Mathematical modeling / English

Heat Equation

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IN THIS VIDEO

01Heat equation definition
02Separated solution ansatz
03Eigenvalue and eigenfunction analogy

FROM THE LEARNING NOTES

\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}

Heat equation definition

SELECTED MOMENT

Choosing sine eigenfunctions

SUGGESTED PREPARATION

Partial derivatives
Ordinary differential equations
Eigenvalues and eigenvectors
Basic separation of variables
Second derivative of sine
Separation of variables
Linearity and superposition for differential equations
Infinite series notation
Second-order ordinary differential equations
Eigenfunctions and eigenvalues for differential operators
Superposition principle for linear PDEs
Basic properties of the sine function
Heat equation on a finite interval
Exponential time factors in PDE solutions
Fourier sine series
Initial and boundary conditions
Separation of variables for the heat equation
Eigenfunctions $\sin(k\pi x)$ on [0,1]
Basic idea of Fourier sine series
Basic partial differential equations
Sine functions on a finite interval
Initial conditions for evolution equations
Fourier series as a method for representing functions

CHAPTERS · TIME REFERENCES

0:00Introduction to the heat equation
0:12Contrast with Laplace's equation and role of time
0:34Separated solution form $u(t,x)=e^{λt}S(x)$
0:56Eigenfunction analogy with ordinary systems
1:14Substitution into the heat equation
1:45Canceling $e^{λt}$ and obtaining an eigenvalue equation
2:00Heat equation and separated form
2:12Choosing sine eigenfunctions
2:27Computing the eigenvalue
3:00General solution by superposition
4:00Board summary: heat equation, separated solutions, eigenfunctions, and general solution
4:20Time dependence: exponential decay and fastest decay for larger k
4:37Slowest-decaying mode: $k=1$ gives $e^{-\pi^2t}$
4:54Physical model: bar with frozen ends and heat escaping
5:34Boundary check: sine eigenfunctions vanish at $x=0$ and $x=1$
6:00Introduce the bar graph and frozen ends
6:20State the initial condition $u(0,x)=1$
6:50Recall the general solution from the upper board
7:10Match the series at $t=0$ to the initial condition
8:00Fourier sine-series interpretation of the coefficients
8:28Writing the explicit solution for $t>0$
9:01Structure of the infinite-sum solution and rapid time decay
9:23Sketch of cooling toward the zero steady state
10:00General heat-equation series on the board
10:05Fourier-series step for finding the coefficients
10:18Why infinitely many solutions are needed for a PDE
10:28Matching the initial function and locating the $B_k$ 's
10:42Copyright and license end card

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This 120-second lecture segment introduces the one-dimensional heat equation $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$ and contrasts it with Laplace's equation by emphasizing that time now appears. The lecturer proposes a separated solution $u(t,x)=e^{\lambda t}S(x)$ , comparing the setup to ordinary systems where $e^{\lambda t}$ multiplies an eigenvector, but here the spatial factor is an eigenfunction. He substitutes the product form into the heat equation, obtains $\lambda e^{\lambda t}S=e^{\lambda t}\frac{\partial^2 S}{\partial x^2}$ , cancels the common exponential factor, and identifies the remaining relation as an eigenvalue equation for $S$ . The final simplified equation is begun verbally but not fully completed within the clip. This 120-second whiteboard segment solves the heat equation by separation of variables. The lecturer starts from $u(t,x)=e^{λt}S(x)$ , chooses sine spatial functions $S(x)=\sin (kπx)$ , and computes their eigenvalue under the second derivative operator as $λ=-k$ ²π². He then uses linearity to build the general solution as an infinite sum over modes k, introducing coefficients $B_k$ and writing the series form $u(t,x)=\sum _{k=1}^\infty B_k e$ ^{-k²π² t} $\sin (kπx)$ . This 120-second lecture segment reviews the displayed one-dimensional heat equation $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$ and its separated product solutions $u(t,x)=e^{\lambda t}S(x)$ . The board lists the sine eigenfunctions $S(x)=\sin(k\pi x)$ with eigenvalues $\lambda=-k^2\pi^2$ , giving the general solution $u(t,x)=\sum_{k=1}^{\infty}B$ _ke^{-k^ $2\pi^2t$ } $S_k$ . The lecturer explains that the time dependence is exponential decay, with larger k decaying faster and $k=1$ decaying most slowly like $e^{-\pi^2t}$ . He interprets the solution physically as heat in a bar whose ends are held at temperature zero, so heat flows out through the frozen ends. He then verifies that the sine factors vanish at $x=0$ and $x=1$ , matching the zero-temperature boundary condition. This clip sets up a heat-equation example on a unit bar with frozen ends and initial temperature $u(0,x)=1$ . The lecturer recalls the general separated solution $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t} S_k(x)$ with $S_k(x)=\sin k\pi x$ , then explains that the unknown coefficients $B_k$ are found by matching the initial condition. He writes $t=0$ and derives the sine-series equation $\sum_{k=1}^{\infty} B_k \sin k\pi x = 1$ , stopping before computing the coefficients explicitly. This 120-second whiteboard segment applies a previously derived heat-equation modal expansion to the specific initial condition $u(0,x)=1$ on $0\le x\le 1$ . The lecturer identifies the coefficient-finding step as a Fourier sine series problem, writes the matching equation $\sum_{k=1}^{\infty}B_k\sin(k\pi x)=1$ at $t=0$ , then reconstructs the solution for $t>0$ as $u(t,x)=\sum_{k=1}^{\infty}B_k e^{-k^2\pi^2 t}\sin(k\pi x)$ . He emphasizes the role of the exponential decay factor and closes with a qualitative sketch showing the temperature profile relaxing toward the zero steady state. This 48-second whiteboard segment reviews the final setup for solving the heat equation by an infinite sine series. The upper board displays the general solution $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}S_k(x)$ together with $S_k(x)=\sin k\pi x$ and $\lambda=-k^2\pi^2$ . The lower board imposes the initial condition $u(0,x)=1$ , producing the matching equation $\sum_{k=1}^{\infty} B_k\sin k\pi x=1$ , and then restates the solution for $t>0$ . The lecturer explains that because the heat equation is a PDE and the initial data is a whole function, infinitely many modes are needed, and Fourier series is the separate step used to determine the coefficients $B_k$ . The clip ends with a copyright and license card.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Introduction to the heat equation 0:12Contrast with Laplace's equation and role of time 0:34Separated solution form

u(t,x)=e^{λt}S(x)

0:56Eigenfunction analogy with ordinary systems 1:14Substitution into the heat equation 1:45Canceling

e^{λt}

and obtaining an eigenvalue equation 2:00Heat equation and separated form 2:12Choosing sine eigenfunctions 2:27Computing the eigenvalue 3:00General solution by superposition 4:00Board summary: heat equation, separated solutions, eigenfunctions, and general solution 4:20Time dependence: exponential decay and fastest decay for larger k 4:37Slowest-decaying mode:

k=1

gives

e^{-\pi^2t}

4:54Physical model: bar with frozen ends and heat escaping 5:34Boundary check: sine eigenfunctions vanish at

x=0

and

x=1

6:00Introduce the bar graph and frozen ends 6:20State the initial condition

u(0,x)=1

6:50Recall the general solution from the upper board 7:10Match the series at

t=0

to the initial condition 8:00Fourier sine-series interpretation of the coefficients 8:28Writing the explicit solution for

t>0

9:01Structure of the infinite-sum solution and rapid time decay 9:23Sketch of cooling toward the zero steady state 10:00General heat-equation series on the board 10:05Fourier-series step for finding the coefficients 10:18Why infinitely many solutions are needed for a PDE 10:28Matching the initial function and locating the

B_k

's 10:42Copyright and license end card

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The segment opens with the lecturer identifying the topic as the heat equation and placing it among the basic partial differential equations. The board shows the central formula $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$ , making clear that the unknown is a function of both time and space.

He then contrasts this equation with Laplace's equation. The key distinction stated in the audio is that Laplace's equation did not involve time, whereas the heat equation does. This motivates the structural feature visible on the board: one time derivative on the left and two space derivatives on the right.

From that structure, the lecturer introduces the dependent variable $u(t,x)$ and emphasizes that the solution depends on both $t$ and $x$ . This sets up the next move: trying to separate the time dependence from the space dependence.

The proposed separated form is written on the board as $u(t,x)=e^{\lambda t}S(x)$ . The lecturer explains this as a product of an exponential factor in time and a function $S$ of space only. The purpose is to reduce the partial differential equation to a simpler relation for the spatial factor.

He explicitly compares this to the method for ordinary systems of differential equations. In that earlier setting, a factor $e^{\lambda t}$ is pulled out with $\lambda$ as the eigenvalue, and the remaining object is an eigenvector. Here, because the remaining factor depends on $x$ , the lecturer calls $S(x)$ an eigenfunction instead.

With the analogy established, he states the plan: substitute the separated form into the heat equation and see what condition determines $S(x)$ . This is the operational core of the segment, turning the ansatz into a derivation.

The substitution is carried out term by term. Differentiating $e^{\lambda t}S(x)$ with respect to $t$ brings down a factor of $\lambda$ , giving $\lambda e^{\lambda t}S$ on the left-hand side. On the spatial side, the lecturer says there are two space derivatives, so the right-hand side becomes $e^{\lambda t}$ times the second derivative of $S$ with respect to $x$ .

The board displays the substituted equation in the form $\lambda e^{\lambda t}S=e^{\lambda t}\frac{\partial S}{\partial x^2}$ . The intended mathematical content, supported by the spoken explanation, is the appearance of the second spatial derivative of $S$ multiplied by the same exponential factor $e^{\lambda t}$ .

The next step is cancellation. Because $e^{\lambda t}$ appears on both sides, the lecturer crosses it out visually and says that this is always possible. The cancellation removes the time-dependent factor from the relation and leaves an equation involving only $\lambda$ and the spatial function $S$ .

He then names the result: an eigenvalue equation. The logical conclusion of the segment is that the separated heat equation has been reduced from a partial differential equation in $u(t,x)$ to an eigenvalue problem for $S(x)$ . The final simplified formula is begun verbally with reference to the second derivative of the function, but the clip ends before the completed expression is fully written on the board.

The segment opens on a blackboard already containing the heat equation $\partial u/\partial t=\partial^2u/\partial x^2$ and the separated ansatz $u(t,x)=e^{\lambda t}S(x)$ . The lecturer’s first task is to interpret $\lambda$ : it is the scalar multiplier that appears when the spatial operator acts on $S$ .

He then chooses the spatial eigenfunctions to be sine functions and writes $S(x)=\sin(k\pi x)$ . This advances the solution method from a generic separated form to a concrete family of candidate modes indexed by $k$ .

To find the associated eigenvalue, he applies two derivatives to $\sin(k\pi x)$ . Each differentiation brings down a factor of $k\pi$ , and the second derivative of sine produces a minus sign, so the result is $-k^2\pi^2\sin(k\pi x)$ . Hence the eigenvalue is $\lambda=-k^2\pi^2$ .

Having identified both the eigenfunctions and eigenvalues, the lecturer shifts from a single mode to the full solution. He states that because the heat equation is linear, known solutions may be combined.

He therefore writes the general solution as an infinite sum over modes: $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$ . The coefficients $B_k$ weight each basic separated solution, and the final remark clarifies that the sine eigenfunction itself carries the index $k$ .

The blackboard presents the heat equation $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$ together with separated solutions $u(t,x)=e^{\lambda t}S(x)$ and a braced list of eigenfunction data. The lecturer identifies the displayed family as eigenfunctions and eigenvalues, then points to the summation as the combination of these solutions.

He clarifies the notation in the general solution $u(t,x)=\sum_{k=1}^{\infty}B_ke^{-k^2\pi^2t}S_k$ by saying that $S_k$ should be written as $\sin(k\pi x)$ . This connects the compact summation notation to the explicit sine eigenfunctions shown beside it.

The next focus is the time factor. Because the eigenvalue is $\lambda=-k^2\pi^2$ , the temporal part becomes $e^{-k^2\pi^2t}$ . The lecturer describes this dependence on $t$ as fast decay, and as $k$ increases the exponent $-k^2\pi^2$ becomes more negative, so later terms in the sum decay more rapidly.

For the slowest-decaying displayed mode, he takes $k=1$ . Substituting $k=1$ into $e^{-k^2\pi^2t}$ gives $e^{-\pi^2t}$ , so the first term has the form involving $B_1$ or, in his speech, $b_1$ , times $e^{-\pi^2t}$ . This establishes that the lowest-index sine mode persists longest.

The lecturer then interprets the mathematics physically. He describes a material bar whose ends are kept at temperature zero, or frozen. Heat initially inside the bar flows through the bar and escapes out the ends, so the whole bar approaches freezing. In the formula, this loss of heat is represented by the exponential decay factors in time.

Finally, he checks why the sine eigenfunctions fit the boundary condition. At $x=0$ , $\sin(k\pi x)=\sin 0=0$ , so the left end is frozen. At $x=1$ , $\sin(k\pi x)=\sin(k\pi)=0$ for integer $k$ , so the right end is also frozen. Thus the displayed sine modes satisfy $u(t,0)=0$ and $u(t,1)=0$ , matching the physical setup of both ends held at temperature zero.

The lecturer begins by turning the abstract heat-equation solution into a concrete picture. On the lower board he writes “Graph” and draws a horizontal bar representing the spatial interval from 0 to 1.

He marks both ends with f and explains verbally that the ends are kept frozen. This establishes the boundary picture: a bar whose endpoints are held fixed while heat can leave through those ends.

Next he specifies the initial state. Above the bar he writes $u(0,x)=1$ and says he is starting with a warm bar, so the temperature at time 0 is 1 across the interval. He also labels the horizontal direction as x and later marks the endpoints as $x=0$ and $x=1$ .

The physical description is refined by saying the sides are insulated, so the heat escapes only out the two ends. This connects the sketch to the standard one-dimensional heat-flow model on a finite rod.

To solve the problem, he points back to the upper board and recalls the previously derived general solution $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t} S_k(x)$ , together with $S_k(x)=\sin k\pi x$ and $\lambda=-k^2\pi^2$ . The key point is that this formula still contains unknown coefficients $B_k$ .

He then states the method for determining those coefficients: match the initial condition. In other words, specialize the general solution at $t=0$ and compare it with the prescribed function $u(0,x)=1$ .

On the lower board he writes $t=0$ and constructs the matching equation step by step. Setting $t=0$ makes the exponential factor equal to 1, so the series reduces to $\sum_{k=1}^{\infty} B_k \sin k\pi x$ .

He equates that reduced series to 1, obtaining $\sum_{k=1}^{\infty} B_k \sin k\pi x = 1$ . This is the central algebraic condition of the clip: the coefficients $B_k$ must be chosen so that a sine series represents the constant initial temperature profile.

The segment ends before the actual Fourier coefficients are computed. The lecturer indicates that once this matching equation is solved for the $B_k$ ’s, one can substitute them back into the general solution to get the final answer for $t>0$ .

The clip opens with the upper board already showing the separated heat-equation solution $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}S_k$ , together with $S_k(x)=\sin(k\pi x)$ and $\lambda=-k^2\pi^2$ . On the lower board, the lecturer points to the initial-condition equation $\sum_{k=1}^{\infty} B_k\sin(k\pi x)=1$ and explains that determining the coefficients is a Fourier sine series problem: only sine modes appear here, not cosine modes.

He then turns to the right side of the lower board and writes the solution for positive time. Starting from the coefficients fixed by the initial condition, he reconstructs each mode by reattaching the exponential factor, producing $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$ . The spoken explanation makes clear that the $B_k$ come from the initial data, while the full time-dependent answer is obtained by multiplying each sine mode by $e^{-k^2\pi^2 t}$ .

After completing the formula, he comments on its structure. Although the answer is an infinite sum, he presents it as a standard and workable PDE solution format: a numerical coefficient $B_k$ , a time factor that decays rapidly, and a spatial factor depending on x. This verbal breakdown corresponds directly to the three visible pieces of each term in the written series.

To interpret the formula geometrically, he returns to the graph labeled $u(0,x)=1$ over $0\le x\le 1$ . He sketches a sequence of curves below the initial flat profile, showing the temperature dropping first near the ends and then overall. The visual progression illustrates the effect of the decaying exponentials, and he concludes that the steady state is the entire bar at temperature zero.

The segment opens on a two-board layout. The upper board states the modal solution form for the heat equation, $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t} S_k(x)$ , and beside it lists the displayed sine mode $S_k(x)=\sin k\pi x$ with eigenvalue $\lambda=-k^2\pi^2$ . The lecturer points to this structure while saying that this is what the solutions to the heat equation look like.

He then identifies the missing computational stage: after writing the infinite series, one still has to determine the coefficients $B_k$ . In his words, this is the Fourier series step of finding the coefficients in the infinite series of solutions. The board itself does not compute those coefficients here; it only exposes where they enter.

Next he explains why the series is infinite rather than finite. Because the problem is a partial differential equation and the initial condition is a whole function of $x$ , a single mode cannot match the data. The solution must combine infinitely many separated modes so that their sum can reproduce the prescribed initial profile.

The lower board makes that matching explicit. It records the initial condition $u(0,x)=1$ and, underneath the label $t=0$ , writes $\sum_{k=1}^{\infty} B_k \sin k\pi x = 1$ . This follows directly from substituting $t=0$ into the general series, since $e^{-k^2\pi^2\cdot 0}=1$ . The lecturer states that Fourier series tells us how to perform this matching and how to find the $B_k$ 's.

To the right, under $t>0$ , the board restores the time-dependent form $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$ . Thus the same sine basis that matches the initial state at $t=0$ evolves for positive time by multiplying each mode by its exponential decay factor. The lecturer closes by calling Fourier series a separate and important question.

The lecture image gives way to a black end card reading © 2015 copyright Gilbert Strang and stating that the work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike International License, with the license URL shown below.

Knowledge cards

Heat equation definition

The lecture defines the one-dimensional heat equation as $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$ . The lecturer stresses that, unlike Laplace's equation, this PDE includes time. The dependent variable is $u(t,x)$ , with one time derivative and two space derivatives.

\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}

Separated solution ansatz

To solve the heat equation, the lecturer looks for product solutions of the form $u(t,x)=e^{\lambda t}S(x)$ . The factor $e^{\lambda t}$ carries the time dependence, while $S(x)$ carries the space dependence.

u(t,x)=e^{\lambda t}S(x)

Eigenvalue and eigenfunction analogy

The lecturer compares this setup to ordinary systems of differential equations, where $e^{\lambda t}$ is paired with an eigenvector and $\lambda$ is the eigenvalue. In the PDE case, the remaining object depends on $x$ , so it is called an eigenfunction rather than an eigenvector.

Substitution into the heat equation

Substituting $u(t,x)=e^{\lambda t}S(x)$ into the heat equation gives a time-derivative side $\lambda e^{\lambda t}S$ and a space-derivative side involving $e^{\lambda t}$ times the second derivative of $S$ . The board shows the substituted relation before cancellation.

\lambda e^{\lambda t}S=e^{\lambda t}\frac{\partial^2 S}{\partial x^2}

Canceling the exponential factor

Because the same factor $e^{\lambda t}$ appears on both sides after substitution, the lecturer cancels it. This step removes the explicit time dependence and leaves an equation for the spatial function $S$ .

Resulting eigenvalue equation

After cancellation, the lecturer identifies the remaining relation as an eigenvalue equation for $S(x)$ . The segment ends as he begins to state the second-derivative form, so the final simplified equation is not fully completed on screen within this clip.

Separated solution form for the heat equation

The lecturer begins from the heat equation $\partial u/\partial t=\partial^2u/\partial x^2$ and uses separated solutions $u(t,x)=e^{\lambda t}S(x)$ . In this form, the time dependence is exponential and the spatial dependence is isolated in $S(x)$ , which is then treated through an eigenvalue problem.

u(t,x)=e^{\lambda t}S(x)

Sine eigenfunction choice

For the spatial factor, the lecturer selects sine functions and writes $S(x)=\sin(k\pi x)$ . This gives a concrete family of candidate eigenfunctions labeled by the mode index $k$ .

S(x)=\sin(k\pi x)

Eigenvalue from two derivatives

Applying the second derivative to $\sin(k\pi x)$ brings down $k\pi$ twice and introduces a minus sign, yielding $-k^2\pi^2\sin(k\pi x)$ . Therefore the associated eigenvalue is $\lambda=-k^2\pi^2$ .

\lambda=-k^2\pi^2

General solution by superposition

Because the heat equation is linear, the lecturer builds the general solution by summing basic separated solutions over all modes. The resulting series is $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$ , where $B_k$ are coefficients for each mode.

u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)

Heat equation shown on the board

The clip centers on the one-dimensional heat equation $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$ , where $u(t,x)$ is the temperature field in a bar. The displayed equation has no extra diffusivity constant, so the spatial second derivative appears with coefficient 1.

\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}

Separated product solutions

Solutions are built in separated form $u(t,x)=e^{\lambda t}S(x)$ , with an exponential time factor and a spatial eigenfunction $S(x)$ . The board also records the eigenvalue relation $\lambda S=S''$ , although some nearby chalk writing is crossed out.

u(t,x)=e^{\lambda t}S(x),\qquad \lambda S=S''

Sine eigenfunctions and eigenvalues

The specific modes listed are $S(x)=\sin(k\pi x)$ with eigenvalues $\lambda=-k^2\pi^2$ for positive integers $k$ . The lecturer explicitly clarifies that the $S_k$ in the summation should be understood as $\sin(k\pi x)$ .

S_k(x)=\sin(k\pi x),\qquad \lambda_k=-k^2\pi^2

General solution as an infinite superposition

The general solution displayed is $u(t,x)=\sum_{k=1}^{\infty}B_ke^{-k^2\pi^2t}S_k$ . It combines all separated sine modes, each multiplied by a coefficient $B_k$ and its own exponential decay factor.

u(t,x)=\sum_{k=1}^{\infty}B_ke^{-k^2\pi^2t}S_k

Why larger k decays faster

The time factor $e^{-k^2\pi^2t}$ becomes more strongly damped as $k$ grows because the negative exponent has magnitude $k^2\pi^2t$ . Therefore later terms in the sum decay more quickly than earlier terms.

e^{-k^2\pi^2t}

Slowest-decaying mode is $k=1$

Among positive integer modes, the smallest decay rate occurs at $k=1$ , giving the temporal factor $e^{-\pi^2t}$ . The lecturer identifies this as the term that decays most slowly, with coefficient spoken as $b_1$ and written on the board as $B_k$ notation.

k=1:\quad e^{-\pi^2t}

Physical meaning: heat escaping a frozen-ended bar

The decay is interpreted as heat leaving a material bar whose ends are held at temperature zero. Heat flows within the bar and escapes through the frozen ends, so the temperature inside the bar tends toward zero over time.

Sine modes satisfy zero-temperature endpoints

The boundary condition is checked directly from the sine eigenfunctions: $\sin(k\pi\cdot 0)=0$ and $\sin(k\pi\cdot 1)=\sin(k\pi)=0$ . Hence each mode vanishes at both endpoints, matching $u(t,0)=0$ and $u(t,1)=0$ .

\sin(k\pi\cdot 0)=0,\qquad \sin(k\pi\cdot 1)=\sin(k\pi)=0

Heat-equation example on a unit bar

The clip introduces a concrete initial-boundary-value problem: a bar from 0 to 1 with both ends kept frozen and initial temperature $u(0,x)=1$ . The sketch on the board records the physical setup that the later series solution must satisfy.

0\le x\le 1,\quad u(0,x)=1

General separated solution recalled from the board

Before solving the specific example, the lecturer points to the previously derived formula for the heat equation on this interval. It expresses the solution as a sum of decaying exponential time factors multiplied by sine spatial modes, with unknown coefficients $B_k$ .

u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t} S_k(x),\quad S_k(x)=\sin k\pi x

Coefficients are found by matching the initial condition

The method emphasized in the clip is to determine the unknown coefficients by imposing the initial data. One sets $t=0$ in the general solution and matches the result to the prescribed initial temperature profile.

\left.u(t,x)\right|_{t=0}=u(0,x)

Derivation of the sine-series matching equation

Substituting $t=0$ into the general solution turns every exponential factor into 1, leaving only the sine series. For the present problem the initial profile is the constant 1, so the coefficients must satisfy a Fourier sine representation of 1 on (0,1).

\sum_{k=1}^{\infty} B_k \sin k\pi x = 1

What is completed and what is left for later

By the end of the clip the lecturer has reduced the problem to the equation that determines the $B_k$ ’s, but he has not yet computed their explicit values. The next mathematical step would be to solve this sine-series identity and substitute the resulting coefficients back into the full solution for $t>0$ .

Fourier sine-series step for the heat equation

The lecturer identifies the coefficient-finding problem as a Fourier sine series because the spatial basis consists only of $\sin(k\pi x)$ terms. At $t=0$ the exponential factors become 1, so matching the initial condition $u(0,x)=1$ gives $\sum_{k=1}^{\infty} B_k\sin(k\pi x)=1$ .

\sum_{k=1}^{\infty} B_k\sin(k\pi x)=1

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 53

$u(t,x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows the solution form $u(t,x) = e^{λt} S(x)$ .

Symbol

$u(t,x)$

Meaning

The unknown function or solution of the heat equation, depending on time t and space x.

Domain

Function of two variables, t and x.

t

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says that time comes into the heat equation and that the solution depends on t.
Formula
Observation
t appears in ∂u/∂t and in $e^{λt}$ .

Symbol

Meaning

Time variable.

Domain

Variable appearing in the time derivative and exponential factor.

x

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the solution depends on x and that x is also a coordinate here.
Formula
Observation
x appears in ∂^2u/∂ $x^2$ and $S(x)$ .

Symbol

Meaning

Space variable or spatial coordinate.

Domain

Variable appearing in the second spatial derivative and in the eigenfunction $S(x)$ .

λ

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer refers to λ as the eigenvalue and says the time derivative brings down a λ.
Formula
Observation
λ appears in $e^{λt}$ and in $λ e^{λt} S = e^{λt}$ ∂S/∂ $x^2$ .

Symbol

Meaning

Eigenvalue associated with the separated time factor $e^{λt}$ .

Domain

Scalar parameter multiplying t in the exponential and appearing as a coefficient after substitution.

$S(x)$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer calls S an eigenfunction because it depends on x.
Formula
Observation
The board shows $S(x)$ in $u(t,x)=e^{λt}S(x)$ and in the substituted equation.

Symbol

$S(x)$

Meaning

Spatial factor in the separated solution; described by the lecturer as an eigenfunction.

Domain

Function of x.

$\frac{\partial u}{\partial t}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The top line on the board reads Heat equation ∂u/∂t = ∂^2u/∂ $x^2$ .

Symbol

$\frac{\partial u}{\partial t}$

Meaning

First partial derivative of u with respect to time t.

Domain

Left-hand side of the heat equation.

$\frac{\partial^2 u}{\partial x^2}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The top line on the board reads Heat equation ∂u/∂t = ∂^2u/∂ $x^2$ .

Symbol

$\frac{\partial^2 u}{\partial x^2}$

Meaning

Second partial derivative of u with respect to space x.

Domain

Right-hand side of the heat equation.

$\lambda e^{\lambda t} S$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The right side of the board shows $λ e^{λt} S = e^{λt}$ ∂S/∂ $x^2$ .

Symbol

$\lambda e^{\lambda t} S$

Meaning

Result of applying the time derivative to $e^{λt}S(x)$ , with S treated as the spatial factor.

Domain

Left-hand side after substituting the separated form into the heat equation.

$e^{\lambda t}\frac{\partial S}{\partial x^2}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The right side of the board shows $λ e^{λt} S = e^{λt}$ ∂S/∂ $x^2$ .

Symbol

$e^{\lambda t}\frac{\partial S}{\partial x^2}$

Meaning

Expression shown on the board for the spatial-derivative side after substitution; the displayed derivative notation is ambiguous relative to the spoken second derivative.

Domain

Right-hand side after substituting the separated form into the heat equation.

$u(t,x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows the heat equation as ∂ $u/t$ = ∂²u/∂x² and the separated solution form $u(t,x) = e^{λt} S(x)$ .

Symbol

$u(t,x)$

Meaning

Temperature or unknown function of time t and space x.

Domain

Function of two variables t and x.

λ

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says that λ is “some number times my function.”
Formula
Observation
The separated solution is written as $e^{λt} S(x)$ , and later λ is identified as -k^ $2π^2$ .

Symbol

Meaning

Eigenvalue associated with the spatial operator d²/dx² in the separated solution.

Domain

Scalar parameter; in this example it becomes negative for sine eigenfunctions.

$S(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $S(x) = \sin kπx$ .
Audio
Observation
The lecturer says he is looking for functions S and that they will be sine functions.

Symbol

$S(x)$

Meaning

Spatial eigenfunction in the separated solution $u(t,x)=e^{λt}S(x)$ .

Domain

Function of x.

Knowledge points · 26

Heat equation definition

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer introduces this as the heat equation video and says it is the second of the three basic partial differential equations.
Formula
Observation
The board states Heat equation ∂ $u/t$ = ∂^2u/∂ $x^2$ .

Definition

Explanation

The segment defines the one-dimensional heat equation as a partial differential equation relating the first time derivative of u to the second space derivative of u. The lecturer contrasts it with Laplace's equation by saying that now time comes into the equation.

Formula

\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}

Conditions

The dependent variable is $u(t,x)$ .
The equation contains one time derivative and two space derivatives.

Separated solution ansatz

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says he hopes to separate the two parts and looks for solutions like $e^{λt}$ times a function of x.
Formula
Observation
The board writes Solutions $u(t,x)=e^{λt}S(x)$ .

Definition

Explanation

The lecturer proposes looking for solutions in product form, with a time-dependent exponential factor and a space-dependent factor $S(x)$ . He explicitly compares this to solving ordinary systems of differential equations, where $e^{λt}$ multiplies an eigenvector; here the spatial factor is called an eigenfunction because it depends on x.

Formula

u(t,x)=e^{\lambda t}S(x)

Conditions

The solution is sought as a product of a function of t and a function of x.
λ is treated as the eigenvalue.
$S(x)$ is treated as the eigenfunction.

Prerequisites

Heat equation definition

Substituting the separated form into the heat equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says he substitutes the proposed solution into the differential equation to discover what determines $S(x)$ .
Formula
Observation
The board shows $λ e^{λt} S = e^{λt}$ ∂ $S/x^2$ .

Uncertainties

The displayed right-hand derivative is written as ∂S/∂ $x^2$ , while the spoken explanation refers to two space derivatives and the second derivative of the function.

Method

Explanation

The method is to insert $u(t,x)=e^{λt}S(x)$ into the heat equation. The time derivative brings down λ, producing $λ e^{λt}S$ on the left. On the spatial side, the lecturer says there are two space derivatives, leading to an expression involving the second derivative of S multiplied by $e^{λt}$ .

Formula

\lambda e^{\lambda t}S=e^{\lambda t}\frac{\partial^2 S}{\partial x^2}

Conditions

Use the product form $u(t,x)=e^{λt}S(x)$ .
Differentiate with respect to t on the left and with respect to x twice on the right.

Prerequisites

Heat equation definition
Separated solution ansatz

Canceling the exponential factor to obtain an eigenvalue equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, 'And always, I cancel e to the lambda t' and then says he has an eigenvalue equation.
Animation
Observation
The lecturer draws diagonal cancellation marks through the $e^{λt}$ factors on both sides of the substituted equation.

Uncertainties

The final simplified equation is not fully written before the clip ends; only the start of the next line, 'So the second derivative of my function is', is audible.

Method

Explanation

After substitution, the common factor $e^{λt}$ appears on both sides and is canceled visually and verbally. The lecturer identifies the remaining relation as an eigenvalue equation for the unknown function $S(x)$ .

Formula

e^{\lambda t}\text{ cancels from both sides, leaving an equation for }S(x)

Conditions

The same nonzero exponential factor $e^{λt}$ must appear on both sides after substitution.

Prerequisites

Substituting the separated form into the heat equation

Separated solutions of the heat equation

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board displays “Heat equation ∂u/∂t = ∂²u/∂x²” and “Solutions $u(t,x)=e^{λt}S(x)$ ”.
Audio
Observation
The lecturer explains that λ is a number multiplying the function and identifies S as a sine function.

Definition

Explanation

For the heat equation, the lecturer uses separated solutions of the form $u(t,x)=e^{λt}S(x)$ . The time dependence is exponential, while the spatial part $S(x)$ is chosen to satisfy an eigenvalue problem for the second derivative operator.

Formula

u(t,x)=e^{\lambda t}S(x)

Conditions

Applied to the heat equation ∂u/∂t = ∂²u/∂x².
The spatial factor $S(x)$ is treated as an eigenfunction of d²/dx².

Sine eigenfunction for the spatial operator

Clear evidence

Shown in the video

Evidence

Formula
Observation
The lecturer writes $S(x)=\sin kπx$ .
Audio
Observation
He says the functions S will be sine functions and asks what the eigenvalue is.

Uncertainties

The boundary conditions that select these sine modes are not stated in this clip.

Formula

Explanation

The spatial eigenfunction used in this example is a sine function with argument kπx. Taking two derivatives returns the same sine function multiplied by a constant, so it is an eigenfunction of the second derivative operator.

Formula

S(x)=\sin(k\pi x)

Conditions

Used as the spatial factor in $u(t,x)=e^{λt}S(x)$ .
The integer index k labels different modes.

Prerequisites

Separated solutions of the heat equation

Eigenvalue produced by two derivatives of $\sin (kπx)$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says that taking two derivatives makes kπ come out twice with a minus sign.
Formula
Observation
The board shows $λ=-k^2π^2$ .

Formula

Explanation

Differentiating $\sin (kπx)$ twice brings down the factor kπ twice and introduces a minus sign, giving the eigenvalue $λ=-k^2π^2$ for the spatial operator.

Formula

\lambda=-k^2\pi^2

Conditions

Applies to $S(x)=\sin (kπx)$ .
Uses the second derivative operator d²/dx².

Prerequisites

Sine eigenfunction for the spatial operator

General solution as a linear combination of separated solutions

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says that if several solutions are known and the equation is linear, one takes combinations of the basic solutions.
Formula
Observation
He writes a general solution as a sum over k from 1 to infinity.

Uncertainties

The full final formula is not completely legible in every frame, but the spoken explanation and visible terms identify the intended series form.

Method

Explanation

Because the heat equation is linear, the lecturer constructs the general solution by summing basic separated solutions over all modes k, with coefficients $B_k$ multiplying each mode.

Formula

u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)

Conditions

Requires linearity of the PDE.
Uses the previously found eigenpairs ( $\sin (kπx)$ , -k²π²).
The sum runs over $k=1$ ,2,3,... in the displayed construction.

Prerequisites

Separated solutions of the heat equation
Eigenvalue produced by two derivatives of $\sin (kπx)$

One-dimensional heat equation

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board title reads "Video 8.3: Heat equation" followed by $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$ .
Audio
Observation
The lecture context is the heat equation solution family.

Definition

Explanation

The clip displays the heat equation as a partial differential equation relating the time derivative of temperature u to its second spatial derivative. In the lecturer's physical interpretation, u represents temperature in a material bar, with heat flowing through the bar and escaping at the ends.

Formula

\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}

Conditions

u is a function of time t and position x
The displayed equation uses unit diffusivity because no constant multiplies $\frac{\partial^2 u}{\partial x^2}$

Separated product solutions

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes "Solutions $u(t,x)=e^{\lambda t}S(x)$ ".
Audio
Observation
The lecturer calls the displayed family eigenfunctions and eigenvalues.

Definition

Explanation

The heat equation is solved by looking for product solutions in which the time dependence and space dependence separate: a temporal exponential factor multiplies a spatial function. The spatial function is treated as an eigenfunction, and the exponent parameter is treated as the corresponding eigenvalue.

Formula

u(t,x)=e^{\lambda t}S(x)

Conditions

S is the spatial part
$\lambda$ is the eigenvalue controlling the time factor

Prerequisites

One-dimensional heat equation

Eigenvalue equation for the spatial part

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $d^2S = \lambda S = \frac{\partial^2 S}{\partial x^2}$ , with parts crossed out.
Audio
Observation
The lecturer refers to the family of eigenfunctions and eigenvalues.

Uncertainties

The first expression is visually ambiguous because of chalk crossings; the readable mathematical content is the eigenvalue relation $\lambda S=S$ ''.

Formula

Explanation

The spatial factor $S(x)$ satisfies an eigenvalue equation for the second-derivative operator. The crossed-out writing on the board indicates revision, but the retained relation identifies $\lambda S$ with the second derivative of S.

Formula

\lambda S=S''

Conditions

S is twice differentiable in x
$\lambda$ is the eigenvalue associated with S

Prerequisites

Separated product solutions

General solution as a superposition

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes "General sol $u(t,x)=\sum_{k=1}^{\infty}B$ _ke^{-k^ $2\pi^2t$ } $S_k$ ".
Audio
Observation
The lecturer says this is the combination of these solutions and calls it the general solution.

Formula

Explanation

The general solution shown is an infinite linear combination of separated solutions. Each term has a coefficient $B_k$ , an exponential time-decay factor $e^{-k^2\pi^2t}$ , and a spatial eigenfunction $S_k$ .

Formula

u(t,x)=\sum_{k=1}^{\infty}B_ke^{-k^2\pi^2t}S_k

Conditions

k runs over positive integers
The formula assumes the listed sine modes form the relevant solution family

Prerequisites

Separated product solutions
Eigenvalue equation for the spatial part

Claims and conditions · 13

Structure of the heat equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, 'Now time comes into the heat equation. We have a time derivative and matching with two space derivatives.'
Formula
Observation
The board equation is ∂u/∂t = ∂^2u/∂ $x^2$ .

Proposition

Statement

The heat equation presented in this segment contains one time derivative on one side and two space derivatives on the other side.

Hypotheses

The equation under discussion is the heat equation written on the board.

Quantifiers

For the displayed function $u(t,x)$ in this lecture segment.

Analogy between PDE separation and ODE eigenvalue solutions

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says this is exactly like solving ordinary systems of differential equations, where $e^{λt}$ was pulled out with λ as the eigenvalue and an eigenvector appeared; here it is an eigenfunction because it depends on x.

Proposition

Statement

Seeking solutions of the heat equation in the form $e^{λt}S(x)$ is analogous to solving ordinary systems of differential equations using $e^{λt}$ times an eigenvector, except that the spatial factor here is an eigenfunction rather than a vector.

Hypotheses

The PDE solution is sought by separating time and space dependence.

Quantifiers

For the comparison made in this lecture segment between ordinary systems and the heat equation.

After cancellation the remaining relation is an eigenvalue equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says he cancels $e^{λt}$ and has an eigenvalue equation.
Animation
Observation
Cancellation marks are drawn through the exponential factors on both sides.

Uncertainties

The exact final simplified formula is not completed on screen within the clip.

Proposition

Statement

After substituting $u(t,x)=e^{λt}S(x)$ into the heat equation and canceling the common factor $e^{λt}$ , the remaining relation is an eigenvalue equation for S.

Hypotheses

The substituted equation contains the same factor $e^{λt}$ on both sides.

Quantifiers

For the separated solution form used in this segment.

$\sin (kπx)$ is an eigenfunction of d²/dx² with eigenvalue -k²π²

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer states that after taking two derivatives one gets back $\sin (kπx)$ , so it is an eigenfunction, and the eigenvalue is -k²π².
Formula
Observation
The board shows $S(x)=\sin kπx$ and $λ=-k^2π^2$ .

Proposition

Statement

For $S(x)=\sin (kπx)$ , applying the second derivative operator gives d² $S/dx$ ² = -k²π² $S(x)$ , so the associated eigenvalue is $λ=-k$ ²π².

Hypotheses

$S(x)=\sin (kπx)$ .
The operator under consideration is d²/dx².

Quantifiers

For each mode index k used in the construction.

Linear combinations of separated solutions give the general solution

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says that if one knows several solutions and has a linear equation, one just takes combinations of those basic solutions.

Uncertainties

The proof of superposition is not shown in this clip; only the principle is stated.

Proposition

Statement

Because the heat equation is linear, sums of known separated solutions are also solutions, and the general solution is formed by taking such combinations over all modes.

Hypotheses

The PDE is linear.
Each term in the sum is a known solution of the heat equation.

Quantifiers

Over the indexed family of basic solutions used in the series.

Higher-index modes decay faster

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the dependence on t is fast decay, and as k gets larger, later terms in the sum decay really fast.
Formula
Observation
The displayed factor is $e^{-k^2\pi^2t}$ .

Proposition

Statement

In the displayed general solution, the temporal factor for the kth mode is $e^{-k^2\pi^2t}$ ; as k increases, the magnitude of the negative exponent increases, so later terms decay faster in time.

Hypotheses

k is a positive integer
$t\ge 0$
The general solution uses the displayed factor $e^{-k^2\pi^2t}$

Quantifiers

For each positive integer k and nonnegative time t, larger k gives a faster exponential decay rate.

The $k=1$ mode is the slowest-decaying displayed mode

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the term that decays most slowly is k equal one, with $a b_1$ and $e^{-\pi^2t}$ .
Formula
Observation
Setting $k=1$ in $e^{-k^2\pi^2t}$ gives $e^{-\pi^2t}$ .

Uncertainties

The spoken coefficient is heard as $b_1$ while the board notation is $B_k$ .

Proposition

Statement

Among the modes indexed by positive integers k in the displayed sum, the slowest decay occurs for $k=1$ , whose temporal factor is $e^{-\pi^2t}$ .

Hypotheses

k ranges over positive integers
The decay factor is $e^{-k^2\pi^2t}$
$t\ge 0$

Quantifiers

For positive integer mode indices k, the minimum decay rate occurs at $k=1$ .

Sine eigenfunctions satisfy the zero endpoint condition

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says when x is zero the sine is zero, and when x is one he has sine of k pi, which is again zero.
Formula
Observation
The board lists $S(x)=\sin k\pi x$ .

Proposition

Statement

For every positive integer k, $\sin(k\pi x)$ equals 0 at $x=0$ and at $x=1$ , so the displayed spatial eigenfunctions match the stated frozen-end boundary condition.

Hypotheses

k is a positive integer
The spatial eigenfunction is $\sin(k\pi x)$

Quantifiers

For all positive integers k, the function $\sin(k\pi x)$ vanishes at both endpoints $x=0$ and $x=1$ .

Physical interpretation of decay as heat leaving the bar

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer describes a material bar with ends kept at temperature zero, heat flowing around in the bar, and heat flowing out the ends as the bar approaches freezing.

Proposition

Statement

In the lecturer's physical description, the exponential decay of the solution modes corresponds to heat initially inside the bar flowing out through ends held at temperature zero, so the bar approaches freezing.

Hypotheses

u represents temperature in a material bar
The ends are kept at temperature zero
Heat is initially present in the bar

Quantifiers

For the described bar with frozen ends, the displayed decaying solution is interpreted as heat escaping over time.

The coefficient-finding step is a Fourier sine series problem

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, 'Whenever I have to find these coefficients, this is a Fourier sine series. I have only sines, not cosines here.'
Formula
Observation
The displayed initial-condition equation contains only $\sin(k\pi x)$ terms.

Proposition

Statement

For this heat-equation example, determining the coefficients $B_k$ from the initial condition is a Fourier sine series problem, not a full cosine-including Fourier series problem.

Hypotheses

The spatial basis functions are $S_k(x)=\sin(k\pi x)$ .
The initial profile is represented on $0<x<1$ by a sine expansion.

Quantifiers

For the displayed example with $u(0,x)=1$ and basis $\sin(k\pi x)$ .

Long-time steady state is zero temperature

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer explicitly says the steady state is the whole thing at temperature zero.
Diagram
Observation
The sketched later-time curves move downward toward the horizontal axis.

Proposition

Statement

In the depicted evolution, the steady-state temperature profile is $u=0$ throughout the bar.

Hypotheses

Each mode carries the factor $e^{-k^2\pi^2 t}$ .
The boundary values shown in the sketch remain zero.

Quantifiers

As $t \to \infty$ for $0 \le x \le 1$ .

A PDE heat problem requires infinitely many modes to match a full initial function

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, 'Once again, we have infinitely many solutions. We're talking about a partial differential equation. We have a whole function to match, so we need all of those.'

Proposition

Statement

Because the heat equation is a partial differential equation and the initial data is a whole function, the solution must use infinitely many separated modes rather than a single one.

Hypotheses

The problem is a partial differential equation
There is a full initial function to match

Quantifiers

For the displayed heat-equation setup, infinitely many solution modes are needed.

Derivations and proofs · 9

Derivation of the eigenvalue equation from the separated heat-equation ansatz

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer explains separation, substitution, cancellation, and the resulting eigenvalue equation.
Formula
Observation
The board shows $u(t,x)=e^{λt}S(x)$ , then $λ e^{λt}S=e^{λt}$ ∂S/∂ $x^2$ , with cancellation marks through $e^{λt}$ .

Uncertainties

The last written simplification is incomplete in the clip; the spoken phrase begins 'So the second derivative of my function is' but does not finish before the segment ends.

Proof

Steps

Expression
$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$
Explanation
Start from the heat equation written on the board.
Justification
Given definition of the heat equation in this lecture segment.
Shown in the video
Expression
$u(t,x)=e^{\lambda t}S(x)$
Explanation
Assume a separated solution with an exponential time factor and a spatial factor $S(x)$ .
Justification
Separation-of-variables ansatz stated by the lecturer.
Shown in the video
Expression
$\frac{\partial}{\partial t}\left(e^{\lambda t}S(x)\right)=\lambda e^{\lambda t}S(x)$
Explanation
Differentiate the assumed solution with respect to time; the exponential derivative brings down λ.
Justification
Standard derivative rule for $e^{λt}$ , as described by the lecturer when he says the time derivative brings down a λ.
Shown in the video
Expression
$\frac{\partial^2}{\partial x^2}\left(e^{\lambda t}S(x)\right)=e^{\lambda t}\frac{\partial^2 S}{\partial x^2}$
Explanation
Differentiate the assumed solution twice with respect to space; the time factor remains multiplied outside the spatial derivative.
Justification
The lecturer says there are two space derivatives and writes the substituted spatial side with $e^{λt}$ multiplying the derivative expression for S.
Derived from the video
Expression
$\lambda e^{\lambda t}S=e^{\lambda t}\frac{\partial^2 S}{\partial x^2}$
Explanation
Equate the time-derivative side and the space-derivative side after substitution into the heat equation.
Justification
Substitution into the heat equation.
Shown in the video
Expression
$\lambda S=\frac{\partial^2 S}{\partial x^2}$
Explanation
Cancel the common factor $e^{λt}$ from both sides to obtain the eigenvalue equation for S.
Justification
The lecturer explicitly cancels $e^{λt}$ and calls the result an eigenvalue equation.
Derived from the video

Conclusion

The separated heat equation reduces to an eigenvalue problem for the spatial function S, with the final simplified equation not fully completed on screen before the clip ends.

Derivation of the eigenvalue for $\sin (kπx)$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, “take two derivatives of this,” then explains that kπ comes out twice with a minus sign.
Formula
Observation
The board shows $S(x)=\sin kπx$ and $λ=-k^2π^2$ .

Proof

Steps

Expression
$S(x)=\sin(k\pi x)$
Explanation
Start from the proposed spatial eigenfunction written on the board.
Justification
Observed directly from the lecturer’s writing and speech.
Shown in the video
Expression
$\frac{d^2}{dx^2}\sin(k\pi x)$
Explanation
Apply the second derivative operator to the sine function.
Justification
This is the eigenvalue test for the spatial operator d²/dx².
Shown in the video
Expression
$-k^2\pi^2\sin(k\pi x)$
Explanation
Two differentiations bring down the factor kπ twice and introduce a minus sign.
Justification
Standard differentiation rule for sine composed with a linear argument, as explained verbally by the lecturer.
Shown in the video
Expression
$\lambda=-k^2\pi^2$
Explanation
Since the result is a constant multiple of the original function, that constant is the eigenvalue.
Justification
Definition of eigenfunction/eigenvalue for the operator d²/dx².
Shown in the video

Conclusion

The function $\sin (kπx)$ is an eigenfunction of d²/dx² with eigenvalue $λ=-k$ ²π².

Construction of the general solution from eigenpairs

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the previous pair was simple and now writes the general solution by taking combinations of basic solutions.
Formula
Observation
The board shows a sum from $k=1$ to infinity with coefficients $B_k$ , exponential decay, and sine factors.

Uncertainties

Some intermediate handwriting is partially obscured while the lecturer writes, but the final spoken formula identifies the intended expression.

Proof

Steps

Expression
$u(t,x)=e^{\lambda t}S(x)$
Explanation
Begin with the separated solution form already on the board.
Justification
Previously established setup for solving the heat equation.
Shown in the video
Expression
$\lambda=-k^2\pi^2,\quad S(x)=\sin(k\pi x)$
Explanation
Substitute the eigenvalue and eigenfunction found for each mode k.
Justification
From the derivation of the sine eigenpair.
Shown in the video
Expression
$u_k(t,x)=e^{-k^2\pi^2 t}\sin(k\pi x)$
Explanation
This gives one basic separated solution for a fixed k.
Justification
Direct substitution into the separated form.
Derived from the video
Expression
$u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$
Explanation
Form the general solution by summing over all modes with coefficients $B_k$ .
Justification
Linearity of the heat equation allows superposition of solutions, as stated by the lecturer.
Shown in the video

Conclusion

The general solution is represented as an infinite linear combination of separated sine-mode solutions with coefficients $B_k$ .

From eigenvalue to exponential decay factor

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $u(t,x)=e^{\lambda t}S(x)$ , $\lambda=-k^2\pi^2$ , and the general solution with $e^{-k^2\pi^2t}$ .
Audio
Observation
The lecturer discusses dependence on t as fast decay and identifies the $k=1$ factor as $e^{-\pi^2t}$ .

Proof

Steps

Expression
$u(t,x)=e^{\lambda t}S(x)$
Explanation
Start from the separated product solution displayed on the board.
Justification
Observed board formula for solutions of the heat equation.
Shown in the video
Expression
$\lambda=-k^2\pi^2$
Explanation
Use the eigenvalue listed for the sine spatial modes.
Justification
Observed board formula pairing $S(x)=\sin k\pi x$ with $\lambda=-k^2\pi^2$ .
Shown in the video
Expression
$e^{\lambda t}=e^{-k^2\pi^2t}$
Explanation
Substitute the eigenvalue into the temporal factor.
Justification
Direct substitution of $\lambda=-k^2\pi^2$ into $e^{\lambda t}$ .
Derived from the video
Expression
$k=1:\quad e^{-\pi^2t}$
Explanation
For the first mode, the decay factor becomes $e^{-\pi^2t}$ .
Justification
Substitute $k=1$ into $e^{-k^2\pi^2t}$ ; this matches the lecturer's spoken example.
Derived from the video

Conclusion

The displayed separated solutions decay in time according to $e^{-k^2\pi^2t}$ , with the $k=1$ mode decaying like $e^{-\pi^2t}$ .

Checking the endpoint values of the sine eigenfunctions

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer evaluates the sine functions at $x=0$ and $x=1$ and states both give zero.
Formula
Observation
The board lists $S(x)=\sin k\pi x$ .

Proof

Steps

Expression
$S_k(x)=\sin(k\pi x)$
Explanation
Take the displayed kth spatial eigenfunction.
Justification
Observed board formula and lecturer's clarification for $S_k$ .
Shown in the video
Expression
$S_k(0)=\sin(k\pi\cdot 0)=\sin 0=0$
Explanation
At the left endpoint $x=0$ , the sine factor is zero.
Justification
Evaluation of the sine function at 0.
Derived from the video
Expression
$S_k(1)=\sin(k\pi\cdot 1)=\sin(k\pi)=0$
Explanation
At the right endpoint $x=1$ , the sine factor is also zero for integer k.
Justification
The sine of an integer multiple of $\pi$ is zero; this matches the lecturer's statement.
Derived from the video

Conclusion

Each displayed sine eigenfunction vanishes at both endpoints, matching the stated zero-temperature boundary condition.

Substitute $t=0$ into the general solution and match $u(0,x)=1$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, "when t is zero, this is one, and I just have the $s_k$ , the sine of k pi x. That has to match the one."
Formula
Observation
Lower board shows $t=0$ followed by $\sum_{k=1}^{\infty} B_k \sin k\pi x = 1$ .

Proof

Steps

Expression
$u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t} S_k(x)$
Explanation
Start from the displayed general solution on the upper board.
Justification
Observed formula on the board.
Shown in the video
Expression
$S_k(x)=\sin k\pi x$
Explanation
Replace $S_k(x)$ by its explicit sine form.
Justification
Observed definition on the upper board.
Shown in the video
Expression
$t=0$
Explanation
Impose the initial time used to match the initial condition.
Justification
Explicitly written on the lower board and stated in the audio.
Shown in the video
Expression
$e^{-k^2\pi^2\cdot 0}=1$
Explanation
At $t=0$ the exponential time factor equals 1.
Justification
Standard evaluation of the exponential at zero; consistent with the lecturer saying "when t is zero, this is one."
Derived from the video
Expression
$\sum_{k=1}^{\infty} B_k \sin k\pi x = 1$
Explanation
After setting $t=0$ , the remaining series must equal the initial temperature profile 1.
Justification
Matching the substituted general solution to the written initial condition $u(0,x)=1$ .
Shown in the video

Conclusion

The coefficients $B_k$ are determined by requiring the sine series $\sum_{k=1}^{\infty} B_k \sin k\pi x$ to represent the constant initial profile 1 on $0<x<1$ .

Deriving the initial-condition equation from the general solution

Clear evidence

Shown in the video

Evidence

Formula
Observation
Top board shows $u(t,x)=\sum B_k e^{-k^2\pi^2 t}\sin(k\pi x)$ ; lower board writes $\sum B_k\sin(k\pi x)=1$ at $t=0$ .
Audio
Observation
The lecturer says he is finding the coefficients so that this will match one, the initial condition.

Proof

Steps

Expression
$u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$
Explanation
Start from the general separated solution displayed on the top board.
Justification
Given formula on the board.
Shown in the video
Expression
$e^{-k^2\pi^2\cdot 0}=1$
Explanation
At the initial time $t=0$ , each exponential factor equals 1.
Justification
Direct substitution $t=0$ into the exponential.
Derived from the video
Expression
$u(0,x)=\sum_{k=1}^{\infty} B_k\sin(k\pi x)$
Explanation
Therefore the initial profile is represented by a pure sine series.
Justification
Substitute $t=0$ into the general solution.
Derived from the video
Expression
$\sum_{k=1}^{\infty} B_k\sin(k\pi x)=1$
Explanation
Matching the prescribed initial condition $u(0,x)=1$ yields the coefficient equation shown on the lower board.
Justification
Equate the series representation of $u(0,x)$ to the given initial value 1.
Shown in the video

Conclusion

The coefficients $B_k$ must be chosen so that the sine series equals 1 on the interval, which is exactly the Fourier sine-series matching step.

Reconstructing the later-time solution after fixing $B_k$

Clear evidence

Shown in the video

Evidence

Formula
Observation
After writing the initial series, the lecturer writes $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$ for $t>0$ .
Audio
Observation
He says, 'and then for t greater than zero, the solution u will be ... the sum of these $b_k$ 's ... times $e^{-k^2\pi^2 t}$ times $\sin (k\pi x)$ .'

Proof

Steps

Expression
$\sum_{k=1}^{\infty} B_k\sin(k\pi x)=1$
Explanation
First determine $B_k$ from the initial condition at $t=0$ .
Justification
Displayed lower-board equation and spoken explanation.
Shown in the video
Expression
$u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$
Explanation
Then reinsert the time-dependent exponential factor into each mode to obtain the solution for $t>0$ .
Justification
Use the general solution structure from the top board with the now-determined coefficients.
Shown in the video

Conclusion

Once the Fourier sine coefficients are known from the initial data, the full heat-equation solution for positive time is the same sine series multiplied mode-by-mode by $e^{-k^2\pi^2 t}$ .

Deriving the initial-condition equation from the general series

Clear evidence

Shown in the video

Evidence

Formula
Observation
Top board gives $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t} S_k(x)$ with $S_k(x)=\sin k\pi x$ .
Formula
Observation
Bottom board writes $u(0,x)=1$ and $\sum_{k=1}^{\infty} B_k \sin k\pi x = 1$ under $t=0$ .

Proof

Steps

Expression
$u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t} S_k(x)$
Explanation
Start from the displayed general solution on the top board.
Justification
Board formula.
Shown in the video
Expression
$S_k(x)=\sin k\pi x$
Explanation
Substitute the displayed spatial mode into the general solution.
Justification
Board formula beside the general solution.
Shown in the video
Expression
$u(0,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2\cdot 0}\sin(k\pi x)$
Explanation
Set $t=0$ in the series.
Justification
Direct substitution of the initial time into the displayed formula.
Derived from the video
Expression
$\sum_{k=1}^{\infty} B_k \sin(k\pi x)=1$
Explanation
Use the initial condition $u(0,x)=1$ and simplify $e^{0}=1$ to obtain the matching equation shown on the board.
Justification
Initial condition plus elementary simplification of the exponential factor.
Shown in the video

Conclusion

The initial condition $u(0,x)=1$ becomes the Fourier sine matching equation $\sum_{k=1}^{\infty} B_k \sin(k\pi x)=1$ on the displayed interval.

Worked examples · 3

Slowest-decaying first mode

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the term that decays most slowly is k equal one, with $a b_1$ and $e^{-\pi^2t}$ .
Formula
Observation
The general solution contains B_ke^{-k^ $2\pi^2t$ } $S_k$ .

Uncertainties

The coefficient is spoken as $b_1$ but written on the board as $B_k$ .

Problem

Identify the temporal factor of the slowest-decaying term in the displayed general solution.

Given

The general solution contains terms B_ke^{-k^ $2\pi^2t$ } $S_k$
k is a positive integer
The lecturer states that larger k gives faster decay

Goal

Find the form of the $k=1$ time-decay factor.

Steps

Expression
$B_ke^{-k^2\pi^2t}S_k$
Explanation
Start from the general kth term shown on the board.
Justification
Observed formula for the general solution.
Shown in the video
Expression
$k=1$
Explanation
Choose the smallest positive integer index because the lecturer identifies it as the slowest-decaying term.
Justification
Audio statement that the term decaying most slowly has k equal one.
Shown in the video
Expression
$B_1e^{-\pi^2t}S_1$
Explanation
Substitute $k=1$ into the displayed kth term.
Justification
Direct substitution into B_ke^{-k^ $2\pi^2t$ } $S_k$ .
Derived from the video

Answer

The first mode has temporal factor $e^{-\pi^2t}$ , with coefficient written on the board as $B_1$ and spoken as $b_1$ .

Verification

This agrees with the lecturer's spoken example: "there'll be $a b_1$ and $e^{-\pi^2t}$ ".

Warm bar with frozen ends and uniform initial temperature

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, "Maybe I start off with a warm bar," "So I have an ordinary heated bar," and "And I put it in the freezer."
Diagram
Observation
Lower board draws a horizontal bar from 0 to 1, labels both ends f, marks $x=0$ and $x=1$ , and writes $u(0,x)=1$ .

Uncertainties

The final explicit values of $B_k$ are not computed within this clip.

Problem

Graph the solution for a bar on $0\le x\le 1$ whose ends are kept frozen and whose initial temperature is $u(0,x)=1$ .

Given

The bar runs from 0 to 1.
Both ends are marked f and described as frozen.
The sides are insulated so heat escapes only out the two ends.
The initial condition is $u(0,x)=1$ .

Goal

Reduce the general heat-equation solution to the equation that determines the coefficients for this specific initial-value problem.

Steps

Expression
$0\le x\le 1$
Explanation
Draw the physical bar as the spatial domain.
Justification
Stated in the audio as "here's my bar from zero to one" and shown on the board.
Shown in the video
Expression
$u(0,x)=1$
Explanation
Assign the initial temperature profile across the whole bar.
Justification
Written on the board and spoken as "So u at zero and x, I'll say be one."
Shown in the video
Expression
$u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t} S_k(x)$
Explanation
Use the previously obtained general solution as the starting formula.
Justification
Referenced from the upper board when the lecturer says he must remember the general solution.
Shown in the video
Expression
$t=0:\ \sum_{k=1}^{\infty} B_k \sin k\pi x = 1$
Explanation
Set $t=0$ and match the series to the initial condition.
Justification
Written on the lower board and explained verbally as matching the initial picture.
Shown in the video

Answer

The worked setup ends with the coefficient-matching equation $\sum_{k=1}^{\infty} B_k \sin k\pi x = 1$ .

Verification

The lecturer states that from this equation he will find the $B_k$ 's and then obtain the final solution for $t>0$ .

Example: bar initially at uniform temperature 1

Clear evidence

Shown in the video

Evidence

Diagram
Observation
Graph labeled $u(0,x)=1$ shows a flat top over $0\le x\le 1$ .
Formula
Observation
Lower board writes $\sum_{k=1}^{\infty} B_k\sin k\pi x=1$ under $t=0$ .
Audio
Observation
The lecturer says the temperature starts out through the whole bar at one.

Uncertainties

The numerical values of $B_k$ are not computed in this clip; the lecturer says a separate Fourier-series video is needed to know what these numbers come out to be.

Problem

Solve the heat equation on $0\le x\le 1$ with initial temperature $u(0,x)=1$ , using the displayed sine-mode expansion.

Given

$u(0,x)=1$ for $0<x<1$
Spatial modes are $\sin(k\pi x)$
General solution form is $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$

Goal

Express the solution as a Fourier sine series whose coefficients are fixed by the initial condition, then write the time-dependent solution for $t>0$ .

Steps

Expression
$u(0,x)=\sum_{k=1}^{\infty} B_k\sin(k\pi x)$
Explanation
Set $t=0$ in the general solution so the exponentials become 1.
Justification
Substitution into the displayed general formula.
Derived from the video
Expression
$\sum_{k=1}^{\infty} B_k\sin(k\pi x)=1$
Explanation
Match the resulting sine series to the constant initial profile 1.
Justification
Initial condition stated on the board and in speech.
Shown in the video
Expression
$u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$
Explanation
Restore the time factor to get the solution for later times.
Justification
General solution structure from the top board.
Shown in the video

Answer

$u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$ , where $B_k$ are the Fourier sine coefficients determined by $\sum_{k=1}^{\infty} B_k\sin(k\pi x)=1$ .

Verification

The answer is verified structurally in the video by checking that at $t=0$ the exponentials equal 1 and the series reproduces the initial condition; explicit numerical verification of $B_k$ is deferred to another lecture.

Visual events · 15

Blackboard layout for the heat equation lecture

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A blackboard fills the background with the title Video 8.3: Heat equation, the equation ∂u/∂t = ∂^2u/∂ $x^2$ , and the solution form $u(t,x)=e^{λt}S(x)$ followed by the substituted equation.

Objects

Blackboard text 'Video 8.3: Heat equation'.
Equation ∂u/∂t = ∂^2u/∂ $x^2$ .
Solution form $u(t,x)=e^{λt}S(x)$ .
Substituted equation $λ e^{λt}S=e^{λt}$ ∂S/∂ $x^2$ .
Lecturer standing at the board.

Changes

The lecturer points to the time derivative and spatial derivative terms while explaining the equation structure.
He gestures to the separated solution form and labels the exponential and $S(x)$ factors as eigenvalue and eigenfunction parts.
He draws cancellation marks through the $e^{λt}$ factors on both sides of the substituted equation.
He begins writing a new line below the substituted equation near the end of the clip.

Invariants

The main heat equation remains visible throughout the clip.
The separated solution form remains visible throughout the clip.

Interpretation

The visual sequence tracks the algebraic reduction from the PDE to a separated eigenvalue equation: define the equation, propose a product solution, substitute, then cancel the common exponential factor.

Visual cancellation of the common exponential factor

Clear evidence

Shown in the video

Evidence

Animation
Observation
Diagonal chalk strokes cross out $e^{λt}$ on both sides of $λ e^{λt}S=e^{λt}$ ∂ $S/x^2$ .

Objects

Term $e^{λt}$ on the left side of the substituted equation.
Term $e^{λt}$ on the right side of the substituted equation.
Diagonal cancellation marks.

Changes

Both exponential factors are marked for removal from the equation.

Invariants

The remaining factors λ, S, and the derivative expression for S stay on the board.

Interpretation

The animation represents division or cancellation by the common nonzero factor $e^{λt}$ , leaving an equation involving only λ and the spatial function S.

Initial blackboard layout

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A blackboard fills the frame with the title “Video 8.3: Heat equation,” the PDE ∂u/∂t = ∂²u/∂x², and the separated solution template $u(t,x)=e^{λt}S(x)$ .

Objects

Blackboard text for the heat equation
Separated solution formula $u(t,x)=e^{λt}S(x)$
Lecturer standing at the board

Changes

The lecturer points to the existing formulas while speaking about λ and S.

Invariants

The heat equation itself remains displayed unchanged.
The separated form $e^{λt}S(x)$ remains the reference structure.

Interpretation

The visual setup establishes that the lesson is solving the heat equation by separating time and space factors.

Writing the sine eigenfunction and its eigenvalue

Clear evidence

Shown in the video

Evidence

Animation
Observation
The lecturer writes $S(x)=\sin kπx$ and then writes $λ=-k^2π^2$ beneath it.

Objects

Formula $S(x)=\sin kπx$
Formula $λ=-k^2π^2$
Chalk in the lecturer’s hand

Changes

A new spatial eigenfunction is added below the earlier solution template.
The corresponding eigenvalue is then written directly underneath.

Invariants

The original heat equation and separated solution form remain visible above.

Interpretation

The board progression shows the concrete eigenpair selected for the spatial part of the solution.

Building the series solution on the board

Clear evidence

Shown in the video

Evidence

Animation
Observation
The lecturer writes “General sol $u(t,x)$ =” and then adds a summation symbol with $k=1$ below and infinity above, followed by $B_k$ , an exponential factor, and a sine factor.

Uncertainties

The exact final handwritten layout is partly obscured during writing, but the spoken formula clarifies the intended expression.

Objects

Label “General sol”
Summation sign $\sum$ with limits $k=1$ to $\infty$
Coefficient $B_k$
Exponential factor $e^{-k^2π^2 t}$
Sine factor $\sin (kπx)$

Changes

The lecture moves from a single eigenpair to an indexed family of solutions.
Coefficients $B_k$ are introduced to weight each mode.

Invariants

The underlying separated structure $e^{λt}S(x)$ continues to determine each term in the sum.

Interpretation

The visual sequence demonstrates superposition: many basic separated solutions are combined into one general solution.

Blackboard organization of the heat-equation solution

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A blackboard fills the background with the title "Video 8.3: Heat equation" and the equation $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$ at the top.
Diagram
Observation
Below are lines for separated solutions, a general solution sum, and a brace listing $S(x)=\sin k\pi x$ and $\lambda=-k^2\pi^2$ .

Objects

Title "Video 8.3: Heat equation"
Equation $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$
Separated solution $u(t,x)=e^{\lambda t}S(x)$
Crossed-out eigenvalue-relation writing
General solution $\sum_{k=1}^{\infty}B$ _ke^{-k^ $2\pi^2t$ } $S_k$
Braced list $S(x)=\sin k\pi x$ and $\lambda=-k^2\pi^2$

Changes

The lecturer points among the eigenfunction/eigenvalue list, the separated solution, and the general solution.
Around 118 seconds, the lower blackboard panel is pushed upward, moving the main writing higher in the frame.

Invariants

The displayed heat equation remains visible throughout the clip.
The general solution and the braced sine/eigenvalue list remain on the board throughout the clip.

Interpretation

The board organizes the lesson from PDE to separated solutions to eigenfunction data to the superposed general solution, supporting the spoken explanation of modal decay and boundary conditions.

Hand demonstration of the physical bar and endpoint evaluation

Clear evidence

Shown in the video

Evidence

Animation
Observation
The lecturer uses his hands to indicate a bar and its ends while describing temperature zero at the ends and heat flowing out.
Animation
Observation
He gestures toward the sine formula while evaluating $x=0$ and $x=1$ .

Objects

Lecturer's hands
Mental image of a material bar
Endpoints corresponding to $x=0$ and $x=1$
Displayed sine eigenfunction $\sin k\pi x$

Changes

His hands move apart and inward to represent the bar and its ends.
He points to the sine expression while discussing endpoint values.

Invariants

The verbal boundary condition remains that both ends are kept at temperature zero.
The spatial eigenfunction under discussion remains $\sin(k\pi x)$ .

Interpretation

The gesture sequence connects the abstract sine zeros at $x=0$ and $x=1$ to the physical picture of frozen ends through which heat escapes.

Physical sketch of the heated bar

Clear evidence

Shown in the video

Evidence

Diagram
Observation
Lower board shows a horizontal line segment with endpoints labeled 0 and 1, both ends marked f, later annotated $x=0$ and $x=1$ , and $u(0,x)=1$ written above.
Audio
Observation
The lecturer says, "here's my bar from zero to one" and "This way is x."

Objects

Horizontal bar segment
Endpoint labels 0 and 1
Frozen-end markers f
Axis label x
Initial condition $u(0,x)=1$

Changes

The lecturer first draws the bar from 0 to 1.
He then marks both ends with f to indicate frozen ends.
He adds the direction label x and the endpoint labels $x=0$ and $x=1$ .
He writes $u(0,x)=1$ above the bar to specify the initial temperature.

Invariants

The spatial interval remains $0\le x\le 1$ throughout the sketch.
Both ends remain labeled as frozen.

Interpretation

The diagram encodes the boundary and initial data for the heat equation: a unit-length bar with fixed cold ends and uniform initial temperature 1.

Reference back to the upper-board general solution

Clear evidence

Shown in the video

Evidence

Diagram
Observation
Around the middle of the clip the lecturer points up to the boxed general solution on the upper board.
Audio
Observation
He says, "let me remember what the general solution looked like."

Objects

Boxed formula $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t} S_k(x)$
Definitions $S_k(x)=\sin k\pi x$ and $\lambda=-k^2\pi^2$

Changes

The lecturer shifts attention from the lower-board sketch to the upper-board formula.
He points to the boxed expression while discussing how to proceed.

Invariants

The upper-board formula remains unchanged during this reference.

Interpretation

The visual gesture links the concrete initial-value problem on the lower board to the previously derived general series solution above.

Persistent reference board with the general solution

Clear evidence

Shown in the video

Evidence

Diagram
Observation
Throughout the clip, the upper board remains visible with General Sol., the boxed series, $S_k(x)=\sin k\pi x$ , and $\lambda=-k^2\pi^2$ .

Objects

Upper chalkboard text
Boxed formula $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t} S_k$
Side notes $S_k(x)=\sin k\pi x$ and $\lambda=-k^2\pi^2$

Invariants

The top-board formulas stay unchanged during the clip.
They serve as the reference structure reused on the lower board.

Interpretation

The upper board supplies the already-derived modal decomposition, while the lower board applies it to the specific initial condition $u(0,x)=1$ .

Initial condition displayed as both graph and series

Clear evidence

Shown in the video

Evidence

Diagram
Observation
Lower-left graph is labeled Graph and $u(0,x)=1$ , with horizontal axis $f\to x$ and marks $x=0$ , $x=1$ .
Formula
Observation
Below it appears $t=0$ and $\sum_{k=1}^{\infty} B_k\sin k\pi x=1$ .

Objects

Rectangular initial profile
Axis labels $x=0$ and $x=1$
Equation $\sum B_k\sin k\pi x=1$

Invariants

The initial profile is constant and equal to 1 across the bar.
The series contains only sine terms.

Interpretation

The visual pairing shows that the constant initial temperature is being represented by a Fourier sine series.

Stepwise construction of the $t>0$ solution

Clear evidence

Shown in the video

Evidence

Animation
Observation
The lecturer turns to the lower-right board and writes $u(t,x)=\sum_{k=1}^{\infty} B_k e^{-k^2\pi^2 t}\sin(k\pi x)$ step by step.
Audio
Observation
He narrates the construction as the sum of $b_k$ 's times $e^{-k^2\pi^2 t}$ times $\sin (k\pi x)$ .

Objects

Handwriting of $u(t,x)$
Summation symbol
Coefficient $B_k$
Exponential factor
Sine factor

Changes

First the left-hand side $u(t,x)$ is written.
Then the summation and coefficient $B_k$ are added.
Next the exponential decay factor $e^{-k^2\pi^2 t}$ is inserted.
Finally the spatial factor $\sin(k\pi x)$ completes each term.

Invariants

The coefficient $B_k$ comes from the initial condition.
The spatial basis remains $\sin(k\pi x)$ .

Interpretation

The animation of writing makes explicit how the later-time solution is assembled from the initial-condition coefficients and the modal decay factor.

Misconceptions · 9

Confusing Laplace's equation with the heat equation regarding time dependence

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, 'We had Laplace's equation, that was time was not there. Now time comes into the heat equation.'

Misconception

One might think the previously discussed Laplace equation and the heat equation have the same variable structure.

Clarification

The lecturer distinguishes them by stating that Laplace's equation did not involve time, whereas the heat equation includes a time derivative.

Calling the spatial factor an eigenvector in the PDE setting

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, 'Here, it's an eigenfunction because it depends on x. We didn't have x before.'

Misconception

One might carry over the ordinary-systems terminology and treat the x-dependent factor as an eigenvector.

Clarification

In this heat-equation setting, the lecturer identifies the x-dependent factor $S(x)$ as an eigenfunction rather than an eigenvector.

One eigenpair is not the whole solution

Clear evidence

Shown in the video

Evidence

Audio
Observation
After finding one eigenpair, the lecturer explicitly contrasts it with the “general solution” and says one must take combinations of basic solutions.

Misconception

Finding a single separated solution such as $e^{λt}S(x)$ for one k gives only one basic mode, not the full solution of the heat equation.

Clarification

The lecturer emphasizes that because the equation is linear, the general solution is built by summing many such basic solutions with coefficients $B_k$ .

$S_k$ is not fully explicit inside the summation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, "And for $S_k I$ should have written in sine k pi x."
Formula
Observation
The general solution uses $S_k$ , while the braced list separately gives $S(x)=\sin k\pi x$ .

Misconception

Reading the general solution as if $S_k$ alone specifies the spatial function could leave the actual sine form implicit.

Clarification

The lecturer explicitly corrects the notation by saying that $S_k$ should be written as $\sin(k\pi x)$ , which is also listed in the braced eigenfunction formula.

Crossed-out board writing should not be read as a separate final formula

Approximate timing

Shown in the video

Evidence

Diagram
Observation
The board contains crossed-out writing near the eigenvalue relation, including $d^2S$ and $\lambda S$ .
Formula
Observation
The readable retained relation is $\lambda S=\frac{\partial^2 S}{\partial x^2}$ .

Uncertainties

The exact intended pre-correction expression is obscured by chalk crossings.

Misconception

The crossed-out symbols near the eigenvalue relation might be mistaken for additional independent equations.

Clarification

The crossings indicate revision on the board; the usable mathematical content is the eigenvalue relation $\lambda S=S$ '' together with the listed sine eigenfunctions and eigenvalues.

Confusing the general solution with the fully determined solution

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, "the numbers are always found by matching the initial condition."
Formula
Observation
He writes $t=0$ and then the series equal to 1.

Misconception

One might think the boxed series already gives the answer without further work.

Clarification

The clip emphasizes that the coefficients $B_k$ are still unknown until the initial condition $u(0,x)=1$ is matched at $t=0$ .

An infinite series solution is not necessarily unusable

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, 'You may think, well, that's a pretty messy solution because it's an infinite sum. But it's not bad for partial differential equations.'

Misconception

One might think that because the heat-equation solution is an infinite sum, it is too messy to be meaningful.

Clarification

The video stresses that for PDEs this structure is normal and useful: the solution separates into numerical coefficients, rapidly decaying time factors, and simple spatial sine modes.

Do not confuse this with a full Fourier series containing cosines

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer explicitly says, 'This is a Fourier sine series. I have only sines, not cosines here.'

Misconception

One might assume that any Fourier-type coefficient calculation includes both sine and cosine terms.

Clarification

In this example the expansion uses only sine modes $\sin(k\pi x)$ , so the coefficient-finding problem is specifically a Fourier sine series problem.

Mistaking the displayed setup for an actual coefficient computation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says this is the step 'which I didn't take' and identifies it as the Fourier series step of finding the coefficients.
Audio
Observation
He closes by calling Fourier series 'a separate and important question.'

Misconception

One might think the board already computes the values of the coefficients $B_k$ .

Clarification

In this clip the lecturer only sets up the matching equation $\sum_{k=1}^{\infty} B_k \sin k\pi x = 1$ and states that Fourier series is the method for finding the $B_k$ 's; the computation itself is deferred.

Concept relations · 25

Heat equation definition → Separated solution ansatz

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer first states the heat equation and then says he hopes to separate the two parts and look for solutions like $e^{λt}S(x)$ .

Prerequisite

Explanation

The separated solution form is introduced specifically as a method for solving the heat equation defined earlier in the segment.

Separated solution ansatz → Substituting the separated form into the heat equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says he substitutes the proposed solution into the differential equation.
Formula
Observation
The board moves from $u(t,x)=e^{λt}S$ (x} to $λ e^{λt}S=e^{λt}$ ∂S/∂ $x^2$ .

Application

Explanation

The product ansatz is applied by direct substitution into the heat equation to derive an equation for S.

Substituting the separated form into the heat equation → Canceling the exponential factor to obtain an eigenvalue equation

Clear evidence

Shown in the video

Evidence

Animation
Observation
Cancellation marks are drawn only after the substituted equation is present on the board.

Prerequisite

Explanation

The cancellation step depends on having first substituted the separated form into the heat equation and obtained a common exponential factor on both sides.

Separated solution ansatz → Analogy between PDE separation and ODE eigenvalue solutions

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer compares the current setup to ordinary systems of differential equations with $e^{λt}$ and eigenvectors, then says here the factor is an eigenfunction.

Contrast

Explanation

The separated PDE solution resembles the ODE eigenvalue solution form, but the spatial object changes from an eigenvector to an eigenfunction.

Separated solutions of the heat equation → Sine eigenfunction for the spatial operator

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board first shows $u(t,x)=e^{λt}S(x)$ , then specializes $S(x)=\sin kπx$ and $λ=-k^2π^2$ .

Contains

Explanation

The general separated form contains the more specific choice of sine eigenfunctions used in this example.

Sine eigenfunction for the spatial operator → Eigenvalue produced by two derivatives of $\sin (kπx)$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer asks for the eigenvalue after writing the sine function and then computes it.
Formula
Observation
$S(x)=\sin kπx$ is paired with $λ=-k^2π^2$ on the board.

Proof dependency

Explanation

The eigenvalue is obtained by applying the second derivative operator to the sine eigenfunction.

Eigenvalue produced by two derivatives of $\sin (kπx)$ → General solution as a linear combination of separated solutions

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the previous result was a simple pair and now forms the general solution by taking combinations.
Formula
Observation
The series uses $e^{-k^2π^2 t}$ and $\sin (kπx)$ , exactly the factors from the eigenpair.

Application

Explanation

The eigenvalue and eigenfunction are substituted into the separated form and then summed to produce the general solution.

Linear combinations of separated solutions give the general solution → General solution as a linear combination of separated solutions

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer explicitly says that if one has a linear equation and several solutions, one takes combinations.

Proof dependency

Explanation

The method of writing an infinite sum relies on the linearity/superposition property stated in the claim.

One-dimensional heat equation → Separated product solutions

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board places the heat equation above the separated solution $u(t,x)=e^{\lambda t}S(x)$ .

Application

Explanation

The separated product form is presented as a method for constructing solutions of the displayed heat equation.

Separated product solutions → Eigenvalue equation for the spatial part

Clear evidence

Shown in the video

Evidence

Formula
Observation
The separated solution contains $S(x)$ , and the adjacent board relation identifies $\lambda S$ with S''.

Proof dependency

Explanation

The spatial factor S in the product solution must satisfy the eigenvalue equation for the second-derivative operator.

Eigenvalue equation for the spatial part → Sine eigenfunctions and negative eigenvalues

Clear evidence

Shown in the video

Evidence

Formula
Observation
The braced list pairs $S(x)=\sin k\pi x$ with $\lambda=-k^2\pi^2$ .

Special case

Explanation

The sine functions with eigenvalues -k^ $2\pi^2$ are the specific eigenfunction-eigenvalue pairs displayed for this heat-equation problem.

Sine eigenfunctions and negative eigenvalues → General solution as a superposition

Clear evidence

Shown in the video

Evidence

Formula
Observation
The general solution sums B_ke^{-k^ $2\pi^2t$ } $S_k$ , and the lecturer clarifies $S_k$ as $\sin(k\pi x)$ .

Contains

Explanation

The general solution is built by superposing the separated modes whose spatial parts are the sine eigenfunctions.

Find an answer · 31

What is the heat equation shown in this lecture segment?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board displays the heat equation at the beginning of the clip.

Knowledge points

Heat equation definition

Why does the lecturer say time comes into the heat equation?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer contrasts Laplace's equation with the heat equation by saying time now comes in.

Knowledge points

Heat equation definition
Confusing Laplace's equation with the heat equation regarding time dependence

What separated form of the solution is proposed for the heat equation?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $u(t,x)=e^{λt}S(x)$ .

Knowledge points

Separated solution ansatz

Why is $S(x)$ called an eigenfunction rather than an eigenvector?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says S is an eigenfunction because it depends on x.

Knowledge points

Separated solution ansatz
Calling the spatial factor an eigenvector in the PDE setting

Why can $e^{λt}$ be canceled after substituting the separated solution?

Clear evidence

Shown in the video

Evidence

Animation
Observation
The lecturer crosses out $e^{λt}$ on both sides of the substituted equation.

Knowledge points

Canceling the exponential factor to obtain an eigenvalue equation
Derivation of the eigenvalue equation from the separated heat-equation ansatz

What equation remains after canceling the exponential factor?

Approximate timing

Shown in the video

Evidence

Audio
Observation
The lecturer says he has an eigenvalue equation and begins to state the second derivative of the function.

Uncertainties

The final simplified equation is not fully completed within the clip.

Knowledge points

Canceling the exponential factor to obtain an eigenvalue equation
Derivation of the eigenvalue equation from the separated heat-equation ansatz

What is λ in the separated heat-equation solution and how is it found?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer asks what the eigenvalue is and then computes it from two derivatives.

Knowledge points

Separated solutions of the heat equation
Eigenvalue produced by two derivatives of $\sin (kπx)$

Why does the lecturer choose sine functions for $S(x)$ ?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the functions S will be sine functions and writes $S(x)=\sin kπx$ .

Knowledge points

Sine eigenfunction for the spatial operator

Why does differentiating $\sin (kπx)$ twice produce a minus sign and k²π²?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer explains that taking two derivatives makes kπ come out twice with a minus sign.

Knowledge points

Eigenvalue produced by two derivatives of $\sin (kπx)$
Derivation of the eigenvalue for $\sin (kπx)$

How is the general solution of the heat equation built from the separated solutions?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says that for a linear equation one takes combinations of basic solutions and writes a sum from $k=1$ to infinity.

Knowledge points

General solution as a linear combination of separated solutions
Linear combinations of separated solutions give the general solution
Construction of the general solution from eigenpairs

What do the coefficients $B_k$ represent in the series solution?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer introduces coefficients called $B_k$ in the series.

Knowledge points

General solution as a linear combination of separated solutions
$B_k$

What heat equation is written on the board in this clip?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Top board line reads $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$ .

Knowledge points

One-dimensional heat equation

Coverage and review notes

Covered · Introduction to the heat equation, its place among basic PDEs, and contrast with Laplace's equation.

Covered · Presentation of the separated solution ansatz $u(t,x)=e^{λt}S(x)$ and analogy with eigenvectors from ordinary systems.

Covered · Substitution of the separated form into the heat equation to obtain an equation involving λ, S, and derivatives of S.

Covered · Cancellation of the common exponential factor and identification of the remaining relation as an eigenvalue equation; the final simplified line is begun but not completed on screen.

Covered · The board shows the heat equation and the separated solution template while the lecturer introduces λ and S.

Covered · The lecturer writes $S(x)=\sin (kπx)$ , asks for the eigenvalue, and derives $λ=-k$ ²π².

Covered · The lecturer summarizes that a family of eigenfunctions and eigenvalues has been found and transitions toward the general solution.

Covered · The lecturer explains superposition for a linear equation and writes the series solution with coefficients $B_k$ .

Covered · Final spoken clarification that the eigenfunction should carry the index k; no new mathematical content beyond the series construction.

Covered · The board displays the heat equation, separated solutions, the general summation, and the sine/eigenvalue list; the lecturer identifies the family of eigenfunctions and eigenvalues and clarifies $S_k$ as sine k pi x.

Covered · The lecturer analyzes the time dependence $e^{-k^2\pi^2t}$ , states that larger k decays faster, and identifies the $k=1$ term as the slowest-decaying displayed mode.

Covered · The lecturer gives the physical interpretation: a material bar with ends kept at temperature zero, heat flowing inside the bar and escaping through the frozen ends.

Covered · The lecturer checks that $\sin(k\pi x)$ vanishes at $x=0$ and $x=1$ , connecting the sine eigenfunctions to the zero-temperature boundary condition; near the end the lower blackboard panel is pushed upward.

Covered · The entire clip presents one continuous worked setup: drawing the bar, stating the initial condition, recalling the general solution, and reducing it to the coefficient-matching equation.

Covered · Lecturer points to the prewritten general solution and explains that finding the coefficients is a Fourier sine series problem matching the initial value 1.

Covered · He writes the explicit $t>0$ solution $u(t,x)=\sum B_k e^{-k^2\pi^2 t}\sin(k\pi x)$ on the lower right board.

Covered · He comments that although the answer is an infinite sum, its structure is manageable because it separates coefficients, rapidly decaying time factors, and x-dependent sine modes.

Covered · He sketches successive temperature profiles decreasing toward zero and states that the steady state is temperature zero.

Covered · Lecture segment: the board shows the heat-equation series solution, the sine eigenpair, the initial-condition matching equation, the positive-time formula, and the lecturer explains that Fourier series determines the coefficients.

Covered · End card with copyright and Creative Commons license text; no mathematical content.

Explore the knowledge in this video

Reviewed subject paths

Heat Equation

The explanation, unpacked.

Chapters

Learning script

Knowledge cards

Heat equation definition

Separated solution ansatz

Eigenvalue and eigenfunction analogy

Substitution into the heat equation

Canceling the exponential factor

Resulting eigenvalue equation

Separated solution form for the heat equation

Sine eigenfunction choice

Eigenvalue from two derivatives

General solution by superposition

Heat equation shown on the board

Separated product solutions

Sine eigenfunctions and eigenvalues

General solution as an infinite superposition

Why larger k decays faster

Slowest-decaying mode is k=1k=1k=1

Physical meaning: heat escaping a frozen-ended bar

Sine modes satisfy zero-temperature endpoints

Heat-equation example on a unit bar

General separated solution recalled from the board

Coefficients are found by matching the initial condition

Derivation of the sine-series matching equation

What is completed and what is left for later

Fourier sine-series step for the heat equation

Detailed learning notes

u(t,x)u(t,x)u(t,x)

t

x

λ

S(x)S(x)S(x)

∂u∂t\frac{\partial u}{\partial t}∂t∂u​

∂2u∂x2\frac{\partial^2 u}{\partial x^2}∂x2∂2u​

λeλtS\lambda e^{\lambda t} SλeλtS

eλt∂S∂x2e^{\lambda t}\frac{\partial S}{\partial x^2}eλt∂x2∂S​

u(t,x)u(t,x)u(t,x)

λ

S(x)S(x)S(x)

Heat equation definition

Separated solution ansatz

Substituting the separated form into the heat equation

Canceling the exponential factor to obtain an eigenvalue equation

Separated solutions of the heat equation

Sine eigenfunction for the spatial operator

Eigenvalue produced by two derivatives of sin⁡(kπx)\sin (kπx)sin(kπx)

General solution as a linear combination of separated solutions

One-dimensional heat equation

Separated product solutions

Eigenvalue equation for the spatial part

General solution as a superposition

Structure of the heat equation

Analogy between PDE separation and ODE eigenvalue solutions

After cancellation the remaining relation is an eigenvalue equation

sin⁡(kπx)\sin (kπx)sin(kπx) is an eigenfunction of d²/dx² with eigenvalue -k²π²

Linear combinations of separated solutions give the general solution

Higher-index modes decay faster

The k=1k=1k=1 mode is the slowest-decaying displayed mode

Sine eigenfunctions satisfy the zero endpoint condition

Physical interpretation of decay as heat leaving the bar

The coefficient-finding step is a Fourier sine series problem

Long-time steady state is zero temperature

A PDE heat problem requires infinitely many modes to match a full initial function

Derivation of the eigenvalue equation from the separated heat-equation ansatz

Derivation of the eigenvalue for sin⁡(kπx)\sin (kπx)sin(kπx)

Construction of the general solution from eigenpairs

From eigenvalue to exponential decay factor

Checking the endpoint values of the sine eigenfunctions

Substitute t=0t=0t=0 into the general solution and match u(0,x)=1u(0,x)=1u(0,x)=1

Deriving the initial-condition equation from the general solution

Reconstructing the later-time solution after fixing BkB_kBk​

Deriving the initial-condition equation from the general series

Slowest-decaying first mode

Warm bar with frozen ends and uniform initial temperature

Example: bar initially at uniform temperature 1

Blackboard layout for the heat equation lecture

Visual cancellation of the common exponential factor

Slowest-decaying mode is $k=1$

$u(t,x)$

$S(x)$

$\frac{\partial u}{\partial t}$

$\frac{\partial^2 u}{\partial x^2}$

$\lambda e^{\lambda t} S$

$e^{\lambda t}\frac{\partial S}{\partial x^2}$

$u(t,x)$

$S(x)$

Eigenvalue produced by two derivatives of $\sin (kπx)$

$\sin (kπx)$ is an eigenfunction of d²/dx² with eigenvalue -k²π²

The $k=1$ mode is the slowest-decaying displayed mode

Derivation of the eigenvalue for $\sin (kπx)$

Substitute $t=0$ into the general solution and match $u(0,x)=1$

Reconstructing the later-time solution after fixing $B_k$

Stepwise construction of the $t>0$ solution

$S_k$ is not fully explicit inside the summation

Sine eigenfunction for the spatial operator → Eigenvalue produced by two derivatives of $\sin (kπx)$

Eigenvalue produced by two derivatives of $\sin (kπx)$ → General solution as a linear combination of separated solutions

Why is $S(x)$ called an eigenfunction rather than an eigenvector?

Why can $e^{λt}$ be canceled after substituting the separated solution?

Why does the lecturer choose sine functions for $S(x)$ ?

Why does differentiating $\sin (kπx)$ twice produce a minus sign and k²π²?

What do the coefficients $B_k$ represent in the series solution?