Method of Characteristics 1: Constant Coefficients

1. Formula
   Observation

   The unknown function is written as u in equations such as au_x + bu_y = 0 and $5u_x + 2u_y = 0$.

2. Audio
   Observation

   The lecturer refers to “the partial derivative of u with respect to x” and “with respect to y.”

u

Unknown scalar function of two variables, implicitly $u(x,y)$, whose first partial derivatives appear in the PDE.

Function of variables x and y; explicit domain not stated in the clip.

1. Formula
   Observation

   Coefficients are written as $a(x,y)$, $b(x,y)$, $c(x,y)$, $f(x,y)$, and the boundary condition uses $u(0,y)=\sin y$.

x, y

Independent variables of the PDE.

1. Formula
   Observation

   $u_x$ appears in all three displayed PDE forms and in $5u_x+2u_y=0$.

2. Audio
   Observation

   The lecturer says “partial derivative of u with respect to x.”

$u_x$

Partial derivative of u with respect to x, equivalently $\partial u/\partial x$.

1. Formula
   Observation

   $u_y$ appears in all three displayed PDE forms and in $5u_x+2u_y=0$.

2. Audio
   Observation

   The lecturer says “partial derivative of u with respect to y.”

$u_y$

Partial derivative of u with respect to y, equivalently $\partial u/\partial y$.

1. Formula
   Observation

   The first case is written as au_x + bu_y = 0, and the lecturer identifies this as the constant-coefficient case.

2. Audio
   Observation

   “where we have constant coefficients.”

a, b

Constant coefficients multiplying $u_x$ and $u_y$ in the first PDE case.

1. Formula
   Observation

   The second and third cases display $a(x,y)u_x + b(x,y)u_y = 0$ and $a(x,y)u_x + b(x,y)u_y + c(x,y)u = f(x,y)$.

2. Audio
   Observation

   “the case where a and b are functions of x and y, so we have variable coefficients.”

$a(x,y)$, $b(x,y)$

Variable coefficient functions multiplying $u_x$ and $u_y$ in the second and third cases.

1. Formula
   Observation

   The third displayed equation contains + $c(x,y)u$.

2. Audio
   Observation

   “some term involving u without any partial derivative.”

$c(x,y)$

Coefficient function multiplying the zeroth-order term u in the general case.

1. Formula
   Observation

   The third displayed equation has right-hand side $f(x,y)$.

2. Audio
   Observation

   “some right hand side which is a function of x and y.”

$f(x,y)$

Inhomogeneous forcing or source term on the right-hand side of the general first-order linear PDE.

1. Formula
   Observation

   The worked example is written as $5u_x + 2u_y = 0$.

2. Audio
   Observation

   “where a is five and b is two.”

$a=5$, $b=2$

Specific constant coefficients chosen for the example PDE.

1. Formula
   Observation

   The second line of the example system is $u(0,y)=\sin y$.

2. Audio
   Observation

   “when x equals zero then the function u along that line is going to be given by sine y.”

$u(0,y)=\sin y$

Initial/boundary condition prescribing the values of u on the line $x=0$.

1. Formula
   Observation

   The rewritten PDE is shown as (5,2)·∇$u = 0$.

2. Audio
   Observation

   The lecturer says this form will allow arguments involving directional derivatives.

1. The clip does not explicitly expand ∇u on screen before it ends.

$\nabla u$

Gradient vector of u, understood from the displayed dot-product form as the vector of first partial derivatives.

1. Formula
   Observation

   The vector (5,2) appears in the rewritten equation (5,2)·∇$u = 0$.

2. Audio
   Observation

   This follows directly from the example coefficients $a=5$ and $b=2$.

(5,2)

Constant vector formed from the coefficients of $u_x$ and $u_y$ in the example PDE.

1. Caption evidence
   Observation

   Title card reads “Method of Characteristics” with subtitle “(A method to solve first order linear PDEs)”.

2. Audio
   Observation

   “it’s the method that we’ve used to solve first order linear partial differential equations.”

The video introduces the method of characteristics as a technique for solving first-order linear partial differential equations. The title card and narration both state this scope explicitly.

1. Applies to first-order linear PDEs.

2. Presented here as a revision/assignment-support series rather than a full course lecture.

1. Formula
   Observation

   Three equations are written in sequence: au_x + bu_y = 0; $a(x,y)u_x + b(x,y)u_y = 0$; $a(x,y)u_x + b(x,y)u_y + c(x,y)u = f(x,y)$.

2. Audio
   Observation

   “I’m going to split it up into three different videos corresponding to these three cases.”

The lecturer organizes the topic into three increasing levels of generality: constant coefficients and homogeneous; variable coefficients and homogeneous; and the most general displayed case with variable coefficients, a zeroth-order term in u, and a nonzero right-hand side.

1. First equation: a and b are constants.

2. Second equation: a and b depend on x and y.

3. Third equation: also includes $c(x,y)u$ and right-hand side $f(x,y)$.

1. Scope of the method of characteristics in this series

1. Formula
   Observation

   The first equation is highlighted and labeled as case 1: au_x + bu_y = 0.

2. Audio
   Observation

   “the first case is going to be this one here where we have constant coefficients.”

Case 1 is the simplest displayed form, where the coefficients multiplying the partial derivatives do not depend on x or y and the equation is set equal to zero.

1. a and b are constants.

2. No u term without derivatives is present.

3. Right-hand side is 0.

1. Three displayed cases of first-order linear PDEs

1. Formula
   Observation

   The board shows the system {$5u_x + 2u_y = 0$; $u(0,y) = \sin y$}.

2. Audio
   Observation

   “our partial differential equation is five times the partial derivative of u with respect to x plus two times the partial derivative of u with respect to y equals zero… when x equals zero then the function u along that line is going to be given by sine y.”

The lecturer selects a concrete instance of case 1 by taking $a=5$ and $b=2$, then pairs the PDE with the condition $u(0,y)=\sin y$. This sets up an initial-value problem on the line $x=0$.

1. This is a specific choice within the constant-coefficient case.

2. The condition prescribes u on the vertical line $x=0$.

1. Constant-coefficient homogeneous first-order linear PDE

1. Formula
   Observation

   Red text appears listing “Linear, first order, homogeneous, constant coefficients.”

2. Audio
   Observation

   The lecturer explains each property in turn: linear because only multiplication by constants and differentiation occur; first order because only first derivatives appear; homogeneous because all u terms are on one side and 0 is on the other; constant coefficients because 5 and 2 are constants.

The example equation is identified as a linear, first-order, homogeneous PDE with constant coefficients. The narration gives the reason for each label directly from the displayed formula.

1. Linearity refers to operations on u and its derivatives.

2. First order means no second or higher derivatives appear.

3. Homogeneous means the right-hand side is 0 after collecting u-terms.

4. Constant coefficients means 5 and 2 do not depend on x or y.

1. Worked example: constant-coefficient PDE with initial condition

1. Formula
   Observation

   Yellow text writes “Rewrite PDE as (5,2)·∇$u = 0$”.

2. Audio
   Observation

   “the first step… is to rewrite it in a particular form. And this will allow us to use arguments involving directional derivatives.”

1. The clip stops before the lecturer carries out the next step using directional derivatives.

The constant-coefficient PDE $5u_x+2u_y=0$ is recast as (5,2)·∇$u=0$. This reformulation packages the coefficients into a fixed vector and prepares the argument for interpreting the PDE through directional derivatives.

1. Used for the displayed example with coefficients 5 and 2.

2. Intended to connect the PDE to directional-derivative reasoning.

1. Worked example: constant-coefficient PDE with initial condition
2. Classification of the example PDE

1. Formula
   Observation

   Top-left board text lists the problem as {$5u_x + 2u_y = 0$, $u(0,y)=\sin y$}.

2. Caption evidence
   Observation

   Top-right red text reads "Linear, first order, homogeneous, constant coefficients."

The clip presents a first-order linear homogeneous PDE with constant coefficients, together with a boundary condition on $x=0$. The classification is shown directly on the board rather than derived.

1. The PDE is linear in $u_x$ and $u_y$.

2. The order is first because only first partial derivatives appear.

3. It is homogeneous because the right-hand side is 0.

4. The coefficients 5 and 2 are constants.

1. Formula
   Observation

   Above ∇u the lecturer writes ($u_x$,$u_y$).

2. Audio
   Observation

   He says ∇u is the vector whose first component is the partial derivative of u with respect to x and whose second component is the partial derivative of u with respect to y.

For a scalar function $u(x,y)$, the gradient is the vector collecting its first partial derivatives. In this clip it is introduced specifically to explain the compact form of the PDE.

1. u must have partial derivatives with respect to x and y.

2. This statement is for a two-variable setting.

1. Classification of the displayed PDE problem

1. Audio
   Observation

   The lecturer explains that the dot product multiplies the first components and adds the second components multiplied by each other.

2. Formula
   Observation

   The board shows (5,2)·∇$u=0$ and the spoken expansion gives $5u_x+2u_y$.

The clip uses the dot product to rewrite the PDE in vector form. Expanding the product recovers the original scalar equation exactly.

1. Both operands are two-component vectors.

2. The equality is algebraic expansion of the dot product.

1. Gradient of a two-variable function

1. Caption evidence
   Observation

   New yellow text appears: "That is, the directional derivative of u in the direction (5,2) is zero."

2. Audio
   Observation

   The lecturer says that if a vector dotted with the gradient gives the directional derivative, then here the directional derivative of u in direction (5,2) equals zero.

Once the PDE is written as (5,2)·∇$u=0$, the lecturer interprets the left-hand side as the directional derivative of u in the direction (5,2). This is the conceptual reason for rewriting the equation.

1. The interpretation applies to the vector (5,2) as the chosen direction.

2. The equation states that this directional derivative is zero.

1. Expanding (5,2)·∇u

1. Caption evidence
   Observation

   Yellow text states: "Therefore u is constant along curves in the direction (5,2)".

2. Audio
   Observation

   The lecturer says that if the derivative in some direction is zero, then u must be constant along any curve in that direction.

3. Audio
   Observation

   Later he calls these curves characteristic curves and says the plotted black lines show them.

1. The video does not spell out the regularity assumptions needed for the constancy claim beyond the displayed PDE context.

The method step is to infer from a zero directional derivative that the solution u stays constant along curves tangent to the direction (5,2). In the lecture these curves are called characteristic curves.

1. The directional derivative of u in the direction (5,2) is zero.

2. The curves considered lie in that direction.

1. Directional derivative reading of the rewritten PDE

1. Diagram
   Observation

   At 123s the board view changes to a 3D surface plot with axes x, y, and u.

2. Audio
   Observation

   The lecturer says he is showing what the actual solution looks like and identifies the picture as a surface plot of u.

1. The exact formula for the plotted surface is not stated in this clip.

The lecturer jumps ahead to a graphical preview of the solution. The surface plot visualizes u as a function of x and y, while black lines on the surface indicate curves in the direction (5,2) along which u remains constant.

1. The plot is a visual aid, not a derivation of the solution formula in this clip.

1. Constancy along curves in direction (5,2)

1. Audio
   Observation

   “In this series of videos I’m going to speak about the method of characteristics… it’s the method that we’ve used to solve first order linear partial differential equations.”

2. Caption evidence
   Observation

   Subtitle on title card: “(A method to solve first order linear PDEs)”.

Within this video series, the method of characteristics is the method being taught for solving first-order linear partial differential equations.

1. The statement is scoped to the lecture series referenced by the speaker.

2. It concerns first-order linear PDEs, not nonlinear or higher-order equations.

For the class of first-order linear PDEs addressed in the series.

1. Formula
   Observation

   Red annotation lists “Linear, first order, homogeneous, constant coefficients.”

2. Audio
   Observation

   The lecturer justifies each descriptor from the displayed equation $5u_x+2u_y=0$.

The equation $5u_x+2u_y=0$ is linear, first order, homogeneous, and has constant coefficients.

1. The equation is read exactly as displayed on the board.

2. Coefficients 5 and 2 are treated as constants.

For the specific displayed example PDE.

1. Formula
   Observation

   Board text: “Rewrite PDE as (5,2)·∇$u = 0$”.

2. Audio
   Observation

   “this will allow us to use arguments involving directional derivatives.”

1. The actual directional-derivative argument is not yet shown in this clip.

Rewriting $5u_x+2u_y=0$ as (5,2)·∇$u=0$ is intended to enable arguments involving directional derivatives.

1. The example coefficients are 5 and 2.

2. ∇u denotes the gradient of u.

For the displayed example at the end of the clip.

1. Audio
   Observation

   The lecturer expands the dot product and says that looking back at the original equation, that is exactly what they had before.

2. Formula
   Observation

   Board shows both $5u_x+2u_y=0$ and (5,2)·∇$u=0$.

For this problem, (5,2)·∇$u=0$ is just a rewriting of $5u_x+2u_y=0$.

1. u is a function of x and y.

2. ∇$u=(u_x,u_y)$.

For the displayed u and the fixed vector (5,2).

1. Caption evidence
   Observation

   The board text says: "Therefore u is constant along curves in the direction (5,2)".

2. Audio
   Observation

   The lecturer says that if we have a derivative in some direction that's zero, then that implies the function itself, u, must be constant along any curve that is in that direction.

1. The clip does not state differentiability or smoothness assumptions explicitly.

2. The argument is presented as a lecture claim rather than a fully formal proof.

If the directional derivative of u in the direction (5,2) is zero, then u is constant along curves in the direction (5,2).

1. (5,2)·∇$u=0$.

2. The curves under consideration lie in the direction (5,2).

For curves in the direction (5,2) in the setting of the displayed PDE.

1. Audio
   Observation

   The speaker says u is constant along each characteristic curve and stays constant if you keep going in the direction 5,2.

2. Caption evidence
   Observation

   The board states: Therefore u is constant along curves in the direction (5,2).

If (5,2)$\cdot \nabla u = 0$, then u is constant along curves in the direction (5,2).

1. The PDE is written as (5,2)$\cdot \nabla u = 0$.

2. One moves along curves in the direction (5,2).

For each characteristic curve in the direction (5,2), u takes a particular constant value.

1. Audio
   Observation

   The speaker says these really are completely parallel lines.

2. Diagram
   Observation

   The contour plot shows evenly spaced parallel diagonal bands.

The characteristic curves shown in the top-down plot are parallel straight lines.

1. The direction is fixed as (5,2).

2. The displayed solution has constant coefficients.

All displayed characteristic lines share the same direction and slope.

1. Formula
   Observation

   The board moves from $5u_x+2u_y=0$ to (5,2)·∇$u=0$.

2. Audio
   Observation

   The lecturer calls this “the first step… to rewrite it in a particular form.”

1. The intermediate expansion of ∇$u=(u_x,u_y)$ is not explicitly written before the clip ends.

1. Expression

   $$5u_x + 2u_y = 0$$
   Explanation

   Start from the example PDE written on the board.

   Justification

   Given problem statement.

   Shown in the video
2. Expression

   $$\nabla u = (u_x,u_y)$$
   Explanation

   Identify the gradient of u in two variables as the vector of its first partial derivatives.

   Justification

   Standard definition of the gradient; not explicitly written on screen in this clip.

   Supplementary explanation
3. Expression

   $$(5,2)\cdot \nabla u = 0$$
   Explanation

   Package the constant coefficients 5 and 2 into a vector and take their dot product with the gradient.

   Justification

   Algebraic rewriting of the left-hand side using the gradient notation.

   Shown in the video

The example PDE is equivalently expressed as (5,2)·∇$u=0$, setting up directional-derivative reasoning.

1. Audio
   Observation

   The lecturer verbally expands the dot product term by term.

2. Formula
   Observation

   The board contains (5,2)·∇$u=0$ and the earlier PDE $5u_x+2u_y=0$.

1. Expression

   $$\nabla u=(u_x,u_y)$$
   Explanation

   Introduce the gradient of u as the vector of its partial derivatives.

   Justification

   Definition of the gradient in two variables, stated by the lecturer.

   Shown in the video
2. Expression

   $$(5,2)\cdot\nabla u=(5,2)\cdot(u_x,u_y)$$
   Explanation

   Substitute the explicit form of ∇u into the dot product.

   Justification

   Direct substitution using the definition above.

   Shown in the video
3. Expression

   $$(5,2)\cdot(u_x,u_y)=5u_x+2u_y$$
   Explanation

   Expand the dot product by multiplying corresponding components and adding.

   Justification

   Definition of the Euclidean dot product.

   Shown in the video
4. Expression

   $$5u_x+2u_y=0$$
   Explanation

   Set the expanded expression equal to zero to recover the original PDE.

   Justification

   The displayed vector equation has right-hand side 0.

   Shown in the video

The compact equation (5,2)·∇$u=0$ is exactly the same as the original PDE $5u_x+2u_y=0$.

1. Caption evidence
   Observation

   Yellow explanatory text appears stating the directional derivative interpretation.

2. Audio
   Observation

   The lecturer explains why the rewriting was useful and connects the dot product with the gradient to the directional derivative.

1. The clip does not separately prove the general directional-derivative identity; it invokes it as known background.

1. Expression

   $$(5,2)\cdot\nabla u=0$$
   Explanation

   Start from the rewritten PDE in vector form.

   Justification

   This is the displayed equation after the previous derivation.

   Shown in the video
2. Expression

   $$D_{(5,2)}u=0$$
   Explanation

   Interpret the dot product of a direction vector with the gradient as the directional derivative of u in that direction.

   Justification

   Standard directional-derivative identity invoked by the lecturer.

   Shown in the video
3. Expression

   $$u\ \text{is constant along curves in the direction }(5,2)$$
   Explanation

   Conclude that u does not change when moving along curves tangent to (5,2).

   Justification

   If the derivative in a given direction is zero, the function is constant along curves in that direction, as stated in the lecture.

   Shown in the video

The PDE implies that u remains constant along characteristic curves in the direction (5,2).

1. Audio
   Observation

   The speaker derives $dy/dx$ from change in y over change in x, then integrates both sides with respect to x.

2. Caption evidence
   Observation

   The board shows $dy/dx = 2/5$ and then $y = 2/5 x + c$.

1. Expression

   $$\frac{dy}{dx} = \frac{2}{5}$$
   Explanation

   The slope of each characteristic line is read from the direction (5,2): 2 units up for every 5 units across.

   Justification

   Observed from the spoken explanation and the annotated triangle on the contour plot.

   Shown in the video
2. Expression

   $$\int \frac{dy}{dx}\,dx = \int \frac{2}{5}\,dx$$
   Explanation

   Both sides of the ODE are integrated with respect to x.

   Justification

   The speaker explicitly says all we need to do is integrate both sides with respect to x.

   Shown in the video
3. Expression

   $$y = \frac{2}{5}x + c$$
   Explanation

   The integration yields the equation of a straight line with slope $2/5$ and arbitrary constant c.

   Justification

   Standard integration of a constant right-hand side; the constant c labels the family of parallel characteristics.

   Shown in the video

The characteristic curves are the straight lines $y = \frac{2}{5}x + c$.

1. Audio
   Observation

   The lecturer says, 'we can just rearrange this equation here to get that c is y minus two fifths of x'.

2. Formula
   Observation

   The term ($y - 2/5 x$) appears inside the function f in the general solution formula.

1. Expression

   $$y = \frac{2}{5}x + c$$
   Explanation

   Start with the equation of the characteristic curves.

   Justification

   Definition of characteristic curves for direction (5,2).

   Shown in the video
2. Expression

   $$c = y - \frac{2}{5}x$$
   Explanation

   Isolate the constant c by subtracting ($2/5$)x from both sides.

   Justification

   Algebraic manipulation.

   Shown in the video

The constant c, which labels the characteristic curve, can be expressed in terms of the coordinates x and y as $c = y - (2/5)x$.

1. Formula
   Observation

   The system {$5u_x + 2u_y = 0$, $u(0,y) = \sin y$} is shown, followed by the step-by-step substitution $u(0,y) = f(y - 2/5 * 0) = f(y) = \sin y$.

2. Audio
   Observation

   The lecturer explains substituting $x=0$ into the general solution and equating it to the boundary condition to find f.

1. Expression

   $$u(x,y) = f\left(y - \frac{2}{5}x\right)$$
   Explanation

   State the general solution derived from the method of characteristics.

   Justification

   General solution form for the PDE.

   Shown in the video
2. Expression

   $$u(0,y) = f\left(y - \frac{2}{5}(0)\right)$$
   Explanation

   Apply the boundary condition by substituting $x = 0$ into the general solution.

   Justification

   Substitution based on the given boundary condition $u(0,y) = \sin y$.

   Shown in the video
3. Expression

   $$u(0,y) = f(y)$$
   Explanation

   Simplify the expression inside the function f.

   Justification

   Arithmetic simplification: ($2/5$)*$0 = 0$.

   Shown in the video
4. Expression

   $$f(y) = \sin y$$
   Explanation

   Equate the result of the substitution to the specified boundary condition.

   Justification

   Given boundary condition $u(0,y) = \sin y$.

   Shown in the video
5. Expression

   $$f(s) = \sin s$$
   Explanation

   Identify the function f itself.

   Justification

   Since $f(y) = \sin y$ for any y, the function f is the sine function.

   Derived from the video

The arbitrary function f in the general solution is determined to be the sine function, i.e., $f(z) = \sin z$ for any argument z.

1. Formula
   Observation

   $u(0,y) = f(y) = \sin y$

2. Formula
   Observation

   $u(x,y) = \sin(y - \frac{2}{5}x)$

1. Expression

   $$u(x,y) = f\left(y - \frac{2}{5}x\right)$$
   Explanation

   Start with the general solution derived from the method of characteristics.

   Justification

   Given in the video as the general form of the solution.

   Shown in the video
2. Expression

   $$u(0,y) = f\left(y - \frac{2}{5}(0)\right) = f(y)$$
   Explanation

   Substitute $x=0$ into the general solution to apply the boundary condition.

   Justification

   Algebraic substitution.

   Derived from the video
3. Expression

   $$f(y) = \sin y$$
   Explanation

   Equate the result to the given boundary condition $u(0,y) = \sin y$ to find the specific function f.

   Justification

   Given boundary condition.

   Shown in the video
4. Expression

   $$u(x,y) = \sin\left(y - \frac{2}{5}x\right)$$
   Explanation

   Substitute the specific function f back into the general solution to obtain the final explicit solution.

   Justification

   Function composition.

   Shown in the video

The unique solution satisfying both the PDE and the boundary condition is $u(x,y) = \sin(y - \frac{2}{5}x)$.

1. Formula
   Observation

   The board displays {$5u_x + 2u_y = 0$; $u(0,y)=\sin y$}.

2. Audio
   Observation

   The lecturer states $a=5$, $b=2$ and explains the boundary condition on $x=0$.

3. Formula
   Observation

   Later the same example is rewritten as (5,2)·∇$u=0$.

1. No final solution $u(x,y)$ is obtained within this clip.

Consider the first-order linear PDE $5u_x+2u_y=0$ together with the condition $u(0,y)=\sin y$.

1. PDE: $5u_x+2u_y=0$.

2. Condition: $u(0,y)=\sin y$.

3. This is a specific instance of the constant-coefficient case with $a=5$ and $b=2$.

Set up the problem and begin the method of characteristics by rewriting the PDE in a form suitable for directional-derivative arguments.

1. Expression

   $$\begin{cases}5u_x + 2u_y = 0\\ u(0,y)=\sin y\end{cases}$$
   Explanation

   Write the PDE and its initial condition as a system.

   Justification

   Directly displayed on the board and described in the audio.

   Shown in the video
2. Expression

   $$\text{Linear, first order, homogeneous, constant coefficients}$$
   Explanation

   Classify the PDE before solving it.

   Justification

   The lecturer explicitly names these properties and justifies them from the formula.

   Shown in the video
3. Expression

   $$(5,2)\cdot \nabla u = 0$$
   Explanation

   Rewrite the differential operator using the coefficient vector and gradient.

   Justification

   Stated as the first step toward using directional derivatives.

   Shown in the video

The clip ends after the setup stage: the example is classified and rewritten as (5,2)·∇$u=0$, but the solution is not completed here.

Verification is limited to consistency checks within the clip: the rewritten vector (5,2) matches the coefficients in $5u_x+2u_y=0$, and the classification matches the displayed equation.

1. Diagram
   Observation

   A 3D surface plot replaces the board view; axes are labeled x, y, and u.

2. Audio
   Observation

   The lecturer says he will show what the actual solution looks like and describes the plot as a surface plot of u.

3. Audio
   Observation

   He says the black plotted curves are characteristic curves along which the solution remains constant.

1. The exact analytic expression for the plotted solution is not given in this clip.

2. The clip cuts off before the lecturer finishes describing the axes.

Show what the solution surface of the PDE looks like and how the constancy claim appears geometrically.

1. The PDE has been rewritten as (5,2)·∇$u=0$.

2. The lecturer claims u is constant along curves in the direction (5,2).

3. A 3D surface plot of $u(x,y)$ is displayed with black curves drawn on it.

Use the plot to visualize the solution and the characteristic curves.

1. Expression

   $$$$
   Explanation

   Switch from the handwritten board to a 3D surface plot of u over the x-y plane.

   Justification

   The lecturer explicitly says he is jumping forward to show the actual solution.

   Shown in the video
2. Expression

   $$$$
   Explanation

   Read the axes: x and y are horizontal coordinates and u is the vertical value of the solution.

   Justification

   The plot labels and the lecturer's description identify the axes.

   Shown in the video
3. Expression

   $$$$
   Explanation

   Observe black curves drawn on the surface; these represent curves in the direction (5,2).

   Justification

   The lecturer says the characteristic curves are plotted in black.

   Shown in the video
4. Expression

   $$$$
   Explanation

   Along each black curve, the surface height does not change, illustrating that u is constant there.

   Justification

   This matches the prior claim that zero directional derivative implies constancy along those curves.

   Derived from the video

The solution is visualized as a wavy surface in (x,y,u)-space, with black characteristic curves indicating directions along which u stays constant.

The visual check is that the black curves lie on level portions of the surface, consistent with the spoken claim of constancy along curves in direction (5,2).

1. Caption evidence
   Observation

   The board displays the system $5u_x + 2u_y = 0$ with $u(0,y) = \sin y$.

2. Diagram
   Observation

   The 3D and contour plots visualize the corresponding solution surface and level bands.

1. The full derivation of the closed-form solution is not shown in this clip; only the characteristic geometry and characteristic equation are derived here.

Analyze the first-order PDE $5u_x + 2u_y = 0$ together with the boundary condition $u(0,y)=\sin y$ by identifying the characteristic direction and characteristic curves.

1. PDE: $5u_x + 2u_y = 0$.

2. Boundary condition: $u(0,y) = \sin y$.

3. The PDE is described on the board as linear, first order, homogeneous, constant coefficients.