PDE | Finite differences: introduction

1. Formula
   Observation

   The symbol f appears in the difference quotient formula and the differential equation example.

f

A function of a single variable x.

Real-valued function of a real variable.

1. Formula
   Observation

   The symbol x appears as the independent variable in the difference quotient and the differential equation.

x

The independent variable of the function f.

Real numbers.

1. Formula
   Observation

   The symbol h appears in the denominator and argument shift of the difference quotient.

h

A small step size used to approximate the derivative.

Small positive real numbers.

1. Formula
   Observation

   The symbol f'(x) appears as the result of the limit definition of the derivative.

f'(x)

The derivative of the function f with respect to x.

Real numbers.

1. Formula
   Observation

   The board repeatedly shows f' = f, $f(0)=1$, and Solution: $f(x)=e^x$.

2. Audio
   Observation

   The speaker distinguishes the actual solution f from the approximate values produced by the finite difference method.

f

The unknown function in the differential equation and its exact solution; later contrasted with discrete approximations.

Function of a continuous variable x.

1. Formula
   Observation

   Visible formulas include f'(x), $f(x+h)$, $f(x)$, and $x_k$ = kh.

x

Continuous independent variable in the differential equation and exact solution.

Continuous variable; grid points are written as $x_k$.

1. Formula
   Observation

   The forward difference formula contains h in the denominator and in $f(x+h)$.

2. Audio
   Observation

   The speaker says to multiply both sides by h and later chooses $h = 0.5$.

h

Step size used in the finite difference approximation and in defining the discrete grid.

Positive step size; in the worked example $h=0.5$.

1. Formula
   Observation

   The board writes $x_k$ = kh and $f_{k+1} = (1+h)f_k$.

2. Audio
   Observation

   The speaker says $x_k$ equals k times h and describes stepping from $f_0$ to $f_1$ to $f_2$.

k

Integer index labeling grid points and sequence terms.

Nonnegative integer index in the recurrence.

1. Formula
   Observation

   The board writes where $x_k$ = kh.

$x_k$

The kth grid point, defined as kh.

Discrete set {0,h,2h,...}.

1. Formula
   Observation

   The board writes $f(x_k) \approx f_k$ and $f_{k+1} = (1+h)f_k$, $f_0 = 1$.

2. Audio
   Observation

   The speaker introduces $f_k$ as an approximation to f at $x_k$.

$f_k$

Discrete approximation to the value of f at the grid point $x_k$.

Sequence indexed by k.

1. Formula
   Observation

   The recurrence block includes $f_0 = 1$.

2. Audio
   Observation

   The speaker says we start off at $f_0$.

$f_0$

Initial term of the discrete approximation sequence.

First sequence value, equal to 1.

1. Formula
   Observation

   The board writes $f_{k+1} = (1+h)f_k$.

$f_{k+1}$

Next term in the discrete approximation sequence.

Defined recursively from $f_k$.

1. Audio
   Observation

   The speaker introduces finite difference methods as a numerical technique for solving differential equations by replacing derivatives with algebraic approximations.

2. Diagram
   Observation

   Text on screen reads 'Finite difference methods' and 'replace derivatives in a differential equation with approximations'.

Finite difference methods are numerical techniques used to solve differential equations. The core idea is to replace every derivative in the differential equation with an algebraic approximation, resulting in a system of purely algebraic equations that can be solved to find an approximate solution.

1. Applicable to differential equations.

2. Produces approximate rather than exact solutions.

1. Audio
   Observation

   The speaker explains that the derivative is defined as the limit of the difference quotient as h approaches zero, and for small h, it can be approximated by the difference quotient itself.

2. Formula
   Observation

   The formula $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f$'(x) is shown, followed by $\frac{f(x+h) - f(x)}{h} \approx f$'(x).

The difference quotient is used to approximate the derivative of a function. For a sufficiently small step size h, the derivative f'(x) is approximately equal to the difference quotient.

1. h must be relatively small.

1. f
2. x
3. h
4. f'(x)

1. Formula
   Observation

   Top right of the board shows ($f(x+h)-f(x)$)/h $\approx f$'(x).

2. Audio
   Observation

   The speaker explains replacing derivatives in a differential equation with approximations and solving the resulting algebraic equations.

The video defines the basic finite-difference idea by approximating the derivative f'(x) with the quotient ($f(x+h)-f(x)$)/h. This converts a differential equation into an algebraic relation involving nearby values of the function.

1. Used to replace derivatives in a differential equation with approximations.

2. h is the step size between neighboring evaluation points.

1. Formula
   Observation

   Left side text reads 'replace derivatives in a differential equation with approximations' and right side text reads 'solve the resulting algebraic equations to get approximate solution'.

2. Audio
   Observation

   The speaker frames the method as turning the differential equation into something solvable incrementally.

The method has two stages shown on the board: first replace derivatives by finite-difference expressions, then solve the resulting algebraic or recurrence equations to obtain an approximate solution.

1. Applies to differential equations when one wants an approximate numerical solution.

2. The replacement step introduces a step size h.

1. Forward difference approximation of a derivative

1. Formula
   Observation

   Lower left shows f' = f and $f(0)=1$, with Solution: $f(x)=e^x$.

2. Audio
   Observation

   The speaker uses this equation as the example being discretized.

The worked example is the initial value problem f'=f with $f(0)=1$. The exact solution displayed on the board is $f(x)=e^x$, which serves as the reference against which the finite-difference approximation is compared.

1. This is the specific differential equation used for demonstration.

1. Formula
   Observation

   After substitution the board shows ($f(x+h)-f(x)$)/h = $f(x)$, then rewrites it as $f(x+h)=(1+h)f(x)$.

2. Audio
   Observation

   The speaker says to multiply both sides by h and add $f(x)$, obtaining $f(x+h)=(1+h)f(x)$.

Substituting the forward difference for f' in f'=f gives ($f(x+h)-f(x)$)/h=$f(x)$. Rearranging yields $f(x+h)=(1+h)f(x)$, a relation that expresses the next value in terms of the current one.

1. Derived from the example ODE f'=f using the forward difference approximation.

1. Forward difference approximation of a derivative
2. Example initial value problem f'=f, $f(0)=1$

1. Audio
   Observation

   The speaker says we can solve the equation incrementally, starting at $f(0)=1$ and then getting $f(0+h)$, $f(h+h)$, etc.

2. Formula
   Observation

   The board later lists $f(0)=1$, $f(h)=1+h$, $f(2h)=(1+h)^2$, etc.

1. The intermediate values are spoken before all are fully written on the board.

Once the equation is rewritten as $f(x+h)=(1+h)f(x)$, values can be generated step by step from the initial condition. Starting from $f(0)=1$ gives $f(h)=1+h$, $f(2h)=(1+h)^2$, and in general f(kh)=($1+h$)^k.

1. Requires the initial value $f(0)=1$.

2. Uses the recurrence $f(x+h)=(1+h)f(x)$.

1. Discretized form of f'=f
2. Example initial value problem f'=f, $f(0)=1$

1. Audio
   Observation

   The speaker says there is a conflict of notation because $f(x)$ denotes the actual solution while $f(0)$, $f(h)$, $f(2h)$ were being used for approximations.

2. Formula
   Observation

   The board still shows Solution: $f(x)=e^x$ on the left while the right side lists approximate values.

The video identifies a notational problem: $f(x)$ is already being used for the exact solution, so writing $f(h)$, $f(2h)$, ... for numerical approximations can be misleading. The speaker explicitly states that $f(h)$ is not equal to $e^h$, only approximately related to it.

1. Arises when comparing the exact solution $f(x)=e^x$ with values produced by the finite difference scheme.

1. Example initial value problem f'=f, $f(0)=1$
2. Incremental solution from the recurrence

1. Formula
   Observation

   The board writes $f(x_k) \approx f_k$ where $x_k$ = kh.

2. Audio
   Observation

   The speaker says instead of calling the approximate values $f(0)$, $f(h)$, etc., we will call them $f_k$, an approximation to f at $x_k$.

To separate exact and approximate quantities, the video introduces grid points $x_k$=kh and writes $f_k$ as the approximation to f at $x_k$. This makes the discrete unknowns a sequence rather than values of the exact function.

1. k indexes the grid points.

2. $f_k$ is an approximation, not the exact value $f(x_k)$.

1. Notation conflict between exact and approximate values

1. Formula
   Observation

   The board replaces the earlier functional form with $f_{k+1}=(1+h)f_k$ and $f_0=1$.

2. Audio
   Observation

   The speaker says the sequence $f_k$ will be defined by the finite difference equation above, determining $f_{k+1}$ from $f_k$.

In the corrected notation, the finite difference method becomes the recurrence $f_{k+1}=(1+h)f_k$ with initial condition $f_0=1$. This defines the whole sequence of approximations from the starting value.

1. Derived from the discretized equation $f(x+h)=(1+h)f(x)$.

2. k is a nonnegative integer index.

1. Discretized form of f'=f
2. Grid-point notation $f_k \approx f(x_k)$

1. Formula
   Observation

   At the end of the clip the board writes h=.5, $f_0=1$, $f_1=1.5$, $f_2=2.25$, $f_3=3.375$.

2. Audio
   Observation

   The speaker says, 'Let's choose h equals 0.5.'

1. The clip ends immediately after introducing the concrete computation, before further explanation.

The video begins a numerical example by choosing $h=0.5$ and listing the first few sequence values generated by the recurrence: $f_0=1$, $f_1=1.5$, $f_2=2.25$, $f_3=3.375$.

1. Uses the recurrence $f_{k+1}=(1+h)f_k$ with $h=0.5$.

1. Recurrence relation for the approximation sequence

1. Formula
   Observation

   Orange text shows $f_{k+1} = (1+h)f_k$ and $f_0 = 1$

2. Audio
   Observation

   Speaker explains plugging in values to get $f_1$, $f_2$, etc.

The video presents a discrete recurrence relation used to approximate the continuous differential equation. By replacing the derivative with a finite difference approximation, the continuous problem is converted into an iterative formula where each new value depends on the previous one.

1. h is the step size

2. $x_k$ = kh

3. $f_k$ approximates $f(x_k)$

1. Audio
   Observation

   The speaker says, 'It's not true that f of h equals e to the h. It's only an approximation. One plus h is approximately equal to e to the h.'

For the example f'=f with $f(0)=1$, the finite-difference value $1+h$ approximates $e^h$ but is not equal to it.

1. The exact solution is $f(x)=e^x$.

2. The finite difference approximation gives the next value as $1+h$ from the initial value 1.

For the chosen step size h in the example.

1. Audio
   Observation

   The speaker sets up the example f' = f with $f(0) = 1$, states the exact solution is $e^x$, and then replaces f' with the difference quotient to form the finite difference equation.

2. Formula
   Observation

   The equations f' = f, $f(0) = 1$, Solution: $f(x) = e^x$, and $\frac{f(x+h) - f(x)}{h} = f(x)$ are written sequentially on the board.

1. Expression

   $$f' = f, \quad f(0) = 1$$
   Explanation

   Start with the given first-order ordinary differential equation and its initial condition.

   Justification

   Problem statement.

   Shown in the video
2. Expression

   $$\frac{f(x+h) - f(x)}{h} = f(x)$$
   Explanation

   Replace the derivative f' with its difference quotient approximation to obtain the finite difference equation.

   Justification

   Definition of the difference quotient approximation.

   Shown in the video

The differential equation is transformed into an algebraic finite difference equation.

1. Audio
   Observation

   The speaker says to multiply both sides by h and add $f(x)$.

2. Formula
   Observation

   The board changes from ($f(x+h)-f(x)$)/h=$f(x)$ to $f(x+h)=(1+h)f(x)$.

1. Expression

   $$\frac{f(x+h)-f(x)}{h}=f(x)$$
   Explanation

   Start from the forward-difference substitution into the ODE f'=f.

   Justification

   Definition of the finite difference approximation applied to the example equation.

   Shown in the video
2. Expression

   $$f(x+h)-f(x)=h f(x)$$
   Explanation

   Multiply both sides by h.

   Justification

   Algebraic equivalence when h is treated as the step size.

   Shown in the video
3. Expression

   $$f(x+h)=(1+h)f(x)$$
   Explanation

   Add $f(x)$ to both sides and factor.

   Justification

   Basic algebraic rearrangement.

   Shown in the video

The discretized equation can be written as $f(x+h)=(1+h)f(x)$.

1. Audio
   Observation

   The speaker says we can start at $f(0)=1$ and use the finite difference equation to get $f(0+h)$, then $f(h+h)$, etc.

2. Formula
   Observation

   The board later displays $f(0)=1$, $f(h)=1+h$, $f(2h)=(1+h)^2$, etc.

1. The spoken derivation precedes some of the written values appearing on the board.

1. Expression

   $$f(0)=1$$
   Explanation

   Use the given initial condition.

   Justification

   Problem statement for the example IVP.

   Shown in the video
2. Expression

   $$f(h)=(1+h)f(0)=1+h$$
   Explanation

   Apply the recurrence at $x=0$.

   Justification

   From $f(x+h)=(1+h)f(x)$.

   Shown in the video
3. Expression

   $$f(2h)=(1+h)f(h)=(1+h)^2$$
   Explanation

   Apply the recurrence again at $x=h$.

   Justification

   Iterating the same recurrence relation.

   Shown in the video
4. Expression

   $$f(kh)=(1+h)^k$$
   Explanation

   Continue the pattern for k steps.

   Justification

   Repeated multiplication by the factor ($1+h$).

   Derived from the video

Starting from $f(0)=1$, the finite difference scheme produces f(kh)=($1+h$)^k.

1. Formula
   Observation

   The board introduces $x_k$=kh and $f(x_k)\approx f_k$, then rewrites the update as $f_{k+1}=(1+h)f_k$ with $f_0=1$.

2. Audio
   Observation

   The speaker says the sequence $f_k$ will be defined by the finite difference equation above.

1. Expression

   $$x_k=kh$$
   Explanation

   Define the grid points used by the method.

   Justification

   Introduction of discrete notation to avoid confusion with the exact function.

   Shown in the video
2. Expression

   $$f(x_k)\approx f_k$$
   Explanation

   Let $f_k$ denote the approximation to f at $x_k$.

   Justification

   New notation stated on the board and in the audio.

   Shown in the video
3. Expression

   $$f_{k+1}=(1+h)f_k,\quad f_0=1$$
   Explanation

   Translate $f(x+h)=(1+h)f(x)$ into recurrence form for the sequence.

   Justification

   Substitute $x=x_k$ and use the definitions of $x_k$ and $f_k$.

   Shown in the video

The finite difference method is equivalently expressed as the recurrence $f_{k+1}=(1+h)f_k$ with $f_0=1$.

1. Audio
   Observation

   Speaker narrates the arithmetic: 'I plug in f zero equals one... times one plus point five... that's one point five'

2. Formula
   Observation

   Green text lists $f_0=1$, $f_1=1.5$, $f_2=2.25$, $f_3=3.375$

1. Expression

   $$f_0 = 1$$
   Explanation

   Start with the given initial condition.

   Justification

   Given in the problem statement.

   Shown in the video
2. Expression

   $$f_1 = (1 + 0.5)f_0 = 1.5(1) = 1.5$$
   Explanation

   Substitute $k=0$ and $h=0.5$ into the recurrence relation.

   Justification

   Definition of the finite difference recurrence.

   Shown in the video
3. Expression

   $$f_2 = (1 + 0.5)f_1 = 1.5(1.5) = 2.25$$
   Explanation

   Substitute $k=1$ and the newly calculated $f_1$.

   Justification

   Recursive application of the formula.

   Shown in the video
4. Expression

   $$f_3 = (1 + 0.5)f_2 = 1.5(2.25) = 3.375$$
   Explanation

   Substitute $k=2$ and the newly calculated $f_2$.

   Justification

   Recursive application of the formula.

   Shown in the video

The sequence of approximate values is generated iteratively: 1, 1.5, 2.25, 3.375.

1. Audio
   Observation

   The speaker walks through the example of converting f' = f into a finite difference equation.

2. Formula
   Observation

   The equations f' = f, $f(0) = 1$, and $\frac{f(x+h) - f(x)}{h} = f(x)$ are displayed.

Convert the differential equation f' = f with initial condition $f(0) = 1$ into a finite difference equation using the difference quotient approximation.

1. Differential equation: f' = f

2. Initial condition: $f(0) = 1$

3. Approximation: f' $\approx \frac{f(x+h) - f(x)}{h}$

Find the corresponding finite difference equation.

1. Expression

   $$f' = f$$
   Explanation

   Identify the derivative term in the differential equation.

   Justification

   Given problem.

   Shown in the video
2. Expression

   $$\frac{f(x+h) - f(x)}{h} = f(x)$$
   Explanation

   Substitute the difference quotient for f'.

   Justification

   Difference quotient approximation.

   Shown in the video

The finite difference equation is $\frac{f(x+h) - f(x)}{h} = f(x)$ with $f(0) = 1$.

The speaker notes that the exact solution to the original ODE is $f(x) = e^x$, which serves as a benchmark for the numerical approximation.

1. Formula
   Observation

   The board writes h=.5, $f_0=1$, $f_1=1.5$, $f_2=2.25$, $f_3=3.375$.

2. Audio
   Observation

   The speaker says, 'I'd like to conclude with a concrete computation for a particular value of h. Let's choose h equals 0.5.'

1. The clip stops before the speaker explains the listed numbers in detail.

Compute the first few finite-difference approximations for f'=f, $f(0)=1$ using $h=0.5$.

1. f'=f

2. $f(0)=1$

3. $h=0.5$

4. $f_{k+1}=(1+h)f_k$

Find $f_0$,$f_1$,$f_2$,$f_3$ for the chosen step size.

1. Expression

   $$f_0=1$$
   Explanation

   Use the initial condition.

   Justification

   Given starting value of the recurrence.

   Shown in the video
2. Expression

   $$f_1=(1+0.5)f_0=1.5$$
   Explanation

   Apply the recurrence once.

   Justification

   From $f_{k+1}=(1+h)f_k$ with $h=0.5$.

   Shown in the video
3. Expression

   $$f_2=(1+0.5)f_1=2.25$$
   Explanation

   Apply the recurrence again.

   Justification

   Iterating the same recurrence relation.

   Shown in the video
4. Expression

   $$f_3=(1+0.5)f_2=3.375$$
   Explanation

   Apply the recurrence a third time.

   Justification

   Iterating the same recurrence relation.

   Shown in the video

$f_0=1$,\ $f_1=1.5$,\ $f_2=2.25$,\ $f_3=3.375$

Each value follows directly from multiplying the previous value by 1.5, matching the displayed board entries.

1. Formula
   Observation

   Left side shows f'=f, $f(0)=1$; Right side shows the discrete calculations and comparison table

2. Audio
   Observation

   Speaker walks through the entire example from setup to error comparison

Approximate the solution to the differential equation f' = f with initial condition $f(0) = 1$ using a step size of $h = 0.5$, and compare the results to the exact solution.

1. Differential equation: f' = f

2. Initial condition: $f(0) = 1$

3. Step size: $h = 0.5$

4. Exact solution: $f(x) = e^x$

Compute discrete approximations $f_k$ and evaluate their accuracy against the true function values.

1. Expression

   $$f_{k+1} = (1+h)f_k$$
   Explanation

   Set up the finite difference recurrence relation based on the forward difference approximation of the derivative.

   Justification

   Standard finite difference discretization for ODEs.

   Shown in the video
2. Expression

   $$f_0 = 1, f_1 = 1.5, f_2 = 2.25, f_3 = 3.375$$
   Explanation

   Calculate the first few terms of the sequence iteratively.

   Justification

   Arithmetic substitution into the recurrence.

   Shown in the video
3. Expression

   $$f(0)=1, f(0.5)≈1.65, f(1)≈2.72, f(1.5)≈4.48$$
   Explanation

   Evaluate the exact solution $e^x$ at the corresponding grid points $x_k = 0$, 0.5, 1, 1.5.

   Justification

   Direct evaluation of the known analytical solution.

   Shown in the video
4. Expression

   $$Compare f_k vs f(x_k)$$
   Explanation

   Observe that the approximate values are consistently lower than the exact values, and the gap widens as x increases.

   Justification

   Visual inspection of the plotted points and the numerical table.

   Shown in the video

The finite difference method produces the sequence {1, 1.5, 2.25, 3.375}, which underestimates the true values of $e^x$ at the respective grid points, demonstrating accumulating truncation error.

The speaker explicitly compares the calculated $f_k$ values to the evaluated $e^{x_k}$ values both numerically in a table and visually on a graph.

1. Animation
   Observation

   Handwritten text and formulas appear sequentially on a black background as the speaker talks.

1. Black background

2. White and colored handwritten text

3. Formulas

1. Text 'Finite difference methods' appears.

2. Text 'replace derivatives...' appears.

3. Arrow and 'solve the resulting...' text appear.

4. Limit definition of derivative appears.

5. Approximation formula appears.

6. Example ODE and solution appear.

7. Finite difference equation for the example appears.

1. The black background remains constant.

2. The handwriting style remains consistent.

The visual progression mirrors the logical flow of the lecture, introducing concepts before applying them to an example.

1. Diagram
   Observation

   The blackboard is organized into a title at top left, a general formula at top right, explanatory text across the middle, and the example ODE with its transformation below.

1. Title 'Finite difference methods'

2. General formula ($f(x+h)-f(x)$)/h $\approx f$'(x)

3. Text 'replace derivatives in a differential equation with approximations'

4. Text 'solve the resulting algebraic equations to get approximate solution'

5. Example ODE f'=f, $f(0)=1$

6. Solution $f(x)=e^x$

1. The lower-right portion is progressively rewritten from functional notation to sequence notation.

2. A concrete numerical list for $h=0.5$ appears near the end.

1. The title and the general forward-difference formula remain visible throughout most of the clip.

2. The left-side exact solution $f(x)=e^x$ remains on the board as a reference.

The layout visually separates the general method, the example differential equation, the exact solution, and the evolving discrete approximation scheme.

1. Animation
   Observation

   The board transitions from $f(x+h)=(1+h)f(x)$ and a list of values to the cleaner notation $f(x_k)\approx f_k$, $x_k$=kh, and then $f_{k+1}=(1+h)f_k$, $f_0=1$.

1. $f(x+h)=(1+h)f(x)$

2. $f(0)=1$, $f(h)=1+h$, $f(2h)=(1+h)^2$, etc.

3. $f(x_k)\approx f_k$

4. $x_k$=kh

5. $f_{k+1}=(1+h)f_k$

6. $f_0=1$

1. The earlier value list is replaced by indexed sequence notation.

2. The recurrence is restated in terms of k rather than x.

1. The multiplicative factor ($1+h$) stays the same.

2. The initial value remains 1.

This visual change encodes the conceptual shift from ambiguous function-value notation to a well-defined discrete sequence of approximations.

1. Animation
   Observation

   A new block appears at the upper right listing h=.5, $f_0=1$, $f_1=1.5$, $f_2=2.25$, $f_3=3.375$.

1. Only the beginning of this numerical example is shown before the clip ends.

1. h=.5

2. $f_0=1$

3. $f_1=1.5$

4. $f_2=2.25$

5. $f_3=3.375$

1. Specific numeric values replace the purely symbolic recurrence for a concrete case.

1. The values are generated by the same recurrence $f_{k+1}=(1+h)f_k$.

The board instantiates the abstract method with a particular step size to show how the sequence is computed in practice.

1. Animation
   Observation

   Axes appear, followed by green 'x' marks for approximate values, then a white curve for the exact solution

2. Audio
   Observation

   Speaker says 'Now I can plot those guys roughly on a graph...'

1. Coordinate axes

2. Green 'x' markers representing ($x_k$, $f_k$)

3. White curve representing $y = e^x$

4. Numerical data tables

1. Axes are drawn

2. Approximate points are plotted sequentially

3. Exact solution curve is overlaid

4. Comparison table appears

1. Step size h remains 0.5

2. Initial condition $f(0)=1$ remains fixed

The visual overlay demonstrates that while the finite difference method tracks the general shape of the exponential growth, it systematically lags behind the true curve, illustrating the concept of numerical error accumulation.

1. Audio
   Observation

   The speaker warns that using $f(h)$, $f(2h)$, ... for approximations conflicts with $f(x)$ as the exact solution and says 'It's not true that f of h equals e to the h.'

One may incorrectly think the values produced by the recurrence, such as $f(h)=1+h$, are exactly the values of the true solution $f(x)=e^x$ at those points.