Consider the partial differential equation (Laplace's equation) (∂^2 u)/(∂x^2)+(∂^2 u)/(∂y^2)=0

Consider the partial differential equation (Laplace's...

Key Concepts

Second Order Linear ODEs with Constant Coefficients

Once the PDE is reduced to an ordinary differential equation through the substitution method, it often takes the form of a second order linear ODE with constant coefficients. This class of differential equations is well-understood, and standard methods, such as solving the characteristic equation, are used to obtain the general solution.

Chain Rule in Multivariable Calculus

The chain rule is a fundamental tool in multivariable calculus used to differentiate composite functions. When a substitution introduces a new variable that depends on the original variables, the chain rule allows one to correctly compute the partial derivatives of the transformed function, enabling the reduction of partial differential equations to ordinary differential equations.

Characteristic Equation

The characteristic equation is derived from a linear ordinary differential equation with constant coefficients by assuming an exponential solution. The resulting algebraic equation, usually quadratic in the case of a second order ODE, provides the roots that determine the behavior of the general solution, including oscillatory or exponential behavior depending on whether the roots are real or complex.

Laplace's Equation

Laplace's equation is a second order partial differential equation that arises in many areas such as physics and engineering, particularly in steady-state heat conduction, potential flow, and electrostatics. Its key property is that the sum of its second spatial derivatives equals zero, and it is widely used as a prototype for solving and understanding more complex PDEs.

Change of Variables (Substitution Method)

In solving partial differential equations, a change of variables or substitution method is often used to simplify the equation by transforming it into a more manageable form, such as an ordinary differential equation. This involves introducing a new variable that combines the original independent variables in a way that takes advantage of symmetries or specific structure in the problem.

Transcript

00:01 Okay, so for this partial differential equation here, we're going to use this substitution here with this here.

00:10 Okay? so then now, we just need to find the partial derivative with respect to x and the partial derivative of respect to y.

00:20 So then first, d squared, d squared u, d x squared, this is first going to be.

00:31 Be called d d x of and then we're going to take the partial derivative of this with respect to x right so remember we're going to need to use product rule here so we're going to take the first times or the derivative of the first times second so first we have the derivative of the first is one over alpha e to the x over alpha and then just times f of xe here.

01:05 Then plus, now the first, which is just e to the x over alpha, and then we need to take the derivative of this with respect to x.

01:16 So we're going to take the chain rule here.

01:18 So that's going to become f prime, f prime of x.

01:25 So that's going to be f prime of xxie times the integral of xxx.

01:34 Or sorry, times the derivative of this with respect to x, which is going to become beta.

01:44 Then we still need to take the derivative again.

01:48 So we're going to use product rule again.

01:52 Okay, so then we have here.

01:55 So the derivative of this with respect to x, we'll take the derivative of this again.

02:01 So 1 over alpha squared, e to the x over alpha times f of c plus and then we're going to do the first, sorry, first times the derivative of second.

02:20 So plus 1 over alpha, e to the x over alpha times f prime of c times beta.

02:30 And then now we're going to take the derivative of this, again, with respect to x.

02:34 So i'm just going to put the beta out front.

02:40 Beta is out front.

02:42 And then we're going to take beta over alpha.

02:51 So i'm going to take the derivative of this first part.

02:54 So beta divided by alpha, e to the x divided by alpha, then f prime, c.

03:01 And then a plus beta squared, e to the x over alpha, then f double prime of c.

03:23 So the first beta comes from this here, but then taking the derivative, the partial derivative of this with respect to x is going to be equal to so f double prime of xe and then times the derivative of c with respect to x, which is a beta again.

03:45 So we're going to get another beta on the outside.

03:48 Okay.

03:50 So now we can simplify this a little bit.

03:53 We'll have 1 over alpha squared, e to the x over alpha times f of c, plus, and then we can combine the two f prime terms.

04:07 So we'll have e to the x, e to the x to the alpha.

04:15 And actually the e to the x to the alpha, i'm going to factor that from all of the terms.

04:22 So we have here 1 over alpha squared f of c, and then plus here we have a beta over alpha, and then another beta over alpha.

04:36 So it becomes 2 plus 2 beta beta, over alpha f prime of c and then plus a beta squared f double prime of c c now we need to find d squared u d y squared so this is going to be equal to so first d d y of now take the derivative of this with respect to to y.

05:16 So the only y term here is this.

05:19 So this is actually going to be a lot easier of a derivative to take.

05:23 We just leave this part as a constant.

05:26 So it just becomes e to the x over a or x over alpha and then f prime of c times negative alpha.

05:37 Okay.

05:37 Then take the derivative again and and again we just get e to the x over alpha and then and f double prime, f double prime, c of alpha squared.

05:52 So i'm going to put the alpha, or yeah, then alpha squared here.

05:58 So we just multiply by another negative alpha.

06:01 So adding all the terms together now, so we substitute this into here, and then this whole thing into here.

06:10 Okay, so we have, from our first part, we have e to the x over alpha, so i'm going to factor out that from every term.

06:22 Okay.

06:23 And then i'm going to rewrite it with the f prime first.

06:26 So we have beta squared, f double prime.

06:30 Actually write it.

06:34 Beta squared, f double prime, f double prime of c.

06:40 And then there's also actually an alpha squared.

06:45 So it should be alpha squared plus beta squared f double prime of c, and then plus our f prime term is going to be 2 beta over alpha.

07:01 So we'll have a 2 -2 -beta over alpha, f -prime of x, and then plus our final term is going to be 1 over alpha, f -prime of x.

07:20 And that is all equal to zero.

07:27 So this alpha squared f double prime, that is our d squared u, d, y squared, and the rest all comes from the d x squared.

07:37 Okay, so now since e to the x over alpha cannot be equal to zero, that reduces just this part right here equal to zero.

07:47 So we have just alpha squared plus beta squared, f double prime of c plus 2 beta over alpha f prime of x plus 1 over alpha f prime of x plus 1 over alpha squared f prime of x is equal to 0 so then we can divide everything by an alpha squared beta squared so we'll get f double prime c plus and then this whole thing divided by alpha squared plus beta squared.

08:27 That was what we defined to be p.

08:29 So we have two p f prime, f prime of c, and then plus one over alpha squared or sorry, this is one over alpha squared plus beta squared.

08:47 We called that q.

08:49 So we're just going to have one over alpha squared and then this whole thing is q.

08:53 So we're going to have q over alpha squared.

08:57 So q over alpha squared.

09:00 And then, oops, sorry, this should be an f of x, not an f prime of x, 0.

09:07 F of, oh, actually that should be of x, f of c, f of c.

09:18 Oops.

09:21 This also up here should be f of c.

09:24 F of c is equal to zero here...