# PDE Heat Equation 2 Dimensions

1. Mar 10, 2016

### RJLiberator

1. The problem statement, all variables and given/known data
Show that if v(x,t) and w(y,t) are solutions of the 1-dimensional heat equation (v_t = k*v_xx and w_t = k*w_yy), then u(x,y,t) = v(x,t)w(y,t) satisfies the 2-dimensional heat equation. Can you generalize to 3 dimensions? Is the same result true for solutions of the wave equation?

2. Relevant equations

3. The attempt at a solution

Honestly, I have no idea what I am doing. This is all very interesting and it SEEMS like it should be answered by a "oh, yes, that's obvious just do this" quick few lines.

But I'm so foreign with my PDE course that this is causing great stress.

I need a start, if you can tell me what I should look at, or where I should start, please do so. I will attack it then.

2. Mar 10, 2016

### Ray Vickson

Step 1: write down the 2D heat equation for a general function u(x,y,t).
Step 2: plug in the suggested form for u(x,y,t) to see if it "works".

3. Mar 10, 2016

### RJLiberator

I agree with your method and can see that this problem isn't that hard... once you understand the heat equation.

I am searching now for more information on the 2D heat equation.

4. Mar 10, 2016

### RJLiberator

A good starting point?

Homogeneous Dirichlet B.C.:

5. Mar 11, 2016

### LCKurtz

Not much of a start I'd say. A better start would be to write down what you are given, then write down what you are to prove. Then you could start working on it...

6. Mar 11, 2016

### RJLiberator

@LCKurtz ,

I have an extension on this assignment.

I am going to go through some intense studying tonight and I will report back with my updated tries for this problem and the other PDE problem I posted.

Thank you for your guidance thus far! Hopefully I come through.

7. Mar 12, 2016

### RJLiberator

@Ray Vickson
The 2D heat equation I am wondering what this means.

I have the 1d heat equation as follows:
u_t =k∇^2u
u(x,t) = k*∇^2u

is the 2d heat equation just
u(x,y,t) = k*(uxx+kuyy)

?

Separation of variable yiels
T'/(K*T) = X''/X = Y''/Y = -v^2

Last edited: Mar 12, 2016
8. Mar 12, 2016

### LCKurtz

You mean $u_t = k(u_{xx}+u_{yy})$, but, yes, that is what it means.

This problem has nothing to do with separation of variables. Start by following my advice in post #5.

9. Mar 12, 2016

### RJLiberator

Okay, what do we know?

Well, we know the 2 dimension heat equation now:
u(x,y,t) = k(U_xx+U_yy)
we know u(x,y,t) = v(x,t)w(y,t) from problem description.

we know
V_t = k*V_xx and w_t=k*w_yy

Do we set:
v(x,t)w(y,t) = k(U_xx+U_yy)

The thing that is confusing me is the left hand side, how do I work with a function of this nature.

10. Mar 13, 2016

### LCKurtz

It is very bad form, not to mention confusing, when you use both u and U and v and V to represent the same thing as you have done below. Don't do that.

I wouldn't call that u(x,y,t) because it suggests it is the same u in the heat equation. Call it something new like$$F(x,y,t) = v(x,t)w(y,t)$$
The question is whether or not $F(x,y,t)$ satisfies the heat equation $u_t = k(u_{xx} + u_{yy})$.

No. You plug $F(x,y,t)$ into the heat equation and see if what you are given makes it work.

Last edited: Mar 14, 2016