# Verification of solution to Heat Equation

1. Oct 8, 2011

### rexasaurus

1. verify that u(t,x,y)=e-λtsin(αt)cos(βt) (for arbitrary α, β and with λ=α22) satisfies the 2-D Heat Equation.

2. ut=Δu

3. I began with:
Δu=uxx+uyy.
note the equation does not contain variable "x"
so uxx=0 i.e. Δu=uyy

uy=e-λtsin(αt){-βsin(βt)}
uyy=e-λtsin(αt){-β2cos(βt)}

next I found ut
ut=cos(βy) {e-λtαcos(αt)+sin(αt)-λe-λt}

I have tried to reduce both equations but don't see how they are equal. I also have tried using the λ=α22 to re-write the eq. Any suggestions?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 8, 2011

### LCKurtz

u(t,x,y)=e-λtsin(αx)cos(βy)

Last edited: Oct 8, 2011
3. Oct 8, 2011

### rexasaurus

The original equation was posted incorrectly.

The actual question states:
u(t,x,y)=e-λtsin(αt)cos(βy)

My apologies

4. Oct 8, 2011

### LawrenceC

The heat equation in 2D is the Fourier equation. I think the answer they want you to check out is:

u(x,y,t) = e**(-lambda*t) * sin(alpha*x) * cos(beta*y)

Take the partial derivatives and plug them back into the original equation and you'll see that the equation is satisfied if lambda = alpha**2 + beta**2. The original equation is the Laplacian of u equals the partial of u with respect to time.

5. Oct 8, 2011

### rexasaurus

I checked the problem statement

The question actually states:
u(t,x,y)=e-λtsin(αt)cos(βy)

Could the instructor possibly have made a typo??

When I set ut=uyy (from simplifying Δu) I was able to substitute: λ=22

αcos(αt)+β2sin(αt)=(α22)sin(αt)

Any thoughts?

6. Oct 8, 2011

### LCKurtz

Of course he could have. That t in the sine term should be x. Then it all works.

7. Oct 8, 2011

### rexasaurus

the last substitution should read λ=α22

8. Oct 8, 2011