# Heat equation in 2dim

1. May 25, 2006

### JohanL

Im trying to solve the heat equation in 2dim on a plate.
0=<x=<L, 0=<y=<L. With homogenous dirichlet conditions on the boundary and the initial condition:
T(x,y)=T0sin(pi*x/L)sin(pi*y/L)

With separation of variables i get the solution
$$T(x,y,t)=\sum_{m=0}^\infty\sum_{n=0}^\infty B_{mn}*exp[\frac{-(m^2+n^2)\pi^2kt}{L^2}]*sin[\frac{m\pi x}{L}]*sin[\frac{m\pi y}{L}]$$

m,n integers and k the constant from the heat equation.

Now the initial condition determine the constants B_mn

$$B_{mn} = \frac{4T_0}{L^2}*\int_{0}^{L}\int_{0}^{L} sin[\frac{\pi x}{L}]*sin[\frac{\pi y}{L}] sin[\frac{m\pi x}{L}]*sin[\frac{n\pi y}{L}] dx dy$$

But an integral like
$$\int_{0}^{L}sin\frac{\pi x}{L}sin\frac{m\pi x}{L}dx$$

is zero for m not equal to 1.
So m=1 and n=1...???
and with this i dont get a fourier series as the solution...
What am i doing wrong ?

Last edited: May 25, 2006
2. May 25, 2006

### HallsofIvy

Staff Emeritus
"Bmn= 0" is not an initial condition! Your initial condition is that
$$T(x,y,0)=\sum_{n=0}^\infty B_{mn}sin[\frac{m\pi x}{L}]*sin[\frac{m\pi y}{L}]= T_0sin(\frac{\pi x}{L})sin(\frac{pi y}{L})$$

Now, apply the Fourier integral to both sides of that.

3. May 25, 2006

### JohanL

I never said that Bmn= 0 is an initial condition, i said that the integral is zero for m NOT equal to one, because of the orthogonality of sin-functions.

Im not sure what you mean by applying the Fourier integral, i have only used that with Fourier transforms.

This is how i get the expression for Bmn:
I have the general solution

$$T(x,y,t)=\sum_{m=0}^\infty\sum_{n=0}^\infty B_{mn}*exp[\frac{-(m^2+n^2)\pi^2kt}{L^2}]*sin[\frac{m\pi x}{L}]*sin[\frac{n\pi y}{L}]$$

Then i expand the initial condition in the eigenfunctions

$$(sin[\frac{m\pi x}{L}]*sin[\frac{n\pi y}{L}])_{m,n=1}^{\infty}=\Psi$$

The initial condition is

$$f(x,y)=T_0sin(\frac{\pi x}{L})sin(\frac{\pi y}{L})$$

$$T(x,y,0)=\sum_{n=0}^\infty\sum_{m=0}^\infty B_{mn}sin[\frac{m\pi x}{L}]*sin[\frac{m\pi y}{L}]= f(x,y)$$

I multiply with the eigenfunctions and integrate over the domin to get an expression for Bmn:

$$B_{mn}=\frac{(f,\Psi)}{(\Psi,\Psi)}$$

where the normalisation integral is

$$(\Psi,\Psi) = \int_{0}^{L}\int_{0}^{L}sin^2[\frac{m\pi x}{L}]*sin^2[\frac{n\pi y}{L}] dxdy=L^2/4$$

and

$$(f,\Psi) = T_0*\int_{0}^{L}\int_{0}^{L} sin[\frac{\pi x}{L}]*sin[\frac{\pi y}{L}]*sin[\frac{m\pi x}{L}]*sin[\frac{n\pi y}{L}]dxdy$$

this integral is

$$(f,\Psi) = T_0\int_{0}^{L}sin[\frac{\pi x}{L}]*sin[\frac{m\pi x}{L}]dx\int_{0}^{L} sin[\frac{\pi y}{L}]*sin[\frac{n\pi y}{L}]dy$$

but for m not equal to 1

$$\int_{0}^{L}sin[\frac{\pi x}{L}]*sin[\frac{m\pi x}{L}]dx=\frac{1}{2}\int_{0}^{L}cos[\frac{(1-m) \pi x}{L}]dx+\int_{0}^{L}cos[\frac{(1+m)\pi x}{L}]dx=\frac{Lm*sin(m\pi)}{\pi-m^2\pi}=0$$

and the same for the integral over y.
Therefor m and n has to be 1 and i dont get a series solution!

Im probably doing some silly misstakes, can you find them?

Last edited: May 26, 2006
4. May 31, 2006

### JohanL

hmm, i was probably right from the beginning.
I got confused because i had never solved a problem before where this happens.

Is it like this? :
When you have a certain "frequency" as the initial condition you cant get a solution for later times as a superposition over other frequencies.
Thats why in this case only the terms with m=1,n=1 survived in the sum.