What am I doing wrong in solving the heat equation in 2 dimensions?

[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
<br /> 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}]<br />

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

Now the initial condition determine the constants B_mn

<br /> 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<br />

But an integral like
<br /> \int_{0}^{L}sin\frac{\pi x}{L}sin\frac{m\pi x}{L}dx<br />

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

 

[HallsofIvy]

"B_mn= 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.

 

[JohanL]

"B_mn= 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.

I never said that B_mn= 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 B_mn:
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<br />

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 B_mn:

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 don't get a series solution!

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

 

[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 can't 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.