Heat equation in the first quadrant.

[A_B]

Homework Statement

Solve the heat equation u_t=u_{xx}+u_{yy} fot t>0 in the first quadrant of \mathbb{R}^2. The boundary conditions are u(0,y,t)=u(x,0,t)=0 and the initial temperature distribution is
<br /> f(x,y)=<br /> \begin{cases}<br /> 1 \;\;\;\; \text{in the square } \; 0&lt;x&lt;1; \; 0&lt;y&lt;1 \\<br /> 0 \;\;\;\; \text{elsewhere}<br /> \end{cases}<br />

The Attempt at a Solution

I can solve problems in finite domains, ie where enough boundary conditions are given to determine the countable infinite set of eigenvalues. This problem is on a semi-infinite domain, with insufficient boundary conditions to do so. I have come to understand that to solve such a problem, one replaces "the sum over eigenvalues" with an integral of the solutions obtained by separation of variables, over the (roots of the) eigenvalues.

This is where I got (detailed derivations not typed out)

Separation of variables (solutions of the form u=X(x)Y(y)T(t) gives ODEs
<br /> \begin{cases}<br /> -\ddot{X} = \lambda_1 \;\;\;\;\;\;\;\;\;\;\;\; X(0)=0\\<br /> -\ddot{Y} = \lambda_2 \;\;\;\;\;\;\;\;\;\;\;\; Y(0)=0\\<br /> \dot{T}=-(\lambda_1+\lambda_2)T<br /> \end{cases}<br />

The "eigenvalues" can be shown to be all positive. So let \lambda_1 = \gamma_1^2 and \lambda_2 = \gamma_2^2. With \gamma_1 and \gamma_2 both positive.

The solutions are
<br /> X(x)=\sin(\gamma_1 x) <br />
<br /> Y(y)=\sin(\gamma_2 y) <br />
<br /> T(t) = e^{-(\gamma_1^2 + \gamma_2^2)t}<br />

So the solution for separated variables is
<br /> u(x,y,t)=\sin(\gamma_1 x)\sin(\gamma_2 y) e^{-(\gamma_1^2 + \gamma_2^2)t}<br />

Since there are solutions for every positive \gamma_1 and \gamma_2, we can't determine eigenvalues like in regular problems. We write the final solution as a weighed integral of the solution for separated variables, with respect to the gamma's.

<br /> u(x,y,t) = \int_0^\infty\int_0^\infty d\gamma_1 d\gamma_2 \alpha(\gamma_1, \gamma_2) \sin(\gamma_1 x)\sin(\gamma_2 y) e^{-(\gamma_1^2 + \gamma_2^2)t}<br />The function \alpha(\gamma_1,\gamma_2) is determined by the initial conditions:

<br /> f(x,y) = u(x,y,0) = \int_0^\infty\int_0^\infty d\gamma_1 d\gamma_2 \alpha(\gamma_1, \gamma_2) \sin(\gamma_1 x)\sin(\gamma_2 y) <br />

And here I'm stuck, how do I determine alpha?Any help is much appreciated,
A_B

 

[Thaakisfox]

Fourier transform ;)

 

[A_B]

I solved it, the last formula expresses f(x,y) as a double Fourier sine transform of α(γ_1, γ_2). So taking the inverse Fourier sine transform of f(x,y) twice gives α.

A_B