# Heat equation in the first quadrant.

1. Dec 21, 2011

### A_B

1. The problem statement, all variables and given/known data
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
$$f(x,y)= \begin{cases} 1 \;\;\;\; \text{in the square } \; 0<x<1; \; 0<y<1 \\ 0 \;\;\;\; \text{elsewhere} \end{cases}$$

3. 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
$$\begin{cases} -\ddot{X} = \lambda_1 \;\;\;\;\;\;\;\;\;\;\;\; X(0)=0\\ -\ddot{Y} = \lambda_2 \;\;\;\;\;\;\;\;\;\;\;\; Y(0)=0\\ \dot{T}=-(\lambda_1+\lambda_2)T \end{cases}$$

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
$$X(x)=\sin(\gamma_1 x)$$
$$Y(y)=\sin(\gamma_2 y)$$
$$T(t) = e^{-(\gamma_1^2 + \gamma_2^2)t}$$

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

Since there are solutions for every positive $\gamma_1$ and $\gamma_2$, we cant 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.

$$u(x,y,t) = \int_0^\infty\int_0^\infty d\gamma_1 d\gamma_2 \alpha(\gamma_1, \gamma_2) \sin(\gamma_1 x)\sin(\gamma_2 y) e^{-(\gamma_1^2 + \gamma_2^2)t}$$

The function $\alpha(\gamma_1,\gamma_2)$ is determined by the initial conditions:

$$f(x,y) = u(x,y,0) = \int_0^\infty\int_0^\infty d\gamma_1 d\gamma_2 \alpha(\gamma_1, \gamma_2) \sin(\gamma_1 x)\sin(\gamma_2 y)$$

And here I'm stuck, how do I determine alpha?

Any help is much appreciated,
A_B

Last edited: Dec 21, 2011
2. Dec 24, 2011

### Thaakisfox

Fourier transform ;)

3. Dec 26, 2011

### 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