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A_B
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Homework Statement
Solve the heat equation [itex]u_t=u_{xx}+u_{yy}[/itex] fot [itex]t>0[/itex] in the first quadrant of [itex]\mathbb{R}^2[/itex]. The boundary conditions are [itex]u(0,y,t)=u(x,0,t)=0[/itex] and the initial temperature distribution is
[tex]
f(x,y)=
\begin{cases}
1 \;\;\;\; \text{in the square } \; 0<x<1; \; 0<y<1 \\
0 \;\;\;\; \text{elsewhere}
\end{cases}
[/tex]
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 [itex]u=X(x)Y(y)T(t)[/itex] gives ODEs
[tex]
\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}
[/tex]
The "eigenvalues" can be shown to be all positive. So let [itex]\lambda_1 = \gamma_1^2[/itex] and [itex]\lambda_2 = \gamma_2^2[/itex]. With [itex]\gamma_1[/itex] and [itex]\gamma_2[/itex] both positive.
The solutions are
[tex]
X(x)=\sin(\gamma_1 x)
[/tex]
[tex]
Y(y)=\sin(\gamma_2 y)
[/tex]
[tex]
T(t) = e^{-(\gamma_1^2 + \gamma_2^2)t}
[/tex]
So the solution for separated variables is
[tex]
u(x,y,t)=\sin(\gamma_1 x)\sin(\gamma_2 y) e^{-(\gamma_1^2 + \gamma_2^2)t}
[/tex]
Since there are solutions for every positive [itex]\gamma_1[/itex] and [itex]\gamma_2[/itex], 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.
[tex]
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}
[/tex]The function [itex]\alpha(\gamma_1,\gamma_2)[/itex] is determined by the initial conditions:
[tex]
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)
[/tex]
And here I'm stuck, how do I determine alpha?Any help is much appreciated,
A_B
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