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## Homework Statement

Consider

[tex] \frac{\partial u}{\partial t} = k\left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) \\

0<x<L\\

y>0 [/tex]

subject to the initial condition

IC: [itex] u(x,y,0) = f(x,y) [/itex]

And solve with the following boundary conditions:

BC1: [itex] \quad u(0,y,t) = 0 [/itex]

BC2: [itex] \quad u(L,y,t) = 0[/itex]

BC3: [itex] \quad u(x,0,t) = 0[/itex]

## Homework Equations

Already stated above:

## The Attempt at a Solution

A solution is given as follows:

You can determine the Fourier sine transform in y:

[tex] \frac{\partial \overline{U}}{\partial t} = k\left( \frac{\partial^2 \overline{U}}{\partial x^2} - \omega^2 \overline{U} \right) [/tex]

Then this eqaution can be solved with Fourier sine series using BC1 and BC2:

[tex] \overline{U} = \sum_{n=1}^{\infty} A_n sin(\frac{n \pi x}{L}) [/tex]

And the previous equation can be used to get an expression for [itex] A_n [/itex]:

[tex] \frac{\partial A_n}{\partial t} = -kA_n[({\frac{n\pi}{L}}^2 + \omega^2] \\

A_n(\omega,t) = A_n(\omega,0) e^{-kt[({\frac{n\pi}{L}}^2 + \omega^2]} [/tex]

With the initial condition IC1 you can get:

[tex] \overline{U}(x,\omega,0) = F(x,\omega) = \sum_{n=1}^{\infty} A_n(\omega,0) sin(\frac{n \pi x}{L}) \\

A_n(\omega,0) = \frac{2}{L} \int_0^L F(x,\omega) sin(\frac{n \pi x}{L}) dx [/tex]

And now you can use the inverse Fourier sine transform to get [itex] u(x,y,t) [/itex]:

[tex] u(x,y,t) = \int_0^\infty \overline{U}(x,\omega,t) sin(\omega y) d\omega [/tex]

With [itex] \overline{U}(x,\omega,t) [/itex]:

[tex] \overline{U}(x,\omega,t) = \sum_{n=1}^{\infty} A_n(\omega,t) sin(\frac{n \pi x}{L}) [/tex]

With [itex] A_n(\omega,t) [/itex]:

[tex] A_n(\omega,t) = A_n(\omega,0) e^{-kt[({\frac{n\pi}{L}}^2 + \omega^2]} [/tex]

With [itex] A_n(\omega,0) [/itex]:

[tex] A_n(\omega,0) = \frac{2}{L} \int_0^L F(x,\omega) sin(\frac{n \pi x}{L}) dx [/tex]

With [itex] F(x,\omega) [/itex]:

[tex] F(x,\omega) = \frac{2}{\pi} \int_0^\infty f(x,y) sin(\omega y) dy [/tex]

I think I understand the solution, but the problem is, I don't see how BC3 comes in to play. I now have to do solve the same problem but with these conditions:

IC: [itex] u(x,y,0) = f(x,y) [/itex]

BC1: [itex] \quad u(0,y,t) = 0 [/itex]

BC2: [itex] \quad u(L,y,t) = 0[/itex]

BC3: [itex] \quad \frac{\partial u}{\partial y}(x,0,t) = 0[/itex]

but in the derivation of the previous problem I only followed how IC1, BC1 and BC2 were used to derive the solution. How does BC3 matter?