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Homework Statement
Consider
\frac{\partial u}{\partial t} = k\left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) \\<br /> 0<x<L\\<br /> y>0
subject to the initial condition
IC: u(x,y,0) = f(x,y)
And solve with the following boundary conditions:
BC1: \quad u(0,y,t) = 0
BC2: \quad u(L,y,t) = 0
BC3: \quad u(x,0,t) = 0
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:
\frac{\partial \overline{U}}{\partial t} = k\left( \frac{\partial^2 \overline{U}}{\partial x^2} - \omega^2 \overline{U} \right)
Then this eqaution can be solved with Fourier sine series using BC1 and BC2:
\overline{U} = \sum_{n=1}^{\infty} A_n sin(\frac{n \pi x}{L})
And the previous equation can be used to get an expression for A_n:
\frac{\partial A_n}{\partial t} = -kA_n[({\frac{n\pi}{L}}^2 + \omega^2] \\<br /> A_n(\omega,t) = A_n(\omega,0) e^{-kt[({\frac{n\pi}{L}}^2 + \omega^2]}
With the initial condition IC1 you can get:
\overline{U}(x,\omega,0) = F(x,\omega) = \sum_{n=1}^{\infty} A_n(\omega,0) sin(\frac{n \pi x}{L}) \\<br /> A_n(\omega,0) = \frac{2}{L} \int_0^L F(x,\omega) sin(\frac{n \pi x}{L}) dx
And now you can use the inverse Fourier sine transform to get u(x,y,t):
u(x,y,t) = \int_0^\infty \overline{U}(x,\omega,t) sin(\omega y) d\omega
With \overline{U}(x,\omega,t):
\overline{U}(x,\omega,t) = \sum_{n=1}^{\infty} A_n(\omega,t) sin(\frac{n \pi x}{L})
With A_n(\omega,t):
A_n(\omega,t) = A_n(\omega,0) e^{-kt[({\frac{n\pi}{L}}^2 + \omega^2]}
With A_n(\omega,0):
A_n(\omega,0) = \frac{2}{L} \int_0^L F(x,\omega) sin(\frac{n \pi x}{L}) dx
With F(x,\omega):
F(x,\omega) = \frac{2}{\pi} \int_0^\infty f(x,y) sin(\omega y) dyI think I understand the solution, but the problem is, I don't see how BC3 comes into play. I now have to do solve the same problem but with these conditions:
IC: u(x,y,0) = f(x,y)
BC1: \quad u(0,y,t) = 0
BC2: \quad u(L,y,t) = 0
BC3: \quad \frac{\partial u}{\partial y}(x,0,t) = 0
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?