MHB Infinite domain to finite plate by a change of variables

[Dustinsfl]

Consider the following solution to the steady state heat diffusion problem on an infinite y domain.
\[
T(x, y) = \sum_{n = 1}^{\infty}c_n\exp\left(-\frac{\pi n}{\ell} y\right)
\sin\left(\frac{\pi n}{\ell}x\right)
\]
How does one obtain the results of finite plate by making the change of variables \(d - y\) for \(y\) and considering the linit as \(d\to\infty\)?

Making that sub we have \(\exp(-\lambda_nd)\exp(\lambda_ny)\). If we take the limit as d goes to inifinity, we get 0. Therefore, \(T(x,y) = 0\). This doesn't seem correct.

 

[Dustinsfl]

Consider \(\exp(d-y) = \sinh(d-y) + \cosh(d-y)\).

Then \(\exp\left(\frac{\pi n}{\ell}(d-y)\right) = \sinh\left(\frac{\pi n}{\ell}(d-y)\right) + \cosh\left(\frac{\pi n}{\ell}(d-y)\right)\).
\[
T(x,y) = \sum_{n = 1}^{\infty}c_n\sin\left(\frac{\pi n}{\ell}x\right)\left[\sinh\left(\frac{\pi n}{\ell}(d-y)\right) + \cosh\left(\frac{\pi n}{\ell}(d-y)\right)\right]
\]
General the solution to the Laplace on a rectangle will only have sinh or cosh, but if it has both, they should have their own Fourier coefficients. I don't see away to re-write this any further though. I still don't understand the \(d\to\infty\) part.

If we break \(T\) up, we have
\[
T_1 = T(x,y) = \sum_{n = 1}^{\infty}c_n\sin\left(\frac{\pi n}{\ell}x\right)\sinh\left(\frac{\pi n}{\ell}(d-y)\right)
\]
which implies a boundary condition of \(T(x, d) = f(x)\), and
\[
T_2 = T(x,y) = \sum_{n = 1}^{\infty}c_n\sin\left(\frac{\pi n}{\ell}x\right)\cosh\left(\frac{\pi n}{\ell}(d-y)\right)
\]
which implies a boundary condition of \(T'(x, d) = g(x)\).
Is there more to say about \(T\)?

If \(d\to\infty\), we are back to the infinite domain problem.

 

[Dustinsfl]

\[
T(x,y) = \sum_{n = 1}^{\infty}c_n\sin\left(\frac{\pi n}{\ell}x\right)\left[\sinh\left(\frac{\pi n}{\ell}(d-y)\right) + \cosh\left(\frac{\pi n}{\ell}(d-y)\right)\right]
\]

Should this have be written automatically as
\[
T(x,y) = \sum_{n = 1}^{\infty}\sin\left(\frac{\pi n}{\ell}x\right)\left[A_n\sinh\left(\frac{\pi n}{\ell}(d-y)\right) + B_n\cosh\left(\frac{\pi n}{\ell}(d-y)\right)\right]
\]
instead?

This would then make more since for the boundary conditions. We would have the first equal to say \(f(x)\) and the second would be the derivative equal to \(g(x)\).