# [SOLVED]Infinite domain to finite plate by a change of variables

#### dwsmith

##### Well-known member
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.

#### dwsmith

##### Well-known member
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.

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#### dwsmith

##### Well-known member
$T(x,y) = \sum_{n = 1}^{\infty}c_n\sin\left(\frac{\pi n}{\ell}x\right)\left[\sinh\left(\frac{\pi n}{\ell}(d-y)\right) + \cosh\left(\frac{\pi n}{\ell}(d-y)\right)\right]$
Should this have be written automatically as
$T(x,y) = \sum_{n = 1}^{\infty}\sin\left(\frac{\pi n}{\ell}x\right)\left[A_n\sinh\left(\frac{\pi n}{\ell}(d-y)\right) + B_n\cosh\left(\frac{\pi n}{\ell}(d-y)\right)\right]$
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 $$g(x)$$.