# Fourier Series - Laplace Equation - Help, I'm lost!

I'm trying to make sense of this solution but I get lost after line 13. Can anyone explain to me what happens here? I have no idea how the cosh fraction results from what came above it or what is happening as the solution progresses from there.

I also don't know where Bn came from or how the e expression integrates to the result on line 17 ## 1 Answer

Edit:

Equations (1-4) are the problem to consider. Equations (5-7) are the reduction to single variable equations to solve. The $$x$$-equation seems to more familiar, which leaves the $$y$$-equation to consider.

Consider the equation $$f'' - p^2 f = 0$$, where $$f=f(y)$$ with the condition $$f'(1)=0$$ and $$p = n \pi$$. The solution, for the general case is, $$f(y) = C \, \cosh(p y) + D \, \sinh(p y).$$ Applying the condition leads to $$f'(y) = p \, (C \, \sinh(p y) + D \, \cosh(p y))$$ and $$f'(1) = 0 = p \, (C \, \sinh(p) + D \, \cosh(p))$$ or $$D = - C \, \frac{\sinh(n \pi)}{\cosh(n \pi)}.$$ Now, $$f(y) = C \, \left(\cosh(n \pi y) - \frac{\sinh(n \pi)}{\cosh(n \pi)} \, \sinh(n \pi y) \right)$$ and now use $$\cosh(x - y) = \cosh(x) \cosh(y) - \sinh(x) \sinh(y)$$ to obtain the form given in the problem (and in this solution).

Original:

Equation (10) gives the form of the solution that satisfies the pde. From (11-13) it is determined that $$A = 0$$, $$\alpha = n \pi$$ (not $$\alpha$$ of the boundary conditions), and $$C \, \sinh(n \pi) = - D \, \cosh(n \pi).$$ Now, The solution to the $$y$$-variable is $$f(y) = C \, \cosh(n \pi y) + D \, \sinh(n \pi y)$$ which can be seen as $$f(y) = C \, \left( \cosh(n \pi y) - \frac{\sinh(n \pi) \sinh( n \pi y)}{\cosh(n \pi)} \right) = \frac{\cosh(n \pi (1- y))}{\cosh(n \pi)}.$$ Since $$A = 0$$ then the solution is seen as a series of sines in the variable $$x$$ and takes the form $$u(x,y) = \sum_{n=1}^{\infty} \frac{B_{n}}{\cosh(n \pi)} \, \sin(n \pi x) \, \cosh(n \pi (1-y)).$$ Applying the final condition, $$u(x,0) = \alpha \, e^{- \beta x}$$ leads to $$\alpha \, e^{- \beta x} = \sum_{n=1}^{\infty} B_{n} \, \sin(n \pi x).$$ The coefficients $$B_{n}$$ is given by \begin{align} B_{n} &= \alpha \, \int_{0}^{1} e^{- \beta x} \, \sin(n \pi x) \, dx \\ &= \frac{n \pi \alpha}{\beta^2 + n^2 \pi^2} \, (1 - (-1)^n \, e^{-\beta}). \end{align}

From all of this the solution then takes the form $$u(x, y) = \sum_{n=1}^{\infty} \frac{n \pi \alpha}{\beta^2 + n^2 \pi^2} \, (1 - (-1)^n \, e^{-\beta}) \, \sin(n \pi x) \, \cosh(n \pi (1-y)).$$

• Thanks for that, but I'm still not seeing how we get the result in the bracket for your second step. I'm sure it's obvious to most people, but this is a big leap from what I've done in the past and I'm just not able to suss out the steps between the first and second line. – C. Wolfe Mar 21 at 11:39
• @C.Wolfe I made an edit that may be more clear of how the $y$-solution takes the desired form. – Leucippus Mar 21 at 19:27
• oooh! Now it's clear! Thank you so much for that! I kept looking at that transposition thinking that it was just $\tanh$ which was really throwing me off! – C. Wolfe Mar 21 at 19:39