Solving modified heat equation

In my Fourier analysis book, the author introduces some basic PDE problems and how one can solve these using Fourier series. I know how to solve basic heat equation problems, but the above one is different from the previous problems I've worked in terms of the boundary conditions. Using ##u(x,t)=v(x,t)e^{-ht}## I can transform the equation into the heat equation, i.e. ##v_t= v_{xx}## , however, the boundary conditions become $$v(0,t)=0,\quad v(\pi,t)=e^{ht}.$$ I don't know how to deal with non-constant boundary conditions...any ideas on how to proceed?

Think about the long-term steady state. Find a function [itex]u_\infty(x)[/itex] which satisfies [tex]u_\infty'' - hu_\infty = 0[/tex] with [itex]u_\infty(0)= 0[/itex] and [itex]u_\infty(\pi) = 1[/itex]. Then [itex]f(x,t) = u(x,t) - u_\infty(x)[/itex] must satisfy [itex]f_t = f_{xx} - hf[/itex] subject to the self-adjoint boundary condition [itex]f(0,t) = f(\pi,t) = 0[/itex] and the initial condition [itex]f(x,0) = -u_\infty(x)[/itex].

Ok. I found $$u_{\infty}(x)=\frac{\sinh \sqrt{h}x}{\sinh \sqrt{h}\pi}.$$ I still think I have to use the trick ##f(x,t)=e^{-ht}v(x,t)## to turn ##f_t = f_{xx} - hf## into ##v_t = v_{xx}##. The solution to the latter will be $$v(x,t)=\sum_{n=1}^\infty b_n e^{-n^2t}\sin nx.$$ But I'm stuck at how to solve for ##b_n## in $$f(x,0)=v(x,0)=\sum_{n=1}^\infty b_n \sin nx=-\frac{\sinh \sqrt{h}x}{\sinh \sqrt{h}\pi}.$$

Do you not know how to determine the coefficients [itex]b_n[/itex]?

pasmith said:

Do you not know how to determine the coefficients [itex]b_n[/itex]?

Right, I don't. I don't see any connection between ##\sinh## and ##\sin## that's useful here.

If [tex] \sum_{n=1}^\infty b_n \sin nx = f(x), \quad x \in [0, \pi][/tex] then [tex]
b_n = \frac{2}{\pi} \int_0^\pi f(x) \sin nx \,dx.[/tex] This should be derived in any decent textbook on fourier series.

Integrals of the form [itex]\int \sin ax \sinh bx\,dx[/itex] can be done by integratin by parts twice or by expressing everything in terms of exponentials.