Solving modified heat equation

[psie]

Homework Statement

Find a solution of the following problem \begin{align} u_t&= u_{xx} - hu,\qquad &0<x<\pi, \ t>0; \\ u(0,t)&=0,u(\pi,t)=1,\qquad &t>0; \\ u(x,0)&=0,\qquad &0<x<\pi.\end{align} Here $h>0$ is a constant.

Relevant Equations

The heat equation in the form $u_t= u_{xx}$ and its solution $u(x,t)=\sum_{n=1}^\infty b_n e^{-n^2t}\sin nx$.

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?

 

[pasmith]

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

 

[psie]

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 $$f(x,0)=v(x,0)=\sum_{n=1}^\infty b_n \sin nx=-\frac{\sinh \sqrt{h}x}{\sinh \sqrt{h}\pi}.$$

 

[pasmith]

Do you not know how to determine the coefficients $b_n$?

 

[psie]

Do you not know how to determine the coefficients $b_n$?

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

 

[pasmith]

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

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