The solution of heat equation $$u_t = \kappa u_{xx}$$

with separation of variables is

$$u(x,t) = \sum_{n=1}^{\infty}b_n e^{-\kappa n^2 \pi^2 t}\sin n \pi x$$

I spare the details as are well known.

For obtaining $b_n$, we use the initial condition, $u(x,t)$, for example, $$ -100x = \sum_{n=1}^{\infty}b_n \sin n \pi x$$ for the initial condition of $u(0,t=0)=0, u(L,t=0)=100$

How can we obtain $b_n$ when the initial condition is zero, for example

IC: $u(x,0)=0$

BC1: $u(0,t)=0$

BC2: $u(L,t)=100$

or a bit more tricky

IC: $u(x,0)=0$

BC1: $u_x(0,t)=0$

BC2: $u(L,t)=100$

When applying the initial condition, the Fourier series is

$$ 0 = \sum_{n=1}^{\infty}b_n \sin n \pi x$$

How do we obtain $b_n$?

  • $\begingroup$ Boundary conditions need to be homogeneous. You need to apply a transform before proceeding $\endgroup$ – DaveNine Jan 26 at 20:52
  • $\begingroup$ By IC, $u(L, 0) = 0$, but by BC2, $u(L, 0) = 100$. $\endgroup$ – Paul Sinclair Jan 26 at 21:02
  • $\begingroup$ @PaulSinclair IC is for $t=0$ and BCs are for $t>0$. $\endgroup$ – Kimia Jan 27 at 1:01
  • $\begingroup$ Kimia - That does not avoid the problem. $u$ has to be singular at $(L, 0)$ $\endgroup$ – Paul Sinclair Jan 27 at 5:50

The problem isn't the the initial condition is zero, but rather your boundary conditions are not homogeneous. Therefore, you can't apply a series solution just yet.

Instead, try breaking up the solution into

$$ u(x,y) = w(x) + v(x,y) $$

where $w(x)$ is the steady-state solution, satisfying

$$ w_{xx} = 0 $$

with B.C.s: $w_x(0) = 0$, $w(L) = 100$. Solving this gives $w(x) = 100$.

Then, you can solve the residual problem

$$ v_t = \kappa v_{xx} $$

with B.C.s: $v_x(0,t) = v(L,t) = 0$

and I.C.: $v(x,0) = u(x,0) - w(x) = -100$

Now you can proceed as usual.


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