# How to solve the heat equation with initial condition of zero?

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 $$b_n$$?

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