Heat equation in one dimension with constant heat supply

[V0ODO0CH1LD]

Homework Statement

A bar of length $L$ has an initial temperature of $0^{\circ}C$ and while one end ($x=0$) is kept at $0^{\circ}C$ the other end ($x=L$) is heated with a constant rate per unit area $H$. Find the distribution of temperature on the bar after a time $t$.

Homework Equations

Heat equation in one dimension:
\alpha^2u_{xx}=u_t
Initial conditions:
u(0,t)=0
u(L,t)=\frac{k}{H}t
u(x,0)=0

The Attempt at a Solution

If $u(x,t)$ can be written as $u(x,t)=X(x)T(t)$ we can separate variables so that the heat equation becomes $\alpha^2X''(x)T(t)=T'(t)X(t)$ which can be written as
\alpha^2\frac{X''(x)}{X(x)}=\frac{T'(t)}{T(t)}
and since each side only depends on one variable the two ratios above are constant (because if, for example, I differentiate $T'/T$ with respect to $x$ I get $0$ so $X''(x)/X(x)$ is constant). So I get
\alpha^2\frac{X''(x)}{X(x)}=\gamma\Rightarrow{}X(x)=c_1cos\left(\frac{\sqrt{\gamma}}{\alpha}x\right)+c_2sin\left(\frac{\sqrt{\gamma}}{\alpha}x\right)
and
\frac{T'(t)}{T(t)}=\gamma\Rightarrow{}T(t)=ce^{\gamma{}t}
Now from the boundary conditions
u(0,t)=X(0)T(t)=ce^{\gamma{t}}\left[c_1cos\left(\frac{\sqrt{\gamma}}{\alpha}0\right)+c_2sin\left(\frac{\sqrt{\gamma}}{\alpha}0\right)\right]=0\Rightarrow{}c_1=0
and now here's where I got into trouble
u(L,t)=X(L)T(t)=ce^{\gamma{t}}\left[c_1\cos\left(\frac{\sqrt{\gamma}}{\alpha}L\right)+c_2\sin\left(\frac{\sqrt{\gamma}}{\alpha}L\right)\right]=\frac{k}{H}t

But what am I supposed to do with
c_2\sin{\left(\frac{\sqrt{\gamma}}{\alpha}L\right)}=\frac {k}{Hce^{\gamma{t}}}t
to keep going to find $u$?

 

[pasmith]

To use separation of variables here, you need constant boundary conditions and a non-zero initial condition, which is not what you have. But you can obtain a problem with constant (indeed zero) boundary conditions and a non-zero initial condition.

Consider f(x,t) = \frac{ktx}{HL} + g(x). This satisfies f(0,t) = 0 and f(L,t) = \frac{kt}{H} provided g(0) = g(L) = 0. If f is to satisfy f_t = \alpha^2 f_{xx} then we need <br /> \frac{kx}{HL} = \alpha^2 g&#039;&#039;. That's a second-order ODE for g and we have two boundary conditions which g must satisfy, so that's OK.

It won't be the case that f(x,0) = g(x) = 0, but we can always set u(x,t) = f(x,t) + v(x,t) where v satisfies <br /> v_t = \alpha^2 v_{xx}<br /> subject to v(0,t) = v(L,t) = 0 and v(x,0) = -g(x).