# Heat equation in one dimension with constant heat supply

## 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##?

Last edited:

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 $$\frac{kx}{HL} = \alpha^2 g''.$$ 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 $$v_t = \alpha^2 v_{xx}$$ subject to $v(0,t) = v(L,t) = 0$ and $v(x,0) = -g(x)$.