# Nonhomog heat equation that's piecewise

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1. Mar 6, 2016

### Panphobia

1. The problem statement, all variables and given/known data

$$u_{t}=u_{xx}+f(x) \\ u(0,t)=50 \\ u(\pi , t)=0 \\ u(x,0)=g(x)$$
$$0<x<\pi \\ t>0$$
$$f(x)=\begin{cases} 50 & 0<x<\frac{\pi}{2} \\ 0 & \frac{\pi}{2}\leq x< \pi \end{cases}$$
$$g(x)=\begin{cases} 0 & 0<x<\frac{\pi}{2} \\ 50 & \frac{\pi}{2}\leq x< \pi \end{cases}$$

So what I tried to do here is use the principle of superposition to split this problem up into two different problems $m(x,t),n(x,t)$.

$$m_t=m_{xx} \\ m(0,t)=50 \\ m(\pi,t)=0 \\ m(x,0) = g(x)$$
and
$$n_t=n_{xx}+f(x) \\ n(0,t)=0 \\ n(\pi,t)=0 \\ n(x,0) =0$$

I know how solve the first PDE easily, but the second one is giving me some trouble. I know that you are supposed to do a change of variables and then solve it that way, but how do you take care of the piecewise function $f(x)$ when you are transforming back from the change of variables? Will you just have a piecewise solution in the end?

2. Mar 6, 2016

### Brian T

The heat equation has the smoothing property which smooths out any discontinuities in the data, so no the solution after t=0 should not be piece wise

3. Mar 6, 2016

### Panphobia

So when you transform the second with a change of variables you get something like $$v_t=v_{xx} \\ v(0,t)=0 \\ v(\pi, t)=0 \\ v(x,0)=\int \int f(x) dx - Ax - B$$. Solving for this problem $v$ isn't too difficult, but when you transform back, by doing $$u(x,t) = v(x,t)-\int \int f(x) dx + Ax + B$$. Then we see that $f(x)$ is piecewise, so wouldn't that make the whole solution piecewise too?

4. Mar 6, 2016

### LCKurtz

As Brian T said, any discontinuities get smoothed out, but, yes, you are correct that the function would still be expressed as a piecewise formula.