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Nonhomog heat equation that's piecewise

  1. Mar 6, 2016 #1
    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. jcsd
  3. Mar 6, 2016 #2
    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
     
  4. Mar 6, 2016 #3
    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?
     
  5. Mar 6, 2016 #4

    LCKurtz

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    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.
     
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