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Exercise 12.6.10 from the book Applied Partial Differential equations (Haberman) seems to be distinctly different from the other exercises. It is formulated thusly:

Solve $\frac{\partial \rho}{\partial t} + t^2 \frac{\partial \rho}{\partial x} = 4 \rho$ for $x>0$ and $t>0$ with $\rho(0,t) = h(t)$ and $\rho(x,0) = 0$. The latter is the condition usually defined as an arbitrary function.

Normally, we would write out the characteristic equations, i.e. $\frac{\partial t}{\partial s} = 1$, $\frac{\partial x}{\partial s} = t^2$ and $\frac{\partial \rho}{\partial s} = 4 \rho$.

We would then parametrise the initial conditions; this is normally $t=0 \to s=0, x=\tau$. And thus $\rho(0, \tau) = 0$. I'm not sure what to do with the second condition that involves h(t). I surmise this is the problem.

Using the parametrised IC, I find $t = s$, $x = \frac{1}{3} s^3 + \tau$.

$\rho$ would be $\rho(s, \tau) = C(\tau) e^{4 s}$.

I thought that applying the value $h(s)$ for $\rho$ on the line $x=\frac{1}{3} s ^3 + \tau = 0$ would be the correct way to go. However, the first initial condition requires that $C(\tau)=0$. So $\rho$ can't have the value $h(s)$ on the mentioned line.

How to proceed? [Preferably a hint, not a solution].

Daimonie
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  • Actually, it turns out $\rho(s, \tau) = C(\tau) e^{4s} + Q(\tau)$. This changes the deal; from the IC we find that $\rho(\tau, 0) = 0$ so C + Q = 0. Then, we have to apply the BC on $\tau = -\frac{s^3}{3}, s$ and we can find an expression for the final answer $\rho(s, \tau)$ and ultimately $\rho(x,t)$. Just to let you know; it was again a simple error. – Daimonie Oct 12 '14 at 14:15
  • On second thought, that can't be. Since that leads to $\frac{\partial \rho}{\partial s} = C(\tau) e^{4s} \cdot 4 = 4\cdot C(\tau)e^{4s} + Q(\tau)$, which only allows for $Q=0$. Too bad; – Daimonie Oct 14 '14 at 06:56

1 Answers1

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it's me again. As you can see, I commented a few times because I had not found it yet.

I just realised my problem: I was thinking of $h(t)$ as entirely arbitrary. But it isn't. $h(t)$ is only defined for $t>0$! That's important, because we can set it to zero elsewhere.

If we do that, the BC that removes $C(\tau)$ is not a problem. Consistency requires that $h(0)=0$, since the BC and IC intersect there.

If we apply the BC first: $\rho(x=0, t) = h(t)$. Of course, $t=s$ and $x = 0 \rightarrow \tau = -\frac{s^3}{3}$. So we look for $\rho(\tau=-\frac{s^3}{3}, s) = h(s)$. This leads to a simple expression for $C\tau)$:

$C(\tau) = h(s) e^{-4s}$. Since $x = 0$, we find that $s=-(3\tau)^\frac{1}{3}$. Thus, $C(\tau)$ is determined.

Finally, since the initial condition is $\rho(\tau=x, s=0) = 0$, we see that we require $h(-(3\tau)^\frac{1}{3})$. Because $x>0$ for the entire problem, it follows that $-(3\tau)^\frac{1}{3} \leq 0$ and consequently $C(\tau) = 0$. The IC is satisfied.

Daimonie
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