Toronto Math Forum
APM3462018S => APM346Lectures => Topic started by: Tristan Fraser on February 09, 2018, 10:19:46 PM

Given:
$$ u_{tt}  c^2 u_{xx} = 0 $$(1) for $ 0<x<\infty $
$$ u _{t=0} = g(x) , u_t  _{t=0} = h(x) $$(2) for $0<x$, and additionally
$$ (\alpha u + \beta u_t) _{x=0} = q(t) $$(3) for $ t> 0 $
We are asked to evaluate and find the general solution for both regions $ x > ct $ and $0<x<ct$
Given that the most general solution under the first condition is $$u(x,t) = \phi (x+ct) + \psi(xct) $$(I), we will focus on the latter case outlined, i.e. $x<ct$
We know from the definitions that:
$$\phi(x) = \frac{1}{2} g(x) + \frac{1}{2c} \int_{0}^{x} h(x')dx' $$(4) and
$$\psi(x) = \frac{1}{2} g(x)  \frac{1}{2c} \int_{0}^{x} h(x')dx' $$(5)
We can begin examining our boundary conditions. As usual, the particular issue is that $\psi(xct)$ is a problem in this region, as the values might be negative, while $\phi(x+ct)$ will not be
Jumping straight into (3), we apply it to (I), to get:
$$ q(t) = \alpha ( \phi(ct) + \psi(ct) ) + \beta c (\phi(ct) '  \psi(ct) ' ) $$(6) and applying the relation $ x = ct $ to (6):
$$ q(\frac{x}{c}) = \alpha(\phi(x) + \psi(x) ) + \beta c (\phi(x)'  \psi(x)') $$(6')
From here, my steps get a bit more uncertain, where I take the (total) derivative of (4) to be :
$$\phi(x) ' = \frac{1}{2} g'(x) + \frac{1}{2c} (h(x)  h(0) ) + \int_{0}^{x} h'(x')dx' $$(7) using Leibnitz's Rule of Integration
plugging into (6') gives me:
$$ q(\frac{x}{c}) = \alpha( \frac{1}{2}g(x)  \frac{1}{2c}\int_{0}^{x} h(x')dx' + \psi(x)) + \beta c (\frac{1}{2} g'(x) + \frac{1}{2c} (h(x)  h(0) ) + \int_{0}^{x} h'(x')dx  \psi(x)') $$(8)
NOTE:
I originally misread the example, as I meant to talk about example 4, but I have modified the Robin condition such that it involves $\alpha u $ and $\beta u_t $ as opposed to the original wording of the problem.
This leaves me two questions:
1. Is this the correct procedure to get the conclusion outlined in the example?
2. What steps do I need to take to get the conclusion outlined (an expression for $\psi $ only in terms of functions of $q,\psi', \phi' $)?
3. Is this procedure the correct one, when considering boundary conditions?

Equations (6) and (6') can be written as ODEs with respect to $ \psi(x) $:
$$ q(t) = \alpha ( \phi(ct) + \psi(ct) ) + \beta c (\phi(ct) '  \psi(ct) ' ) $$
$$ \alpha \psi(ct)  \beta c (\psi(ct))' = q(t) \alpha \phi(ct)  \beta c (\phi(ct))' $$
and you can relate the above equation to the general $ y + y' = p(t) $ equation from ODEs to solve.

Tristan, your modification of example trivializes it. Indeed, you just have an ODE as $t=0$, which allows you to find $u_{x=0}$ (with continuity of $u$ condition) and you arrive to the standard $u_{t=0}$ case.