### Author Topic: Stuck on Example 4 of 2.6 (modified)  (Read 1004 times)

#### Tristan Fraser

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##### Stuck on Example 4 of 2.6 (modified)
« 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 $$u(x,t) = \phi (x+ct) + \psi(x-ct)$$(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 $\psi(x-ct)$ 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:

$$q(\frac{-x}{c}) = \alpha( \frac{1}{2}g(-x) - \frac{1}{2c}\int_{0}^{-x} h(x')dx' + \psi(x)) + \beta c (\frac{1}{2} g'(-x) + \frac{1}{2c} (h(-x) - h(0) ) + \int_{0}^{-x} h'(x')dx - \psi(x)')$$(

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?

« Last Edit: February 09, 2018, 10:28:36 PM by Tristan Fraser »

#### Ioana Nedelcu

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##### Re: Stuck on Example 3 of 2.6
« Reply #1 on: February 09, 2018, 10:36:47 PM »
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.

#### Victor Ivrii

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.