16

**Quiz-3 / Re: Q3-T5102**

« **on:**February 11, 2018, 01:35:32 AM »

Given:

$$ u_{tt} - c^2u_{xx} = 0 $$ with $ t > 0, x> 0 $

$$u|_{t=0} = g(x) , u_t|_{t=0} = cg'(x) $$ with $ x > 0 $

$$ (u_x + \alpha u_t) |_{x=0} = 0 $$ for $ t> 0 $

Generally:

$$ u(x,t) = \phi(x+ct) + \psi(x-ct) $$

The initial conditions lead to:

$$ \phi(x) = g(x) - \frac{g(0)}{2} $$

$$ \psi(x) = \frac{g(0)}{2} $$

It follows for the nicer region ($x > ct $) we have:

$$u(x,t) = \phi(x+ct) + \psi(x-ct) = g(x+ct) -\frac{g(0)}{2} + \frac{g(0)}{2} = g(x+ct) $$

While for the case of ($0<x<ct$) means that we have to investigate the behaviour of \psi($\gamma $) with $\gamma = x - ct < 0 $

Applying the boundary condition of $ (u_x + \alpha u_t) |_{x=0} = 0 $

Resulting in :

$$ 0 = \phi'(ct) +\psi'(-ct) + \alpha (c \phi'(ct) - c\psi'(-ct)) $$

Rearranging, and applying the substitution of $ x = -ct $ this gives us:

$$ \psi ' (x) = \phi ' (-x) \frac{ (1 + \alpha)}{\alpha - 1 } $$

This is where I usually have trouble with these sorts of problems, where I don't know where/if I am allowed to integrate both sides/ use the original expressions for $\phi$ and $\psi (x) $ above. Although I do know that I should try to manipulate it into the form of an ODE.

Tentatively, I'd start by taking $ \phi'(x) - \psi'(x) = g'(x) $ rearranging to:

$$ \phi'(x) = g'(x) + \psi'(x) $$

the issue is that substituting this expression into the above results in results in an expression of $\psi'(x)$ and $\psi'(-x)$ , so clearly this tentative step is incorrect, but I am unsure of what to do next.

b. Will be discussed later

$$ u_{tt} - c^2u_{xx} = 0 $$ with $ t > 0, x> 0 $

$$u|_{t=0} = g(x) , u_t|_{t=0} = cg'(x) $$ with $ x > 0 $

$$ (u_x + \alpha u_t) |_{x=0} = 0 $$ for $ t> 0 $

Generally:

$$ u(x,t) = \phi(x+ct) + \psi(x-ct) $$

The initial conditions lead to:

$$ \phi(x) = g(x) - \frac{g(0)}{2} $$

$$ \psi(x) = \frac{g(0)}{2} $$

It follows for the nicer region ($x > ct $) we have:

$$u(x,t) = \phi(x+ct) + \psi(x-ct) = g(x+ct) -\frac{g(0)}{2} + \frac{g(0)}{2} = g(x+ct) $$

While for the case of ($0<x<ct$) means that we have to investigate the behaviour of \psi($\gamma $) with $\gamma = x - ct < 0 $

Applying the boundary condition of $ (u_x + \alpha u_t) |_{x=0} = 0 $

Resulting in :

$$ 0 = \phi'(ct) +\psi'(-ct) + \alpha (c \phi'(ct) - c\psi'(-ct)) $$

Rearranging, and applying the substitution of $ x = -ct $ this gives us:

$$ \psi ' (x) = \phi ' (-x) \frac{ (1 + \alpha)}{\alpha - 1 } $$

This is where I usually have trouble with these sorts of problems, where I don't know where/if I am allowed to integrate both sides/ use the original expressions for $\phi$ and $\psi (x) $ above. Although I do know that I should try to manipulate it into the form of an ODE.

Tentatively, I'd start by taking $ \phi'(x) - \psi'(x) = g'(x) $ rearranging to:

$$ \phi'(x) = g'(x) + \psi'(x) $$

the issue is that substituting this expression into the above results in results in an expression of $\psi'(x)$ and $\psi'(-x)$ , so clearly this tentative step is incorrect, but I am unsure of what to do next.

b. Will be discussed later