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APM346-2018S => APM346--Tests => Final Exam => Topic started by: Victor Ivrii on April 11, 2018, 02:35:31 PM

Title: FE-P1
Post by: Victor Ivrii on April 11, 2018, 02:35:31 PM
Solve by the method of characteristics the BVP for a wave equation
&u_{tt}-  16 u_{xx}=0,\qquad 0<x<\infty , \; t>0\label{1-1}\\[2pt]
& u(x,0)=f(x),\label{1-2}\\[2pt]
& u_t(x,0)=g(x),\label{1-3}\\[2pt]
& (u_x  -u )(0,t)= h(t)\label{1-4}
with $f(x)=4e^{-2x}$,   $g(x)=16e^{-2x}$ and  $h(t)=e^{-8t}$. You need to find a continuous solution.
Title: Re: FE-P1
Post by: Tristan Fraser on April 11, 2018, 05:55:16 PM
General solution of

$u =  \phi (x+4t) + \psi (x-4t)$
$u_x =  \phi'(x+4t) + \psi'(x-4t)$
$u_t = 4(\phi' (x+4t) - \psi' (x-4t) $
For $x>4t$:

$4e^{-2x} =  \phi(x) + \psi(x)$
$4e^{-2x} = \phi'(x) - \psi'(x)$
integrate and rearrange to get:

$\phi(x) =  e^{-2x}$ and $\psi(x) = 3e^{-2x}$

$$u = e^{-2(x+4t)} + 3e^{-2(x-4t)} \qquad x>4t$$

For the case of $0<x<4t$:
$u_x =  \phi'(x+4t) + \psi'(x-4t)$ and $u$ are used in BC:

$\phi'(4t) + \psi'(-4t) - \phi(4t) -\psi(-4t) = e^{-8t}$

Then let $x = -4t$ and move the $\psi(x)$ to one side:

$\psi'(x) - \psi(x) = e^{2x} - \phi'(-x) + \phi(-x) $
We know two things: first the reason we're looking at $\psi$ instead of $\phi$ is that it has a chance of being negative, while the other does not, and thus, the second thing is that we can also plug in our previously used functions for $\phi$. The other nice thing is that you can rearrange to ODE:

$(e^{-x} \psi(x))' = e^{x} + (-2e^{x} + e^{x})$ integrate both sides:

$e^{-x} \psi(x) = C$
$\psi(x) =  Ce^{x}$

$u =  e^{-2(x+4t)} + Ce^{x-4t}$

Then check continuity, plug in for $x = 4t$ for both u(x,t), and make sure they match up:

$u = e^{-16t} + C =  e^{-16t} + 3$

Therefore $C= 3$, ensuring continuity.

EDIT: There is a chance I might have misplaced a minus sign, please let me know if I have made such an error, as I will aim to correct it promptly.
Title: Re: FE-P1
Post by: Jingxuan Zhang on April 11, 2018, 07:20:42 PM

I think
$$(e^{-x} \psi(x))' = e^{x} + 2e^{x} + e^{x},$$
not minus.
Title: Re: FE-P1
Post by: Tristan Fraser on April 11, 2018, 08:13:38 PM
@Jingxuan: Thank you!

The adjusted solution gives us;

$(e^{-x} \psi(x)) = 4e^{x} + C $

and then:

$\psi(x) =  4e^{2x} + Ce^{x}$

Now, this gives us:

$u(x,t) =  4e^{2(x-4t)} + Ce^{x-4t} + e^{-2(x+4t)}$

Then checking continuity gives us (x=4t):

$u(x,t) =  4+ C +e^{-2(8t)} = 3+ e^{-16t}$ implying that $C= -1$.

$$u(x,t) =  4e^{2(x-4t)} - e^{x-4t} + e^{-2(x+4t)}\qquad 0<x<4t$$
Title: Re: FE-P1
Post by: Victor Ivrii on April 12, 2018, 05:04:43 AM
Solution is correct but should be written in the form
& ...... && \text{case 1},\\
& ...... && \text{case 2}
where "case 1" and "case 2" (or may be more cases?) should be described.

So that anyone could easily compare what he or she got with yours. I actually highlighted cases in your posts.

From  (\ref{1-1}):
u(x,t)=\phi (x+4t)+\psi (x-4t).
Plugging to (\ref{1-2})--(\ref{1-3}),   we conclude that for $x>0$  $\phi(x) +\psi(x)=4e^{-2x}$, $4 \phi'(x)-4\psi'(x)=16e^{-2x}$; integrating the second equation we get  $\phi (x)-\psi(x)=-2e^{-2x}$ (since we can select constant equal to $0$ here) and finally
\phi(x)=e^{-2x}, \quad \psi(x)=3e^{-2x}\qquad \text{for  } x>0.
We need to find $\psi(x)$ for $x<0$. Plugging (\ref{1-5}) into (\ref{1-4}) we conclude that
&&&\phi'(4t)-\phi(4t)+\psi'(-4t)-\psi(-4t)= e^{-8t}\qquad\text{for }t>0,\iff\\
&\text{and plugging $x=-4t$ we see that}\\
&&&\psi'(x)-\psi(x)=4e^{2x}\qquad\text{for  }x<0,\\
&\text{and solving this ODE}\\
&&&(\psi e^{-x})'= 4e^{x}\implies \psi(x) e^{-x}=4e^{x}+C
and therefore $\psi(x)=4e^{2x}+Ce^{x}$ for $x<0$. Since $\psi(+0)=3$, $\psi(-0)=4+C$ we need for continuity $C=-1$. So
 \psi(x)=4e^{2x}-e^{x}\qquad \text{for } x>0.
u(x,t)= \left\{\begin{aligned}
&e^{-2x-8t}+3e^{-2x+8t} \qquad &&x>4t>0,\\
&e^{-2x-8t} + 4e^{2x-8t}-e^{x-4t}  &&0<x< 4t.

Almost everybody got correctly $\phi(x),\psi(x)$ for $x>0$ and $u(x,t)$ for $x>4t$.
Most correctly wrote equation $\psi' -\psi =4e^{2x}$ for $x<0$.
But then correctly finding one solution $\psi= 4e^{2x}$ (mostly by the method of undetermined coefficient), they wrote $\psi=4e^{2x}+C$ rather than $\psi=4e^{2x}+Ce^{x}$ (WTH). Which in turn yielded term $-1$ instead of correct one $-e^{x-4t}$ in the very end for $0<x<4t$.