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**Home Assignment 2 / Re: Problem 1**

« **on:**October 14, 2012, 11:31:13 PM »

same thing for part C, how would we decide condition c is the necessary initial condition for unique solution

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same thing for part C, how would we decide condition c is the necessary initial condition for unique solution

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is there any clear solution for part c of question four. thank you?

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A)The problem as is has a unique solution. No extra consitions are necessary. Since $x>3t$, we are confident that $x-2t$ is always positive. I.e. initial value functions are defined everywhere in domain of $u(t,x)$. Using d'Alembert's formula we write:

\begin{equation*}

u(t,x)=\frac{1}{2}\Bigl[e^{-(x+2t)}+e^{-(x-2t)}\Bigr]+\frac{1}{4}\int_{x-2t}^{x+2t}e^{-s}\mathrm{d}s

\end{equation*}

$$= \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$

Stated in this way, $u(t,x)$ is determined uniquely in its domain. OKB)In this case, we need an extra boundary condition at $u_{|x=t}=0$ to find the unique solution:

For $x>2t$ general formula for $u$ is as part (A):

$$u(t,x) = \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$

For $t<x<2t$, story is different:

$$u(t,x)= \phi(x+2t)+\psi(x-2t)$$

where $\phi(x+2t)= \frac{1}{4}\bigl[e^{-(x+2t)}+1\bigr]$ is determined by initial conditions at $t=0$. To find $\psi(x-2t)$, we impose $u_{|x=t}=0$ to solution:

$$u_{|x=t}=\phi(3t)+\psi(-t)=0$$

$$\Rightarrow \psi(s)=-\phi(-3s)$$

$$=\frac{-1}{4}\bigl[e^{3s}+1\bigr]$$

Hence the general solution for $t<x<2t$ is:

$$u(t,x)= \frac{1}{4}\bigl[e^{-(x+2t)}+1\bigr]-\frac{1}{4}\Bigl[e^{3x-6t}+1\Bigr]$$

Note that we would not be able to determine $\psi$ if we did not have the extra condition $u|_{x=t}$. Also note that $u_x|_{x=t}=\frac{1}{2}(e^{-3t}) \neq 0$. This means the problem would have been overdetermined, without any solution, if we considered boundary condition $u|_{x=t}=u_x|_{x=t}=0$.C)In this case we need to impose the strongest boundary condition to get the unique solution:

Case $x>2t$ is identical to part (A) and (B):

$$u(t,x) = \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$

We find the solution in region $-3t<x<-2t$ by imposing boundary conditions $u|_{x=-3t}=u_x|_{x=-3t}=0$ to $u(t,x)= \phi(x+2t)+\psi(x-2t)$. This gives

$$\phi(-t)+\psi(-5t)=0$$

$$\phi'(-t)+\psi'(-5t)=0$$

Differentiating first equation and adding to the second we get $-4\psi'(-5t)=0$. Therefore $\psi(s)=C$, $\phi(s)=-C$ and $u(t,x)$ is identically zero.Note that we would not be able to find $\phi$ and $\psi$ uniquely, if we did not have both boundary conditions at $x=-3t$.

From continuety of $u$ in $t>0$, $x>-3t$, we conclude $u|_{x=-2t}=0$. This helps us to find solution for $-2t<x<2t$. Analogous to part (B) we write:

$$u(t,x)= \phi(x+2t)+\psi(x-2t)$$

Impose $u|_{x=-2t}=0$ to $u$ to get $\psi(-4t)=-\phi(0)=C$. Therefore solution here is:

$$u(t,x)=\frac{1}{4}e^{-(x+2t)}+C$$

By continuity at $x=2t$, we get $C=\frac{3}{4}$. General solution for part (C) can be explicitly formulated as

\begin{equation*}

u(x,y)=

\left\{\begin{aligned}[h]

&\frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}, & x>2t\\

&\frac{1}{4}e^{-(x+2t)}+\frac{3}{4}, & -2t<x<2t\\

&0, & -3t<x<-2t\\

\end{aligned}

\right.

\end{equation*}

sorry, but I am still confused about part B, I understand when x<2t will fail the general solution of u. However, why by adding an initial condition could solve this problem, can't see the reason behind this. Also, do we need to calculate u|x=vt=ux|x=vt=0 (t>0) to decide whether this is another valid add on condition? Appreciate your help

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