Author Topic: Problem 1  (Read 2334 times)

Djirar

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Problem 1
« on: September 25, 2012, 12:06:23 PM »
my solution to problem 1.

Victor Ivrii

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Re: Problem 1
« Reply #1 on: September 25, 2012, 03:08:21 PM »
So, what are the answers for (c), (d)? Especially (d). Formulate them explicitly

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Rouhollah Ramezani

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Re: Problem 1
« Reply #2 on: September 25, 2012, 04:28:16 PM »
a,b) This is a first order linear PDE with constant coefficient. We begin by writing  equation of characteristic lines:
\begin{equation*}
\frac{dy}{-3}=dx
\Rightarrow y=-3x+C
\end{equation*}
This implies $u(x,y)=\phi(3x+y)$, for some arbitrary $\phi$. Letting $u|_{x=0}=e^{-y^2}$ we get:
$$\phi(y)=e^{-y^2}
$$
Hence solution to IVP is$$
u(x,y)=e^{-(3x+y)^2}$$
c) With initial condition $u|_{x=0}=y$ we have $\phi(y)=y$, $y>0$. Hence the solution is $$u(x,y)=3x+y$$
Since $3x+y>0$ $\forall x,y>0$, this solution is valid everywhere in the domain of $u(x,y)$.
$$$$
d) In this case, the solution is defined only where $3x+y>0$ i.e. where $y>|3x|$. To find solution on the whole domain, we need to impose another condition e.g. at $y=0$. Letting $u|_{y=0}=x$ we get:
$$\phi(3x)=x, (x<0) \rightarrow \phi(x)=\frac{x}{3}, (x<0)$$ $$\Rightarrow u(x,y)=x+ \frac{y}{3}$$
Above equation is constrained to $x+ \frac{y}{3}<0$.
Final solution would be:
\begin{equation*}
u(x,y)= \left\{
\begin{aligned}
&3x+y,&y>|3x|\\
&x+ \frac{y}{3}, &y<|3x|
\end{aligned}
 \right.
\end{equation*}