# Toronto Math Forum

## APM346-2015S => APM346--Tests => Test 1 => Topic started by: Victor Ivrii on February 12, 2015, 07:23:35 PM

Title: TT1 Problem 1
Post by: Victor Ivrii on February 12, 2015, 07:23:35 PM
Consider the first order equation:
\begin{equation}
u_t + t x u_x = 0.
\label{eq-1}
\end{equation}
a. Find the characteristic curves and sketch them in the $(x,t)$ plane.

b. Write the general solution.

c. Solve  equation (\ref{eq-1})  with the initial condition $u(x,0)= e^{-x^2}$. Explain why the solution is fully  determined by the initial condition.

d. Describe domain in which solution of
\begin{equation}
u_t + t x^2 u_x = 0, \qquad x>0
\label{eq-2}
\end{equation}
is fully determined by the initial condition $u(x,0)=g(x)$ ($x>0$)?
Title: Re: TT1 Problem 1
Post by: Yiyun Liu on February 12, 2015, 10:49:23 PM
\begin{array}{l}
b)\\
\frac{{dt}}{1} = \frac{{dx}}{{tx}} = \frac{{du}}{0}\quad .o.d.e\\
ln(x) = \frac{{{t^2}}}{2} + C\\
\end{array}
Title: Re: TT1 Problem 1
Post by: Yiyun Liu on February 12, 2015, 11:22:11 PM
\begin{array}{l}
c)\\
u(x,0) = {e^{ - {x^2}}}\\
f(lnx) = {e^{ - {x^2}}},\\
f(x) = {e^{ - {e^{2x}}}}\\
u(x,t) = {e^{ - {e^{2lnx - {t^2}}}}} = {e^{ - {x^2}{e^{ - {t^2}}}}}
\end{array}
Since the arbitrary function f is uniquely determined by the initial condition, the solution is fully determined and the problem is well-posed.
Title: Re: TT1 Problem 1
Post by: Victor Ivrii on February 14, 2015, 05:51:54 AM
OK. Please write \ln , \sin etc with leading backslash (and trailing space).

Also what about sketching integral curves?

a) Characteristic curves are $x=Ce^{-t^2/2}$ and sketch is attached;
b) Respectively the general solution is $u(x,t)=f(xe^{-t^2/2})$;
c) Solution satisfying initial condition is $u(x,t)=\exp (-x^2 e^{-t^2})$;
d) is a different equation with characteristics $x= 2/(C-t^2)$ and characteristics passing through $\{(x,0): x>0\}$ cover exactly the right half-plane (see the second picture).