Toronto Math Forum
APM3462015S => APM346Tests => Test 1 => Topic started 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{eq1}
\end{equation}
a. Find the characteristic curves and sketch them in the $(x,t)$ plane.
b. Write the general solution.
c. Solve equation (\ref{eq1}) 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{eq2}
\end{equation}
is fully determined by the initial condition $u(x,0)=g(x)$ ($x>0$)?

\begin{array}{l}
b)\\
\frac{{dt}}{1} = \frac{{dx}}{{tx}} = \frac{{du}}{0}\quad .o.d.e\\
ln(x) = \frac{{{t^2}}}{2} + C\\
u(x,t) = f(C) = f(lnx  \frac{{{t^2}}}{2}),\quad f\quad arbitary
\end{array}

\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 wellposed.

OK. Please write \ln , \sin etc with leading backslash (and trailing space).
Also what about sketching integral curves?
I graded this problem.
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 $x= 2/(Ct^2)$ and characteristics passing through $\{(x,0): x>0\}$ cover exactly the right halfplane (see the second picture).