Toronto Math Forum
APM3462016F => APM346Tests => TT1 => Topic started by: Victor Ivrii on October 19, 2016, 10:24:30 PM

Consider the first order equation:
\begin{equation}
u_t + xt u_x =  u.
\label{eq11}
\end{equation}
(a) Find the characteristic curves and sketch them in the $(x,t)$ plane.
(b) Write the general solution.
(c) Solve equation (\ref{eq11}) with the initial condition $u(x,0)= (x^2+1)^{1}$.
Explain why the solution is fully determined by the initial condition.

a) Characteristic Equation
\begin{equation} \frac{dt}{1} = \frac{dx}{xt} = \frac{du}{u} \end{equation}
From $ \frac{dt}{1} = \frac{dx}{xt} $, $\frac{t^2}{2} + \ln c = \ln x$, thus $ x = ce^\frac{t^2}{2}$
b) General Solution
From $ \frac{dt}{1} = \frac{du}{u}$, $t + \ln k = \ln u$
So $u = ke^{t} = \phi(xe^{\frac{t^2}{2}})e^{t} $
c) Since $u_{t=0} = \frac{1}{1+x^2}$,
$\phi(x) = \frac{1}{1+x^2} $
Therefore, \begin{equation} u(x,t) = \frac{1}{1+x^2e^{t^2}}e^{t}\end{equation}

Just to add onto Roro Sihui Yap's solution for (c), the solution is fully determined by condition at $t=0$ because all characteristics intersect the $x$ axis and do not intersect with each other.

:D