Toronto Math Forum
APM3462016F => APM346Misc => Topic started by: Victor Ivrii on September 22, 2016, 07:33:09 AM

Post solution (without screenshots)
Find the general solution $u=u(x,y,t)$
\begin{equation}
u_t+2 tu_x + xu_y=0
\end{equation}
and also a solution satisfying initial condition $u_{t=0}=\sin (x+y)$.

First, we find the characteristics along which $u$ is constant:
\begin{equation}
dt=\frac{dx}{2t}=\frac{dy}{x}
\end{equation}
Combining the first and second, we have $t^2=x+C_1$. So $x=t^2C_1$. Subbing this in the last fraction and then combining with the first we obtain $t^3/3C_1t=y+C_2$, or subbing in for $C_1$, $2t^3/3+xty=C_2$. So our curve in $\mathbb{R}^3$ is described by the intersection of two surfaces, namely $t^2=x+C_1$ and $2t^3/3+xty=C_2$. Since $u$ is constant along this curve, the general solution takes on the form of $u(x,y,t)=\phi(C_1,C_2)=\phi(t^2x,2t^3/3+xty)$. Now using the initial condition, we have $u_{t=0}=\sin(x+y)=\phi(x,y)$. Thus we have $\phi(x,y)=\sin(xy)$. So we have the solution:
\begin{equation}
u(x,y,t)=\sin(xt^2+2t^3/3xt+y)
\end{equation}
A quick check shows that this is indeed a solution to the ODE and satisfies the initial conditions.