Toronto Math Forum

Consider problem: $U_x+3U_y=xy, U(0,y)=0$, I proceed as follow:
$$\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}$$
Integrate on first two terms:
$$x-\frac{1}{3}y=C \qquad \color{red}{(*)}$$
Integrate on $\frac{du}{xy} = \frac{dx}{1}$: Error! You must remember that in the equation of characteristics $x$ and $y$ are not independent but  connected by (*).
$$U =x^3-\frac{C}{2}x^2+D = \frac{3}{2}x^3+\frac{1}{2}yx^2+D $$
D must be constant along the integral curve hence $D=\phi(x-1/3y)$
Hence, the general solution is $U =\frac{3}{2}x^3+\frac{1}{2}yx^2+\phi(x-1/3y)$.
Impose initial condition: $U(0,y)=0$ we have $\phi(-1/3y)=0$.

Update:
Hi professor, I have fixed the integration part, but I still have question about the constant D when integrating on U. Should I include this $D=\phi(x-\frac{1}{3}y)$ in the solution to IVBP?

In solving the problem $2U_t+3U_x=0, U(0,x)=f(x)$, tutorial gave answer $U=f(x-\frac{3}{2}t)$ while I have $U=f(3t-2x)$. Are both answers valid? We can use different form of C as long as it represent the same family of characteristics right?

I was a little confused about the region marked in red in this example. For the region on the right we get $\phi(x) = \sin(3x)$ and $\psi(x) = 3\sin(3x)$,  and from the boundary conditions we can get $\phi(x) = \psi(x) = 0$ for $t > 0$, so we have $u = 0$ for the region on the left. For the middle region in red, since $\psi(x - \frac{t}{3})$ is undefined, we can just have it as $\psi(x) = 0$ instead?  The middle region is like a mix of the other two regions so I am a little uncertain.

\begin{gather}
u_t+uu_x= 0;\\[2pt]
\end{gather}

Hi Professor, I don't know what the operator for this equation. Does it have an operator? Or is the operator equal to

\begin{gather}

L= \partial_t + u\cdot\partial_x
\end{gather}

\begin{align}
&\left\{\begin{aligned}
&u_{xy}=0,\\[2pt]
&u_{xz}=0;
\end{aligned}\right.\\[2pt]
\end{align}

For this problem I'm not sure if my approach was correct.
\begin{align}
u_{x}=f(x,z)  \textrm{   from the first equation}
\end{align}
\begin{align}
&u_{x}=g(x,y)\textrm{   from the second equation}
\end{align}
\begin{align}
\textrm{Therefore we must have   }\textrm{   }u_{x} = h(x)
\end{align}
\begin{align}
&u=H(x) + m(y,z)\textrm{ } \textrm{where}\textrm{ } H'(x)=h(x)
\end{align}