Solve the following Cauchy problems for a first order PDE: $$(x_2 + x_2^5)v_{x_1} + x_1 v_{x_2} = 0, v(x_1, 0)$$ $$v_0(x_1, 0) = v_0(s)$$
Attempt at the solution:
The characteristic equations are: $$\frac{dx_1}{dt} = x_2 + x_2^5, \frac{dx_2}{dt} = x_1, \frac{dx_3}{dt} = 0$$ $$x_1(s, 0) = s, x_2(s, 0) = 0, x_3(s, 0) = v_0(s)$$
Solving for $x_3$, we get: $$x_3 = v_0(s).$$ Since $dx_2/dt = x_1$ we have: $$\frac{dx^2_2}{dt^2} = x_2+x_2^5 \implies \frac{dx_2}{dt} = \frac{e^{t + c}}{(1 - e^{4(t+c)})^{\frac{1}{4}}} = x_1.$$
I'm unsure how to proceed from this point since integrating further ends up being very complicated. Am I heading in the right direction or is there a trick I'm missing?
Thanks.
$$(x_2 + x_2^5)v_{x_1} + x_1 v_{x_2} = 0$$ System of ODEs for the characteristic curves : $\quad \frac{dx_1}{x_2 + x_2^5}=\frac{dx_2}{x_1}=\frac{dv}{0}$
A first family of characteristic curves comes from $\quad dv=0\quad\to\quad v=c_1$
A second family of characteristic curves comes from $\quad \frac{dx_1}{x_2 + x_2^5}=\frac{dx_2}{x_1} \quad\to\quad x_1^2-x_2^2-\frac{1}{3}x_2^6=c_2$
$c_1=F(c_2)$ any $F$. The general solution of the PDE is : $$v(x_1,x_2)=F(x_1^2-x_2^2+\frac{1}{3}x_2^6) $$ with any differentiable function $F$.
The function $F$ has to be determined according to a given boundary condition.
For example, if the condition is $v(x_1,0)=v_0(x_1)$
$v_0(x_1)=F(x_1^2)\quad\to\quad F(X)=v_0(\sqrt{X}) \quad$ where $\begin{cases} X=x_1^2 \quad\text{on the boundary}\\ X=x_1^2-x_2^2+\frac{1}{3}x_2^6 \quad\text{everywhere} \end{cases}$ $$v(x_1,x_2)=v_0\left(\sqrt{x_1^2-x_2^2+\frac{1}{3}x_2^6}\right) $$