I'm starting to learn about the method of characteristics and tried the problem below, but ran into some problems:
$\begin{align*} v_t + \sin^2(x)v_x &= 0, \ t > 0, \ x \in(-\pi, \pi)\\ v(x,0) &= g(x) \end{align*}$
$v, g$ are continuously differentiable functions in all variables.
Consider the ODE
$\begin{align*} X'(t) &= \sin^2(x), \ x \in (-\pi,\pi)\\ X(0) &= x_0 \end{align*}$.
The solution is $X(t) = \sin^2(x)t + c$, where $c$ is a constant.
Since $X(0) = x_0 = 0 + c$, we get $x_0 = c$.
Further, we can check that $v$ is constant along the characteristic curves $(X(t),t)$ defined by the solutions of the ODE above. Thus,
$v(X(t),t) = v(x_0,0) = g(x_0)$.
We have that $x_0 = x - \sin^2(x)t$, hence $v(x,t) = g(x_0) = g(x-\sin^2(x)t)$.
My problem is that, when I try to differentiate the solution, I can't get the result to satisfy the original DE. Also, I'm not used to working on an interval such as $(-\pi, \pi)$, but rather on all of $\mathbb{R}$. Perhaps I need to specify the solution as a piecewise defined function, somehow?
$$v_t+\sin^2(x) v_x=0$$ The Charpit-Lagrange system of characteristic ODEs is : $$\frac{dt}{1}=\frac{dx}{\sin^2(x)}=\frac{dv}{0}$$ A first characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{\sin^2(x)}$
$t=\int\frac{dx}{\sin^2(x)}=-\cot(x)+$constant. $$t+\cot(x)=c_1$$ A second characteristic equation comes from finite $\frac{dv}{0}\implies dv=0$ $$v=c_2$$ The general solution of the PDE expressed on the form of implicite equation $\Phi(c_1,c_2)=0$ is : $$\Phi(t+\cot(x)\:,\:v)=0$$ where $\Phi$ is an arbitrary function of two variables,
or equivalently on explicite form : $$v=F\big(t+\cot(x)\big)$$ $F$ is an arbitrary function, to be determined according to an initial condition.
Condition : $v(x,0)=g(x)$ with a given function $g$.
$v(x,0)=F\big(0+\cot(x)\big)=g(x)$ $$F\big(\cot(x)\big)=g(x)$$
Let $X=\cot(x)\quad\implies\quad x=\cot^{-1}(X)$ $$F(X)=g\big(\cot^{-1}(X)\big)$$ Now the function $F$ is determined. We put it into the above general solution where $X=t+\cot(x)$ and $F\big(t+\cot(x)\big)=g\big(\cot^{-1}(t+\cot(x))\big) $
This leads to the particular solution which satisfies the initial condition :
$$v(x,t)=g\big(\cot^{-1}(t+\cot(x))\big)$$