I was trying to solve the following PDE:
$$\ln(x_2)\partial_1U+x_2U\partial_2U-U=0$$
with the initial condition $$U(t+1,e)=1$$
I managed to get to the following system:
$$\begin{cases} \dot{x}_1(s)=ln(x_2)\\\dot{x}_2(s)=Ux_2 \\\dot{U}(s)=p_1\ln(x_2)+p_2x_2U \\ \dot{p_1}(s)=p_1-p_2p_1x_2 \\ \dot{p_2}(s)=p_2-\frac{p_1}{x_2}-Up_2-p^2_2x_2 \end{cases} $$
where $x_1>0,\quad x_2>0$ and $p_1=\partial_{x_1}U$ and $p_2=\partial_{x_2}U$.
But, I am not able to solve this system of ODEs. Please help!
Best Answer
The characteristic ODEs are
$$\frac{dx_{1}}{\ln x_{2}} = \frac{dx_{2}}{x_{2} u} = \frac{du}{u}$$
The last equality implies ln u $$u - \ln x_{2} = C_{1}$$
The first and third ratios imply
\begin{align} d x_{1} &= \frac{(u - C_{1}) du}{u} \\ \implies u - C_{1} \ln u - x_{1} &= C_{2} \\ \implies u - C_{1} + C_{1} - C_{1} \ln u - x_{1} &= C_{2} \\ \implies \ln x_{2} + (u - \ln x_{2}) (1 - \ln u) - x_{1} &= C_{2} \\ \end{align}
and hence
$$\ln x_{2} + (u - \ln x_{2})(1 - \ln u) - x_{1} = f(u - \ln x_{2})$$
for some arbitrary differentiable function $f$. You can check this satisfies the PDE by differentiation
\begin{align} u_{x_{1}}(1 - \ln u) - (u - \ln x_{2}) u_{x_{1}} u^{-1} - 1 &= u_{x_{1}} f' \\ \implies u_{x_{1}} \left(1 - \ln u - (u - \ln x_{2}) u^{-1} - f' \right) &= 1 \\ \implies u_{x_{1}} &= \frac{1}{\left(1 - \ln u - (u - \ln x_{2}) u^{-1} - f' \right)} \\ x_{2}^{-1} + (u_{x_{2}} - x_{2}^{-1})(1 - \ln u) - (u - \ln x_{2}) u_{x_{2}} u^{-1} &= (u_{x_{2}} - x_{2}^{-1}) f' \\ \implies u_{x_{2}} \left(1 - \ln u - (u - \ln x_{2}) u^{-1} - f' \right) &= - x_{2}^{-1} (\ln u + f') \\ \implies u_{x_{2}} &= \frac{- x_{2}^{-1} (\ln u + f')}{\left(1 - \ln u - (u - \ln x_{2}) u^{-1} - f' \right)} \end{align}
so that
\begin{align} \implies \ln x_{2} u_{x_{1}} + x_{2} u u_{x_{2}} &= \frac{\ln x_{2} - u (\ln u + f')}{\left(1 - \ln u - (u - \ln x_{2}) u^{-1} - f' \right)} \\ &= \frac{u (\ln x_{2} - u (\ln u + f'))}{\left(u - u \ln u - (u - \ln x_{2}) - u f' \right)} \\ &= \frac{u (- u \ln u + \ln x_{2} - u f')}{\left(- u \ln u + \ln x_{2} - u f' \right)} \\ &= u \end{align}
I leave the domain of convergence and initial condition to you (hint for initial condition; the solution to the PDE can also be written as $$u - \ln x_{2} = g(\ln x_{2} + (u - \ln x_{2})(1 - \ln u) - x_{1})$$ for some arbitrary differentiable function $g$, then note that evaluating both sides at $(x_{1}, x_{2}, u) = (t + 1, e, 1)$ makes the LHS vanish identically).