Unsolvable characteristic system ODE as a part of PDE solution?

Question

I'm trying to solve the following PDE: $$F(x_1,x_2,u,p_1,p_2)=\text{ln}(x_2)p_1+x_2up_2-u=0 \ \ \ \ \ \ p_i=\partial_iu(x_1,x_2)$$ Where the initial conditions are: $$\begin{cases}x_1(t)=t+1 \\x_2(t)=e \\u(t)=1 \\ p_1(t)=0 \\ p_2(t)=\frac{1}{e}\end{cases}$$ The first step in solving this type of nonlinear PDE is to solve the characteristic system ODE: $$\begin{cases} \dot{x_1}(s)&=\partial_{p_1}F&= \text{ln}(x_2)\\ \dot{x_2}(s)&=\partial_{p_2}F&= x_2u\\ \dot{u}(s)&=p_1\partial_{p_1}F+p_2\partial_{p_2}F&=p_1\text{ln}(x_2)+p_2x_2u \\ \dot{p_1}(s)&=-\partial_{x_1}F-p_1\partial_{u}F&=p_1-p_1p_2x_2 \\ \dot{p_2}(s)&=-\partial_{x_2}F-p_2\partial_uF&=-\frac{p_1}{x_2}-up_2+p_2-x_2p_2^2 \\ \end{cases}$$ And i can't seem to find a way to solve it. Am I missing aclever way to solve it or am I making a mistake deriving it? The solution of the PDE has to be: $$u(x_1,x_2)=\text{ln}(x_2)$$ Here is the original problem:

Answer 1

Let $x_1\equiv x$ and $x_2\equiv y$ $$\ln(y)\frac{\partial u}{\partial x}+yu\frac{\partial u}{\partial y}=u$$ The Charpit-Lagrange characteristic ODEs are : $$\frac{dx}{\ln(y)}=\frac{dy}{yu}=\frac{du}{u}$$ A first characteristic equation comes from solving $\frac{dy}{yu}=\frac{du}{u}$ : $$u-\ln(y)=c_1$$ A second characteristic equation comes from solving $\frac{dx}{\ln(y)}=\frac{dy}{yu}$ :

$\frac{dx}{\ln(y)}=\frac{dy}{y(\ln(y)+c_1)}\quad\implies\quad \int dx=\int\frac{\ln(y)}{y(\ln(y)+c_1)}=\ln(y)-c_1\ln(c_1+\ln(y))$+constant. $$x-\ln(y)+c_1\ln(c_1+\ln(y))=c_2$$ The general solution of the PDE on implicit form $c_1=F(c_2)$ is : $$\big(u-\ln(y)\big)=F\Bigg(x-\ln(y)+c_1\ln\Big(c_1+\ln(y)\Big)\Bigg)$$ $F$ is an arbitrary function (to be determined according to the boundary condition). $$\boxed{\big(u-\ln(y)\big)=F\Big(x-\ln(y)+\left(u-\ln(y)\right)\ln\left(u\right)\Big)}$$

Condition : $x(t)=t+1\quad;\quad y(t)=e \quad;\quad u(t)=1.$ $$\big(1-\ln(e)\big)=F\Big((t+1)-\ln(e)+\left(1-\ln(e)\right)\ln\left(1\right)\Big)$$ After simplification : $$0=F(t)$$ The function $F$ is determined : function null. We put it into the above general solution. $$u-\ln(y)=0$$ The particular solution of the PDE which satisfies the boundary condition is : $$u(x,y)=\ln(y)$$ This shows without ambiguity that the particular solution is unique and doesn't depend on $x$.

NOTE :

The conditions $\frac{du}{dx}=0 \quad;\quad \frac{du}{dy}=\frac{1}{e}$ (joint to the other conditions on $y=e$ ) are overabundant. One check that they don't introduce contradiction.