# Unsolvable characteristic system ODE as a part of PDE solution?

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:

• See my answer in math.stackexchange.com/questions/4936784/…. Commented Jun 24 at 0:33
• @Gonçalo How can you be so sure that the solution shouldn't depend on $x_1$? yes the derivative $p_1$ is zero and $u=1$ but that is only for $x_2=e$. How does that give us confidence that for other $x_2$ it wont start depending on $x_1$? Commented Jun 24 at 0:55
• I'm not so sure. That's why I wrote "(...) seems to imply that $U(x_1,x_2)$ does not depend on $x_1$" and "I make no claim that the solution is unique." Commented Jun 24 at 1:03

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.

• Hey that is a great answer. I didn't even notice that this PDE is quasi-linear and just assumed from the getgo that it's nonlinear. Thats why I found the conditions for $p_i$. Can I ask you though, why doesn't the more general method for any nonlinear PDE work here? It seems that the system is still unsolvable. My intuition tells me that it should still be solvable even with the more general nonlinear method. Commented Jun 24 at 9:41
• @Krum Kutsarov. Sorry I don't understand your question. What is the "system" that you call "unsolvable" ? Commented Jun 24 at 12:06
• I understood where is my confusion. I was wondering why the $\dot{p_i}$ part of the system is unsolvable but I thought about it and the only reason we would need to sove this part is if those functions are part of the first three equations which isnt the case here. Commented Jun 24 at 12:11
• OK. All's well that ends well. Commented Jun 24 at 12:23