# Finding solution of semilinear PDE using method of characteristics

Using Evans' book, I'm trying to find an explicit solution of the following problem:

Consinder the boundary value problem $$x_1^2u_{x_1}(x_1,x_2)-x_2^2u_{x_2}(x_1,x_2)=u^2(x_1,x_2)\qquad (x_1,x_2)\in\mathbb{R}^2,$$ $$u=1\qquad \text{on }\Gamma,$$ where $$\Gamma=\{(x_1,x_2)\in\mathbb{R}^2\mid x_1>0,x_2=2x_1\}$$.

Verify that the boundary value problem is locally solvable and use the method of characteristics to find an explicit solution.

So far, I don't know how to show the first part, should I show that the triplet $$(p^0,z^0,x^0)$$ is noncharacteristic?

For the second part, I have done this, trying to follow Example 2 in $$\S$$3.2.2.b. in Evans' book and also using that notation :

We can rewrite the PDE to look like this: $$\begin{pmatrix} x_1^2\\-x_2^2 \end{pmatrix}p(x_1,x_2)-z^2(x_1,x_2).$$ Which gives the system $$\begin{cases} \dot{x^1}(s)=(x^2)^2,\qquad \dot{x^2}(s)=-(x^2)^2\\ \dot{z}(s)=z^2.\end{cases}$$ When trying to continue know, I think I need to get a system where we integrated the previous, giving $$\begin{cases}x^1(s)=\frac{x^0}{1-sx^0},\qquad x^2(s)=\frac{2x^0}{1-s(2x^0)}\\z(s)=\frac{z^0}{1-sz^0}=\frac{g(x^0)}{1-sg(x^0)}=\frac{1}{1-s}.\end{cases}$$

But from here I don't know how to continue, for which $$x_0,s$$ do I get $$(x_1,x_2)=(x^1(s),x^2(s))$$? I would say after this, we can conclude $$u(x_1,x_2)=u(x^1(s),x^2(s))=z(s)=\frac{1}{1-s}$$, with the appropriate value for $$s$$.

I don't know if this approach is correct and whether the steps are valid. So any advice/tips would be very much welcome. Also anything that could help we solve the first part would be welcome!

In my opinion, in this case it's simpler to solve the parametrization invariant Lagrange-Charpit equations $$(x=x_1, y=x_2)$$ $$\frac{dx}{x^2}=-\frac{dy}{y^2}=\frac{du}{u^2}. \tag{1}$$ The solution to $$\frac{dx}{x^2}=-\frac{dy}{y^2}$$ yields $$\frac{1}{x}+\frac{1}{y}=C_1$$. Similarly, the solution to $$\frac{dx}{x^2}=\frac{du}{u^2}$$ yields $$\frac{1}{u}=\frac{1}{x}+C_2$$. From these solutions it follows that $$u(x,y)=\left(\frac{1}{x}+f\left(\frac{1}{x}+\frac{1}{y}\right)\right)^{-1}. \tag{2}$$ One can determine the the function $$f(\cdot)$$ using the boundary condition $$u(x,2x)=1$$: $$1=\left(\frac{1}{x}+f\left(\frac{1}{x}+\frac{1}{2x}\right)\right)^{-1} =\left(\frac{1}{x}+f\left(\frac{3}{2x}\right)\right)^{-1}$$ $$\implies f\left(\frac{3}{2x}\right)=1-\frac{1}{x} \implies f(t)=1-\frac{2t}{3}, \tag{3}$$ hence $$u(x,y)=\left(\frac{1}{x}+1-\frac{2}{3}\left(\frac{1}{x}+\frac{1}{y}\right)\right)^{-1}=\frac{3xy}{3xy+y-2x}. \tag{4}$$
• Answer: $u \left( x,y \right) =\,{\frac {3xy}{3\,xy-2\,x+y}}$ Jan 5 at 13:37
• @AleksasDomarkas I get $u(x,y)=\frac{3xy}{5y+2x-2xy}$, could you perhaps elaborate? Jan 5 at 16:21
• @Gonçalo I see now, I found the wrong value for $f(t)$, which lead to a different outcome, thank you for taking the time to further work it out! Jan 6 at 9:24