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!


1 Answer 1


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 $$ u(x,y)=\left(\frac{1}{x}+1-\frac{2}{3}\left(\frac{1}{x}+\frac{1}{y}\right)\right)^{-1}=\frac{3xy}{3xy+y-2x}. \tag{4} $$

  • 2
    $\begingroup$ Answer: $u \left( x,y \right) =\,{\frac {3xy}{3\,xy-2\,x+y}}$ $\endgroup$ Jan 5 at 13:37
  • $\begingroup$ @AleksasDomarkas I get $u(x,y)=\frac{3xy}{5y+2x-2xy}$, could you perhaps elaborate? $\endgroup$ Jan 5 at 16:21
  • 1
    $\begingroup$ I included in my answer a derivation of the solution given by @AleksasDomarkas. $\endgroup$
    – Gonçalo
    Jan 5 at 16:59
  • 1
    $\begingroup$ @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! $\endgroup$ Jan 6 at 9:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .