$$\frac{\partial u}{\partial x} \frac{\partial u}{\partial y}=1 \qquad \qquad u=0 \; \text{ when } \; x+y=1$$

Find all possible solutions and state where each one exists.


Using the method of characteristics (Charpit's equations), we end up with

$$x=s \pm t \;,\; y=1-s \pm t \;,\; \frac{\partial u}{\partial x} = \pm 1 \;,\; \frac{\partial u}{\partial y}= \pm 1 \;,\; u=2t$$

The set of all solutions is then

$$u(x,y) = \pm(x+y-1)$$

However, how do you know where these solution does/doesn't "exist"?

There seems to be no problems as far as the characteristic projections are concerned (they are a set of straight lines perpendicular to the initial data), and the Jacobian is $J= \pm 2$ which is non-zero.

So is $u$ just supposed to be a multi-valued function in this case?

  • 1
    $\begingroup$ I don't get what you are trying to say? Imposing $u(x,1-x)=0$ leaves us with exactly two solutions namely $u(x,y)= \pm (x+y-1)$. But the question is, is there a reason why one of the solutions should not exist at a particular region in the $xy$-plane? $\endgroup$ Oct 25, 2018 at 13:16
  • $\begingroup$ Maybe you can tell is the boom ir the lesson where the problem is. $\endgroup$ Oct 25, 2018 at 19:26


You must log in to answer this question.