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This is a general question about the case when the characteristic equations are such that are not (easily) solvable. The following is just an example; in particular, the right-hand side need not be zero.

Suppose we have an equation like $yu_x(x,y)+xu_y=0$ subject to some appropriate condition (e.g. $u(0,x)=h(x)$ for some function $h$).

As this is a quasilinear first-order PDE, we can try and solve it using the method of characteristics. So, we set $x=x(s,t)$, $y=y(s,t)$, and $z(s,t)=u(x(s,t),y(s,t))$. We have to solve $$\frac{dx}{dt}=y,\quad \frac{dy}{dt}=x,\quad \frac{dz}{dt}=0$$ subject to some conditions. The third equation gives $z(s,t)=A(s)$ for some function $A(s)$ to be found using the conditions. However, the first two equations lead nowhere because to solve, say, the first one, you have to have an expression for $y$ in terms of $s$ and $t$. But you don't have this because in order to solve the second equation you have to have an expression for $x$ in terms of $s$ and $t$.

(We could perhaps write $y=\int x\,dt + C(s)$ for some $C$ and then substitute in $$\frac{dx}{dt}=\int x\,dt+C(s)$$ but then how do solve for $x$ from this?)

So my question is: how does the method of characteristics apply in this situation? Is it perhaps that we can solve these equations but I just don't see why?

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    $\begingroup$ Notice that $\frac{dx}{dt} = y$ and $\frac{dy}{dt} = x \implies \frac{d^2x}{dt^2} = x$. After solving for $x(t)$ you can substitute this in the $y$-equation and solve this. Thus you end up with nice ODE's for all the characteristic functions $(x(t),y(t),z(t))$. $\endgroup$
    – Winther
    Feb 26, 2016 at 0:27

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There is formal way to find eigenvalues of coefficent matrix, and get the solution. There is a short way. Notice that $$x\frac{dx}{dt} -y \frac{dy}{dt} = 0$$ Furthermore, $$ \frac{d}{dt}(x^2 - y^2) = 0 $$ so $x^2 - y^2 = C = Constant$

If C > 0 or x > y, $x = \sqrt{C}cosh t$, $y = \sqrt{C}sinh t$; If C = 0 or x = y, $x = y = t$; If C < 0 or x < y $x = \sqrt{|C|} sinh t$, $ y = \sqrt{|C|} cosh t$.

Edit to address the general question.

Between solving characteristics equation explicitly and numerical solution, there is a half-way approach as I demonstrated for this question: find a function f(x,y) that would be constant along the characteristics. It is kind of "level set" approach to describe the soultion of PDE.

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  • $\begingroup$ This is a nice trick for this particular problem, however if you read the first line the question is about the general case: "This is a general question about the case when the characteristic equations are such that are not (easily) solvable. The following is just an example" $\endgroup$
    – Winther
    Feb 26, 2016 at 0:30
  • $\begingroup$ @Winter, assume the charatristic exists, and use numerical method to trace through characteristic equations. If you cannot solve, you have to prove they exists for some domain, and use numerical methods to solve it. $\endgroup$
    – runaround
    Feb 26, 2016 at 0:35
  • $\begingroup$ @runaround Thank you for the reply. I don't really see though how the displayed equation implies that $x^2-y^2=\mathrm{constant}$. Can you elaborate a bit? $\endgroup$
    – Zugzwang14
    Feb 26, 2016 at 0:47
  • $\begingroup$ @Ryuki, I added missing step in the answer, The derivative wrt t of $x^2 - y^2$ is 0 along the characteristc. $\endgroup$
    – runaround
    Feb 26, 2016 at 0:54

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