# Solving a Quasilinear First-Order PDE with the method of characteristics

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?

• 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 $(x(t),y(t),z(t))$. Feb 26, 2016 at 0:27

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 $x = \sqrt{|C|} sinh t$, $y = \sqrt{|C|} cosh t$.
• @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? Feb 26, 2016 at 0:47
• @Ryuki, I added missing step in the answer, The derivative wrt t of $x^2 - y^2$ is 0 along the characteristc. Feb 26, 2016 at 0:54