Consider the PDE


with the condition $u(x,0)=f(x)$. I want to solve this using the method of characteristics. The first thing I've done was to solve the characteristic system. In that case the system is

$$\dfrac{dx}{dt}=u, \quad \dfrac{dy}{dt}=1, \quad \dfrac{du}{dt}=0.$$

In that case we get, with $c_1,c_2,c_3$ constants, $u=c_3$, $x=c_1+c_3t$ and $y=t+c_2$.

Now as I understand, the initial condition can be thought of as a curve $\sigma(s)=(s, 0, f(s))$ which cuts the characteristics. The idea then is to write $u$ as a function os $s$ and $t$ that is $s$ picks a characteristic and $t$ the value of $u$ at some point of it.

In that case if we imagine that give $s$ the characteristic is intercepted at $t=0$ we find $c_2=0$, $c_1=s$ and $c_3=f(s)$.

Now for any $t$ on this characteristic $u$ is the same so $u(s,t)=f(s)$. Now we have to express $s$ and $t$ in terms of $x,y$. The point is that we easily get $t=y$, but $s=x-yf(s)$. In that case we would have

$$u(x,y) = f(x-yf(s))$$

which is obviously wrong.

What is wrong here? How is the right way to do this?

  • 1
    $\begingroup$ The "calculus" tag wasn't really appropriate; of course everything in analysis is based on calculus, but that's not a reason to tag. Consider that some users interested in answering PDE questions may be ignoring calculus tag. $\endgroup$ – user147263 Nov 17 '15 at 5:37

Your solution is correct up to and including $s=x-yf(s)$, but you didn't recognize this as an equation for $s$, to be solved in terms of $x,y$. Suppose you can solve it as $s = g(x,y)$. Then plug the solution into $u(x,y)=f(s)$ and obtain $u(x,y)=f(g(x,y))$.

Very simple example: $f(s)=2s$. Then $s = x-2ys$ has solution $s= x/(1+2y)$, and therefore $u(x,y) = 2x/(1+2y)$.

Things get more complicated when the equation $s=x-yf(s)$ has multiple solutions, meaning that characteristics want to intersect each other. This is where you start considering shock waves and their propagation.

Also, it may happen that $s=x-yf(s)$ has no solutions, meaning that $(x,y)$ is either outside of the domain where solution makes sense, or within a rarefaction wave.


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