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I recently came across the method of characteristics (before that I had only used separation of variables) and decided to read up on it. I tried reading the wikipedia article on it, but I'm still not clear how to get the solution! They all only seem to give the characteristics... I thought I'd start with a very simple form of their Eq.(1) as:

$$\frac{\partial u}{\partial x}+a(x,y,u) \frac{\partial u}{\partial y}=0$$

I'm not clear as to how to solve this, as it starts with "suppose a solution $u$ is known" (which I don't know!), and with $z=u(x,y)$, I end up with

$$\begin{align} dx/dt&=1\\ dy/dt&=a(x,y,z)\\ dz/dt&=0 \end{align} $$

How do I proceed from here? How do I get $u(x,y)$? I know I need a boundary condition for a D.E., but I don't know what to choose... is $u(0,y)=u_0$ a sufficient BC?

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The solution is already given in the linked document – Nana Jun 26 '11 at 19:33
@Nana: Do you mean John Fritz' PDE book? If so, could you please post the solution? I've ordered the book today and it should arrive in a week, – user7815 Jun 26 '11 at 20:09
That would be Fritz John, not John Fritz. – Corey Jun 26 '11 at 20:18
@Corey: oops, you're right. My mind automatically read that as J.F because Fritz is a common last name! thanks for correcting. – user7815 Jun 26 '11 at 20:20
@doob: I actually meant the Wikipedia article. You can however find additional info in this document: – Nana Jun 26 '11 at 20:25
up vote 3 down vote accepted

I don't like the Wikipedia article either. Try looking at Walter Strauss' PDE book (that's where I learned this method from--and I think he explains it much better[but my professor explained it better than both!]).

Firstly, in that example from wikipedia, I don't see the initial condition anywhere: $u(x_0,0) = f(x_0)$. Also I'm going to make $a$ a constant; you'll be able to see how to work from there.

To solve this pde, you want a function $u$ such that when we take a derivative of $u$ with respect to some variable $s$, we get both sides of the equation we want (both 0 and $u_x + au_y$)

Remember from Calculus 3 the chain rule: $\dfrac{du}{ds} = \dfrac{\partial{u}}{\partial{x}}\dfrac{dx}{ds} + \dfrac{\partial{u}}{\partial{y}}\dfrac{dy}{ds} = 0$, with zero being what we want the RHS of our equation to look like (it could be something else).

Now we see something! If we set $\dfrac{dx}{ds} = a, \dfrac{dy}{ds} = 1, \dfrac{du}{ds} = 0$, we recover what we were looking for!

Now we have three ODEs in a single variable. We integrate, and we need some initial conditions (which [kinda] have). Now we have the characteristics name: Along this made up parameterization if we let $x(0) = x, y(0) = 0$, we recover our original initial condition:

$u(x(0),y(0)) = u(x_0,0) = f(x_0)$!

So let's integrate our 3 ODEs:

$x(s) = as + c_1$

$y(s) = s + c_2$

$u(s) = c_3$

use our initial conditions:

$x(0) = x_0 \Rightarrow c_1 = x_0, x(s) = as + x_0$

$y(0) = 0 \Rightarrow c_2 = 0, y(s) = s$

$u(0) = f(x_0) \Rightarrow c_3 = f(x_0), u(s) = f(x_0)$

Now we must solve for $x_0$, since we have $u(s) = u(x(s),y(s))$ but that does us no good.

Look at our solved ODE in x! $x_0 = as-x$ and now our second ODE! $s = y$! Therefore

$u(x(s),y(s)) = f(x_0) = f(ay-x)$.

This means the solution to the PDE $au_x + u_y = 0$ with initial condition $u(x_0, 0) = f(x_0)$ is solved by changing $f(x_0)$ to $f(ay-x)$ (ex: $f(x) = 3x$ becomes $f(ay-x) = 3(ay-x)$)

Hope this helped. Please let me know if I can help more. I left out some of the geometry of why this exactly works, but now you should be able to solve them.

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this is a great answer! Maybe you could also comment on this (unanswered) question? @mathmath8128… – GregVoit Jul 12 '15 at 12:18

You can still follow the method in

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

The remaining is $\dfrac{dy}{dt}=a(x,y,u)=a(t,y,u_0)$ with letting $y(0)=f(u_0)$

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