I am learning about solving p.d.e.s by the method of characteristics at the moment. I was given an "algorithm" to solve these problems but I want to know also what is going on, how it works and what it is doing intuitively/physically/graphically. It would be great if someone could either explain it or provide a good link To reference material. Thank you.


The basic idea is that we look for parametric curves along which the PDE tells us how the function changes. Suppose you have a smooth parametric curve in the $x-y$ plane: $x = X(t)$, $y = Y(t)$. Consider how a smooth function $u(x,y)$ changes as you move along the curve. The chain rule says $$\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dX}{dt} + \frac{\partial u}{\partial y} \frac{dY}{dt}$$ Now this looks rather like the left side of a first-order PDE $$a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} = c(x,y)$$ In fact, if you can find $X(t)$ and $Y(t)$ such that $\frac{dX}{dt} = a(X(t),Y(t))$ and $\frac{dY}{dt} = b(X(t),Y(t))$, this tells you that along that curve $\frac{du}{dt} = c(X(t), Y(t))$, so that if you know $u(X(0), Y(0))$ you can get $$u(X(s),Y(s)) = u(X(0), Y(0)) + \int_0^s c(X(t),Y(t))\ dt$$

Now, for any point $(x_1, y_1)$ in the plane, suppose you want to find $u(x_1, y_1)$, where $u(x,y)$ satisfies the PDE $a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} = c(x,y)$ plus some boundary condition. Then you want to find a curve $x = X(t)$, $y = Y(t)$ satisfying the system of ordinary differential equations $\frac{dX}{dt} = a(X(t),Y(t))$ and $\frac{dY}{dt} = b(X(t),Y(t))$ that passes through the given point $(x_1, y_1)$, follow that curve to some $(x_0, y_0)$ where your boundary condition tells you the value of $u$, and then you can get $u(x_1, y_1)$ by an integral as above.

  • $\begingroup$ I am confused about the transformation used in method of characteristics. To have change of coordinates, we use the slope dy/dx (=y/1 say) and we get eta=xy. But for the second one, in most cases, it is assumed that zeta=x out of the blue. My question is, how we arrived at zeta=x? any thoughts? $\endgroup$ – zhk Mar 19 '17 at 13:54
  • $\begingroup$ What do you mean by eta and zeta? $\endgroup$ – Robert Israel Mar 19 '17 at 17:34
  • $\begingroup$ These are the new coordinates, eta=eta(x,y) and zeta=zeta(x,y). For more details, please see section 2.3 page 8-9. web.math.ucsb.edu/~grigoryan/124A.pdf $\endgroup$ – zhk Mar 19 '17 at 17:44
  • $\begingroup$ Any comment on my question? $\endgroup$ – zhk Mar 23 '17 at 13:20

This is an intuitive take which is intended to be as simple as possible.

Let's say you're measuring some system for two variables $x$ and $t$, and you discover that you can describe the system by $u_t+a(x,t)u_x=0$.

Okay, but what if $x$ and $t$ just superficially look like independent variables, but in fact $x$ is a function of $t$?

If it were so, then $$\frac{d}{dt}u(x(t),t)=u_t+x'(t)u_x$$ which means that if we can find a function $x(t)$ with the property that $x'(t)=a(x,t)$, then $u(x(t),t)$ would have the property that $$\frac{d}{dt}u(x(t),t)=u_t+a(x,t)u_x$$ which is exactly the type of function we are looking for.

So if it really is the case that our measured variable $x$ actually is a function of the underlying variable $t$, we only need to find a function $x(t)$ such that $x'(t)=a(x(t),t)$ with $x(0)=x_0$ and solve the much easier ODE $$\frac{d}{dt}u(x(t),t)=0$$ and then $u(x(t),t)$ satisfies the PDE we started with.

Usually these problems come with some boundary conditions like $u(x,0)=\phi(x)$, so we solve our easy ODE: $$u(x(t),t)=C$$

If we set $C=\phi(x_0)$ we see that $u(x(t),t)=u(x(0),0)=u(x_0,0)=\phi(x_0)$ (indeed, $u$ is constant wrt $t$)

So $$u(x(t),t)=\phi(x_0)$$

is a solution to the PDE provided we can find an $x(t)$ such that $x'(t)=a(x(t),t)$ with $x(0)=x_0$, which we can. When you find $x(t)$, you can solve it for $x_0$ and substitute so you get a solution function consisting of $x$'s and $t$'s, which you can then differentiate to check that it satisfies the PDE.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.