Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Reading about characteristic curves for second-order equations, in particular semi-linear equations of second order with two independent variables:

$a(x,y)u_{xx}+2b(x,y)u_{xy}+c(x,y)u_{y,y}=f(x,y,u,u_{x},u_{y})$ $ $ $ $ $(1)$

My book reference, define characteristic curve to $(1)$, as plane curves along which the PDE can be written in a form containing only total derivatives of $u_{x}$ and $u_{y}$.

I do not understand this definition (only total derivatives of $u_{x}$ and $u_{y}$??? ), I do not know how to see it this, for example, in equation


I read some questions about characteristic curves here, but did not help.

Can anyone help me? Thank you.

share|cite|improve this question
In general, the method of characteristics is only used for first order equations. However, if you're going to use it for higher order ones, then naturally higher order derivatives should turn up. – Ray Yang Feb 14 '13 at 22:59
Sorry @Ray Yang, I do not see how that helps me in my question.Thanks anyway. – Anderson Lima Feb 15 '13 at 14:12

They just mean that the PDE can factored into something of the form $\prod_i\left(a_i\partial_t^{\alpha_i} +b_i\partial_x^{\beta_i}+c_i \right)$, where $\beta_i,\alpha_i\in\mathbb{N}$. So for example, the heat equation is not of this form, but the wave equation is. And once it's in this form, you just solve this nested sequence of characteristic problems.

If you want an example, see how Evans derives the solution for the easiest form of wave equation.

Since you haven't accepted, here are all the details. The example PDE you posted can be factored as $$(\partial_x+\partial_y)(x\partial_x+x\partial_y-1)u=0$$ Solve it by letting $v(x,y):=(x\partial_x+x\partial_y-1)u$. Then we have $v_x+v_y=0$ Solve this for $v$, and then solve $v=(x\partial_x+x\partial_y-1)u$ for u.

share|cite|improve this answer

Another definition: Let $$\sum_{i,j=1}^na_{ij}(x)U_{x_ix_j}+\sum_{i,j=1}^nb_i(x)U_{x_i}+c(x)U=f(x)$$ $$x=(x_1\dots x_n)$$

A smooth surface $S:\Phi(x)=0$ with the domain of $\Phi$ being a subset of the domain of the coefficients in the equation is called a characteristic if $\forall x \in S$ $$\sum_{i,j=1}^na_{ij}(x){\partial\Phi\over\partial x_i}(x){\partial\Phi\over\partial x_j}(x)=0$$ For n=2 we talk of characteristic curves.

Under a twice smooth bijective change of variables $y=y(x)$, the characteristic of the transformed equation is $\Phi((x(y))=0$, where x(y) is the inverse change.

share|cite|improve this answer
Sorry @George, I do not see how that helps me in my question.Thanks anyway. – Anderson Lima Feb 15 '13 at 14:14
see this – Student Feb 16 '13 at 13:32
@George Maybe you can explain more please? It looks interesting but I don't understand. Why aren't the $b_i$ and $c$ coefficients involved in determining whether the surface is characteristic? – StuartHa Mar 3 '13 at 6:53
Because they are the coefficients in front of lower order derivatives, I guess. That is how we defined it in class, @Stuart. I never questioned why, but the answer is that it leads to nice results. – Student Mar 3 '13 at 17:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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