# Method of characteristic for second order pde

Can I use the method of characteristic to solve second order pdes? For instance I canconsider the equation $$u_t+u_x=u_{xx}$$

• The 'standard' method does not work (atleast not for this equation) as one needs a first order (quasi) linear PDE for it to apply. But there do exist generalisations though. These are a bit more messy to apply (one generally needs to perform some change of coordinates). See for example here (page 10+) for how to do it for second order PDEs. – Winther Sep 29 '14 at 22:11
• @thanasissdr That is exactly the generalized method I was talking about. – Winther Sep 29 '14 at 22:22
• @Winther Sorry, didn't check your link. – thanasissdr Sep 29 '14 at 22:54

For the equation you specify, consider the change of coordinates

$$\tau = t \qquad y = x - t$$

we have

$$\partial_t = \partial_\tau - \partial_y$$

and

$$\partial_x = \partial_y$$

from the change of variables formula. So

$$u_t + u_x = u_{xx} \implies u_\tau = u_{yy}$$

In other words, your equation is in fact a variable-transformed heat equation and for this method of characteristics will not work (well).

• Where the heck is the stupid downvote from? +1 – Jack Sep 25 '17 at 0:58