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

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  • $\begingroup$ 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. $\endgroup$ – Winther Sep 29 '14 at 22:11
  • $\begingroup$ @thanasissdr That is exactly the generalized method I was talking about. $\endgroup$ – Winther Sep 29 '14 at 22:22
  • $\begingroup$ @Winther Sorry, didn't check your link. $\endgroup$ – thanasissdr Sep 29 '14 at 22:54
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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).

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

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