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I'm looking at this second order quasilinear PDE:

$\alpha_{xx} - \alpha_{yy} - m\alpha^2 = 0$

Attempted Strategies:

  • Fourier Transform resulted in convolution due to the $\alpha^2$ term, I don't think that would be pretty to work on.
  • Separation of Variables resulted in not being able to separate the variables.
  • I tried substituting $u=\alpha_x$ and $v=\alpha_y$ but I'm left with $(\int u_x dx)^2$ or $(\int u_y dy)^2$ in my equation afterwords due to substituting $u$ and $v$ back in, which seems like a dead end.

Do you guys have any suggestions for an analytical solution method?

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Did you try the change of variables used to solve the wave equation: $s=x+y$ and $t=x-y$? Not sure where that leads, but it'll give you something to do. –  AppliedSide Jan 29 '11 at 22:20
    
This is not a quasilinear PDE (well, technically it is, since a linear PDE is also quasilinear). This is better called a semi-linear PDE: the terms with all highest derivatives are linear. –  Willie Wong Jan 29 '11 at 23:38
    
Also, are you trying to solve the initial value problem? If so, which of the two variables is "time"? –  Willie Wong Jan 29 '11 at 23:42

2 Answers 2

up vote 1 down vote accepted

Letting $t=x+y$ and $s=x-y$ as in the standard treatment of the 1-D wave equation, the equation $$\alpha_{st}=\frac{m}{4} \alpha^2.$$ If $m>0$, trying the trick of separation of variables does, indeed, lead to an explicit solution (i.e., assume $\alpha(s,t)=A(s)B(t)$).

As Willie Wong mentions below, this method had no hope from the start of producing a set of particular solutions which could be used to write a general solution, since a linear combination of particular solutions will not be a solution, in general.

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Yes, but that is not going to recover the general solution. For linear equations, you can hope that through separation of variables you can decompose any solution as a superposition of these "special" solutions. The nonlinearity however prevents superposition. –  Willie Wong Jan 30 '11 at 0:36
1  
I read the OP as asking for an explicit solution, or a method to find one. I never meant to suggest this was a method for finding a solution to any given boundary problem. –  AppliedSide Jan 30 '11 at 3:38
    
Thanks for the tip. –  user6390 Feb 1 '11 at 16:13

I think it's pretty hopeless to get an explicit formula for solutions for this equation (although specific examples of solutions might be known). Your equation is called the nonlinear wave equation, and it is an active area of research (to get a glimpse of the variety of research results, find the DispersiveWiki website, although this will not help in your case). In general it is not even true that an initial value problem (that is, prescribing the solution for $x=0$, here I'm interpreting your $x$ as time) has a global solution; often one has local solutions (that is, for small $x$) which can blow up in finite time.

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I don't think the OP will find much useful things on Dispersive Wiki: the wave equation in 1 spatial and 1 temporal dimension is not dispersive: wave packets maintain size and shape over time, they just travel. –  Willie Wong Jan 29 '11 at 23:50
    
But Florian's description is generally correct: depending on the sign of the parameter $m$ and which of $x,y$ you take as the time parameter, the initial value problem either admits global solution with energy decay or blows up in finite time. –  Willie Wong Jan 30 '11 at 0:03
    
I put in the reference to the DispersiveWiki to support my claim that nonlinear wave equations are hard, but I agree that it is potentially confusing.But global existence by energy arguments works only for nonlinearities of type $|\alpha|^{p-1}\alpha$ and not for $\alpha^2$ –  Florian Jan 30 '11 at 1:25
    
And I don't quite believe that "wave packets maintain size and shape over time", what exactly do you mean by this? –  Florian Jan 30 '11 at 1:40
    
@Florian: the solution to the linear wave equation $u_{tt} - u_{xx} = 0$ is a travelling wave packet, a fact known since D'Alembert in the 18th century. This simply follows from the method of characteristics after writing the equation as $(\partial_t + \partial_x) (\partial_t - \partial_x) u = 0$. Your previous comment is duly noted: I was being a bit sloppy. –  Willie Wong Jan 30 '11 at 10:57

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