Nonlinear PDE Help: Solving \alpha(uu_x)_x = u_t | Initial Value Problem Tips"

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Hello. I was wondering if anyone here had come across an equation similar to this one:
\alpha(uu_x)_x= u_t

Any info regarding this equation or tips on how to solve this would be appreciated :)

I came across these solutions: http://eqworld.ipmnet.ru/en/solutions/npde/npde1201.pdf, but how do I choose which one to use? I am looking at a initial value problem. And u > 0.
 
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maka89 said:
Hello. I was wondering if anyone here had come across an equation similar to this one:
\alpha(uu_x)_x= u_t

Any info regarding this equation or tips on how to solve this would be appreciated :)

I came across these solutions: http://eqworld.ipmnet.ru/en/solutions/npde/npde1201.pdf, but how do I choose which one to use? I am looking at a initial value problem. And u > 0.
I don't think the equation given in the link you posted will be helpful. The equation there is equivalent to ##a(w^m w_x)_x = w_t##.
The only thing that comes to mind in your equation is to take the partial w.r.t x of the left side (using the product rule). That would leave you with ##\alpha[(u_x)^2 + uu_{xx}] = u_t##, although I'm not sure that gets you anywhere.
 
maka89 said:
Hello. I was wondering if anyone here had come across an equation similar to this one:
\alpha(uu_x)_x= u_t

Any info regarding this equation or tips on how to solve this would be appreciated :)

For \alpha &gt; 0 this is a one-dimensional diffusion equation <br /> u_t - (Du_x)_x = 0<br /> where the diffusivity D is not constant, but is instead proportional to the density u of the diffused quantity: D = \alpha u.

I came across these solutions: http://eqworld.ipmnet.ru/en/solutions/npde/npde1201.pdf, but how do I choose which one to use? I am looking at a initial value problem. And u > 0.

You seem to be dealing with the case m = 1. I don't think the given analytic solutions will help you, except for particular special cases of the initial condition. For generic initial conditions you must fall back on numerical methods.
 
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