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I recently encountered a viscous Burgers' equation type PDE, but with the addition of a derivative-squared nonlinear term (in dimensionless form):

$u_t - u_{xx} + uu_x - u_x^2 = 0\,,$

where the boundary conditions require the solution to vanish at +/- infinity. My question is: does anyone know if this equation has a name or at least if some information is available. Can one use a Cole-Hopf transformation of some kind to transform this to the heat equation, as is the case in the regular Burgers' equation?

Does it have some physical origin or can it be derived from e.g. the Navier-Stokes equation?

Thanks!

PS: I'm sorry if my typesetting or language is not correct, I am new to the online forum stuff.

JonasB
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    There wouldn't be a 'nice' transformation, if one exists. Letting $u=f(v(x,t))$ $$\implies v_{t}f'+v_{x}ff'\color{red}{-v_{x}^{2}f'^{2}}-v_{x}^{2}f''-v_{xx}f'=0$$ where $f'=df/dv$ and the red term comes from the introduced nonlinearity to the ordinary Burgers equation. The Hopf-Cole transform comes about from solving $$v_{x}ff'-v_{x}^{2}f''=0$$ for $f$, which is easily doable. In order to reduce this new problem to the heat equation, we need to solve $$v_{x}ff'-v_{x}^{2}f'^{2}-v_{x}^{2}f''=0$$ but that $f'^{2}$ term in the middle makes life very difficult. – Matthew Cassell Jun 28 '19 at 07:26
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    Multiplying by $e^u$, we obtain:

    $$(e^u)t - (e^u){xx} + u(e^u)_x = 0$$

    Substituting:

    $$e^u=v$$

    $$v_t-v_{xx}+v_x \log v=0$$

    Not very pretty but maybe there's some way to attack this new equation.

     – Yuriy S Sep 14 '19 at 10:57
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    The non linearity is very inusual. Where does this come from ? – Héhéhé Sep 15 '19 at 05:19

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