# Separable solution to a nonlinear parabolic PDE

I seek a separable solution to the nonlinear parabolic partial differential equation,

$\frac{\partial u}{\partial t} = u \frac{\partial u}{\partial x^2} + u^2.$

The physics of the problem allow either homogeneous Neumann or homogeneous Dirichlet boundary conditions at prescribed points, say at $x=\pm1$. The initial condition is a prescribed function $u(x,t=0) = f(x)$.

Assuming a form $u(x,t) = f(x)g(t)$ results in

$\frac{g^{\prime}}{g^2} = f^{\prime\prime} + f = \lambda,$

for seperation constant $\lambda$.

The PDE therefore reduces to two ODEs. Their solutions are

$g(t) = \frac{1}{c_0 - \lambda t}$ and $f(x) = c_1\sin x + c_2 \cos x + \lambda.$

I am not sure what to do next. The solutions for the functions $f(x)$ and $g(t)$ don't have the same form as the typical eigenfunction expansions that arise, for example, with linear heat and wave equations. Is there are fundamental idea that I am missing? How do I enforce (either type of) boundary conditions and the initial condition?

• Are you quoting the PDE correctly? Where are you getting that second derivative with respect to $x$? There's no derivative with respect to $x$ at all in the equation you gave. Commented Feb 12, 2016 at 1:48
• Sorry about that. There was a typo but it's fixed now. Commented Feb 12, 2016 at 1:53
• Out of curiosity: how does this equation arise, i.e. what is the underlying physical problem? Commented Feb 15, 2016 at 18:02
• The problem arises from the study of the frictional sliding of thin film on a rigid substrate. The terms have more physical meaning when written as $\dot u/u - u = u''$. $u$ is the sliding velocity. The two terms on the left represent two different frictional processes and the term on the right represents a flexure-type elastic operator. Commented Feb 16, 2016 at 18:54

EDIT: Well, one thing you can try is a series solution in powers of $t$.
Let $u(x,t) = \sum_{n=0}^\infty u_n(x) t^n$, where $u_0(x) = u(x,0) = f(x)$. Substituting this into your PDE and expanding in powers of $t$, the coefficient of $t^n$ will give you an equation for $u_{n+1}$ in terms of the previous $u_j$ and their derivatives.
\eqalign{ u_1(x) &= u_0(x) u_0''(x) + u_0(x)^2\cr 2 u_2(x) &= u_0(x) u_1''(x) + u_1(x) u_0''(x) + 2 u_0(x) u_1(x)\cr 3 u_3(x) &= u_0(x) u_2''(x) + u_1(x) u_1''(x) + u_2(x) u_0''(x) + 2 u_0(x) u_2(x) + u_1(x)^2\cr etc &\cr}