# Classic results for porous medium equation

### Selfsimilar solutions

$\newcommand{\R}{\mathbb{R}} \newcommand{\dx}{\mathrm{d}}$ Let us seek for solutions of $\partial \varrho = \Delta \varrho^m$ satisfying the scaling hypothesis

where $s: \R_+ \to \R_+$ and $\tau:\R_+ \to \R_+$ are reparametrization of time and $u:\R_+\times \R^N \to \R$. The prefactor $\frac{1}{s(t)^N}$ ensures that $u$ conserves mass, i.e.

The time derivative has to satisfy

Further, let us calculate the gradient of $\varrho^m$

and the Laplacian evaluates to

Hence, the function $u$ solves the equation

We want the coefficients to be time-independent and a comparison results in the condition

where $c_1, c_2,c_3>0$ are constants, which can be specified later. From the first equality, we obtain $\tau' = \frac{c_2}{c_1}\frac{s'}{s}$, which is solved by $\tau(t) = \tfrac{c_2}{c_1}\log s(t)$. The second equality, leads to

and integrates to $s(t)^{N(m-1)+2} = \frac{c_3}{c_2} (N(m-1)+2)t+c_4$, where $c_4\in \R$ is a further constant. Hence, we find the scaling relation

We are still free to choose the constants $c_1,c_2,c_3>0$ and $c_3\in \R$. A particular nice choice is given bys $c_1=1$, $c_2=\frac{1}{\alpha}$, $c_3=1$ and $c_4=0$, then we obtain the result: If $\varrho(x,t)$ is a solution of the PME, then $u$ solves

### Equilibrium solutions

From the self similar rescaled solution $u$, we can derive the equilibrium solution. Stationary solutions are given by function $\hat\varrho:\R^N\to \R_+$ satisfying

Hence, by setting the flux inside of the divergence equal to zero

Hence, $\hat\varrho\equiv 0$ is a trivial solution and in the case $m=1$ it is easy to check that $\hat\varrho=e^{-\frac{\alpha}{2} y^2}$ is a solution (Compare this with the Ornstein-Uhlenbeck process, which is a special case of the Fokker-Planck equation}. Therefore, let us assume, that $m\ne 1$. Then, we have

which can be rewritten as

which determines $\hat\varrho^{m-1}$ up to a constant $\tilde \lambda\in \R$

We can only take the power $\frac{1}{m-1}$ if the right hand side is non-zero, hence we set

hereby $(x)_+ := \max\{0, x\}$ denotes the positive part of $x$. The constant $\tilde\lambda$ is chosen such that