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
\[
(x,t)\mapsto (y,\tau):=\left(\frac{x}{s(t)},\tau(t)\right) \qquad \varrho(x,t) = \frac{1}{s(t)^N} u\left(\frac{x}{s(t)},\tau(t)\right),
\]
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.
\[
1 = \int_{\R^N} \varrho(x,t) \; \dx{x} \stackrel{x\mapsto ys}{=} \int_{\R^N} \frac{1}{s(t)^N} \varrho(y s(t),t)\;\dx{y} = \int_{\R^N} u(y,\tau(t)) \; \dx{y}.
\]
The time derivative has to satisfy
\[
\partial_t \varrho = \frac{\tau’}{s^N} \partial_\tau u – \frac{N s’}{s^{N+1}} u – \frac{s’}{s^{N+2}} x \cdot \nabla_y u = \frac{\tau’}{s} \partial_\tau u – \frac{s’}{s^{N+1}} \nabla_y \cdot ( y u) .
\]
Further, let us calculate the gradient of \(\varrho^m\)
\[
\nabla_x \varrho(x,t)^m = \nabla_y\left(\frac{1}{s^{Nm}} u(y,\tau)^m \right) \frac{1}{s}
\]
and the Laplacian evaluates to
\[
\Delta \varrho(x,t)^m = \frac{1}{s^{Nm+2}} \Delta_y u(y,\tau)^m .
\]
Hence, the function \(u\) solves the equation
\[ \label{equ:rescale}
\frac{\tau’}{s^N} \partial_\tau u(y,\tau) – \frac{s’}{s^{N+1}} \nabla_y \cdot \bigl(y\; u(y,\tau)\bigr) = \frac{1}{s^{Nm+2}} \Delta_y u(y,\tau)^m .
\]
We want the coefficients to be time-independent and a comparison results in the condition
\[
c_1 \frac{\tau’}{s^N} = c_2 \frac{s’}{s^{N+1}} = c_3 \frac{1}{s^{Nm+2}} = 1,
\]
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
\[
c_3 = s’ c_1 s^{m(N-1)+1} = c_2 \frac{\dx}{\dx t} \frac{s(t)^{N(m-1)+2} }{N(m-1)+2}
\]
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
\[
s(t) = \left(\frac{c_3}{c_2\alpha} t+c_4\right)^{\alpha} \quad\text{and}\quad \tau(t) = \frac{c_2}{c_1} \log s(t) , \qquad\text{where}\qquad \alpha:= \frac{1}{N(m-1)+2} .
\]
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
\[
\partial_\tau u(y,\tau) = \Delta_y u^m(y,\tau) + \alpha \nabla_y\cdot (y \; u(y,\tau)) , \quad\text{with}\quad \tau(t)=\log t \quad\text{and}\quad s(t) = t^\alpha ,
\]
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
\[
\nabla_y \cdot\left(\nabla_y \hat\varrho^m + \alpha y \hat\varrho\right) = 0 .
\]
Hence, by setting the flux inside of the divergence equal to zero
\[
m \hat\varrho^{m-1} \nabla_y \hat\varrho + \alpha \;y\; \hat\varrho = 0 .
\]
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
\[
m \hat\varrho^{m-2} \nabla_y \hat\varrho +\alpha \;y = 0 ,
\]
which can be rewritten as
\[
\nabla_y \left(\frac{m}{m-1} \hat\varrho^{m-1}\right) = \nabla_y\left(- \frac{\alpha}{2} y^2\right),
\]
which determines \(\hat\varrho^{m-1}\) up to a constant \(\tilde \lambda\in \R\)
\[
\hat\varrho^{m-1} = -\frac{m-1}{m} \frac{\alpha}{2}y^2 + \tilde \lambda
\]
We can only take the power \(\frac{1}{m-1}\) if the right hand side is non-zero, hence we set
\[
\hat\varrho(y) := \begin{cases}
\left(\tilde \lambda- \frac{m-1}{m} \frac{\alpha}{2} y^2\right)_+^{\frac{1}{m-1}} &, \text{ for } m>1 \\
\exp\left(\tilde \lambda – \frac{\alpha}{2} y^2 \right) &, \text{ for } m = 1 \\
\left(\tilde \lambda – \frac{m-1}{m} \frac{\alpha}{2} y^2\right)^{\frac{1}{m-1}} &, \text{ for } m \end{cases}
\]
hereby \((x)_+ := \max\{0, x\}\) denotes the positive part of \(x\). The constant \(\tilde\lambda\) is chosen such that
\[
\int_{\R^N} \hat\varrho(y) \; \dx{y} = 1.
\]