# 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 $(x,t)\mapsto (y,\tau):=\left(\frac{x}{s(t)},\tau(t)\right) \qquad \varrho(x,t) = \frac{1}{s(t)^N} u\left(\frac{x}{s(t)},\tau(t)\right),$ 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 $\int_{\R^N} \hat\varrho(y) \; \dx{y} = 1.$

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