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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