Classic results for porous medium equation

Selfsimilar solutions


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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