Subcriticality and supercriticality

Consider the equation

$\displaystyle \Delta u=u^p\text{ on }\mathbb{R}^n$

usually we call the equation is subcritical when ${p<\frac{n+2}{n-2}}$, supercritical when ${p>\frac{n+2}{n-2}}$. The reason comes from the scalling the solution. Suppose ${u(x)}$ is a solution of the equation, then ${u^\lambda(x)=\lambda^{\frac{2}{p-1}}u(\lambda x)}$ is another solution. Consider the energy possessed by ${u}$ around any point ${x_0}$ of radius ${{\lambda}}$ can be bounded

$\displaystyle \int_{B_{\lambda}(x_0)}|\nabla u(x)|^2dx\leq E$

when ${\lambda\rightarrow 0}$, we scale ${B_\lambda(x_0)}$ to ${B_1(x_0)}$, then ${u}$ will become ${u^\lambda}$ in order to be a solution and ${u^\lambda}$ lives on ${B_1(x_0)}$. While the energy will be

$\displaystyle \int_{B_{1}(x_0)}|\nabla u^\lambda(x)|^2dx=\lambda^{\frac{4}{p-1}+2-n}\int_{B_{\lambda}(x_0)}|\nabla u(y)|^2dy$

If the ${\delta=\frac{4}{p-1}+2-n<0}$, which is ${p> \frac{n+2}{n-2}}$, the energy bound of ${u^\lambda}$ will become ${\lambda^\delta E}$. Since ${\lambda\rightarrow 0}$, the bound deteriorates by ‘zooming in’. In this case, we call the equation is supercritical. The solution looks more singular at this time.

Remark: The energy should include ${\int_{B_{\lambda}(x_0)}u^2dx}$, but somehow this term scale differently with ${\int_{B_{\lambda}(x_0)}|\nabla u(x)|^2dx}$ and can not give one the critical exponent exactly.