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.

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