# Ngô Quốc Anh

## May 21, 2010

### How to remember the definition of sup- and super-solutions?

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 17:10

I am frequently confused the definition of sup- and super-solutions so yesterday I tried to figure out a way to remember those things. Fortunately, I think I got a very simple way to remember. The point is, just denote by $\underline u$ and $\overline u$ the sup- and super-solutions respectively to the very simple PDE $-\Delta u = f(x,u)$,

which one is true $\displaystyle -\Delta \overline u \leqslant f(x,\overline u)$

or $\displaystyle -\Delta \overline u \geqslant f(x,\overline u)$

and similarly to $\underline u$.

Let us consider a general case. Assume we are working with a general second-order elliptic operator in non-divergence form, say $\displaystyle L=-a^{ij}\partial_i\partial_j + b^k\partial_k + c$

where $a^{ij}$ are positive coefficients.  Besides, coefficients of $L$ are assumed to satisfy several conditions like symmetry, etc. but it is not considered here.

Concerning the following PDE $L(u)=f(x,u)$,

we say

function $\overline u$ (resp. $\underline u$) is said to be a super-solution (resp. sub-solution) to the PDE if $L(\overline u) \geqslant f(x,\overline u)$ (resp. $L(\underline u) \leqslant f(x,\underline u)$).

Observe that $^{ij}$ is positive lets us think that $L$ has positive spectrum. So once $\overline u$ is a super-solution, the inequality should be ${\rm LHS} \geqslant {\rm RHS}$ where the left hand side should be an operator with positive spectrum. Similarly, concerning the sub-solutions, we need ${\rm LHS} \leqslant {\rm RHS}$.

Let us go back to the Laplacian operator $\Delta = \partial^i\partial_i$. So $\Delta$ has negative spectrum, this yields $\displaystyle \Delta \overline u \leqslant -f(x,\overline u)$

and $\displaystyle \Delta \underline u \geqslant -f(x,\underline u)$.

Equivalently, $\displaystyle -\Delta \overline u \geqslant f(x,\overline u)$

and $\displaystyle -\Delta \underline u \leqslant f(x,\underline u)$

since $-\Delta$ has positive spectrum.

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