Multiply

Δ u + q u \leq 0

ζ^{2} / u_{ε}

and integrate. Since

ζ

is compactly supported, integration by parts and Young's inequality give

\int q \frac{u}{u _{ε}} ζ^{2} \leq - \int \frac{Δ u}{u _{ε}} ζ^{2} = 2 \int \frac{ζ D u \cdot D ζ}{u _{ε}} - \int \frac{ζ ^{2} ∣ D u ∣ ^{2}}{u _{ε}^{2}} \leq \int ∣ D ζ ∣^{2} .

ε ↓ 0

u / u_{ε}

converges to the indicator of

{u > 0}

\int_{{u > 0}} q ζ^{2} \leq \int ∣ D ζ ∣^{2} .

ii

Solution

Put

v = lo g u_{ε}

, so

D v = D u / u_{ε}

. Choose a cutoff

η

which equals one on

B_{ρ} (z) \cap B_{1}

, is supported in a comparable ball inside

B_{2}

, and satisfies

∣ Dη ∣ \leq C / ρ

. Repeating the calculation from part (i) with

η^{2}

and using

∣ q ∣ \leq μ

\int η^{2} ∣ D v ∣^{2} \leq C \int ∣ Dη ∣^{2} + C μ \int η^{2} \leq C (n, μ) ρ^{n - 2} .

The same construction works for balls meeting

\partial B_{1}

because the cutoff is supported in

B_{2}

. Hence

ρ^{2 - n} \int_{B_{ρ} (z) \cap B_{1}} \frac{∣ D u _{ε} ∣ ^{2}}{u _{ε}^{2}} \leq K (n, μ) .

There are no discussions about this article yet.

There are currently no matching articles.