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单词 ProofOfWeakMaximumPrincipleForRealDomains
释义

proof of weak maximum principle for real domains


First, we show that, if Δf>0 (where Δ denotes the LaplacianDlmfPlanetmath on d) on K, then f cannot attain a maximum on the interior of K. Assume, to the contrary, that f did attain a maximum at a point p located on the interior of K. By the second derivative testMathworldPlanetmath, the matrix of second partial derivativesMathworldPlanetmath of f at p would have to be negative semi-definite. This would imply that the trace of the matrix is negative. But the trace of this matrix is the Laplacian, which was assumed to be strictly positive on K, so it is impossible for f to attain a maximum on the interior of K.

Next, suppose that Δf=0 on K but that f does not attain its maximum on the boundary of K. Since K is compact, f must attain its maximum somewhere, and hence there exists a point p located in the interior of K at which f does attain its maximum. Since K is compact, the boundary of K is also compact, and hence the image of the boundary of K under f is also compact. Since every element of this image is strictly smaller than f(p), there must exist a constant C such that f(x)<C<f(p) whenever x lies on the boundary of K. Furthermore Since K is a compact subset of d, it is bounded. Hence, there exists a constant R>0 so that |x-p|<R for all xK.

Consider the functionMathworldPlanetmath g defined as

g(x)=f(x)+(f(p)-C)|x-p|2R2

At any point xK,

g(x)<f(x)+f(p)-C

In particular, if x lies on the boundary of K, this implies that

g(x)<f(p)

Since g(p)=f(p) this inequalityMathworldPlanetmath implies that g cannot attain a maximum on the boundary of K.

This leads to a contradiction. Note that, since Δf=0 on K,

Δg=d(f(p)-C)R2>0

which implies that g cannot attain a maximum on the interior of K. However, since K is compact, g must attain a maximum somewhere on K. Since we have ruled out both the possibility that this maximum occurs in the interior and the possibility that it occurs on the boundary, we have a contradiction. The only way out of this contradiction is to conclude that f does attain its maximum on the boundary of K.

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更新时间:2025/5/4 19:15:33