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Harnack Inequality

Harnack Inequality - If FunctionU(M)=U(xone,...,xk) {\ displaystyle U (M) = U (x_ {1}, ..., x_ {k})} U (M) = U (x_ {1}, ..., x_ {k}) harmonic ink {\ displaystyle k} k -dimensional ballQR {\ displaystyle Q_ {R}} Q_ {R} radiusR {\ displaystyle R} R centered at some pointM0 {\ displaystyle M_ {0}} M_ {0} is non-negative in this ball, then for its values ​​at pointsM {\ displaystyle M} M the following inequalities are valid inside the ball in question:Rk-2R-r(R+r)k-oneU(M0)⩽U(M)⩽Rk-2R+r(R-r)k-oneU(M0) {\ displaystyle R ^ {k-2} {\ frac {Rr} {(R + r) ^ {k-1}}} U (M_ {0}) \ leqslant U (M) \ leqslant R ^ {k- 2} {\ frac {R + r} {(Rr) ^ {k-1}}} U (M_ {0})} R ^ {{k-2}} {\ frac {Rr} {(R + r) ^ {{k-1}}}} U (M_ {0}) \ leqslant U (M) \ leqslant R ^ {{ k-2}} {\ frac {R + r} {(Rr) ^ {{k-1}}}} U (M_ {0}) wherer=ρ(M0,M)<R {\ displaystyle r = \ rho (M_ {0}, M) <R} r = \ rho (M_ {0}, M) <R .

Proof

By the Poisson formula for pointsM {\ displaystyle M} M inside the ballQR′(R′<R) {\ displaystyle Q_ {R '} (R' <R)} Q_{{R'}}(R'<R) we haveU(M)=Γ(k/2)2πk/2R′∫γ(QR′)U(N)R′2-r2(R′2+r2-2R′rcos⁡θ)r/2dσ {\ displaystyle U (M) = {\ frac {\ Gamma (k / 2)} {2 \ pi ^ {k / 2} R '}} \ int \ limits _ {\ gamma} (Q_ {R'}) U (N) {\ frac {{R '} ^ {2} -r ^ {2}} {({R'} ^ {2} + r ^ {2} -2R'r \ cos \ theta) ^ { r / 2}}} d \ sigma} U(M)={\frac  {\Gamma (k/2)}{2\pi ^{{k/2}}R'}}\int \limits _{{\gamma }}(Q_{{R'}})U(N){\frac  {{R'}^{{2}}-r^{2}}{({R'}^{2}+r^{2}-2R'r\cos \theta )^{{r/2}}}}d\sigma . Given the inequalities(R′-r)2⩽R′2+r2-2R′rcos⁡θ⩽(R′+r)2 {\ displaystyle (R'-r) ^ {2} \ leqslant {R '} ^ {2} + r ^ {2} -2R'r \ cos \ theta \ leqslant (R' + r) ^ {2}} (R'-r)^{2}\leqslant {R'}^{2}+r^{2}-2R'r\cos \theta \leqslant (R'+r)^{2} due to the conditionU(N)⩾0 {\ displaystyle U (N) \ geqslant 0} U(N)\geqslant 0 get from here thatR′k-2R′2-r2(R′+r)kΓ(k/2)2πk/2R′k-one∫γ(QR′)U(N)dσ⩽U(M)⩽R′k-2R′2-r2(R′-r)kΓ(k/2)2πk/2R′k-one∫γ(QR′)U(N)dσ {\ displaystyle {R '} ^ {k-2} {\ frac {{R'} ^ {2} -r ^ {2}} {(R '+ r) ^ {k}}} {\ frac {\ Gamma (k / 2)} {2 \ pi ^ {k / 2} {R '} ^ {k-1}}} \ int \ limits _ {\ gamma} (Q_ {R'}) U (N) d \ sigma \ leqslant U (M) \ leqslant {R '} ^ {k-2} {\ frac {{R'} ^ {2} -r ^ {2}} {(R'-r) ^ {k} }} {\ frac {\ Gamma (k / 2)} {2 \ pi ^ {k / 2} {R '} ^ {k-1}}} \ int \ limits _ {\ gamma} (Q_ {R' }) U (N) d \ sigma} {R'}^{{k-2}}{\frac  {{R'}^{2}-r^{2}}{(R'+r)^{k}}}{\frac  {\Gamma (k/2)}{2\pi ^{{k/2}}{R'}^{{k-1}}}}\int \limits _{\gamma }(Q_{{R'}})U(N)d\sigma \leqslant U(M)\leqslant {R'}^{{k-2}}{\frac  {{R'}^{2}-r^{2}}{(R'-r)^{k}}}{\frac  {\Gamma (k/2)}{2\pi ^{{k/2}}{R'}^{{k-1}}}}\int \limits _{\gamma }(Q_{{R'}})U(N)d\sigma , or, applying the Gauss theoremR′k-2R′2-r2(R′+r)kU(M0)⩽U(M)⩽R′k-2R′2-r2(R′-r)kU(M0) {\ displaystyle {R '} ^ {k-2} {\ frac {{R'} ^ {2} -r ^ {2}} {(R '+ r) ^ {k}}} U (M_ {0 }) \ leqslant U (M) \ leqslant {R '} ^ {k-2} {\ frac {{R'} ^ {2} -r ^ {2}} {(R'-r) ^ {k} }} U (M_ {0})} {R'}^{{k-2}}{\frac  {{R'}^{2}-r^{2}}{(R'+r)^{k}}}U(M_{0})\leqslant U(M)\leqslant {R'}^{{k-2}}{\frac  {{R'}^{2}-r^{2}}{(R'-r)^{k}}}U(M_{0}) . Thus, passing to the limit atR′→R {\ displaystyle R '\ rightarrow R} R'\rightarrow R , we obtain the Harnack inequalityRk-2R-r(R+r)k-oneU(M0)⩽U(M)⩽Rk-2R+r(R-r)k-oneU(M0) {\ displaystyle R ^ {k-2} {\ frac {Rr} {(R + r) ^ {k-1}}} U (M_ {0}) \ leqslant U (M) \ leqslant R ^ {k- 2} {\ frac {R + r} {(Rr) ^ {k-1}}} U (M_ {0})} R^{{k-2}}{\frac  {R-r}{(R+r)^{{k-1}}}}U(M_{0})\leqslant U(M)\leqslant R^{{k-2}}{\frac  {R+r}{(R-r)^{{k-1}}}}U(M_{0}) .

Literature

  • Timan A.F., Trofimov V.N. Introduction to the theory of harmonic functions, M., Nauka , 1968, 206 pp., Shooting gallery 39500 copies.


Source - https://ru.wikipedia.org/w/index.php?title=Garnak Inequality&oldid = 81890642


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Clever Geek | 2019