Green Functions, Potentials, and the Dirichlet Problem for the Heat Equation
Green Functions, Potentials, and the Dirichlet Problem for the Heat Equation
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热方程的格林函数、势和狄利克雷问题
DOI：
10.1112/plms/s3-33.2.251
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发表时间：
1976
影响因子：
1.8
通讯作者：
N. Watson
中科院分区：
文献类型：
--
作者：
N. Watson
Miroslav Dont has kindly pointed out to me that the result of A. Friedman, which was quoted in Lemma 2, is false. A correct form is given by J. Moser [1], along with an example in which Friedman's version breaks down (p. 103).This means that all the results in §5 have to be corrected, but once this is achieved only minor alterations to the proofs of two subsequent results are necessary. Results in the original paper should be replaced by those with similar headings below.LEMMA 2.Let0 < ρ′ < ρ, 0 < τ1‐ < τ2‐ < τ+< τ,and putR= {(x,t): |xi| < ρ fori= 1,…,n, 0 <t< τ},R−= {(x,t): |xi| < ρ′ fori= 1,…,n, τ1‐ <t< τ2‐},R+= {(x,t): |xi| < ρ′ fori= 1,…,n, τ+<t< τ}.Then there is a positive constant k such that, for every non‐negative temperature u on R,max{u(p):pɛR−} ⩽kinf{u(p):pɛR+}.This is Moser's Theorem 1.COROLLARY.LetR−and R be as above, and let q0= (0,τ). Then there is a positive constant k such that for every non‐negative temperature u on Ru(p⩽ kliminfu(q)q→q0,qɛRfor all pɛR−.Proof. Choose τ+such that τ2‐ < τ+< τ. IfR+is as above, with this τ+, there is a constantksuch thatmax{u(p):pɛR−} ⩽kinf{u(p):pɛR+} ⩽ kliminfu(q)q→q0,qɛRfor every non‐negative temperatureuonR.THEOREM 3.Let p0be a point in an open set E, and let K be a compact subset of∧(p0,E).Then there exists a positive constant k such that, for every non‐negative temperature u on∧(p0,E),u(p⩽ kliminfq→0,q∈Λ(p0,E)u(q)whenever pɛK.Proof. Letu(p0) denote the lower limit in (2). Suppose that no such constantkexists. Then for each positive integerj, there is a non‐negative temperatureujon ∧(p0,E) and a pointqjεKsuch thatuj(qj)>juj(p0).SinceKis compact, there exists a subsequence of (qj) which converges to a pointqɛK. We may suppose that (qj) itself has this property.SinceqɛK⊆ ∧(p0,E), we can joinqtop0with a polygonal line Γ inEsuch thattis strictly increasing fromqtop0. Since Γ is compact andEis open, the distance between Γ and Rn+1\Eis positive. Hence there exists δ > 0 such that, for eachp= (x,t) ɛ Γ, the closure of the rectangleR(p) = {(y,s): 0 <t−s< δ, |yi−xi| < δ}lies inE. Since γ has finite length and eachR(p) covers a portion of Γ of length at least δ, there are finitely many rectanglesR(P0),…,R(pm) whose union covers Γ\{p0}. PutRi=R(pi) fori= 0,…,m; thusΓ\{p0} ⊆R0∪…∪Rm⊆ ∧(p0,E).For eachi, with l ⩽i⩽m,piɛ Γ\({p0}∪Ri), so that there isRl, with 0 ⩽ l ⩽mandl≠i, such thatpiɛRl. Enclosepiin a rectangle Rl‐ whose close lies inRl, and apply the corollary of Lemma 2 onRl, thus obtaining a constantki> 0 such thatuj(pi) ⩽kiuj(pl)for allj. Sinceqɛ Γ\{p0}, we haveqɛRhfor someh, where 0 ⩽h⩽m, so that we can encloseqin a rectangle Rh‐, whose closure lies inRh, and apply the above corollary onRhto obtain a constantk′ > 0 such thatu(p) ⩽k′uj(ph)for allpɛ Rh‐ and allj. Repeated application of (4), together with (5) gives us a constantk> 0 such thatuj(p) ⩽kuj(p0)for allpɛ Rh‐ and allj. But someqɛ Rh‐, (3) implies that we can findJsuch tahtqJɛ Rh‐ anduJ(qj) >JuJ(p0) >kuJ(p0), which contradicts (6). Hence the assumption that there is no such constankkis incorrect.The Original versions of the above results were applied directly only in Lemma 11 and Theorem 15. These remain valid, but in the proof of Lemma 11 the open setVmust be chosen such that V―∧(p0,E), and in the proof of Theorem 15 we must choosep1ɛU1\U― such that ∧(p1,U1)U―.We take this opportunity to correct a minor error on p. 280. The upper limit of the integral on the ninth line of the example should be min{1,t}.
Miroslav Dont has kindly pointed out to me that the result of A. Friedman, which was quoted in Lemma 2, is false. A correct form is given by J. Moser [1], along with an example in which Friedman's version breaks down (p. 103).This means that all the results in §5 have to be corrected, but once this is achieved only minor alterations to the proofs of two subsequent results are necessary. Results in the original paper should be replaced by those with similar headings below.LEMMA 2.Let0 < ρ′ < ρ, 0 < τ1‐ < τ2‐ < τ+< τ,and putR= {(x,t): |xi| < ρ fori= 1,…,n, 0 <t< τ},R−= {(x,t): |xi| < ρ′ fori= 1,…,n, τ1‐ <t< τ2‐},R+= {(x,t): |xi| < ρ′ fori= 1,…,n, τ+<t< τ}.Then there is a positive constant k such that, for every non‐negative temperature u on R,max{u(p):pɛR−} ⩽kinf{u(p):pɛR+}.This is Moser's Theorem 1.COROLLARY.LetR−and R be as above, and let q0= (0,τ). Then there is a positive constant k such that for every non‐negative temperature u on Ru(p⩽ kliminfu(q)q→q0,qɛRfor all pɛR−.Proof. Choose τ+such that τ2‐ < τ+< τ. IfR+is as above, with this τ+, there is a constantksuch thatmax{u(p):pɛR−} ⩽kinf{u(p):pɛR+} ⩽ kliminfu(q)q→q0,qɛRfor every non‐negative temperatureuonR.THEOREM 3.Let p0be a point in an open set E, and let K be a compact subset of∧(p0,E).Then there exists a positive constant k such that, for every non‐negative temperature u on∧(p0,E),u(p⩽ kliminfq→0,q∈Λ(p0,E)u(q)whenever pɛK.Proof. Letu(p0) denote the lower limit in (2). Suppose that no such constantkexists. Then for each positive integerj, there is a non‐negative temperatureujon ∧(p0,E) and a pointqjεKsuch thatuj(qj)>juj(p0).SinceKis compact, there exists a subsequence of (qj) which converges to a pointqɛK. We may suppose that (qj) itself has this property.SinceqɛK⊆ ∧(p0,E), we can joinqtop0with a polygonal line Γ inEsuch thattis strictly increasing fromqtop0. Since Γ is compact andEis open, the distance between Γ and Rn+1\Eis positive. Hence there exists δ > 0 such that, for eachp= (x,t) ɛ Γ, the closure of the rectangleR(p) = {(y,s): 0 <t−s< δ, |yi−xi| < δ}lies inE. Since γ has finite length and eachR(p) covers a portion of Γ of length at least δ, there are finitely many rectanglesR(P0),…,R(pm) whose union covers Γ\{p0}. PutRi=R(pi) fori= 0,…,m; thusΓ\{p0} ⊆R0∪…∪Rm⊆ ∧(p0,E).For eachi, with l ⩽i⩽m,piɛ Γ\({p0}∪Ri), so that there isRl, with 0 ⩽ l ⩽mandl≠i, such thatpiɛRl. Enclosepiin a rectangle Rl‐ whose close lies inRl, and apply the corollary of Lemma 2 onRl, thus obtaining a constantki> 0 such thatuj(pi) ⩽kiuj(pl)for allj. Sinceqɛ Γ\{p0}, we haveqɛRhfor someh, where 0 ⩽h⩽m, so that we can encloseqin a rectangle Rh‐, whose closure lies inRh, and apply the above corollary onRhto obtain a constantk′ > 0 such thatu(p) ⩽k′uj(ph)for allpɛ Rh‐ and allj. Repeated application of (4), together with (5) gives us a constantk> 0 such thatuj(p) ⩽kuj(p0)for allpɛ Rh‐ and allj. But someqɛ Rh‐, (3) implies that we can findJsuch tahtqJɛ Rh‐ anduJ(qj) >JuJ(p0) >kuJ(p0), which contradicts (6). Hence the assumption that there is no such constankkis incorrect.The Original versions of the above results were applied directly only in Lemma 11 and Theorem 15. These remain valid, but in the proof of Lemma 11 the open setVmust be chosen such that V―∧(p0,E), and in the proof of Theorem 15 we must choosep1ɛU1\U― such that ∧(p1,U1)U―.We take this opportunity to correct a minor error on p. 280. The upper limit of the integral on the ninth line of the example should be min{1,t}.