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Proof of The Viscous Hamilton-Jacobi Operator is Continuous, Strictly Proper and Satisfies the Structure Condition

propositionprop:viscous-hamilton-jacobi-operator-2026a
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· 9,744 chars · 15 deps · depth 24 Reason: First publication. Proof for the viscous Hamilton-Jacobi operator: strict properness from the monotonicity of the trace, the structure condition by reduction to the example of an elliptic operator with a continuous inhomogeneity, and continuity by an explicit estimate of the four terms.

Strict properness reduces to the monotonicity of the trace for the semidefinite ordering, and the structure condition follows by applying the example of an elliptic operator with a continuous inhomogeneity. Continuity is checked directly, by bounding each of the four terms of FF separately: a trace norm bound, the Cauchy-Schwarz inequality for the quadratic term, and the continuity of ff.

Proof

Conventions. The order \le and the arithmetic of R\mathbb{R} are those of the ordered field of real numbers. Multiplication by a nonnegative real number preserves \le: if aba\le b and 0λ0\le\lambda, then either a=ba=b, and the two products are equal, or a<ba<b, and then λaλb\lambda a\le\lambda b by claim 10 of Elementary Order Arithmetic in an Ordered Field when 0<λ0<\lambda, while λ=0\lambda=0 makes both products 00. Adding a fixed real number to both sides of an inequality also preserves it, by the compatibility of \le with addition. Finally, 0<210<2^{-1} by claims 8 and 7 of Elementary Order Arithmetic in an Ordered Field, so κ2=κ21\tfrac{\kappa}{2}=\kappa\cdot2^{-1} is nonnegative and 12=121\tfrac{1}{2}=1\cdot 2^{-1} is positive.

Proof of claim 2. We first check degenerate ellipticity. Let xΩx\in\Omega, rRr\in\mathbb{R}, pRnp\in\mathbb{R}^{n} and X,YS(n)X,Y\in\mathcal{S}(n) with XYX\preceq Y. By The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound §monotone, tr(X)tr(Y)\operatorname{tr}(X)\le\operatorname{tr}(Y). Multiplying by the nonnegative κ2\tfrac{\kappa}{2} gives κ2tr(X)κ2tr(Y)\tfrac{\kappa}{2}\operatorname{tr}(X)\le\tfrac{\kappa}{2}\operatorname{tr}(Y), and claim 4 of Elementary Order Arithmetic in an Ordered Field then gives

κ2tr(Y)κ2tr(X).-\tfrac{\kappa}{2}\operatorname{tr}(Y)\le-\tfrac{\kappa}{2}\operatorname{tr}(X).

Adding γr+12p2f(x)\gamma r+\tfrac{1}{2}\lVert p\rVert^{2}-f(x) to both sides yields F(x,r,p,Y)F(x,r,p,X)F(x,r,p,Y)\le F(x,r,p,X). As xx, rr, pp, XX and YY were arbitrary, FF is degenerate elliptic, which is condition 1 of Strictly Proper Second-Order Equation Operator.

For condition 2, let xΩx\in\Omega, pRnp\in\mathbb{R}^{n}, XS(n)X\in\mathcal{S}(n) and r,sRr,s\in\mathbb{R} with srs\le r. The three terms κ2tr(X)-\tfrac{\kappa}{2}\operatorname{tr}(X), 12p2\tfrac{1}{2}\lVert p\rVert^{2} and f(x)-f(x) are the same in F(x,r,p,X)F(x,r,p,X) and in F(x,s,p,X)F(x,s,p,X), so

F(x,r,p,X)F(x,s,p,X)=γrγs=γ(rs)F(x,r,p,X)-F(x,s,p,X)=\gamma r-\gamma s=\gamma(r-s)

by distributivity. In particular γ(rs)F(x,r,p,X)F(x,s,p,X)\gamma(r-s)\le F(x,r,p,X)-F(x,s,p,X), and claim 2 is proved.

Proof of claim 3. Define G:R×Rn×S(n)RG:\mathbb{R}\times\mathbb{R}^{n}\times\mathcal{S}(n)\to\mathbb{R} by

G(r,p,X)=γrκ2tr(X)+12p2,G(r,p,X)=\gamma r-\tfrac{\kappa}{2}\operatorname{tr}(X)+\tfrac{1}{2}\lVert p\rVert^{2},

so that F(x,r,p,X)=G(r,p,X)f(x)F(x,r,p,X)=G(r,p,X)-f(x) for every xΩx\in\Omega, rRr\in\mathbb{R}, pRnp\in\mathbb{R}^{n} and XS(n)X\in\mathcal{S}(n). The computation carried out in the proof of claim 2, with the term f(x)-f(x) omitted throughout, shows that G(r,p,Y)G(r,p,X)G(r,p,Y)\le G(r,p,X) whenever rRr\in\mathbb{R}, pRnp\in\mathbb{R}^{n} and X,YS(n)X,Y\in\mathcal{S}(n) satisfy XYX\preceq Y; that is, GG is degenerate elliptic in the matrix variable in the sense required by Structure Condition: a Degenerate Elliptic Operator with a Continuous Inhomogeneity. Since ff and ω\omega are as required there, that example applies to this GG, this ff and this ω\omega, and its conclusion Structure Condition: a Degenerate Elliptic Operator with a Continuous Inhomogeneity §structure is exactly claim 3.

Proof of claim 1. Fix x0Ωx_{0}\in\Omega, r0Rr_{0}\in\mathbb{R}, p0Rnp_{0}\in\mathbb{R}^{n} and X0S(n)X_{0}\in\mathcal{S}(n), and let εR\varepsilon\in\mathbb{R} be positive. Put 4=2+24=2+2; then 0<20<2 by claim 8 of Elementary Order Arithmetic in an Ordered Field and hence 0<40<4 by claim 3 there, so 414^{-1} exists and is positive by claim 7 there, and ε=ε41\varepsilon'=\varepsilon\cdot4^{-1} is positive by claim 5 there. Moreover ε+ε+ε+ε=(1+1+1+1)ε=4(ε41)=ε\varepsilon'+\varepsilon'+\varepsilon'+\varepsilon'=(1+1+1+1)\varepsilon'=4\bigl(\varepsilon\cdot4^{-1}\bigr)=\varepsilon by distributivity, commutativity and the defining property of the multiplicative inverse.

Four bounds. We record four estimates.

(i) For sRs\in\mathbb{R}, distributivity and claim 4 of Properties of the Absolute Value in an Ordered Field give γsγr0=γ(sr0)=γsr0=γsr0|\gamma s-\gamma r_{0}|=|\gamma(s-r_{0})|=|\gamma||s-r_{0}|=\gamma|s-r_{0}|, the last step because γ=γ|\gamma|=\gamma by the definition of the absolute value, γ\gamma being nonnegative.

(ii) Let YS(n)Y\in\mathcal{S}(n). Then YX0S(n)Y-X_{0}\in\mathcal{S}(n) by Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation §symmetric, and since the entries of YX0Y-X_{0} are the differences of the corresponding entries by the definition of a matrix difference, claim 1 of Basic Properties of the Trace, applied with the coefficients 11 and 1-1, gives tr(Y)tr(X0)=tr(YX0)\operatorname{tr}(Y)-\operatorname{tr}(X_{0})=\operatorname{tr}(Y-X_{0}). Hence, by The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound §norm-bound and Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation §norm,

tr(Y)tr(X0)βnYX0=βndS(n)(Y,X0),\bigl|\operatorname{tr}(Y)-\operatorname{tr}(X_{0})\bigr|\le\beta_{n}\lVert Y-X_{0}\rVert=\beta_{n}\,d_{\mathcal{S}(n)}(Y,X_{0}),

where βn=k=1n1\beta_{n}=\sum_{k=1}^{n}1 is as in The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound.

(iii) Let qRnq\in\mathbb{R}^{n} and put h=qp0h=q-p_{0}, so that q=p0+hq=p_{0}+h. By claim 1 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n and claims 1, 2 and 5 of Bilinearity and Symmetry of the Dot Product on Rn\mathbb{R}^n,

q2=qq=p0p0+(p0h)+(hp0)+hh=p02+2(p0h)+h2,\lVert q\rVert^{2}=q\cdot q=p_{0}\cdot p_{0}+(p_{0}\cdot h)+(h\cdot p_{0})+h\cdot h=\lVert p_{0}\rVert^{2}+2(p_{0}\cdot h)+\lVert h\rVert^{2},

where 2(p0h)2(p_{0}\cdot h) abbreviates (p0h)+(p0h)(p_{0}\cdot h)+(p_{0}\cdot h), the two middle terms being equal by the symmetry of the dot product. Hence q2p02=2(p0h)+h2\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}=2(p_{0}\cdot h)+\lVert h\rVert^{2}, and by claims 4 and 5 of Properties of the Absolute Value in an Ordered Field, Cauchy-Schwarz Inequality for the Euclidean Dot Product and the nonnegativity of norms recorded in claim 1 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n,

q2p022p0h+h22p0h+h2.\bigl|\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}\bigr|\le2|p_{0}\cdot h|+\lVert h\rVert^{2}\le2\lVert p_{0}\rVert\lVert h\rVert+\lVert h\rVert^{2}.

If moreover h1\lVert h\rVert\le1, then multiplying by the nonnegative h\lVert h\rVert gives h2h\lVert h\rVert^{2}\le\lVert h\rVert, so that by distributivity

q2p02(2p0+1)h.\bigl|\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}\bigr|\le\bigl(2\lVert p_{0}\rVert+1\bigr)\lVert h\rVert .

(iv) Since ΩΩ\Omega\subseteq\overline{\Omega} and ff is continuous at x0x_{0} relative to Ω\overline{\Omega}, there is a positive δfR\delta_{f}\in\mathbb{R} such that every yΩy\in\overline{\Omega} with dE(x0,y)<δfd_{E}(x_{0},y)<\delta_{f} satisfies f(y)f(x0)<ε|f(y)-f(x_{0})|<\varepsilon'; in particular this holds for every yΩy\in\Omega.

Choice of δ\delta. Put

c1=γ,c2=κ2βn+1,c3=12(2p0+1)+1.c_{1}=\gamma,\qquad c_{2}=\tfrac{\kappa}{2}\beta_{n}+1,\qquad c_{3}=\tfrac{1}{2}\bigl(2\lVert p_{0}\rVert+1\bigr)+1 .

Each cic_{i} is positive: c1=γc_{1}=\gamma is positive by hypothesis, while κ2βn\tfrac{\kappa}{2}\beta_{n} and 12(2p0+1)\tfrac{1}{2}(2\lVert p_{0}\rVert+1) are nonnegative, being products of nonnegative numbers — for βn\beta_{n} this is recorded in The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound, where 1βn1\le\beta_{n} — so c2c_{2} and c3c_{3} are at least 11 and 0<10<1 by claim 6 of Elementary Order Arithmetic in an Ordered Field. By claim 7 there each ci1c_{i}^{-1} exists and is positive, and by claim 5 there each εci1\varepsilon'c_{i}^{-1} is positive. Applying claim 9 of Elementary Order Arithmetic in an Ordered Field repeatedly to the five positive numbers 11, δf\delta_{f}, εc11\varepsilon'c_{1}^{-1}, εc21\varepsilon'c_{2}^{-1} and εc31\varepsilon'c_{3}^{-1} produces a real number δ\delta that is at most each of them and equal to one of them, hence positive.

Conclusion. Let yΩy\in\Omega, sRs\in\mathbb{R}, qRnq\in\mathbb{R}^{n} and YS(n)Y\in\mathcal{S}(n) satisfy

dE(y,x0)<δ,sr0<δ,qp0<δ,dS(n)(Y,X0)<δ,d_{E}(y,x_{0})<\delta,\qquad|s-r_{0}|<\delta,\qquad\lVert q-p_{0}\rVert<\delta,\qquad d_{\mathcal{S}(n)}(Y,X_{0})<\delta,

and put h=qp0h=q-p_{0}, so that h<δ1\lVert h\rVert<\delta\le1. Grouping the four terms of FF,

F(y,s,q,Y)F(x0,r0,p0,X0)=(γsγr0)κ2(tr(Y)tr(X0))+12(q2p02)(f(y)f(x0)).F(y,s,q,Y)-F(x_{0},r_{0},p_{0},X_{0})=\bigl(\gamma s-\gamma r_{0}\bigr)-\tfrac{\kappa}{2}\bigl(\operatorname{tr}(Y)-\operatorname{tr}(X_{0})\bigr)+\tfrac{1}{2}\bigl(\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}\bigr)-\bigl(f(y)-f(x_{0})\bigr).

By claims 2, 4 and 5 of Properties of the Absolute Value in an Ordered Field, the absolute value of the left-hand side is at most the sum of

γsr0,κ2tr(Y)tr(X0),12q2p02,f(y)f(x0).\gamma|s-r_{0}|,\qquad\tfrac{\kappa}{2}\bigl|\operatorname{tr}(Y)-\operatorname{tr}(X_{0})\bigr|,\qquad\tfrac{1}{2}\bigl|\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}\bigr|,\qquad|f(y)-f(x_{0})| .

We bound each summand by ε\varepsilon'. By (i) and claim 10 of Elementary Order Arithmetic in an Ordered Field, γsr0<γδ\gamma|s-r_{0}|<\gamma\delta, and multiplying δεc11\delta\le\varepsilon'c_{1}^{-1} by the nonnegative γ=c1\gamma=c_{1} gives γδε\gamma\delta\le\varepsilon'; hence the first summand is less than ε\varepsilon'. By (ii), κ2βnc2\tfrac{\kappa}{2}\beta_{n}\le c_{2}, so multiplying by the nonnegative dS(n)(Y,X0)d_{\mathcal{S}(n)}(Y,X_{0}) and using (ii) again gives κ2tr(Y)tr(X0)c2dS(n)(Y,X0)<c2δε\tfrac{\kappa}{2}|\operatorname{tr}(Y)-\operatorname{tr}(X_{0})|\le c_{2}\,d_{\mathcal{S}(n)}(Y,X_{0})<c_{2}\delta\le\varepsilon', the last two steps by claim 10 and by multiplying δεc21\delta\le\varepsilon'c_{2}^{-1} by c2c_{2}. By (iii), and since h1\lVert h\rVert\le1, the same reasoning with c3c_{3} gives 12q2p02c3h<c3δε\tfrac{1}{2}|\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}|\le c_{3}\lVert h\rVert<c_{3}\delta\le\varepsilon'. By (iv) and dE(y,x0)<δδfd_{E}(y,x_{0})<\delta\le\delta_{f}, the fourth summand is less than ε\varepsilon'.

Adding the four strict inequalities by claim 3 of Elementary Order Arithmetic in an Ordered Field, applied repeatedly, and using claim 2 there,

F(y,s,q,Y)F(x0,r0,p0,X0)<ε+ε+ε+ε=ε.\bigl|F(y,s,q,Y)-F(x_{0},r_{0},p_{0},X_{0})\bigr|<\varepsilon'+\varepsilon'+\varepsilon'+\varepsilon'=\varepsilon .

Thus FF is continuous at (x0,r0,p0,X0)(x_{0},r_{0},p_{0},X_{0}) in the sense of Continuity of a Second-Order Equation Operator §at-point. Since the quadruple was arbitrary, FF is continuous, which is claim 1.

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