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Rescaling a geometric Schrödinger equation without treating its potential as forcing

Statement

In a coordinate ball $B_{r} (x_{0})$ let $- div (A \nabla u) + q u = 0$ , with $A$ symmetric and $A \geq λ I$ for some $λ > 0$ . Define $v (y) = u (x_{0} + r y)$ on $B_{1}$ . Then $- {div}_{y} (\tilde{A} \nabla_{y} v) + \tilde{q} v = 0$ , where $\tilde{A} (y) = A (x_{0} + r y) / λ$ and $\tilde{q} (y) = r^{2} q (x_{0} + r y) / λ$ . Thus $\tilde{A} \geq I$ , $∥ \tilde{A} ∥ \leq ∥ A ∥ / λ$ and $∥ \tilde{q} ∥_{\infty} \leq r^{2} ∥ q ∥_{\infty} / λ$ .

Facts & Assumptions

Given: The divergence and weak forms, matrix $A$ and potential coefficient $q$ are those of Coordinate and weak forms of the geometric Schrödinger equation. Here $∥ A ∥$ is its operator-norm supremum. This is a homogeneous equation for an operator that includes the potential; $q u$ is not taken as an independently controlled forcing.

[F1]

The coordinate supplier gives $u \in C_{l o c}^{2}$ and smooth $A, q$ , with the weak identity for every real smooth test compactly supported in the interior coordinate ball. In this regularity all test integrands are continuous and compactly supported; the chart integrals in the calculation below may be evaluated as compactly supported Riemann integrals.

[F2]

The total-derivative chain rule The chain rule for total derivatives: D(g∘ f)(a)=Dg(f(a))∘ Df(a) applies to the affine maps $T (y) = x_{0} + r y$ and $T^{- 1} (x) = (x - x_{0}) / r$ , and to the differentiable functions composed with them. Here $r > 0$ , $D T = r I$ and $D T^{- 1} = r^{- 1} I$ ; these affine derivatives follow directly from their exact linear increments.

[F3]

The compact-support substitution formula A compactly supported Riemann integrand admits the global change-of-variables formula from a diffeomorphism near the relevant compact preimage applies to an injective $C^{1}$ map with invertible derivative and a compactly supported Riemann integrable function whose support lies in its image. Here $T : B_{1} \to B_{r} (x_{0})$ is a smooth diffeomorphism, $∣ \det D T ∣ = r^{n}$ , and the integrands, extended by zero, are continuous and compactly supported in $B_{r} (x_{0})$ . Their restrictions to any nondegenerate bounding rectangle are Riemann integrable by Every continuous function on a closed nondegenerate rectangle in ℝᵐ is Riemann integrable. This uses the compact-support Riemann formula, not a change-of-variables assertion for arbitrary $L^{1}$ data.

Proof

Given: The hypotheses and data in the Statement and Facts & Assumptions.

Goal: Establish the conclusions of the Statement.

Proof technique: direct.

1.1F1F2F3givenalgebra

Fix a real $η \in C_{c}^{\infty} (B_{1})$ and set $φ = η \circ T^{- 1}$ . Its support is $T (supp η)$ , a compact subset of $B_{r} (x_{0})$ , so it is an admissible test in the coordinate weak identity. The chain rule gives $\nabla_{y} v = r (\nabla_{x} u) \circ T$ and $(\nabla_{x} φ) \circ T = r^{- 1} \nabla_{y} η$ . The functions $A \nabla u \cdot \nabla φ$ and $q u φ$ , extended by zero outside the ball, are continuous compactly supported Riemann integrands. Apply compact-support substitution separately to these functions using $∣ \det D T ∣ = r^{n}$ to obtain $r^{n - 2} \int_{B_{1}} A (T (y)) \nabla_{y} v \cdot \nabla_{y} η d y + r^{n} \int_{B_{1}} q (T (y)) v η d y = 0$ . Dividing by $λ r^{n - 2} > 0$ proves the claimed homogeneous weak equation for every such $η$ . The transformed coefficients are smooth and $v$ is $C^{2}$ , so the weak identity has the classical interpretation as well, by the same interior integration-by-parts and continuous-residual argument supplied by the coordinate lemma.

2.1step 1.1givenalgebra∎

For every vector $ξ$ , $ξ^{T} \tilde{A} (y) ξ = λ^{- 1} ξ^{T} A (T (y)) ξ \geq ∣ ξ ∣^{2}$ . Since $T$ maps $B_{1}$ bijectively onto $B_{r} (x_{0})$ , composition preserves the coefficient suprema. Thus $∥ \tilde{A} ∥ \leq ∥ A ∥ / λ$ and $∥ \tilde{q} ∥_{\infty} \leq r^{2} ∥ q ∥_{\infty} / λ$ . These statements do not estimate $q u$ by an additive source norm.

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