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Dirichlet eigenvalues, simplicity and the nodal theorem on an interval

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ and let $V \in C ([0, L]; R)$ , with $V_{\min} : = \min_{[0, L]} V$ and $V_{\max} : = \max_{[0, L]} V$ . For $E \in R$ let $u (\cdot, E)$ be the solution of $- u^{''} + V u = E u$ , $u (0) = 0$ , $u^{'} (0) = 1$ , and let $θ (\cdot, E)$ be its Prufer angle, as in Prufer angle for the interval equation and strict monotonicity in the energy. Call $E$ a Dirichlet eigenvalue when there is a nonzero $w \in H^{2} (0, L)$ (One-dimensional Sobolev spaces on intervals) with $w^{*} (0) = w^{*} (L) = 0$ satisfying $- w^{''} + V w = E w$ , $w^{*}$ the continuous representative of Absolutely continuous representative of an L2 function with L2 weak derivative. Then:

(a) $E$ is a Dirichlet eigenvalue if and only if $θ (L, E) = k π$ for some integer $k \geq 1$ ;

(b) for every $k \geq 1$ the equation $θ (L, E) = k π$ has exactly one solution $E_{k}$ ; the sequence $(E_{k})_{k \geq 1}$ is strictly increasing, $E_{k} \to \infty$ , and it is exactly the sequence of Dirichlet eigenvalues, each with one-dimensional eigenspace: $\ker (- \frac{d^{2}}{d x^{2}} + V - E_{k}) = C u (\cdot, E_{k})$ ;

(d) nodal count: for every $E$ , the number of zeros of $u (\cdot, E)$ in $(0, L]$ equals $⌊ θ (L, E) / π ⌋$ ; consequently the $k$ -th eigenfunction $u (\cdot, E_{k})$ has exactly $k - 1$ zeros in $(0, L)$ , and the same holds for every nonzero eigenfunction with eigenvalue $E_{k}$ .

Facts & Assumptions

Given: $0 < L < \infty$ ; real continuous $V$ ; the solution family $u (\cdot, E)$ and Prufer angle $θ (\cdot, E)$ of Prufer angle for the interval equation and strict monotonicity in the energy; the spaces One-dimensional Sobolev spaces on intervals.

[F1]

$u (\cdot, E)$ is the unique $C^{2}$ solution of its initial value problem, $r : = (u^{2} + u^{' 2})^{1 / 2} > 0$ , $u = r \sin θ$ , $u^{'} = r \cos θ$ , $θ (0, E) = 0$ , $θ$ is $C^{1}$ in $x$ with $θ^{'} = \cos^{2} θ + (E - V) \sin^{2} θ$ , $E \mapsto θ (L, E)$ is continuous and strictly increasing on $R$ , $θ (x, E) > 0$ for $x > 0$ , and every level $k π$ in $(0, θ (L, E))$ is attained exactly once (Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence, Prufer angle for the interval equation and strict monotonicity in the energy). For the explicit free solution in step 1.4 we use that $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , $\sin 0 = 0$ , $\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine) and the zero sets $\sin x = 0 \Leftrightarrow x \in π Z$ , $\cos x = 0 \Leftrightarrow x \in (Z + \frac{1}{2}) π$ (The zero sets of sine and cosine and the least positive common period 2 pi).

[F2]

If $w \in H^{2} (0, L)$ and $- w^{''} + V w = E w$ with $w^{*} (0) = w^{*} (L) = 0$ and $w \neq 0$ , then $w^{'} (0) \neq 0$ : otherwise $w$ and the zero function are two $C^{2}$ solutions (by Absolutely continuous representative of an L2 function with L2 weak derivative, the class $w$ has a $C^{2}$ representative because its second weak derivative is the continuous function $(V - E) w^{*}$ , and the representative solves the same initial value problem) with the same initial data, contradicting uniqueness in [F1]. Consequently the $C^{2}$ representative of $w$ equals $w^{'} (0) u (\cdot, E)$ .

[F3]

Comparison lemma for scalar ODEs, a direct consequence of Gronwall's inequality (Gronwall's integral inequality with variable and constant coefficients): let $g : R \to R$ be Lipschitz with constant $K$ , and let $φ, ψ$ be continuous and piecewise $C^{1}$ on $[0, L]$ with $φ (0) = ψ (0)$ , $φ^{'} = g (φ)$ and $ψ^{'} \leq g (ψ)$ wherever differentiable. Then $ψ \leq φ$ . Indeed, for $w : = ψ - φ$ the inequality $w^{'} \leq g (ψ) - g (φ) \leq K w$ holds on every interval where $w > 0$ ; if $w (x_{1}) > 0$ and $x_{0} : = \max {x \leq x_{1} : w (x) = 0}$ , then $w (x_{0}) = 0$ and $w > 0$ on $(x_{0}, x_{1}]$ , so $(w e^{- K (x - x_{0})})^{'} \leq 0$ there with value $0$ at $x_{0}$ , forcing $w (x_{1}) \leq 0$ , a contradiction. The reversed inequality $ψ^{'} \geq g (ψ)$ likewise gives $ψ \geq φ$ .

[F4]

Intermediate value theorem for continuous real functions on $[0, L]$ (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ); the integral of a continuous function is continuous (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

Proof

1.1F1F2algebra

[Every eigenvalue is a root] Let $E$ be a Dirichlet eigenvalue with nonzero eigenfunction $w$ . By [F2], the $C^{2}$ representative of $w$ equals $w^{'} (0) u (\cdot, E)$ with $w^{'} (0) \neq 0$ . Since $w^{*} (L) = 0$ , $u (L, E) = 0$ , and $u (L, E) = r (L, E) \sin θ (L, E)$ with $r (L, E) > 0$ , so $\sin θ (L, E) = 0$ and $θ (L, E) \in π Z$ ; by [F1] $θ (L, E) > 0$ , hence $θ (L, E) = k π$ for some integer $k \geq 1$ . This also shows that the eigenspace at $E$ is contained in $C u (\cdot, E)$ , hence is one-dimensional.

1.2F1algebra

[Every root is an eigenvalue] Conversely, suppose $θ (L, E) = k π$ , $k \geq 1$ . Then $u (L, E) = r \sin θ (L, E) = 0$ and $u (0, E) = 0$ , while $u (\cdot, E)$ is a nonzero $C^{2}$ solution of $- u^{''} + V u = E u$ . Its class lies in $H^{2} (0, L)$ because $u$ is $C^{2}$ with bounded derivatives on the compact interval, and its continuous representative has the endpoint values $0$ . Hence $E$ is a Dirichlet eigenvalue.

1.3F1F3

[Free comparison] Let $β \in R$ and let $v_{β}$ be the solution of $v^{''} + β v = 0$ , $v (0) = 0$ , $v^{'} (0) = 1$ , with Prufer angle $φ_{β}$ (so $v_{β} = r_{β} \sin φ_{β}$ , $v_{β}^{'} = r_{β} \cos φ_{β}$ , $φ_{β} (0) = 0$ , $φ_{β}^{'} = \cos^{2} φ_{β} + β \sin^{2} φ_{β}$ by [F1] applied with $V \equiv 0$ ). If $a : = E - V (x) \leq β$ on $[0, L]$ , then $θ (x, E) \leq φ_{β} (x)$ for all $x$ ; if $a \geq β$ , then $θ (x, E) \geq φ_{β} (x)$ . Indeed, in the first case $θ^{'} = \cos^{2} θ + a \sin^{2} θ \leq \cos^{2} θ + β \sin^{2} θ = g (θ)$ and $φ_{β}^{'} = g (φ_{β})$ with $g (ϑ) = \cos^{2} ϑ + β \sin^{2} ϑ$ Lipschitz; [F3] with $ψ = θ$ (so $ψ^{'} \leq g (ψ)$ ) and $φ = φ_{β}$ gives $φ_{β} \geq θ$ , and the second case is symmetric.

1.4F1algebra

[The free angle and its zero count] For $β > 0$ the solution is $v_{β} (x) = β^{- 1 / 2} \sin (β^{1 / 2} x)$ , whose zeros are $j π / β^{1 / 2}$ , $j \in N$ , all simple. By [F1] every level $j π$ below $φ_{β} (L)$ is attained exactly once, so with $m : = ⌊ L β^{1 / 2} / π ⌋$ the number of zeros of $v_{β}$ in $(0, L]$ equals $m$ and $m π \leq φ_{β} (L) < (m + 1) π$ ; when $L β^{1 / 2} / π = k \in N$ the $k$ -th zero is exactly at $L$ , so $φ_{β} (L) = k π$ . In particular $φ_{β} (L) \geq k π$ whenever $L β^{1 / 2} \geq k π$ . For $β \leq 0$ the solution satisfies $v_{β}^{''} = - β v_{β} \geq 0$ wherever $v_{β} > 0$ ; since $v_{β} (0) = 0$ , $v_{β}^{'} (0) = 1$ and $v_{β}^{''} \geq 0$ on $[0, L]$ , we get $v_{β}^{'} > 0$ and $v_{β} > 0$ on $(0, L]$ , so $v_{β}$ has no zeros there and $0 \leq φ_{β} (L) < π / 2$ .

2.1step 1.3step 1.4algebra

[Endpoint limits of $θ (L, \cdot)$ ] If $E \leq V_{\min}$ , then $a = E - V (x) \leq 0$ , so by step 1.3 with $β = 0$ , $0 < θ (L, E) \leq φ_{0} (L) < π / 2$ . If $E \geq V_{\max} + (k π / L)^{2}$ , then $a = E - V (x) \geq E - V_{\max} = : β \geq (k π / L)^{2}$ and step 1.3 gives $θ (L, E) \geq φ_{β} (L) \geq k π$ by step 1.4.

3.1step 1.3step 1.4step 2.1F1F4algebra

[Existence, uniqueness and bounds for $E_{k}$ ] Let $k \geq 1$ . Put $E^{-} : = V_{\min} - 1 \leq V_{\min}$ and $E^{+} : = V_{\max} + (k π / L)^{2}$ . By step 2.1, $θ (L, E^{-}) < π / 2 < k π \leq θ (L, E^{+})$ . Since $E \mapsto θ (L, E)$ is continuous and strictly increasing [F1], [F4] gives exactly one $E_{k} \in (E^{-}, E^{+}]$ with $θ (L, E_{k}) = k π$ . Since $k \mapsto k π$ is strictly increasing and $E \mapsto θ (L, E)$ is strictly increasing, the sequence $(E_{k})$ is strictly increasing; and $E_{k} \to \infty$ because $θ (L, E_{k}) = k π \to \infty$ . For the bounds, if $E = V_{\min} + (k π / L)^{2}$ then $a = E - V (x) \leq (k π / L)^{2} = : β$ and step 1.3 with step 1.4 gives $θ (L, E) \leq φ_{β} (L) = k π$ (the free solution has its $k$ -th zero exactly at $L$ ); by strict monotonicity this forces $E \leq E_{k}$ , i.e. $E_{k} \geq V_{\min} + (k π / L)^{2}$ . Similarly, at $E = V_{\max} + (k π / L)^{2}$ one has $θ (L, E) \geq φ_{β} (L) = k π$ with the same $β$ , so $E \geq E_{k}$ , i.e. $E_{k} \leq V_{\max} + (k π / L)^{2}$ .

4.1step 1.1step 1.2step 3.1F1∎

[Zero count and nodal theorem] For fixed $E$ , by [F1] the zeros of $u (\cdot, E) = r \sin θ (\cdot, E)$ in $(0, L]$ are exactly the points where $θ (\cdot, E)$ crosses a level $j π$ , $j \geq 1$ ; each level $j π < θ (L, E)$ is attained exactly once, no level $> θ (L, E)$ is attained, and if $θ (L, E) = m π$ then $L$ itself is a zero. Hence $# {x \in (0, L] : u (x, E) = 0} = ⌊ θ (L, E) / π ⌋ .$ In particular, for $E = E_{k}$ with $θ (L, E_{k}) = k π$ , the solution $u (\cdot, E_{k})$ has exactly $k$ zeros in $(0, L]$ , one of which is $L$ , hence exactly $k - 1$ zeros in $(0, L)$ ; by step 1.1 every nonzero eigenfunction at $E_{k}$ is a nonzero multiple of $u (\cdot, E_{k})$ and has the same zeros. This is (d), and steps 1.1, 1.2 and 3.1 give (a), (b) and (c).

Dirichlet eigenvalues, simplicity and the nodal theorem on an interval

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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