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Spherical Hankel functions: explicit solutions, constant Wronskian and exact Dirichlet-to-Neumann bounds

Statement

For $l \in N_{0}$ and $z > 0$ set

$h_{l} (z) = (- i)^{l + 1} \frac{e^{i z}}{z} \sum_{j = 0}^{l} \frac{i^{j} (l + j)!}{j! (l - j)! (2 z)^{j}} .$

Then $h_{l}$ is a complex-valued $C^{2}$ function of $z > 0$ solving

$h_{l}^{''} (z) + \frac{2}{z} h_{l}^{'} (z) + [1 - \frac{l (l + 1)}{z^{2}}] h_{l} (z) = 0,$

it never vanishes on $(0, \infty)$ , and its Wronskian-type identity $Im (\overline{h_{l} (z)} h_{l}^{'} (z)) = z^{- 2}$ holds. For $k > 0$ and $R > 0$ the spherical Dirichlet-to-Neumann coefficient

$t_{l} = \frac{k h_{l}^{'} (k R)}{h_{l} (k R)}$

satisfies the exact bounds

$Re t_{l} \leq - \frac{1}{R}, ∣ Re t_{l} ∣ \leq \frac{l + 1}{R}, 0 < Im t_{l} \leq k .$

These are exact identities and inequalities for the displayed explicit functions, not asymptotic formulas; the argument uses only elementary differentiation and a constant Wronskian, so no large- $l$ asymptotic input is required.

Facts & Assumptions

Given: An integer $l \in N_{0}$ , the explicit function $h_{l}$ of the statement on $(0, \infty)$ , and reals $k > 0$ , $R > 0$ . Complex-valued functions of the real variable $z$ are handled componentwise under the identification $C ≅ R^{2}$ of $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves, with derivative and integral as in The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral.

[L1]

Differentiation. The complex exponential is entire and its complex derivative is itself: the complex exponential is entire with $\exp^{'} = \exp$ . Linearity, product, reciprocal, and quotient rules for complex derivatives: sums, scalar multiples, products and (where the denominator is nonzero) quotients of complex differentiable functions are complex differentiable with the usual rules; in particular $z \mapsto e^{i z} P (z)$ and $z \mapsto e^{i z} P (z) / z$ are holomorphic on their domains for polynomial $P$ , hence $C^{2}$ on $(0, \infty)$ along the real axis. Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : the same rules hold for real-valued functions, and (componentwise) for the real and imaginary parts of a complex-valued differentiable function: if $y = u + i v$ with $u, v$ real differentiable on an interval, then $y^{'} = u^{'} + i v^{'}$ , $Re y^{'} = (Re y)^{'}$ , $Im y^{'} = (Im y)^{'}$ .

[L2]

Constants from a vanishing derivative. A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant: a real-valued function continuous on an order-convex set, with derivative zero at every interior point, is constant; and two such functions with equal derivatives differ by a constant.

Proof

1.1L1algebra

Write $P (z) = \sum_{j = 0}^{l} c_{j} z^{- j}$ with $c_{j} = \frac{i^{j} (l + j)!}{j! (l - j)! 2^{j}}$ , so that $h_{l} = (- i)^{l + 1} e^{i z} P (z) / z$ , and put $y = z h_{l} = (- i)^{l + 1} e^{i z} P (z)$ . Substituting $y = e^{i z} P$ into the equation $y^{''} + (1 - l (l + 1) z^{- 2}) y = 0$ , which is equivalent to the stated radial equation because $h = y / z$ gives $h^{''} + \frac{2}{z} h^{'} + (1 - l (l + 1) z^{- 2}) h = \frac{1}{z} (y^{''} + (1 - l (l + 1) z^{- 2}) y)$ , produces by the product rule [L1] the equation $P^{''} + 2 i P^{'} - l (l + 1) z^{- 2} P = 0$ for $P$ . Inserting the finite expansion $P = \sum_{j} c_{j} z^{- j}$ and matching the coefficient of $z^{- m - 2}$ gives $[m (m + 1) - l (l + 1)] c_{m} - 2 i (m + 1) c_{m + 1} = 0$ , that is $c_{m + 1} = \frac{i (l - m) (l + m + 1)}{2 (m + 1)} c_{m}$ , and the displayed factorials satisfy this recursion because $\frac{c_{m + 1}}{c_{m}} = \frac{i (l - m) (l + m + 1)}{2 (m + 1)}$ ; the recursion terminates with $c_{l + 1} = 0$ , so $P$ is exactly this polynomial and $y$ , hence $h_{l}$ , satisfies the stated equation and is $C^{2}$ on $(0, \infty)$ . The leading term is $(- i)^{l + 1} e^{i z} / z$ with remainder $O (z^{- 2})$ .

2.1step 1.1L1L2algebra

With $y = u + i v$ , $u = Re y$ , $v = Im y$ , put $W = u v^{'} - v u^{'} = Im (\overline{y} y^{'})$ . The componentwise real and imaginary parts of the equation of step 1.1 give $u^{''} + (1 - l (l + 1) z^{- 2}) u = 0$ and $v^{''} + (1 - l (l + 1) z^{- 2}) v = 0$ ; the product rule [L1] gives $W^{'} = u v^{''} - v u^{''} = u [- (1 - l (l + 1) z^{- 2}) v] - v [- (1 - l (l + 1) z^{- 2}) u] = 0$ on $(0, \infty)$ , so $W$ is constant by [L2]. Since $P (z) \to c_{0} = 1$ and $P^{'} (z) \to 0$ as $z \to \infty$ , $y = C_{l} e^{i z} (1 + o (1))$ and $y^{'} = i C_{l} e^{i z} (1 + o (1))$ , where $C_{l} = (- i)^{l + 1}$ has modulus one, so $\overline{y} y^{'} = ∣ y ∣^{2} (i + o (1))$ and $W \to Im i = 1$ ; hence $W \equiv 1$ . Consequently $Im (\overline{h_{l}} h_{l}^{'}) = Im ((\overline{y} / z) (y^{'} / z - y / z^{2})) = W / z^{2} - ∣ y ∣^{2} Im (1 / z^{3}) = z^{- 2}$ , which is nonzero for every $z > 0$ ; therefore $h_{l} (z) \neq 0$ on $(0, \infty)$ .

3.1step 1.1step 2.1L1

Put $F = ∣ y ∣^{2}$ and $c (z) = 1 - l (l + 1) z^{- 2}$ . By step 2.1 the complex quantity $A = \overline{y} y^{'}$ has $Im A = 1$ , so $A^{'} = Re (A)^{'} = \frac{1}{2} F^{''}$ is real (componentwise product rule [L1]); on the other hand $A^{'} = ∣ \overline{y^{'}} ∣^{2} + \overline{y} y^{''} = ∣ y^{'} ∣^{2} - c F$ . Differentiating $\frac{1}{2} F^{''} = ∣ y^{'} ∣^{2} - c F$ gives $\frac{1}{2} F^{'''} = 2 Re (\overline{y^{'}} y^{''}) - c^{'} F - c F^{'}$ ; since $y^{''} = - c y$ , $Re (\overline{y^{'}} y^{''}) = - c Re (\overline{y^{'}} y) = - c \frac{F^{'}}{2}$ , and with $c^{'} = 2 l (l + 1) z^{- 3}$ this yields the linear third-order equation $F^{'''} + 4 (1 - l (l + 1) z^{- 2}) F^{'} + 4 l (l + 1) z^{- 3} F = 0$ . Since $P = \sum_{j} c_{j} z^{- j}$ with $c_{j} = i^{j} (real)$ , all cross terms of $∣ P ∣^{2} = \sum_{j, k} c_{j} \overline{c_{k}} z^{- j - k}$ with $j - k$ odd cancel in the sum over $(j, k)$ and $(k, j)$ , so $F = ∣ P ∣^{2}$ is a finite even polynomial in $z^{- 1}$ , $F = \sum_{j \geq 0} a_{j} z^{- 2 j}$ with real $a_{j}$ and $a_{0} = 1$ ; matching the coefficient of $z^{- 2 m - 3}$ in the third-order equation gives $- 4 m (m + 1) (2 m + 1) a_{m} + 8 m l (l + 1) a_{m} + 4 l (l + 1) a_{m} - 8 (m + 1) a_{m + 1} = 0$ , i.e. $a_{m + 1} = \frac{(2 m + 1) [l (l + 1) - m (m + 1)]}{2 (m + 1)} a_{m}$ , whose positive solution is $a_{j} = \frac{(2 j)! (l + j)!}{2^{2 j} (j!)^{2} (l - j)!}$ for $0 \leq j \leq l$ and $a_{j} = 0$ for $j > l$ , because the factor $[l (l + 1) - j (j + 1)]$ vanishes at $j = l$ . In particular $a_{j} > 0$ for $j \leq l$ , $a_{0} = 1$ , and $∣ h_{l} (z) ∣^{2} = F (z) / z^{2} = \sum_{j = 0}^{l} a_{j} z^{- 2 j - 2} \geq z^{- 2}$ .

4.1step 2.1step 3.1algebra

On $(0, \infty)$ one has $\frac{h_{l}^{'}}{h_{l}} = \frac{\overline{h_{l}} h_{l}^{'}}{∣ h_{l} ∣^{2}}$ and, from step 3.1 and $\overline{y} y^{'} = \frac{F^{'}}{2} + i$ (step 2.1), $Re (\overline{h_{l}} h_{l}^{'}) = Re (\frac{\overline{y} y^{'}}{z^{2}} - \frac{∣ y ∣^{2}}{z^{3}}) = \frac{F^{'}}{2 z^{2}} - \frac{F}{z^{3}} = - \frac{1}{z^{3}} \sum_{j = 0}^{l} (1 + j) a_{j} z^{- 2 j}$ and $Im (\overline{h_{l}} h_{l}^{'}) = z^{- 2}$ ; dividing by $∣ h_{l} ∣^{2} = F (z) / z^{2}$ gives $Re \frac{h_{l}^{'} (z)}{h_{l} (z)} = - \frac{1}{z} \frac{\sum_{j = 0}^{l} (1 + j) a_{j} z^{- 2 j}}{\sum_{j = 0}^{l} a_{j} z^{- 2 j}}$ and $Im \frac{h_{l}^{'} (z)}{h_{l} (z)} = \frac{1}{F (z)}$ . With $z = k R$ and weights $a_{j} (k R)^{- 2 j} \geq 0$ , the real part of $t_{l} = k h_{l}^{'} (k R) / h_{l} (k R)$ is $- \frac{1}{R}$ times a weighted average of the numbers $1 + j$ , $j = 0, \dots, l$ , so $- (l + 1) / R \leq Re t_{l} \leq - 1 / R$ ; the imaginary part is $k / F (k R)$ , and since $F (k R) = \sum_{j \geq 0} a_{j} (k R)^{- 2 j} \geq a_{0} = 1$ and $F < \infty$ , one has $0 < Im t_{l} \leq k$ . This proves all the stated bounds, and the bounds are exact identities for the displayed $h_{l}$ .

5.1step 1.1step 2.1step 3.1step 4.1∎

Collecting: step 1.1 gives the $C^{2}$ solution property and the explicit leading term; step 2.1 gives the constant Wronskian $Im (\overline{h_{l}} h_{l}^{'}) = z^{- 2}$ and nonvanishing; steps 3.1 and 4.1 give the positive coefficient representation $∣ h_{l} ∣^{2} = z^{- 2} \sum_{j} a_{j} z^{- 2 j} \geq z^{- 2}$ and the exact bounds $Re t_{l} \leq - 1 / R$ , $∣ Re t_{l} ∣ \leq (l + 1) / R$ , $0 < Im t_{l} \leq k$ for the Dirichlet-to-Neumann coefficient.

Depends on

Dependency tree · 2 levels

Sources

NIST Digital Library of Mathematical Functions, Sec. 10.47 Standard Solutions of the Spherical Bessel Equation (Eqs. 10.47.1-10.47.9), with Sec. 10.50 checked for Wronskian identities