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The separated sine and cosine rectangle profiles: eigen-identities, boundary traces, orthogonality and Dirichlet integrals

Statement

Let $L_{1}, L_{2} > 0$ and $R : = (0, L_{1}) \times (0, L_{2})$ . For integers $m, n \geq 0$ put $h_{m n} (x_{1}, x_{2}) : = \cos (\frac{m π x_{1}}{L_{1}}) \cos (\frac{n π x_{2}}{L_{2}})$ and $k_{m n}^{2} : = (\frac{m π}{L_{1}})^{2} + (\frac{n π}{L_{2}})^{2}$ ; for integers $m, n \geq 1$ put $e_{m n} (x_{1}, x_{2}) : = \sin (\frac{m π x_{1}}{L_{1}}) \sin (\frac{n π x_{2}}{L_{2}})$ . Then, with $Δ = \partial_{1}^{2} + \partial_{2}^{2}$ , $\nabla = (\partial_{1}, \partial_{2})$ and the Riemann integrals over the rectangle of Axis-parallel rectangles in $R^{m}$ and their volume: (a) $Δ h_{m n} = - k_{m n}^{2} h_{m n}$ on $R$ and $\partial_{1} h_{m n} = 0$ on both faces $x_{1} \in {0, L_{1}}$ , $\partial_{2} h_{m n} = 0$ on both faces $x_{2} \in {0, L_{2}}$ ; hence $\partial_{n} h_{m n} = 0$ on every open face of $\partial R$ ; (b) $Δ e_{m n} = - k_{m n}^{2} e_{m n}$ on $R$ and $e_{m n} = 0$ on $\partial R$ for $m, n \geq 1$ ; (c) [orthogonality and norms] $\int_{R} h_{m n} h_{m^{'} n^{'}} = 0$ unless $(m, n) = (m^{'}, n^{'})$ , and $\int_{R} h_{m n}^{2} = \frac{L_{1} L_{2}}{4} (1 + δ_{m 0}) (1 + δ_{n 0})$ ; $\int_{R} e_{m n} e_{m^{'} n^{'}} = 0$ unless $(m, n) = (m^{'}, n^{'})$ , and $\int_{R} e_{m n}^{2} = \frac{L_{1} L_{2}}{4}$ for $m, n \geq 1$ ; the mixed integrals are instead $\int_{R} h_{m n} e_{p q} = J_{m p} (L_{1}) J_{n q} (L_{2}), J_{m p} (L) : = \int_{0}^{L} \cos (m π x / L) \sin (p π x / L) d x,$ for $m, n \geq 0$ , $p, q \geq 1$ . Here $J_{m p} (L) = 0$ when $m, p$ have matching parity (including $p = m$ ), and $J_{m p} (L) = 2 p L / [π (p^{2} - m^{2})]$ when their parities differ. In particular $\int_{R} h_{00} e_{11} = 4 L_{1} L_{2} / π^{2}$ , so the sine and cosine systems are not mutually orthogonal; (d) [Dirichlet-type identities] $\int_{R} ∣ \nabla h_{m n} ∣^{2} = k_{m n}^{2} \int_{R} h_{m n}^{2}$ for all $m, n \geq 0$ and $\int_{R} ∣ \nabla e_{m n} ∣^{2} = k_{m n}^{2} \int_{R} e_{m n}^{2}$ for $m, n \geq 1$ .

Facts & Assumptions

Given: Positive reals $L_{1}, L_{2}$ , the rectangle $R = (0, L_{1}) \times (0, L_{2})$ , and the profiles $h_{m n}$ ( $m, n \geq 0$ ), $e_{m n}$ ( $m, n \geq 1$ ) of the Statement.

[F1]

For an open $U \subseteq R^{2}$ and $f : U \to R$ , $\partial_{j} f (a) = D_{e_{j}} f (a)$ is the $j$ th partial derivative, computed by restricting to coordinate lines (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ), and for $f \in C^{2}$ the Laplacian is $Δ f = div \nabla f = \partial_{1} \partial_{1} f + \partial_{2} \partial_{2} f$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F2]

$[a, b] = {x : a_{j} \leq x_{j} \leq b_{j}}$ with $vol [a, b] = \prod_{j} (b_{j} - a_{j})$ , and $[a, b]$ is nondegenerate when every $a_{j} < b_{j}$ (Axis-parallel rectangles in $R^{m}$ and their volume); every continuous real function on a closed nondegenerate rectangle is Riemann integrable (Every continuous function on a closed nondegenerate rectangle in $R^{m}$ is Riemann integrable); and for $f$ Riemann integrable on a product $A \times B$ , the section-integral functions are integrable and $\int_{A \times B} f = \int_{A} \int_{B} f_{x}$ whenever the inner sections are integrable (Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections).

[F3]

If $G : [a, b] \to R$ is differentiable on $[a, b]$ and $G^{'}$ is integrable there, then $\int_{a}^{b} G^{'} = G (b) - G (a)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[F4]

$(\sin x)^{'} = \cos x$ and $(\cos x)^{'} = - \sin x$ on $R$ , with $\sin 0 = 0$ , $\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine); derivatives of sums, scalar multiples and products obey the sum, scalar and product rules (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 $g (c) \neq 0$ ); and composites of totally differentiable maps obey the chain rule (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ), which applies to the affine maps $x_{j} \mapsto \frac{m π x_{j}}{L_{j}}$ and $x_{j} \mapsto \frac{n π x_{j}}{L_{j}}$ .

[F5]

$\sin (k π) = 0$ and $\cos (k π) = (- 1)^{k}$ for every integer $k$ , and $\sin 0 = 0$ , $\cos 0 = 1$ : the zero sets, period and special values of sine and cosine (The zero sets of sine and cosine and the least positive common period 2 pi, Quarter-turn values and shifts by pi/2 and pi).

[F6]

For all real $A, B$ , $\sin (A + B) = \sin A \cos B + \cos A \sin B$ and $\cos (A + B) = \cos A \cos B - \sin A \sin B$ (The addition formulas for sine and cosine), hence the product-to-sum identities $\cos A \cos B = \frac{1}{2} [\cos (A - B) + \cos (A + B)], \sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)], \sin A \cos B = \frac{1}{2} [\sin (A + B) + \sin (A - B)] .$

Proof

technique · direct: differentiate the products, then integrate the resulting continuous one-variable functions by product-to-sum, the fundamental theorem and Fubini

1.1givenF1F4F5algebra

[First derivatives and traces] Let $m, n \geq 0$ and $x \in R$ . The partial derivatives of the cosine products are $\partial_{1} h_{m n} (x) = - \frac{m π}{L_{1}} \sin (\frac{m π x_{1}}{L_{1}}) \cos (\frac{n π x_{2}}{L_{2}})$ and $\partial_{2} h_{m n} (x) = - \frac{n π}{L_{2}} \cos (\frac{m π x_{1}}{L_{1}}) \sin (\frac{n π x_{2}}{L_{2}})$ : fix the other coordinate and apply the chain rule to $x_{j} \mapsto \cos (\frac{m π x_{j}}{L_{j}})$ composed with the affine map, i.e. the derivative $- \sin \cdot \frac{m π}{L_{j}}$ , together with the product rule. Evaluating with [F5] gives $\partial_{1} h_{m n} (0, x_{2}) = \partial_{1} h_{m n} (L_{1}, x_{2}) = 0$ and $\partial_{2} h_{m n} (x_{1}, 0) = \partial_{2} h_{m n} (x_{1}, L_{2}) = 0$ . Similarly, for $m, n \geq 1$ , $\partial_{1} e_{m n} (x) = \frac{m π}{L_{1}} \cos (\frac{m π x_{1}}{L_{1}}) \sin (\frac{n π x_{2}}{L_{2}})$ , $\partial_{2} e_{m n} (x) = \frac{n π}{L_{2}} \sin (\frac{m π x_{1}}{L_{1}}) \cos (\frac{n π x_{2}}{L_{2}})$ , and $e_{m n} = 0$ on $\partial R$ because a sine factor vanishes whenever $x_{1} \in {0, L_{1}}$ or $x_{2} \in {0, L_{2}}$ .

1.2F3F5F6algebra

[One-dimensional orthogonality and norms] Let $m, m^{'} \geq 0$ and $p, p^{'} \geq 1$ be integers. By [F6], $\cos (m π x / L) \cos (m^{'} π x / L) = \frac{1}{2} [\cos (\frac{(m - m^{'}) π x}{L}) + \cos (\frac{(m + m^{'}) π x}{L})]$ , so the second fundamental theorem [F3] applied to the continuous antiderivatives, together with the integer values [F5], gives $\int_{0}^{L} \cos \frac{m π x}{L} \cos \frac{m^{'} π x}{L} d x = 0$ for $m \neq m^{'}$ , $= \frac{L}{2}$ for $m = m^{'} \geq 1$ and $= L$ for $m = m^{'} = 0$ ; uniformly $\int_{0}^{L} \cos \frac{m π x}{L} \cos \frac{m^{'} π x}{L} d x = δ_{m m^{'}} \frac{L}{2} (1 + δ_{m 0})$ . Likewise $\int_{0}^{L} \sin \frac{p π x}{L} \sin \frac{p^{'} π x}{L} d x = \frac{L}{2} δ_{p p^{'}}$ , because $\sin \frac{p π x}{L} \sin \frac{p^{'} π x}{L} = \frac{1}{2} [\cos \frac{(p - p^{'}) π x}{L} - \cos \frac{(p + p^{'}) π x}{L}]$ and every nonzero integer multiple of $π$ is a zero of sine (and the case $p = p^{'}$ leaves the constant $1$ in the first term).

2.1step 1.1F1F4algebra

[Eigen-identities] Differentiating the formulas of step 1.1 once more in the same coordinate gives $\partial_{1}^{2} h_{m n} = - (\frac{m π}{L_{1}})^{2} h_{m n}$ and $\partial_{2}^{2} h_{m n} = - (\frac{n π}{L_{2}})^{2} h_{m n}$ on $R$ , so $Δ h_{m n} = - k_{m n}^{2} h_{m n}$ by [F1]; the same computation gives $\partial_{1}^{2} e_{m n} = - (\frac{m π}{L_{1}})^{2} e_{m n}$ , $\partial_{2}^{2} e_{m n} = - (\frac{n π}{L_{2}})^{2} e_{m n}$ and $Δ e_{m n} = - k_{m n}^{2} e_{m n}$ . Since each open face of $\partial R$ carries one of the normals $\pm e_{1}, \pm e_{2}$ and the corresponding normal derivative $\partial_{1} h_{m n}$ or $\partial_{2} h_{m n}$ vanishes on it by step 1.1, clause (a) and clause (b) hold.

2.2step 1.2F2algebra

[Rectangle orthogonality, norms and mixed integrals] Each of the functions $h_{m n} h_{m^{'} n^{'}}$ , $e_{m n} e_{m^{'} n^{'}}$ and $h_{m n} e_{p q}$ is continuous on the closed rectangle $\overline{R} = [0, L_{1}] \times [0, L_{2}]$ , hence Riemann integrable there by [F2]; their sections in each coordinate are continuous, hence integrable. Fubini's theorem [F2] therefore factorizes each integral into a product of one-dimensional integrals. By step 1.2, $\int_{R} h_{m n} h_{m^{'} n^{'}} = \frac{L_{1} L_{2}}{4} (1 + δ_{m 0}) (1 + δ_{n 0})$ if $(m, n) = (m^{'}, n^{'})$ and $= 0$ otherwise, and $\int_{R} e_{m n} e_{m^{'} n^{'}} = \frac{L_{1} L_{2}}{4}$ if $(m, n) = (m^{'}, n^{'})$ and $= 0$ otherwise. For the mixed integral, $\int_{R} h_{m n} e_{p q} = J_{m p} (L_{1}) J_{n q} (L_{2})$ with the stated $J_{m p}$ .

3.1step 2.2F3F5F6algebra

[Evaluation of $J$ ] Let $m \geq 0$ , $p \geq 1$ and $L > 0$ . By [F6], $\cos \frac{m π x}{L} \sin \frac{p π x}{L} = \frac{1}{2} [\sin \frac{(p + m) π x}{L} + \sin \frac{(p - m) π x}{L}]$ . If $p = m$ , the second sine vanishes identically and the integral of the first over $[0, L]$ is $\frac{L}{2 π m} (1 - \cos (2 m π)) = 0$ by the fundamental theorem [F3] and [F5] (here $m = p \geq 1$ ); thus $J_{m m} = 0$ . If $p \neq m$ , the fundamental theorem gives $J_{m p} = \frac{L}{2 π} [\frac{1 - \cos ((p + m) π)}{p + m} + \frac{1 - \cos ((p - m) π)}{p - m}] = \frac{L}{2 π} (1 - (- 1)^{p + m}) [\frac{1}{p + m} + \frac{1}{p - m}] = \frac{p L}{π (p^{2} - m^{2})} (1 - (- 1)^{p + m})$ , using $\cos (k π) = (- 1)^{k}$ and $\frac{1}{p + m} + \frac{1}{p - m} = \frac{2 p}{p^{2} - m^{2}}$ . Hence $J_{m p} = 0$ when $m, p$ have matching parity (the case $p = m$ included) and $J_{m p} = \frac{2 p L}{π (p^{2} - m^{2})}$ when their parities differ. In particular $J_{01} (L) = \frac{2 L}{π}$ for $L = L_{1}, L_{2}$ , so $\int_{R} h_{00} e_{11} = J_{01} (L_{1}) J_{01} (L_{2}) = \frac{4 L_{1} L_{2}}{π^{2}}$ .

4.1step 1.1step 2.1F2F3F4algebra∎

[Dirichlet-type identities] Fix $x_{2} \in (0, L_{2})$ and put $G_{1} (x_{1}) : = h_{m n} (x_{1}, x_{2}) \partial_{1} h_{m n} (x_{1}, x_{2})$ on $[0, L_{1}]$ . By the product rule [F4], $G_{1}$ is differentiable on $[0, L_{1}]$ with $G_{1}^{'} = (\partial_{1} h_{m n})^{2} + h_{m n} \partial_{1}^{2} h_{m n}$ , a continuous hence integrable function, and $G_{1} (0) = G_{1} (L_{1}) = 0$ by step 1.1; the fundamental theorem [F3] gives $\int_{0}^{L_{1}} [(\partial_{1} h_{m n})^{2} + h_{m n} \partial_{1}^{2} h_{m n}] d x_{1} = 0$ . The same argument in the second coordinate, and step 2.1 (respectively step 1.1 for the vanishing trace), give $\int_{0}^{L_{2}} [(\partial_{2} h_{m n})^{2} + h_{m n} \partial_{2}^{2} h_{m n}] d x_{2} = 0$ , and the identical argument with $e_{m n}$ in place of $h_{m n}$ gives $\int_{0}^{L_{1}} [(\partial_{1} e_{m n})^{2} + e_{m n} \partial_{1}^{2} e_{m n}] d x_{1} = 0$ and its second-coordinate counterpart, the boundary terms vanishing because $e_{m n}$ is zero on $\partial R$ . Integrating the resulting pointwise identities over the rectangle against the other coordinate and applying Fubini [F2] (all integrands are continuous on $\overline{R}$ ), then substituting $\partial_{1}^{2} h_{m n} = - (\frac{m π}{L_{1}})^{2} h_{m n}$ , $\partial_{2}^{2} h_{m n} = - (\frac{n π}{L_{2}})^{2} h_{m n}$ and the analogous two identities for $e_{m n}$ from step 2.1, yields both statements of clause (d).

The separated sine and cosine rectangle profiles: eigen-identities, boundary traces, orthogonality and Dirichlet integrals

Statement

Facts & Assumptions

Proof

Depends on

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