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Investigation of coefficient expressions arising in the homogenization process

Here we investigate some of the expressions that arise in the coefficients of the homogenized equations. A key question is whether a given expression vanishes for all periodic functions. We have observed that certain expressions seem to vanish if and only if the periodic function has matching derivatives:

f(j)​(0)\displaystyle f^{(j)}(0) =f(j)​(1)\displaystyle=f^{(j)}(1) (1)

for all whole numbers jj. In order to prove this, we may try to write the expression as a linear combination of the differences f(j)​(0)−f(j)​(1)f^{(j)}(0)-f^{(j)}(1). To make this more straightforward, we can consider the case that ff is a trigonometric polynomial, in the hope of extending the proof to analytic functions by using Fourier series.

0.1 Useful trig identities

cos⁡(a)​sin⁡(b)\displaystyle\cos(a)\sin(b) =12​(sin⁡(a+b)+sin⁡(a−b))\displaystyle=\frac{1}{2}(\sin(a+b)+\sin(a-b)) (2)
cos⁡(a)​cos⁡(b)\displaystyle\cos(a)\cos(b) =12​(cos⁡(a+b)+cos⁡(a−b))\displaystyle=\frac{1}{2}(\cos(a+b)+\cos(a-b)) (3)

For simplicity we consider just the function

f​(y)\displaystyle f(y) =∑j=0saj​cos⁡(2​π​j​y).\displaystyle=\sum_{j=0}^{s}a_{j}\cos(2\pi jy). (5)

Then we find

⟦f⟧(y)\displaystyle\llbracket f\rrbracket(y) =∑j=1saj​sin⁡(2​π​j​y)2​π​j.\displaystyle=\sum_{j=1}^{s}a_{j}\frac{\sin(2\pi jy)}{2\pi j}. (6)

0.2 Showing that ⟨f(y)⟦f⟧(y)⟩=0\braket{f(y)\llbracket f\rrbracket(y)}=0

We have

⟨f(y)⟦f⟧(y)⟩\displaystyle\braket{f(y)\llbracket f\rrbracket(y)} =∑j=1s∑k=0saj​ak2​π​j​∫01cos⁡(2​π​y​k)​sin⁡(2​π​y​j)​𝑑y\displaystyle=\sum_{j=1}^{s}\sum_{k=0}^{s}\frac{a_{j}a_{k}}{2\pi j}\int_{0}^{1}\cos(2\pi yk)\sin(2\pi yj)dy (7)
=∑j=1s∑k=0saj​ak4​π​j​∫01(sin⁡(2​π​y​(k+j))+sin⁡(2​π​y​(k−j)))​𝑑y\displaystyle=\sum_{j=1}^{s}\sum_{k=0}^{s}\frac{a_{j}a_{k}}{4\pi j}\int_{0}^{1}\left(\sin(2\pi y(k+j))+\sin(2\pi y(k-j))\right)dy (8)
=0.\displaystyle=0. (9)

Where we have used (2).

0.3 Showing that ⟨(f(y))2⟦f⟧(y)⟩=0\braket{(f(y))^{2}\llbracket f\rrbracket(y)}=0

We have

⟨(f(y))2⟦f⟧(y)⟩\displaystyle\braket{(f(y))^{2}\llbracket f\rrbracket(y)} =∑j=1s∑k=0s∑ℓ=0saj​ak​aℓ2​π​j​∫01sin⁡(2​π​y​j)​cos⁡(2​π​y​k)​cos⁡(2​π​y​ℓ)​𝑑y.\displaystyle=\sum_{j=1}^{s}\sum_{k=0}^{s}\sum_{\ell=0}^{s}\frac{a_{j}a_{k}a_{\ell}}{2\pi j}\int_{0}^{1}\sin(2\pi yj)\cos(2\pi yk)\cos(2\pi y\ell)dy. (10)

Let us define the shorthand

S​(n)\displaystyle S(n) =sin⁡(2​π​y​n)\displaystyle=\sin(2\pi yn) (11)
C​(n)\displaystyle C(n) =cos⁡(2​π​y​n).\displaystyle=\cos(2\pi yn). (12)

Using (3) and (2), the integrand can be expressed as a sum of 4 sines:

4​sin⁡(2​π​y​j)​cos⁡(2​π​y​k)​cos⁡(2​π​y​ℓ)=S​(j+k+ℓ)+S​(j−k−ℓ)+S​(j−k+ℓ)+S​(j+k−ℓ).4\sin(2\pi yj)\cos(2\pi yk)\cos(2\pi y\ell)=S(j+k+\ell)+S(j-k-\ell)+S(j-k+\ell)+S(j+k-\ell).

Thus the integral vanishes, so the entire expression vanishes.

0.4 Showing that ⟨f(y)⟦⟦f⟧⟧(y)⟩≠0\braket{f(y)\llbracket\llbracket f\rrbracket\rrbracket(y)}\neq 0

We have

⟦⟦f⟧⟧[y]\displaystyle\llbracket\llbracket f\rrbracket\rrbracket[y] =∑j=1s−aj(2​π​j)2​cos⁡(2​π​j​y).\displaystyle=\sum_{j=1}^{s}-\frac{a_{j}}{(2\pi j)^{2}}\cos(2\pi jy). (13)

Thus

⟨f(y)⟦⟦f⟧⟧(y)⟩\displaystyle\braket{f(y)\llbracket\llbracket f\rrbracket\rrbracket(y)} =∑j=1s∑k=0s−aj​ak(2​π​j)2​∫01C​(j)​C​(k)​𝑑y\displaystyle=\sum_{j=1}^{s}\sum_{k=0}^{s}-\frac{a_{j}a_{k}}{(2\pi j)^{2}}\int_{0}^{1}C(j)C(k)dy (14)
=∑j=1s∑k=0s−aj​ak2​(2​π​j)2​∫01C​(k+j)+C​(k−j)​d​y\displaystyle=\sum_{j=1}^{s}\sum_{k=0}^{s}-\frac{a_{j}a_{k}}{2(2\pi j)^{2}}\int_{0}^{1}C(k+j)+C(k-j)dy (15)
=∑j=1s−aj22​(2​π​j)2\displaystyle=\sum_{j=1}^{s}-\frac{a_{j}^{2}}{2(2\pi j)^{2}} (16)
≠0.\displaystyle\neq 0. (17)

0.5 Infinite Fourier Series

When the Fourier series of f​(y)f(y) does not terminate, then some of the results above will depend on whether we can interchange the integral and infinite sums. E.g., we would have

⟨(f(y))2⟦f⟧(y)⟩\displaystyle\braket{(f(y))^{2}\llbracket f\rrbracket(y)} =limn→∞∫01∑j=1n∑k=0n∑ℓ=0naj​ak​aℓ2​π​j​sin⁡(2​π​y​j)​cos⁡(2​π​y​k)​cos⁡(2​π​y​ℓ)​d​y.\displaystyle=\lim_{n\to\infty}\int_{0}^{1}\sum_{j=1}^{n}\sum_{k=0}^{n}\sum_{\ell=0}^{n}\frac{a_{j}a_{k}a_{\ell}}{2\pi j}\sin(2\pi yj)\cos(2\pi yk)\cos(2\pi y\ell)dy. (18)

Under what conditions can we change the order of the integral and sums? The dominated convergence theorem suggests that we need some function g​(y)g(y) such that

|∑j=1n∑k=0n∑ℓ=0naj​ak​aℓ2​π​j​sin⁡(2​π​y​j)​cos⁡(2​π​y​k)​cos⁡(2​π​y​ℓ)|≤g​(y)\displaystyle\left|\sum_{j=1}^{n}\sum_{k=0}^{n}\sum_{\ell=0}^{n}\frac{a_{j}a_{k}a_{\ell}}{2\pi j}\sin(2\pi yj)\cos(2\pi yk)\cos(2\pi y\ell)\right|\leq g(y) (19)

for all n,yn,y.

1 Old material using polynomials

1.1 Polynomial derivative differences

Let

p​(x)\displaystyle p(x) =∑k=0sak​xk\displaystyle=\sum_{k=0}^{s}a_{k}x^{k} (20)

Then

p(j)​(1)−p(j)​(0)\displaystyle p^{(j)}(1)-p^{(j)}(0) =∑k=j+1sk!(k−j)!ak=:dj(p).\displaystyle=\sum_{k=j+1}^{s}\frac{k!}{(k-j)!}a_{k}=:d_{j}(p). (21)

1.2 Showing that ⟨f(y)⟦f⟧(y)⟩=0\braket{f(y)\llbracket f\rrbracket(y)}=0

This property is already known, but here we prove it for polynomials. In this case, the expression vanishes regardless of whether it is periodic.

Taking the double-bracket of a generic polynomial (20), we find

⟦p⟧(y)\displaystyle\llbracket p\rrbracket(y) =∑k=0sakk+1​(yk+1−y+k2​(k+2)).\displaystyle=\sum_{k=0}^{s}\frac{a_{k}}{k+1}\left(y^{k+1}-y+\frac{k}{2(k+2)}\right). (22)

Then

⟨p(y)⟦p⟧(y)⟩\displaystyle\braket{p(y)\llbracket p\rrbracket(y)} =∑j,k=0saj​ak​j​k​(j−k)2​(k+1)​(k+2)​(j+1)​(j+2)​(j+k+2)=∑j,k=0sCj​k​aj​ak.\displaystyle=\sum_{j,k=0}^{s}a_{j}a_{k}\frac{jk(j-k)}{2(k+1)(k+2)(j+1)(j+2)(j+k+2)}=\sum_{j,k=0}^{s}C_{jk}a_{j}a_{k}. (23)

Notice that

Cj​k={0j=k−Ck​jj≠k.C_{jk}=\begin{cases}0&j=k\\ -C_{kj}&j\neq k.\end{cases}

Because of this antisymmetry, the sum (23) vanishes.

1.3 ⟨f(y)2⟦f⟧(y)⟩\braket{f(y)^{2}\llbracket f\rrbracket(y)}

From (22) we have

(p(y))2⟦p⟧(y)\displaystyle(p(y))^{2}\llbracket p\rrbracket(y) =∑i,j,k=0sai​aj​akk+1​(yi+j+k+1−yi+j+1+k2​(k+2)​yi+j),\displaystyle=\sum_{i,j,k=0}^{s}\frac{a_{i}a_{j}a_{k}}{k+1}\left(y^{i+j+k+1}-y^{i+j+1}+\frac{k}{2(k+2)}y^{i+j}\right), (24)

and

⟨(p(y))2⟦p⟧(y)⟩\displaystyle\braket{(p(y))^{2}\llbracket p\rrbracket(y)} =∑i,j,k=0sai​aj​akk+1​(1i+j+k+2−1i+j+2+k2​(k+2)​(i+j+1))\displaystyle=\sum_{i,j,k=0}^{s}\frac{a_{i}a_{j}a_{k}}{k+1}\left(\frac{1}{i+j+k+2}-\frac{1}{i+j+2}+\frac{k}{2(k+2)(i+j+1)}\right)
=∑i,j,k=0sai​aj​ak​k​(i+j)​(i+j−k)2​(k+1)​(k+2)​(i+j+1)​(i+j+2)​(i+j+k+2)\displaystyle=\sum_{i,j,k=0}^{s}a_{i}a_{j}a_{k}\frac{k(i+j)(i+j-k)}{2(k+1)(k+2)(i+j+1)(i+j+2)(i+j+k+2)}
=∑k=1s∑j=0s∑ℓ=js+jak​aj​aℓ−j​k​ℓ​(ℓ−k)2​(k+1)​(k+2)​(ℓ+1)​(ℓ+2)​(ℓ+k+2)\displaystyle=\sum_{k=1}^{s}\sum_{j=0}^{s}\sum_{\ell=j}^{s+j}a_{k}a_{j}a_{\ell-j}\frac{k\ell(\ell-k)}{2(k+1)(k+2)(\ell+1)(\ell+2)(\ell+k+2)}
=∑k=1s∑j=0s∑ℓ=js+jakajaℓ−jNℓ,kDℓ,k=:S.\displaystyle=\sum_{k=1}^{s}\sum_{j=0}^{s}\sum_{\ell=j}^{s+j}a_{k}a_{j}a_{\ell-j}\frac{N_{\ell,k}}{D_{\ell,k}}=:S.

We can see immediately that all terms involving a0a_{0} vanish because Nℓ,k=−Nk,ℓN_{\ell,k}=-N_{k,\ell} and Dℓ,k=Dk,ℓD_{\ell,k}=D_{k,\ell}.

How can we show that this vanishes for a smoothly periodic function?

1.4 ⟨f(y)⟦⟦f⟧⟧(y)⟩\braket{f(y)\llbracket\llbracket f\rrbracket\rrbracket(y)}

This expression does not vanish, even if ff is smoothly periodic. Direct computation yields

⟦⟦p⟧⟧(y)\displaystyle\llbracket\llbracket p\rrbracket\rrbracket(y) =∑j,k=0sCj​k​aj​ak\displaystyle=\sum_{j,k=0}^{s}C_{jk}a_{j}a_{k} (25)

where

Cj​k=j​k​(j−1)​(k−1)​(j+k+15)+24​(j+k)12​(j+1)​(j+2)​(j+3)​(k+1)​(k+2)​(k+3)​(j+k+3)\displaystyle C_{jk}=jk\frac{(j-1)(k-1)(j+k+15)+24(j+k)}{12(j+1)(j+2)(j+3)(k+1)(k+2)(k+3)(j+k+3)} (26)

Note that in this case Cj​k=Ck​jC_{jk}=C_{kj}.