马丢函数

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马丢函数（Équation de Mathieu）是1868年法国数学家以米里迂·拉·马丢因研究数学物理所推得的特殊函数，下列马丢方程的解析解：

$$\frac{d^2y}{dx^2}+[a-2q\cos (2x) ]y=0.$$

马丢方程有两个线性无关的解：

奇数解

MathieuCE(n, q, x)，或记为$w_{I}(n,q,x)$,

偶数解

MathieuSE(n, q, x).或记为$w_{II}(n,q,x)$ 称为基本解

## 周期性

马丢函数 MathieuC(a,q,z) 或 MathieuS(a,q,z) 只有一个是周期为 $\pi$ 或$2\pi$的周期解，另一个不是。

马丢函数 MathieuC(a,q,z) 和 MathieuS(a,q,z) 两者都有是周期为$2n\pi$（n≥2)的周期函数。

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## 正交性

- $$\int_{0}^{2*\pi}\! ce_m(x,q)*ce_n(x,q)\,dx=0$$
- $$\int_{0}^{2*\pi}\! ce_m(x,q)*se_n(x,q)\,dx=0$$
- $$\int_{0}^{2*\pi}\! se_m(x,q)*se_n(x,q)\,dx=0$$

## 特征方程

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马丢方程的特征方程是

$$cos(\pi*v)=w_I(a,q,\pi)$$

$$cos(\pi*v)=w_{II}(b,q,\pi)$$

对于给定的v，q, 上列特征方程给出无穷多个a、b解称为特征值。

### 特征值的展开

马丢函数体特征值可展开成级数：

$$a_0(q)= {-(1/2)*z^2+(7/128)*z^4-(29/2304)*z^6+(68687/18874368)*z^8+\mathcal{O}(z^{10})}$$ $$a_1(q)= {1+z-(1/8)*z^2-(1/64)*z^3-(1/1536)*z^4+(11/36864)*z^5+(49/589824)*z^6+(55/9437184)*z^7-(83/35389440)*z^8-(12121/15099494400)*z^9+\mathcal{O}(z^{10})}$$ $$a_2(q)={4+(5/12)*z^2-(763/13824)*z^4+(1002401/79626240)*z^6-(1669068401/458647142400)*z^8+\mathcal{O}(z^{10})}$$ $$a_3(q)= {9+(1/16)*z^2+(1/64)*z^3+(13/20480)*z^4-(5/16384)*z^5-(1961/23592960)*z^6-(609/104857600)*z^7+(4957199/2113929216000)*z^8+(872713/1087163596800)*z^9+\mathcal{O}(z^{10})}$$

$$b_1(q)={1-z-(1/8)*z^2+(1/64)*z^3-(1/1536)*z^4-(11/36864)*z^5+(49/589824)*z^6-(55/9437184)*z^7-(83/35389440)*z^8+(12121/15099494400)*z^9+\mathcal{O}(z^{10})}$$ $$b_2(q)={4-(1/12)*z^2+(5/13824)*z^4-(289/79626240)*z^6+(21391/458647142400)*z^8+\mathcal{O}(z^{10})}$$ $$b_3(q)= {9+(1/16)*z^2-(1/64)*z^3+(13/20480)*z^4+(5/16384)*z^5-(1961/23592960)*z^6+(609/104857600)*z^7+(4957199/2113929216000)*z^8-(872713/1087163596800)*z^9+\mathcal{O}(z^{10})}$$ $$b_4(q)= {16+(1/30)*z^2-(317/864000)*z^4+(10049/2721600000)*z^6-(93824197/2006581248000000)*z^8+\mathcal{O}(z^{10})}$$ $$b_5(q)= {25+(1/48)*z^2+(11/774144)*z^4-(1/147456)*z^5+(37/891813888)*z^6+(7/339738624)*z^7+(63439/201364441399296)*z^8+(1/2130840649728)*z^9+\mathcal{O}(z^{10})}$$

## 级数展开

马丢函数ce,se的级数展开

$$ce_0(z,q)= {1-(1/2)*cos(2*z)*q+(-1/16+(1/32)*cos(4*z))*q^2+((11/128)*cos(2*z)-(1/1152)*cos(6*z))*q^3+O(q^4)}$$ $$ce_1(z,q)= {cos(z)-(1/8)*cos(3*z)*q+(-(1/128)*cos(z)-(1/64)*cos(3*z)+(1/192)*cos(5*z))*q^2+(-(1/512)*cos(z)+(1/3072)*cos(3*z)+(1/1152)*cos(5*z)-(1/9216)*cos(7*z))*q^3+O(q^4)}$$ $$ce_2(z,q)= {cos(2*z)+(1/4-(1/12)*cos(4*z))*q+(-(19/288)*cos(2*z)+(1/384)*cos(6*z))*q^2+(-49/1152+(11/4608)*cos(4*z)-(1/23040)*cos(8*z))*q^3+O(q^4)}$$ $$ce_3(z,q)= {cos(3*z)+((1/8)*cos(z)-(1/16)*cos(5*z))*q+(-(5/512)*cos(3*z)+(1/64)*cos(z)+(1/640)*cos(7*z))*q^2+(-(1/512)*cos(3*z)-(1/4096)*cos(z)+(11/40960)*cos(5*z)-(1/46080)*cos(9*z))*q^3+O(q^4)}$$ $$ce_4(z,q)= {cos(4*z)+((1/12)*cos(2*z)-(1/20)*cos(6*z))*q+(-(17/3600)*cos(4*z)+1/192+(1/960)*cos(8*z))*q^2+((7/28800)*cos(2*z)+(29/288000)*cos(6*z)-(1/80640)*cos(10*z))*q^3+O(q^4)}$$

$$se_1(z,q)= {sin(z)-(1/8)*sin(3*z)*q+(-(1/128)*sin(z)+(1/64)*sin(3*z)+(1/192)*sin(5*z))*q^2+((1/512)*sin(z)+(1/3072)*sin(3*z)-(1/1152)*sin(5*z)-(1/9216)*sin(7*z))*q^3+O(q^4)}$$ $$se_2(z,q)= {sin(2*z)-(1/12)*sin(4*z)*q+(-(1/288)*sin(2*z)+(1/384)*sin(6*z))*q^2+((1/1536)*sin(4*z)-(1/23040)*sin(8*z))*q^3+O(q^4)}$$ $$se_3(z,q)= {sin(3*z)+((1/8)*sin(z)-(1/16)*sin(5*z))*q+(-(5/512)*sin(3*z)-(1/64)*sin(z)+(1/640)*sin(7*z))*q^2+((1/512)*sin(3*z)-(1/4096)*sin(z)+(11/40960)*sin(5*z)-(1/46080)*sin(9*z))*q^3+O(q^4)}$$ $$se_4(z,q)= {sin(4*z)+((1/12)*sin(2*z)-(1/20)*sin(6*z))*q+(-(17/3600)*sin(4*z)+(1/960)*sin(8*z))*q^2+(-(1/1600)*sin(2*z)+(29/288000)*sin(6*z)-(1/80640)*sin(10*z))*q^3+O(q^4)}$$ $$se_5(z,q)= {sin(5*z)+((1/16)*sin(3*z)-(1/24)*sin(7*z))*q+(-(13/4608)*sin(5*z)+(1/384)*sin(z)+(1/1344)*sin(9*z))*q^2+(-(7/73728)*sin(3*z)+(13/258048)*sin(7*z)-(1/9216)*sin(z)-(1/129024)*sin(11*z))*q^3+O(q^4)}$$

## 傅立叶展开式

马丢函数的傅立叶展开：

- $$MathieuCE(2n, q, x)=\sum_{m=0}^{\infty}A_{2m}^{2n}(q)cos(2mx)$$
- $$MathieuCE(2n+1, q, x)=\sum_{m=0}^{\infty}A_{2m+1}^{2n+1}(q)cos[(2m+1)x]$$
- $$MathieuSE(2n+1,q,x)=\sum_{m=0}^{\infty}B_{2m+1}^{2n+1}(q)sin[(2m+1)x]$$
- $$MathieuSE(2n+2,q,x)=\sum_{m=0}^{\infty}B_{2m+2}^{2n+2}(q)sin[(2m+2)x]$$

其中系数A,B满足下列递归关系：

$$aA_0=qA_2$$

$$(a-4)A_2=q(2A_0+A_4)$$

$$(a-4m^2)A_{2m}=q(A_{2m-2}+A_{2m+2})$$

$$(a-1+q)B_1=qB_3$$

$$(a-(2m+1)^2)B_{2m+1}=q(B_{2m-1}+B_{2m+1})$$

## 关系式

马丢方程的基本解$W_I W_{II}$满足下列关系：:

$\begin{vmatrix} w_I(n,q,0) & w_{II}(n,q,0)\\w_{i}^'(n,q,0) & w_{II}^'(n,q,0) \end{vmatrix}$= $\begin{vmatrix} 1 & 0\\0 & 1 \end{vmatrix}$

郎斯基行列式：

$$W[w_I,w_{II}]=1$$

$$w_I(a,q,z+\pi)=w_I(a,q,\pi)*w_I(a,q,z)+w_I^'(a,q,\pi)*w_{II}(a,q,z)$$ $$w_I(a,q,z-\pi)=w_I(a,q,\pi)*w_I(a,q,z)-w_I^'(a,q,\pi)*w_{II}(a,q,z)$$ $$w_{II}(a,q,z+\pi)=w_{II}(a,q,\pi)*w_{II}(a,q,z)+w_{II}^'(a,q,\pi)*w_{II}(a,q,z)$$ $$w_{II}(a,q,z-\pi)=w_{II}(a,q,\pi)*w_{II}(a,q,z)-w_I^'(a,q,\pi)*w_{II}(a,q,z)$$

$$w_I(-z)=w_I(z)$$ $$w_{II}(-z)=-w_{II}(z)$$

## 特例

- $$CE(a,0,z)=cos(az)$$
- $$SE(a,0,z)=sin(az)$$
- $$MathieuA(1,0)=1$$
- $$MathieuA(a,0)=a^2$$
- $$MathieuB(a,0)=a^2$$
- $$MathieuFloquet(a,0,z)=exp(I*sqrt(a)*z)$$

## 夫洛开解

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马丢函数中，如果$f(x)$ 是一个周期为$\omega$的解，并满足下列条件

$f(x+\omega)=\sigma*f(x)$,其中$\sigma$与x 无关，则此解称为夫洛开解。

级数展开

$$MF(1, 1, z) = {.7992-.5734*I+(-.9134+.6553*I)*z+(.3996-.2867*I)*z^2+(-.1523+.1092*I)*z^3+(-.2331+.1673*I)*z^4+O(z^5)}$$ $$MF(1, 2, z) = {.7643-.4526*I+(-1.167+.6910*I)*z+(1.146-.6789*I)*z^2+(-.5835+.3455*I)*z^3+(-.2229+.1320*I)*z^4+O(z^5)}$$ $$MF(1, 3, z) = {.6841-.3703*I+(-1.318+.7135*I)*z+(1.710-.9258*I)*z^2+(-1.098+.5946*I)*z^3+(0.2851e-1-0.1543e-1*I)*z^4+O(z^5)}$$

## 参考文献

- 王竹溪 郭敦仁 《特殊函数概论》第十二章 马丢函数 北京大学出版社 2000
- Frank J Oliver NIST Handbook of Mathematical Functions,Cambridge University PRESS, 2010