MathieuCE 3D
MathieuSE 3D
**马丟函数**（法語：）是1868年法國數學家以米里迂·拉·馬丢（法语：）因研究数学物理所推得的特殊函數，下列马丟方程的解析解：

{\displaystyle {\frac {d^{2}y}{dx^{2}}}+[a-2q\cos(2x)]y=0.}
马丟方程有两个线性无关的解：

奇数解
MathieuCE(n, q, x)，或记为{\displaystyle w\_{I}(n,q,x)},

偶数解
MathieuSE(n, q, x).或记为{\displaystyle w\_{II}(n,q,x)}
称为基本解

周期性
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马丟函数 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。

马丟函数 MathieuC(a,q,z) 和 MathieuS(a,q,z) 两者都有是周期为{\displaystyle 2n\pi }（n≥2)的周期函数。

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正交性
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* {\displaystyle \int \_{0}^{2\*\pi }\!ce\_{m}(x,q)\*ce\_{n}(x,q)\,dx=0}
* {\displaystyle \int \_{0}^{2\*\pi }\!ce\_{m}(x,q)\*se\_{n}(x,q)\,dx=0}
* {\displaystyle \int \_{0}^{2\*\pi }\!se\_{m}(x,q)\*se\_{n}(x,q)\,dx=0}

特征方程
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Mathieu Eigen value a(n,q)
Mathieu eigenvalue b(n,q)
马丟方程的特征方程是

{\displaystyle cos(\pi \*v)=w\_{I}(a,q,\pi )}

{\displaystyle 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

马丟函数体特征值可展开成级数：

{\displaystyle a\_{0}(q)={-(1/2)\*z^{2}+(7/128)\*z^{4}-(29/2304)\*z^{6}+(68687/18874368)\*z^{8}+O(z^{1}0)}}
{\displaystyle 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}+O(z^{1}0)}}
{\displaystyle a\_{2}(q)={4+(5/12)\*z^{2}-(763/13824)\*z^{4}+(1002401/79626240)\*z^{6}-(1669068401/458647142400)\*z^{8}+O(z^{1}0)}}
{\displaystyle 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}+O(z^{1}0)}}

{\displaystyle 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}+O(z^{1}0)}}
{\displaystyle b\_{2}(q)={4-(1/12)\*z^{2}+(5/13824)\*z^{4}-(289/79626240)\*z^{6}+(21391/458647142400)\*z^{8}+O(z^{1}0)}}
{\displaystyle 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}+O(z^{1}0)}}
{\displaystyle b\_{4}(q)={16+(1/30)\*z^{2}-(317/864000)\*z^{4}+(10049/2721600000)\*z^{6}-(93824197/2006581248000000)\*z^{8}+O(z^{1}0)}}
{\displaystyle 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}+O(z^{1}0)}}

级数展开
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马丟函数ce,se的级数展开

{\displaystyle 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})}}
{\displaystyle 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})}}
{\displaystyle 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})}}
{\displaystyle 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})}}
{\displaystyle 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})}}

{\displaystyle 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})}}
{\displaystyle 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})}}
{\displaystyle 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})}}
{\displaystyle 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})}}
{\displaystyle 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
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马丟函数的傅立叶展开：

* {\displaystyle MathieuCE(2n,q,x)=\sum \_{m=0}^{\infty }A\_{2m}^{2n}(q)cos2mx}
* {\displaystyle MathieuCE(2n+1,q,x)=\sum \_{m=0}^{\infty }A\_{2m+1}^{2n+1}(q)cos(2m+1)x}
* {\displaystyle MathieuSE(2n+1,q,x)=\sum \_{m=0}^{\infty }B\_{2m+1}^{2n+1}(q)sin(2m+1)x}
* {\displaystyle 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满足下列归递关系：

  

{\displaystyle aA\_{0}=qA\_{2}}

{\displaystyle (a-4)A\_{2}=q(2A\_{0}+A\_{4})}

{\displaystyle (a-4m^{2})A\_{2m}=q(A\_{2m-2}+A\_{2m+2}}

{\displaystyle (a-1+q)B\_{1}=qB\_{3}}

{\displaystyle (a-(2m+1)^{2})B\_{2m+1}=q(B\_{2m-1}+B\_{2m+1}}

关系式
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马丟方程的基本解{\displaystyle W\_{I}W\_{II}}满足下列关系：:

{\displaystyle {\begin{vmatrix}w\_{I}(n,q,0)&w\_{II}(n,q,0)\\w\_{i}^{'}(n,q,0)&w\_{II}^{'}(n,q,0)\end{vmatrix}}}= {\displaystyle {\begin{vmatrix}1&0\\0&1\end{vmatrix}}}
郎斯基行列式：
{\displaystyle W[w\_{I},w\_{II}]=1}

{\displaystyle 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)}
{\displaystyle 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)}
{\displaystyle 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)}
{\displaystyle 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)}

  

{\displaystyle w\_{I}(-z)=w\_{I}(z)}
{\displaystyle w\_{II}(-z)=-w\_{II}(z)}
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }

特例
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* {\displaystyle CE(a,0,z)=cos(az)}
* {\displaystyle SE(a,0,z)=sin(az)}
* {\displaystyle MathieuA(1,0)=1}
* {\displaystyle MathieuA(a,0)=a^{2}}
* {\displaystyle MathieuB(a,0)=a^{2}}
* {\displaystyle MathieuFloquet(a,0,z)=exp(I\*sqrt(a)\*z)}
* {\displaystyle }

夫洛开解
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Mathieu Floquet
马丟函数中，如果f(x) 是一个周期为\omega 的解，并满足下列条件

{\displaystyle 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
{\displaystyle 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})}}
{\displaystyle 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})}}
{\displaystyle 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
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1. 1 2 3  王竹溪 郭敦仁 603
2. ↑  Frank p659
3. 1 2 3 4  Frank p660

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