马丢函数

马丟函数 MathieuC(a,q,z) 或 MathieuS(a,q,z) 只有一个是周期为 ${\displaystyle \pi }$ 或${\displaystyle 2\pi }$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。 [1]

• ${\displaystyle \int _{0}^{2*\pi }\!ce_{m}(x,q)*ce_{n}(x,q)\,dx=0}$
• ${\displaystyle \int _{0}^{2*\pi }\!ce_{m}(x,q)*se_{n}(x,q)\,dx=0}$
• ${\displaystyle \int _{0}^{2*\pi }\!se_{m}(x,q)*se_{n}(x,q)\,dx=0}$

Mathieu Eigen value a(n,q)

Mathieu eigenvalue b(n,q)

马丟方程的特征方程是[1]

${\displaystyle cos(\pi *v)=w_{I}(a,q,\pi )}$

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

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。

马丟函数ce,se的级数展开[3]

${\displaystyle ce_{0}(z,q)={1-(1/2)*cos(2*z)*q+(-1/16+(1/32)*cos(4*z))*q^{2}+((11/128)*cos(2*z)-(1/1152)*cos(6*z))*q^{3}+O(q^{4})}}$ ${\displaystyle ce_{1}(z,q)={cos(z)-(1/8)*cos(3*z)*q+(-(1/128)*cos(z)-(1/64)*cos(3*z)+(1/192)*cos(5*z))*q^{2}+(-(1/512)*cos(z)+(1/3072)*cos(3*z)+(1/1152)*cos(5*z)-(1/9216)*cos(7*z))*q^{3}+O(q^{4})}}$ ${\displaystyle ce_{2}(z,q)={cos(2*z)+(1/4-(1/12)*cos(4*z))*q+(-(19/288)*cos(2*z)+(1/384)*cos(6*z))*q^{2}+(-49/1152+(11/4608)*cos(4*z)-(1/23040)*cos(8*z))*q^{3}+O(q^{4})}}$ ${\displaystyle ce_{3}(z,q)={cos(3*z)+((1/8)*cos(z)-(1/16)*cos(5*z))*q+(-(5/512)*cos(3*z)+(1/64)*cos(z)+(1/640)*cos(7*z))*q^{2}+(-(1/512)*cos(3*z)-(1/4096)*cos(z)+(11/40960)*cos(5*z)-(1/46080)*cos(9*z))*q^{3}+O(q^{4})}}$ ${\displaystyle ce_{4}(z,q)={cos(4*z)+((1/12)*cos(2*z)-(1/20)*cos(6*z))*q+(-(17/3600)*cos(4*z)+1/192+(1/960)*cos(8*z))*q^{2}+((7/28800)*cos(2*z)+(29/288000)*cos(6*z)-(1/80640)*cos(10*z))*q^{3}+O(q^{4})}}$

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

马丟函数的傅立叶展开：[3]

• ${\displaystyle MathieuCE(2n,q,x)=\sum _{m=0}^{\infty }A_{2m}^{2n}(q)cos2mx}$
• ${\displaystyle MathieuCE(2n+1,q,x)=\sum _{m=0}^{\infty }A_{2m+1}^{2n+1}(q)cos(2m+1)x}$
• ${\displaystyle MathieuSE(2n+1,q,x)=\sum _{m=0}^{\infty }B_{2m+1}^{2n+1}(q)sin(2m+1)x}$
• ${\displaystyle MathieuSE(2n+2,q,x)=\sum _{m=0}^{\infty }B_{2m+2}^{2n+2}(q)sin(2m+2)x}$

其中系数A,B满足下列归递关系：[3]

${\displaystyle aA_{0}=qA_{2}}$

${\displaystyle (a-4)A_{2}=q(2A_{0}+A_{4})}$

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

${\displaystyle (a-1+q)B_{1}=qB_{3}}$

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

马丟方程的基本解${\displaystyle W_{I}W_{II}}$满足下列关系：[3]:

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

郎斯基行列式： ${\displaystyle W[w_{I},w_{II}]=1}$

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

${\displaystyle w_{I}(-z)=w_{I}(z)}$ ${\displaystyle w_{II}(-z)=-w_{II}(z)}$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$

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

Mathieu Floquet

马丟函数中，如果${\displaystyle f(x)}$ 是一个周期为${\displaystyle \omega }$的解，并满足下列条件

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

级数展开

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

1. 王竹溪 郭敦仁 603
2. Frank p659
3. 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