复变函数中的解析函数已知:v(x
v(x,y)=2x+y^2-x^2为调和函数,所以是解析函数的实部或虚部。
先求一个解析函数
g(z)=f(z)/i=-iu(x,y)+v(x,y)
记C(z)为z的共轭
2x+y^2-x^2=
=2[C(z)+z]/2-{[C(z)+z]/2}^2+{[-C(z)+z]/(2i)}^2=
=z-z^2+C(z-z^2)=[g(z)+C(g(z))]/2
==>
g(z)=2[z-z^2]=f(z)/i
==>
f(z)=2[z-z^2]i=-2y+4xy+ [2x+y^2-x^2]i
==>
u(x,y)=-2y+4xy
修改:
2x+y^2-x^2=
=2[C(z)+z]/2-{[...全部
v(x,y)=2x+y^2-x^2为调和函数,所以是解析函数的实部或虚部。
先求一个解析函数
g(z)=f(z)/i=-iu(x,y)+v(x,y)
记C(z)为z的共轭
2x+y^2-x^2=
=2[C(z)+z]/2-{[C(z)+z]/2}^2+{[-C(z)+z]/(2i)}^2=
=z-z^2+C(z-z^2)=[g(z)+C(g(z))]/2
==>
g(z)=2[z-z^2]=f(z)/i
==>
f(z)=2[z-z^2]i=-2y+4xy+ [2x+y^2-x^2]i
==>
u(x,y)=-2y+4xy
修改:
2x+y^2-x^2=
=2[C(z)+z]/2-{[C(z)+z]/2}^2+{[-C(z)+z]/(2i)}^2=
=z-z^2/2+C(z-z^2/2)=[g(z)+C(g(z))]/2
==>
g(z)=2[z-z^2/2]=f(z)/i
==>
f(z)=2[z-z^2/2]i=-2y+2xy+ [2x+y^2-x^2]i
==>
u(x,y)=-2y+2xy
。收起