UA MATH524 复变函数验证一个函数是否为调和函数

称u(x,y)u(x,y)u(x,y)为调和函数，如果它满足Laplace方程：
Δu=0\Delta u=0Δu=0

其中Δ\DeltaΔ为Laplace算子，Δ=∂2∂x2+∂2∂y2\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}Δ=∂x2∂2+∂y2∂2。

假设f=u+ivf=u+ivf=u+iv是解析函数，则它的实部uuu与虚部vvv都是调和函数，称u,vu,vu,v调和共轭(harmonic conjugate)，调和共轭的两个函数之间的关系可以用Cauchy-Riemann方程描述：
∂u∂x=∂v∂y,∂u∂y=−∂v∂x\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}∂x∂u=∂y∂v,∂y∂u=−∂x∂v

例题：验证下面的函数是否为调和函数，并计算它的调和共轭（z=x+iyz=x+iyz=x+iy）

1）ln⁡(x2+y2)\ln(x^2+y^2)ln(x2+y2)，是调和函数，是log⁡z\log zlogz的实部，调和共轭为arg⁡z\arg zargz
2）excos⁡ye^x\cos yexcosy，是调和函数，是eze^zez的实部，调和共轭为exsin⁡ye^x\sin yexsiny
3）u(x,y)=2xyu(x,y)=2xyu(x,y)=2xy，计算
Δu=∂2u∂x2+∂2u∂y2=0+0=0\begin{aligned} \Delta u = \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} =0+0=0\end{aligned}Δu=∂x2∂2u+∂y2∂2u=0+0=0 所以uuu是调和函数，根据Cauchy-Riemann方程，∂v∂y=∂u∂x=2y⇒v(x,y)=y2+p(x)+C1⏟与y无关的积分常量∂v∂x=−∂u∂y=−2x=p′(x)⇒p(x)=−x2+C2\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=2y \Rightarrow v(x,y)=y^2+\underbrace{p(x)+C_1 }_{与y无关的积分常量}\\ \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=-2x = p'(x) \Rightarrow p(x)=-x^2+C_2∂y∂v=∂x∂u=2y⇒v(x,y)=y2+与y无关的积分常量p(x)+C1∂x∂v=−∂y∂u=−2x=p′(x)⇒p(x)=−x2+C2 记c=C1+C2c=C_1+C_2c=C1+C2为积分常数，则uuu的调和共轭为v=y2−x2+cv=y^2-x^2+cv=y2−x2+c

4）u(x,y)=x4−6x2y2+y4u(x,y)=x^4-6x^2y^2+y^4u(x,y)=x4−6x2y2+y4，
Δu=∂2u∂x2+∂2u∂y2=12x2−12y2+(−12x2+12y2)=0\Delta u = \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=12x^2-12y^2+(-12x^2+12y^2)=0Δu=∂x2∂2u+∂y2∂2u=12x2−12y2+(−12x2+12y2)=0所以uuu是调和函数，根据Cauchy-Riemann方程，
∂v∂y=∂u∂x=4x3−12xy2⇒v(x,y)=4x3y−4xy3+p(x)+C1∂v∂x=−∂u∂y=12x2y−4y3=12x2y−4y3+p′(x)⇒p(x)=C2\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=4x^3-12xy^2 \Rightarrow v(x,y)=4x^3y-4xy^3+p(x)+C_1 \\ \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=12x^2y-4y^3=12x^2y-4y^3+p'(x) \Rightarrow p(x)=C_2∂y∂v=∂x∂u=4x3−12xy2⇒v(x,y)=4x3y−4xy3+p(x)+C1∂x∂v=−∂y∂u=12x2y−4y3=12x2y−4y3+p′(x)⇒p(x)=C2 记c=C1+C2c=C_1+C_2c=C1+C2为积分常数，则uuu的调和共轭为v=4x3−12xy2+cv=4x^3-12xy^2+cv=4x3−12xy2+c

5）u(x,y)=x2x2+y2+1u(x,y)=\frac{x^2}{x^2+y^2+1}u(x,y)=x2+y2+1x2
Δu=∂2u∂x2+∂2u∂y2=2x(x2+y2+1)−2x3(x2+y2+1)2+−2x2y(x2+y2+1)2=2xy2+2x−2x2y(x2+y2+1)2\Delta u = \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=\frac{2x(x^2+y^2+1)-2x^3}{(x^2+y^2+1)^2}+\frac{-2x^2y}{(x^2+y^2+1)^2}=\frac{2xy^2+2x-2x^2y}{(x^2+y^2+1)^2}Δu=∂x2∂2u+∂y2∂2u=(x2+y2+1)22x(x2+y2+1)−2x3+(x2+y2+1)2−2x2y=(x2+y2+1)22xy2+2x−2x2y所以uuu不是调和函数

6）u(x,y)=cos⁡(2xy)ex2−y2u(x,y)=\cos(2xy)e^{x^2-y^2}u(x,y)=cos(2xy)ex2−y2
Δu=∂2u∂x2+∂2u∂y2=−2ex2−y2[4xysin⁡(2xy)+(−2x2+2y2−1)cos⁡(2xy)]+2ex2−y2[4xysin⁡(2xy)+(−2x2+2y2−1)cos⁡(2xy)]=0\Delta u = \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} \\ = -2e^{x^2-y^2}[4xy\sin(2xy)+(-2x^2+2y^2-1)\cos(2xy) ] \\ +2e^{x^2-y^2}[4xy\sin(2xy)+(-2x^2+2y^2-1)\cos(2xy) ]=0Δu=∂x2∂2u+∂y2∂2u=−2ex2−y2[4xysin(2xy)+(−2x2+2y2−1)cos(2xy)]+2ex2−y2[4xysin(2xy)+(−2x2+2y2−1)cos(2xy)]=0

所以uuu是调和函数；