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Thursday, 1 October 2015

Harmonic conjugate and analyticity

a) For which values of a,b,c,d is the function u(x,y)=ax^3+bx^2y+cxy^2+dy^3 harmonic?
b) Find a harmonic conjugate v(x,y) of u(x,y) in this case.
c) Find an analytic function f(z)=u(x,y)+iv(x,y), where z=x+iy.



a) Use the Laplace equation for u to get (\partial_x^2+\partial_y^2)u(x,y)=(6a+2c)x+(2b+6d)y=0,
thus c=-3a,b=-3d.

b) u(x,y)=ax^3-3dx^2y-3axy^2+dy^3.
Harmonic conjugate satisfies the Cauchy-Riemann condition v_y=u_x=3a(x^2-y^2)-6dxy,\\ v_x=-u_y=3d(x^2-y^2)+6axy.
Integrating these, we have v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.
Alternatively, we can put f(z)=2u(\dfrac{z}{2},\dfrac{z}{2i})-u(0,0).
Then \begin{align}f(z)&=2\bigg(a(\dfrac{z}{2})^3-3d(\dfrac{z}{2})^2(\dfrac{z}{2i})-3a(\dfrac{z}{2})(\dfrac{z}{2i})^2+d(\dfrac{z}{2i})^3\bigg)\\&=(a+3id+3a+id)\dfrac{z^3}{4}\\&=(a+id)z^3\\&=(a+id)(x+iy)^3\end{align}
The imaginary part is v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.
c) \begin{align}f(z)&=u(x,y)+iv(x,y)\\&=ax^3-3dx^2y-3axy^2+dy^3+i(dx^3+3ax^2y-3dxy^2-ay^3)\\&=a(x^3+3ix^2y-3xy^2-iy^3)+d(ix^3-3x^2y-3ixy^2+y^3)\\&=a(x+iy)^3+id(x+iy)^3\\&=(a+id)z^3\end{align}
Remark: The coefficients 1,3,3,1 hint the use of (?+?)^3. We can also substitute y=0,x=z to get (a+id)z^3 immediately!

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