Question #220631
harmonic conjugate of u(x,y)=e^y cosx is
1
Expert's answer
2021-07-26T16:34:35-0400

By using one of the Cauchy Riemann equations we have vx=uy=eycosx,\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=-e^y\cos x, and partially integrating with respect to xx gives v=eysinx+C(y).v=-e^y\sin x+C(y). Partially differentiating this expression and using the otherCauchy Riemann equation gives vy=eysinx+C(x)=ux=eysinx.\frac{\partial v}{\partial y}=-e^y\sin x+C'(x)=\frac{\partial u}{\partial x}=-e^y\sin x. It follows that C(x)=0,C'(x)=0, and hence C(x)=CC(x)=C is constant. We conclude that harmonic conjugate of u(x,y)=eycosxu(x,y)=e^y \cos x is v=eysinx+C.v=-e^y\sin x+C.


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