Answer to Question #286629 in Differential Equations for Potti

Question

Show that e^x Cosy is Harmonic, find its Conjugate harmonic function.

Answer 1

$$u=e^{x}\cos y$$

$$\dfrac{\partial u}{\partial x}=e^{x}\cos y$$

$$\dfrac{\partial^2 u}{\partial x^2}=e^{x}\cos y$$

$$\dfrac{\partial u}{\partial y}=-e^{x}\sin y$$

$$\dfrac{\partial^2 u}{\partial y^2}=-e^{x}\cos y$$

$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=e^{x}\cos y-e^{x}\cos y$$

Hence

$$\nabla^2u=\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=0, x, y\in \R$$

Therefore the function $u=e^{x}\cos y$ is harmonic.

We find the harmonic conjugate of $u,$ denoted by $v,$ which satisfies the Cauchy-Riemann equations. It follows from 

$$\dfrac{\partial u}{\partial x}=e^{x}\cos y=\dfrac{\partial v}{\partial y}$$

that $v(x, y) = −e ^x \cos y + φ(x),$ where $φ(x)$ is a differentiable real-valued function of $x.$ Also, from

$$\dfrac{\partial u}{\partial y}=-e^{x}\sin y=\dfrac{\partial v}{\partial x}$$

we have $e ^x \cos y = e ^x \cos y −φ'(x)=>φ(x)$ i.e., $φ(x)$ is constant. 

Therefore $v(x, y) = −e ^x \cos y + c, c\in \Complex,$ is the harmonic conjugate of $u.$

Hence

$$f(z)=u+iv=e ^x(\cos y+i \sin y)+ic$$

$$=e^{x+iy} +ic=e^z+ic$$