Question

Obtain the harmonic conjugate v of the function u=2x(1-y)

Answer 1

Answer on Question #84500 – Math – Complex Analysis

Question

Obtain the harmonic conjugate $\mathbf{v}$ of the function $u = 2x(1 - y)$.

Solution

Given that $u = 2x(1 - y)$

a) Prove that $u$ is harmonic:

$$\frac{\partial u}{\partial x} = 2(1 - y)$$ $$\frac{\partial u}{\partial y} = -2x$$ $$\frac{\partial^2 u}{\partial x^2} = 0$$ $$\frac{\partial^2 u}{\partial y^2} = 0$$ $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 + 0 = 0$$

So $u$ is harmonic.

b) Find the harmonic conjugate $\mathbf{v}$ of the function $u$. Obtain $\mathbf{v}$ such that $u, \mathbf{v}$ satisfy the Cauchy-Riemann conditions:

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}; \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$

We get $\frac{\partial v}{\partial y} = 2(1 - y)$; $\frac{\partial v}{\partial x} = 2x$

From the first condition

$$v(x, y) = \int 2(1 - y)\,dy = \int (2 - 2y)\,dy = 2y - y^2 + \varphi(x);$$

The partial derivative of $x$

$$\frac{\partial v}{\partial x} = \varphi'(x)$$

Substitute this in the second condition $\frac{\partial v}{\partial x} = 2x$,

we get $\varphi'(x) = 2x$; $\varphi(x) = x^2 + C$, $C \in R$; $v = 2y - y^2 + x^2 + C$, $C \in R$;

Answer: $v = 2y - y^2 + x^2 + C$, $C \in R$;

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