A harmonic function is analytic if it satisfies the Laplace equation.

Ifu(x,y) = 2x² — 2y² + 4xy is a harmonic function, then its conjugate harmonic function v(x,y) is

The correct answer is: 4xy - 2x² + 2y² + constant

To find the conjugate harmonic function $v(x,y)$ of the given harmonic function $u(x,y)=2x^2-2y^2+4xy$, we use the Cauchy-Riemann equations:

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$

and

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

Step 1: Compute the partial derivatives of $u(x,y)$

Differentiating $u$ with respect to $x$:

$\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(2x^2-2y^2+4xy)=4x+4y$

Differentiating $u$ with respect to $y$:

$\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}(2x^2-2y^2+4xy)=-4y+4x$

Step 2: Set up the relations for $v(x,y)$ using Cauchy-Riemann equations

From $\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}$, we have:

$\frac{\partial v}{\partial y}=4x+4y$

From $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$, we have:

$\frac{\partial v}{\partial x}=-(4x-4y)=-4x+4y$

Step 3: Integrate to find $v(x,y)$

Integrating $\frac{\partial v}{\partial y}$ with respect to $y$:

$v(x,y)=\int (4x+4y)dy=4xy+2y^2+f\left(x\right)$

where $f\left(x\right)$ is an arbitrary function of $x$ acting as the constant of integration.

Now, differentiate this expression for $v(x,y)$ with respect to $x$:

$\frac{\partial v}{\partial x}=4y+f^{\prime }\left(x\right)$

Comparing this with the expression obtained earlier for $\frac{\partial v}{\partial x}$ from the Cauchy-Riemann equations:

$4y+f^{\prime }\left(x\right)=-4x+4y$

Subtracting $4y$ from both sides yields:

$f^{\prime }\left(x\right)=-4x$

Integrating $f^{\prime }\left(x\right)$ with respect to $x$:

$f\left(x\right)=-2x^2+\text{constant}$

Step 4: Formulate the final conjugate harmonic function

Substituting $f\left(x\right)$ back into the expression for $v(x,y)$:

$v(x,y)=4xy+2y^2-2x^2+\text{constant}$

Rearranging the terms gives:

$v(x,y)=4xy-2x^2+2y^2+\text{constant}$