Show that the function u = x^3 - 3xy^2 + 3x is harmonic and determine its corresponding analytic function.

Steps to Solve

1. Determine the Partial Derivatives Calculate the first-order partial derivatives of the function $$ u = x^3 - 3xy^2 + 3x $$

The partial derivatives are:

• With respect to $x$: $$ u_x = \frac{\partial u}{\partial x} = 3x^2 - 3y^2 + 3 $$
• With respect to $y$: $$ u_y = \frac{\partial u}{\partial y} = -6xy $$

2. Calculate the Second Partial Derivatives Next, compute the second-order partial derivatives:

• With respect to $x$ (twice): $$ u_{xx} = \frac{\partial^2 u}{\partial x^2} = 6x $$
• With respect to $y$ (twice): $$ u_{yy} = \frac{\partial^2 u}{\partial y^2} = -6x $$

3. Apply the Laplace Equation For $u$ to be harmonic, it must satisfy the Laplace equation: $$ u_{xx} + u_{yy} = 0 $$

Substituting the second derivatives: $$ 6x - 6x = 0 $$

This confirms that $u$ is harmonic.

4. Determine the Corresponding Analytic Function To find the analytic function, we can use the Cauchy-Riemann equations:

• The general form of the analytic function $f(z) = u + iv$ implies that: $$ v_y = u_x $$ $$ v_x = -u_y $$

From $u_x$ and $u_y$, set up the equations:

• Integrate $u_x = 3x^2 - 3y^2 + 3$ with respect to $y$: $$ v = \int (3x^2 - 3y^2 + 3) , dy = 3x^2y - y^3 + 3y + g(x) $$ where $g(x)$ is a function of $x$.

• Differentiate $v$ with respect to $x$ to find $g'(x)$: $$ v_x = 6xy + g'(x) $$

Set $v_x = -u_y = 6xy$ and solve for $g'(x)$: $$ g'(x) = 0 \Rightarrow g(x) = C $$

5. Final Form of the Analytic Function Substituting back, we get: $$ f(z) = (x^3 - 3xy^2 + 3x) + i(3x^2y - y^3 + 3y + C) $$

Thus, the corresponding analytic function is: $$ f(z) = u + iv $$