if $u(x, y) = x^3 - 3 x y^2$ find $f(z)$, where $f(z)$ is an analytical function

Question

if $u(x, y) = x^3 - 3 x y^2$ find $f(z)$, where $f(z)$ is an analytical function

How can the function can be solve from Cauchy-Riemann equation or other way. I have to find $v(x, y)$ at first. how can I find $v(x, y)$ and How can I find the funtion?

Here is the Cauchy-Riemann

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

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

Answer 1

Hint:

As mentioned in the comments, we have CR equations as:

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

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

Using the first we have:

$$ \frac{\partial(v)}{\partial y}=\frac{\partial(u)}{\partial x}=3x^2-3y^2$$

$$ \frac{\partial(v)}{\partial y}=3x^2-3y^2$$

When you integrate this partially w.r.t $y$ you will get the integrated value of the above and some function of $x$, say $f(x)$ (think why?):

$$v(x,y)=3yx^2-y^3+f(x)$$

Now use the second CR equation on this to find $f(x)$. Can you do it?

You should finally get: $\boxed{v(x,y)=3yx^2-y^3+C}$ where $C$ is an arbitrary constant.

Answer 2

Well, if $v(x, y)$ is the conjugate of $u(x, y)$, so that

$f(z) = f(x, y) = u(x, y) + iv(x, y) \tag 1$

is holomorphic, then $u(x, y)$ and $v(x, y)$ must satisfy the Cauchy-Riemann equations

$u_x = v_y, \tag 2$

$u_y = -v_x, \tag 3$

where we use subscript notation for partial derivatives,

$u_x = \dfrac{\partial u}{\partial x}, \tag 4$

and so forth. If we know $u(x, y)$, we can find $\nabla v(x, y)$ from (2)-(3):

$\nabla v = (v_x, v_y) = (-u_y, u_x); \tag 5$

in the present case

$u(x, y) = x^3 - 3xy^2, \tag 6$

so

$u_x = 3x^2 - 3y^2, \; u_y = -6xy; \tag 7$

therefore,

$v_x = 6xy, \tag 8$

$v_y = 3x^3 - 3y^2; \tag 9$

so now we have the gradient of $v$:

$\nabla v = (6xy, 3x^2 - 3y^2); \tag{10}$

we may express $v(x, y)$ in terms of $\nabla v$ by means of a path integral; suppose $\gamma(t)$ is a continuously differentiable curve in $\Bbb R^2$ such that

$\gamma(0) = (0, 0), \; \gamma(1) = (x, y); \tag{11}$

then

$v(x, y) - v(0, 0) = \displaystyle \int_0^1 \dfrac{dv(\gamma(t))}{dt} \; dt = \int_0^1 \nabla v(\gamma(t)) \cdot \dot \gamma(t) \; dt; \tag{12}$

the trick now is to pick $\gamma(t)$ to make (12) easy to evaluate; this works since the integral is in fact independent of the specific $\gamma(t)$ we choose; therefore we might try

$\gamma(t) = (tx, ty) \tag{13}$

where for the moment we regard $x$ and $y$ as fixed; then

$\dot \gamma(t) = (x, y), \tag{14}$

and

$\nabla v(\gamma(t)) = (6xyt^2, 3(x^2 - y^2)t^2); \tag{15}$

thus,

$\nabla v(\gamma(t)) \cdot \dot \gamma(t) = (6xyt^2, 3(x^2 - y^2)t^2) \cdot (x, y) = (6x^2y + 3x^2y - 3y^3) t^2; \tag{16}$

therefore, by (12)

$v(x, y) - v(0, 0) = \displaystyle \int_0^1 \nabla v(\gamma(t)) \cdot \dot \gamma(t) \; dt$ $= \displaystyle \int_0^1 (6x^2y + 3x^2y - 3y^3) t^2 \; dt = (6x^2y + 3x^2y - 3y^3) \int_0^1 t^2 \; dt$ $= \dfrac{1}{3}(6x^2y + 3x^2y - 3y^3) = 2x^2y + x^2 y - y^3 = 3x^2y - y^3; \tag{17}$

we may thus write

$v(x, y) = 3x^2 y - y^3 + v(0, 0); \tag{18}$

it is easily checked that such $v(x, y)$ satisfies the CR equations with the given $u(x, y)$; thus

$f(z) = f(x, y) = u(x, y) + iv(x, y) = (x^3 = 3x^2y) + i(3x^2 - y^3 + v(0, 0)) \tag{19}$

is a holomorphic function of $z = x + iy$.

Of course, there is an easier way to do this; observing that the holomorphic function

$f(z) = z^3 = (x + iy) = x^3 + 3ix^2y - 3xy^2 - iy^3 = (x^3 - 3x^2y) + i(3x^2 y - y^3), \tag{20}$

we immediately read off that $3x^2 y - y^3$ is the conjugate of $x^3 - 3xy^2$, up to an additive constant.

Answer 3

Using the method proposed by Milne Thomson (see here: Milne Thomson method for determining an analytic function from its real part) one immediately obtains $$f(z)=z^3+ c,\qquad c\in i{\mathbb R}\ ,$$ as we all have guessed anyway.