Construction of Analytic Functions
Finding \( f(z) = u + iv \) when either \( u \) (Real part) or \( v \) (Imaginary part) is known.
Method I: Substitution Method
Suppose \( u(x,y) \) is given. We use the identity involving the conjugate.
- We know \( f(z) = u + iv \). Also, \( \overline{f(\bar{z})} \) is analytic if \( f \) is analytic.
- Consider the identity: \( f(z) + \overline{f(\bar{z})} = 2u(x,y) \).
- In terms of variables, this leads to \( f(x+iy) + \bar{f}(x-iy) = 2u(x,y) \).
- The Trick: Set \( x = z/2 \) and \( y = z/2i \).
- Then \( x+iy = z \) and \( x-iy = 0 \).
Final Formula:
\[ f(z) = 2u\left(\frac{z}{2}, \frac{z}{2i}\right) – u(0,0) + ic \]
(Where \( c \) is a real constant)
Example
Given \( u(x,y) = x^2 – y^2 \). Find \( f(z) \).
Using the formula: \[ \begin{align} f(z) &= 2 \left[ \left(\frac{z}{2}\right)^2 – \left(\frac{z}{2i}\right)^2 \right] – u(0,0) + ic \\ &= 2 \left[ \frac{z^2}{4} – \frac{z^2}{4i^2} \right] – 0 + ic \\ &= 2 \left[ \frac{z^2}{4} – \frac{z^2}{-4} \right] + ic \\ &= 2 \left[ \frac{z^2}{4} + \frac{z^2}{4} \right] + ic \\ &= 2 \left[ \frac{2z^2}{4} \right] + ic \\ &= \mathbf{z^2 + ic} \end{align} \]
Using the formula: \[ \begin{align} f(z) &= 2 \left[ \left(\frac{z}{2}\right)^2 – \left(\frac{z}{2i}\right)^2 \right] – u(0,0) + ic \\ &= 2 \left[ \frac{z^2}{4} – \frac{z^2}{4i^2} \right] – 0 + ic \\ &= 2 \left[ \frac{z^2}{4} – \frac{z^2}{-4} \right] + ic \\ &= 2 \left[ \frac{z^2}{4} + \frac{z^2}{4} \right] + ic \\ &= 2 \left[ \frac{2z^2}{4} \right] + ic \\ &= \mathbf{z^2 + ic} \end{align} \]
Method II: Milne-Thomson Method
This method uses the derivative \( f'(z) \) directly expressed in terms of \( z \).
- We know \( f'(z) = u_x + i v_x \).
- By C-R Equations (\( v_x = -u_y \)): \( f'(z) = u_x – i u_y \).
- Replace \( x \to z \) and \( y \to 0 \) in the partial derivatives.
Formula:
\[ f(z) = \int [ \phi_1(z) – i \phi_2(z) ] \, dz + C \]
Where:
\( \phi_1(z) = u_x(z,0) \) and \( \phi_2(z) = u_y(z,0) \)
Note: Here \( C \) is a complex constant.
Note: Here \( C \) is a complex constant.
Example
Given \( u = x^2 – y^2 + x \). Find \( f(z) \).
Step 1: Partial Derivatives
\( u_x = 2x + 1 \)
\( u_y = -2y \)
Step 2: Apply Milne-Thomson (\( x=z, y=0 \))
\( \phi_1(z) = u_x(z,0) = 2z + 1 \)
\( \phi_2(z) = u_y(z,0) = -2(0) = 0 \)
Step 3: Integrate
\[ f'(z) = (2z+1) – i(0) = 2z+1 \\ f(z) = \int (2z+1) dz = z^2 + z + C \] Since \( u \) was given, the real part of \( C \) is fixed (usually 0 if satisfied).
\( f(z) = z^2 + z + ib \) (where \( b \in \mathbb{R} \))
Step 1: Partial Derivatives
\( u_x = 2x + 1 \)
\( u_y = -2y \)
Step 2: Apply Milne-Thomson (\( x=z, y=0 \))
\( \phi_1(z) = u_x(z,0) = 2z + 1 \)
\( \phi_2(z) = u_y(z,0) = -2(0) = 0 \)
Step 3: Integrate
\[ f'(z) = (2z+1) – i(0) = 2z+1 \\ f(z) = \int (2z+1) dz = z^2 + z + C \] Since \( u \) was given, the real part of \( C \) is fixed (usually 0 if satisfied).
\( f(z) = z^2 + z + ib \) (where \( b \in \mathbb{R} \))
Method III: Exact Differential (Finding Conjugate)
If \( u \) is known, we find \( v \) directly using the total differential \( dv \).
- \( dv = v_x dx + v_y dy \)
- Using C-R Equations: \( v_x = -u_y \) and \( v_y = u_x \).
- Substitute: \( dv = (-u_y) dx + (u_x) dy \).
- This is of the form \( dv = M dx + N dy \). Since \( f \) is analytic, this equation is Exact.
Integration Rule for Exact Equations:
\[ v = \int_{y \text{ const}} M \, dx + \int (\text{terms of } N \text{ free from } x) \, dy + C \]
Example
Given \( u = 2xy + x \). Find \( v \).
1. Calculate Derivatives:
\( u_x = 2y + 1 \)
\( u_y = 2x \)
2. Form \( dv \):
\( dv = (-u_y) dx + (u_x) dy \)
\( dv = \underbrace{(-2x)}_{M} dx + \underbrace{(2y+1)}_{N} dy \)
3. Integrate:
\( \int M dx = \int -2x \, dx = -x^2 \)
\( \int (\text{N terms free of } x) dy = \int (2y+1) dy = y^2 + y \)
Result: \( v = -x^2 + y^2 + y + C \)
So, \( f(z) = (2xy+x) + i(-x^2+y^2+y+C) \).
1. Calculate Derivatives:
\( u_x = 2y + 1 \)
\( u_y = 2x \)
2. Form \( dv \):
\( dv = (-u_y) dx + (u_x) dy \)
\( dv = \underbrace{(-2x)}_{M} dx + \underbrace{(2y+1)}_{N} dy \)
3. Integrate:
\( \int M dx = \int -2x \, dx = -x^2 \)
\( \int (\text{N terms free of } x) dy = \int (2y+1) dy = y^2 + y \)
Result: \( v = -x^2 + y^2 + y + C \)
So, \( f(z) = (2xy+x) + i(-x^2+y^2+y+C) \).
Proofs: Construction of Analytic Functions
Method I: Substitution Method (Derivation)
Goal: Derive the formula to find \( f(z) \) given \( \text{Re}(f) = u(x,y) \).
Step 1: Use the Conjugate Property
Let \( f(z) = u(x,y) + i v(x,y) \). Then the conjugate is:
\[ \overline{f(z)} = u(x,y) – i v(x,y) \]
Since \( f \) is analytic, \( \overline{f(\bar{z})} \) is also analytic. We can define a function \( g(z) = \overline{f(\bar{z})} \).
Step 2: Form the Identity for Real Part u
Adding \( f(z) \) and its conjugate form:
\[ f(z) + \overline{f(z)} = (u+iv) + (u-iv) = 2u(x,y) \]
Replacing \( \overline{f(z)} \) with the function notation \( \bar{f}(\bar{z}) \):
\[ f(z) + \bar{f}(\bar{z}) = 2u(x,y) \]
In terms of variables \( z \) and \( \bar{z} \):
\[
x = \frac{z + \bar{z}}{2}, \quad y = \frac{z – \bar{z}}{2i}
\]
Equation (A):
\[ f(z) + \bar{f}(\bar{z}) = 2u\left( \frac{z + \bar{z}}{2}, \frac{z – \bar{z}}{2i} \right) \quad \dots(A) \]
Step 3: Apply Formal Identity (The “Trick”)
To eliminate \( \bar{z} \), we treat (A) as a formal identity valid for all independent complex variables. We substitute \( z \) for \( z \) and \( 0 \) for \( \bar{z} \) (effectively setting \( \bar{z} = 0 \), or equivalently in the source text, setting \( x = z/2 \) and \( y = z/2i \)).
Putting \( \bar{z} = 0 \) in (A): \[ f(z) + \bar{f}(0) = 2u\left( \frac{z + 0}{2}, \frac{z – 0}{2i} \right) \] \[ f(z) + \bar{f}(0) = 2u\left( \frac{z}{2}, \frac{z}{2i} \right) \]
Putting \( \bar{z} = 0 \) in (A): \[ f(z) + \bar{f}(0) = 2u\left( \frac{z + 0}{2}, \frac{z – 0}{2i} \right) \] \[ f(z) + \bar{f}(0) = 2u\left( \frac{z}{2}, \frac{z}{2i} \right) \]
Step 4: Resolve the Constant
Rearranging the terms:
\[ f(z) = 2u\left( \frac{z}{2}, \frac{z}{2i} \right) – \bar{f}(0) \]
Consider the constant term \( \bar{f}(0) \).
\[ \bar{f}(0) = u(0,0) – i v(0,0) \]
Let \( v(0,0) = c \) (an arbitrary real constant). Then \( \bar{f}(0) = u(0,0) – ic \).
Substituting this back: \[ f(z) = 2u\left( \frac{z}{2}, \frac{z}{2i} \right) – (u(0,0) – ic) \]
Substituting this back: \[ f(z) = 2u\left( \frac{z}{2}, \frac{z}{2i} \right) – (u(0,0) – ic) \]
Result:
\[ f(z) = 2u\left( \frac{z}{2}, \frac{z}{2i} \right) – u(0,0) + ic \]
Method II: Milne-Thomson (Proof)
Goal: Express \( f'(z) \) using only partial derivatives of \( u \).
We know \( f'(z) = u_x + i v_x \).
Using Cauchy-Riemann equations \( v_x = -u_y \):
\[ f'(z) = u_x(x,y) – i u_y(x,y) \]
Since this holds for any \( z = x+iy \), we can set \( y=0 \) (restrict to real axis) to find the functional form in terms of \( z \) (setting \( x=z \)):
\[ f'(z) = u_x(z,0) – i u_y(z,0) \]
Let \( \phi_1(z) = u_x(z,0) \) and \( \phi_2(z) = -u_y(z,0) \).
\[ f'(z) = \phi_1(z) + i \phi_2(z) \]
Integrating with respect to \( z \):
\[ f(z) = \int (\phi_1(z) + i \phi_2(z)) \, dz + C \]
Method III: Exact Differential (Proof of Validity)
Goal: Prove that the differential equation for \( v \) is always solvable (Exact).
Step 1: Form the Differential
Since \( v = v(x,y) \):
\[ dv = \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy \]
Using C-R Equations (\( v_x = -u_y, v_y = u_x \)):
\[ dv = -u_y dx + u_x dy \]
This is of the form \( dv = M dx + N dy \), where:
\[ M = -u_y \quad \text{and} \quad N = u_x \]
Step 2: Condition for Exactness
For the integral \( \int (M dx + N dy) \) to exist, the differential must be Exact.
The condition for exactness is:
\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]
Step 3: Verification
Calculate derivatives of M and N:
\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(-u_y) = -u_{yy} \]
\[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(u_x) = u_{xx} \]
We need to check if \( -u_{yy} = u_{xx} \).
Rearranging, this asks if \( u_{xx} + u_{yy} = 0 \).
Rearranging, this asks if \( u_{xx} + u_{yy} = 0 \).
Conclusion:
Since \( u \) is the real part of an analytic function, it is Harmonic.
Therefore, it satisfies Laplace’s Equation:
\[ \nabla^2 u = u_{xx} + u_{yy} = 0 \implies u_{xx} = -u_{yy} \]
Thus, \( \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} \).
The equation is Exact. The integration is valid.
The equation is Exact. The integration is valid.
v given :
u-v given :
u+v given :