Complex Form of Laplace Operator
Question 1: Derivation
Prove that the complex form of the Laplacian operator is given by:
\[ \nabla^2 = 4 \frac{\partial^2}{\partial z \partial \bar{z}} \]
Step 1: Coordinate Relations
Recall the relationship between Cartesian coordinates \((x,y)\) and complex coordinates \((z, \bar{z})\): \[ z = x + iy \quad \text{and} \quad \bar{z} = x – iy \] Solving for \(x\) and \(y\): \[ x = \frac{z + \bar{z}}{2}, \quad y = \frac{z – \bar{z}}{2i} \]
Recall the relationship between Cartesian coordinates \((x,y)\) and complex coordinates \((z, \bar{z})\): \[ z = x + iy \quad \text{and} \quad \bar{z} = x – iy \] Solving for \(x\) and \(y\): \[ x = \frac{z + \bar{z}}{2}, \quad y = \frac{z – \bar{z}}{2i} \]
Step 2: First Derivative (\(\partial/\partial z\))
Using the Chain Rule for a function \( u(x,y) = u(z, \bar{z}) \): \[ \frac{\partial u}{\partial z} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial z} \] Substituting partials of \(x\) and \(y\): \[ \frac{\partial x}{\partial z} = \frac{1}{2}, \quad \frac{\partial y}{\partial z} = \frac{1}{2i} = -\frac{i}{2} \] \[ \implies \frac{\partial}{\partial z} = \frac{1}{2} \left( \frac{\partial}{\partial x} – i \frac{\partial}{\partial y} \right) \]
Using the Chain Rule for a function \( u(x,y) = u(z, \bar{z}) \): \[ \frac{\partial u}{\partial z} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial z} \] Substituting partials of \(x\) and \(y\): \[ \frac{\partial x}{\partial z} = \frac{1}{2}, \quad \frac{\partial y}{\partial z} = \frac{1}{2i} = -\frac{i}{2} \] \[ \implies \frac{\partial}{\partial z} = \frac{1}{2} \left( \frac{\partial}{\partial x} – i \frac{\partial}{\partial y} \right) \]
Step 3: Mixed Second Derivative (\(\partial^2 / \partial \bar{z} \partial z\))
Now apply \( \frac{\partial}{\partial \bar{z}} \) to the result from Step 2. Recall \( \frac{\partial}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) \) (derived similarly).
Now apply \( \frac{\partial}{\partial \bar{z}} \) to the result from Step 2. Recall \( \frac{\partial}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) \) (derived similarly).
\[
\begin{align}
\frac{\partial^2 u}{\partial \bar{z} \partial z} &= \frac{\partial}{\partial \bar{z}} \left[ \frac{1}{2} (u_x – i u_y) \right] \\
&= \frac{1}{2} \left[ \frac{\partial u_x}{\partial \bar{z}} – i \frac{\partial u_y}{\partial \bar{z}} \right]
\end{align}
\]
Apply Chain Rule again for \(\partial / \partial \bar{z}\):
\[
\begin{align}
&= \frac{1}{2} \left[ \left( \frac{\partial u_x}{\partial x}\frac{\partial x}{\partial \bar{z}} + \frac{\partial u_x}{\partial y}\frac{\partial y}{\partial \bar{z}} \right) – i \left( \frac{\partial u_y}{\partial x}\frac{\partial x}{\partial \bar{z}} + \frac{\partial u_y}{\partial y}\frac{\partial y}{\partial \bar{z}} \right) \right]
\end{align}
\]
Substituting \( \frac{\partial x}{\partial \bar{z}} = \frac{1}{2} \) and \( \frac{\partial y}{\partial \bar{z}} = \frac{-1}{2i} = \frac{i}{2} \):
\[
\begin{align}
&= \frac{1}{2} \left[ \left( \frac{1}{2} u_{xx} + \frac{i}{2} u_{xy} \right) – i \left( \frac{1}{2} u_{yx} + \frac{i}{2} u_{yy} \right) \right] \\
&= \frac{1}{4} \left[ u_{xx} + i u_{xy} – i u_{yx} – i^2 u_{yy} \right]
\end{align}
\]
Step 4: Conclusion
Assuming continuity of second derivatives, \( u_{xy} = u_{yx} \), so the middle terms cancel. Also \( -i^2 = +1 \). \[ \frac{\partial^2 u}{\partial \bar{z} \partial z} = \frac{1}{4} [ u_{xx} + u_{yy} ] = \frac{1}{4} \nabla^2 u \] Rearranging gives the final result:
Assuming continuity of second derivatives, \( u_{xy} = u_{yx} \), so the middle terms cancel. Also \( -i^2 = +1 \). \[ \frac{\partial^2 u}{\partial \bar{z} \partial z} = \frac{1}{4} [ u_{xx} + u_{yy} ] = \frac{1}{4} \nabla^2 u \] Rearranging gives the final result:
\[ \boxed{ \nabla^2 = 4 \frac{\partial^2}{\partial z \partial \bar{z}} } \]
Question 2: Application
Let \( f(z) = u + iv \) be an analytic function. Show that:
\[ \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) |f(z)|^2 = 4 |f'(z)|^2 \]
Solution:
The operator on the Left Hand Side is simply the Laplacian \( \nabla^2 \). Using the result from Question 1, we rewrite the problem as: \[ \nabla^2 |f(z)|^2 = 4 \frac{\partial^2}{\partial z \partial \bar{z}} |f(z)|^2 \]
The operator on the Left Hand Side is simply the Laplacian \( \nabla^2 \). Using the result from Question 1, we rewrite the problem as: \[ \nabla^2 |f(z)|^2 = 4 \frac{\partial^2}{\partial z \partial \bar{z}} |f(z)|^2 \]
Properties of Analytic Functions:
If \( f(z) \) is analytic, it depends only on \( z \) (i.e., independent of \( \bar{z} \)).
Then \( \overline{f(z)} \) depends only on \( \bar{z} \). Let’s denote this as \( \bar{f}(\bar{z}) \).
Rewrite the modulus squared:
\[ |f(z)|^2 = f(z) \cdot \overline{f(z)} = f(z) \cdot \bar{f}(\bar{z}) \]
Then \( \overline{f(z)} \) depends only on \( \bar{z} \). Let’s denote this as \( \bar{f}(\bar{z}) \).
Differentiation:
Substitute this into our operator: \[ 4 \frac{\partial^2}{\partial z \partial \bar{z}} \left( f(z) \cdot \bar{f}(\bar{z}) \right) \] First, differentiate with respect to \( \bar{z} \) (treating \( f(z) \) as constant): \[ = 4 \frac{\partial}{\partial z} \left( f(z) \cdot \frac{\partial}{\partial \bar{z}} \bar{f}(\bar{z}) \right) \] \[ = 4 \frac{\partial}{\partial z} \left( f(z) \cdot \bar{f}'(\bar{z}) \right) \] Next, differentiate with respect to \( z \) (treating \( \bar{f}'(\bar{z}) \) as constant): \[ = 4 \left( \frac{\partial}{\partial z} f(z) \right) \cdot \bar{f}'(\bar{z}) \] \[ = 4 f'(z) \cdot \bar{f}'(\bar{z}) \]
Substitute this into our operator: \[ 4 \frac{\partial^2}{\partial z \partial \bar{z}} \left( f(z) \cdot \bar{f}(\bar{z}) \right) \] First, differentiate with respect to \( \bar{z} \) (treating \( f(z) \) as constant): \[ = 4 \frac{\partial}{\partial z} \left( f(z) \cdot \frac{\partial}{\partial \bar{z}} \bar{f}(\bar{z}) \right) \] \[ = 4 \frac{\partial}{\partial z} \left( f(z) \cdot \bar{f}'(\bar{z}) \right) \] Next, differentiate with respect to \( z \) (treating \( \bar{f}'(\bar{z}) \) as constant): \[ = 4 \left( \frac{\partial}{\partial z} f(z) \right) \cdot \bar{f}'(\bar{z}) \] \[ = 4 f'(z) \cdot \bar{f}'(\bar{z}) \]
Final Result:
Note that \( \bar{f}'(\bar{z}) = \overline{f'(z)} \). \[ 4 f'(z) \cdot \overline{f'(z)} = 4 |f'(z)|^2 \]
Note that \( \bar{f}'(\bar{z}) = \overline{f'(z)} \). \[ 4 f'(z) \cdot \overline{f'(z)} = 4 |f'(z)|^2 \]
\[ \boxed{ \nabla^2 |f(z)|^2 = 4 |f'(z)|^2 } \]
Harmonic Functions
Definition
Let \( u: S \subseteq \mathbb{R}^2 \rightarrow \mathbb{R} \) be a real-valued function of two variables \( u(x,y) \).
We say \( u \) is Harmonic on \( S \) if it satisfies the Laplace Equation: \[ \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \quad \forall (x,y) \in S \] where \( \nabla^2 \) is the Laplacian Operator.
We say \( u \) is Harmonic on \( S \) if it satisfies the Laplace Equation: \[ \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \quad \forall (x,y) \in S \] where \( \nabla^2 \) is the Laplacian Operator.
Example 1: Polynomial Function
Given: \( u(x,y) = x^2 – y^2 \)
Step 1: Calculate partial derivatives w.r.t \( x \)
\[ u_x = \frac{\partial u}{\partial x} = 2x \]
\[ u_{xx} = \frac{\partial^2 u}{\partial x^2} = 2 \]
Step 2: Calculate partial derivatives w.r.t \( y \)
\[ u_y = \frac{\partial u}{\partial y} = -2y \]
\[ u_{yy} = \frac{\partial^2 u}{\partial y^2} = -2 \]
Step 3: Verification
\[ \nabla^2 u = u_{xx} + u_{yy} = 2 + (-2) = 0 \]
Conclusion: Since \( \nabla^2 u = 0 \) for all \( (x,y) \in \mathbb{R}^2 \), the function is Harmonic on \( \mathbb{R}^2 \).
Example 2: Logarithmic Function
Given: \( u(x,y) = \log(x^2 + y^2) \)
Domain: \( \mathbb{R}^2 \setminus \{(0,0)\} \) (Since log is undefined at 0)
Step 1: Derivatives w.r.t \( x \)
\[ u_x = \frac{1}{x^2+y^2} \cdot (2x) = \frac{2x}{x^2+y^2} \]
Using Quotient Rule for second derivative:
\[ u_{xx} = \frac{2(x^2+y^2) – 2x(2x)}{(x^2+y^2)^2} = \frac{2y^2 – 2x^2}{(x^2+y^2)^2} \]
Step 2: Derivatives w.r.t \( y \) (By Symmetry)
\[ u_{yy} = \frac{2x^2 – 2y^2}{(x^2+y^2)^2} \]
Step 3: Summation
\[ \nabla^2 u = \frac{(2y^2 – 2x^2) + (2x^2 – 2y^2)}{(x^2+y^2)^2} = \frac{0}{(x^2+y^2)^2} = 0 \]
Conclusion: \( u \) is Harmonic on its domain \( \mathbb{R}^2 \setminus \{(0,0)\} \).
Example 3: Non-Harmonic Function
Given: \( u(x,y) = x^2 + y^2 \)
Step 1: Derivatives
\[ u_x = 2x \implies u_{xx} = 2 \]
\[ u_y = 2y \implies u_{yy} = 2 \]
Step 2: Verification
\[ \nabla^2 u = u_{xx} + u_{yy} = 2 + 2 = 4 \]
Conclusion: Since \( \nabla^2 u = 4 \neq 0 \), the function is NOT Harmonic anywhere in \( \mathbb{R}^2 \).
Advanced Harmonic Function Problems
Question 1: Radial Harmonic Functions
Problem: Find all the harmonic functions of the form \( u(x,y) = \phi(\sqrt{x^2+y^2}) \) which are not constant.
Step 1: Setup Coordinates
Let \( r = \sqrt{x^2+y^2} \). The function depends only on \( r \):
\[ u(x,y) = \phi(r) \]
Derivatives of \( r \):
\[ \frac{\partial r}{\partial x} = \frac{1}{2\sqrt{x^2+y^2}}(2x) = \frac{x}{r} \]
Similarly, \( \frac{\partial r}{\partial y} = \frac{y}{r} \).
Step 2: First Derivative
Using the Chain Rule:
\[ u_x = \phi'(r) \frac{\partial r}{\partial x} = \phi'(r) \cdot \frac{x}{r} \]
Step 3: Second Derivative (Product Rule)
We need to differentiate \( u_x = x \cdot \frac{\phi'(r)}{r} \) with respect to \( x \).
\[ u_{xx} = 1 \cdot \frac{\phi'(r)}{r} + x \cdot \frac{\partial}{\partial x} \left( \frac{\phi'(r)}{r} \right) \]
Applying Chain Rule to the second term:
\[ \frac{\partial}{\partial x} \left( \frac{\phi'(r)}{r} \right) = \frac{r \phi”(r) r_x – \phi'(r) r_x}{r^2} = \frac{x}{r} \left( \frac{r\phi” – \phi’}{r^2} \right) \]
Substituting back:
\[
\begin{align}
u_{xx} &= \frac{\phi’}{r} + x \left[ \frac{x}{r} \frac{r\phi” – \phi’}{r^2} \right] \\
u_{xx} &= \frac{\phi’}{r} + \frac{x^2}{r^2} \phi” – \frac{x^2}{r^3} \phi’
\end{align}
\]
Step 4: Laplacian Condition
By symmetry, \( u_{yy} \) is the same with \( y \) replacing \( x \):
\[ u_{yy} = \frac{\phi’}{r} + \frac{y^2}{r^2} \phi” – \frac{y^2}{r^3} \phi’ \]
Summing them for \( \nabla^2 u = 0 \):
\[ u_{xx} + u_{yy} = \frac{2\phi’}{r} + \frac{(x^2+y^2)}{r^2}\phi” – \frac{(x^2+y^2)}{r^3}\phi’ = 0 \]
Since \( x^2+y^2 = r^2 \):
\[ \frac{2\phi’}{r} + \phi” – \frac{1}{r}\phi’ = 0 \]
Resulting ODE:
\[ \phi”(r) + \frac{1}{r}\phi'(r) = 0 \]
Step 5: Solving the ODE
Let \( v = \phi’ \). Then \( \frac{dv}{dr} = -\frac{v}{r} \).
\[ \int \frac{dv}{v} = – \int \frac{dr}{r} \implies \ln v = -\ln r + \ln A \implies v = \frac{A}{r} \]
Now integrate \( \phi'(r) = \frac{A}{r} \):
\[ \phi(r) = A \ln r + B \]
Final Solution:
\[ u(x,y) = A \ln(x^2+y^2) + B \]
or more commonly written as \( u = a \ln r + b \).
Note: As \( (x,y) \to (0,0) \), \( |u| \to \infty \). This is the fundamental singularity of the 2D Laplacian.
Question 2: Homogeneous Polynomials
Problem: Find the most general form of a homogeneous polynomial in \( x \) & \( y \) of degree 3 which is Harmonic.
Step 1: General Form
A homogeneous polynomial of degree 3 contains terms where powers sum to 3:
\[ u(x,y) = ax^3 + bx^2y + cxy^2 + dy^3 \]
(We must find conditions on constants \( a, b, c, d \) to minimize unknowns).
Step 2: Calculate Derivatives
First Derivatives:
\[ u_x = 3ax^2 + 2bxy + cy^2 \]
\[ u_y = bx^2 + 2cxy + 3dy^2 \]
Second Derivatives:
\[ u_{xx} = 6ax + 2by \]
\[ u_{yy} = 2cx + 6dy \]
Step 3: Apply Harmonic Condition
\[ \nabla^2 u = u_{xx} + u_{yy} = 0 \]
\[ (6ax + 2by) + (2cx + 6dy) = 0 \]
Grouping by variable:
\[ (6a + 2c)x + (2b + 6d)y = 0 \]
For this to hold for all \( x, y \), the coefficients must vanish:
\[
\begin{cases}
6a + 2c = 0 \implies c = -3a \\
2b + 6d = 0 \implies b = -3d
\end{cases}
\]
Step 4: Substitute Back
Replace \( c \) and \( b \) in the original equation:
\[ u(x,y) = ax^3 + (-3d)x^2y + (-3a)xy^2 + dy^3 \]
Grouping terms by the remaining constants \( a \) and \( d \):
Most General Form:
\[ u(x,y) = a(x^3 – 3xy^2) + d(y^3 – 3x^2y) \]
This represents two linearly independent harmonic polynomials of degree 3.
Properties of Analytic Functions
1. Harmonic Nature of u and v
Theorem: Let \( f: D \subset \mathbb{C} \to \mathbb{C} \) defined by \( w = f(z) = u(x,y) + i v(x,y) \).
If \( f \) is analytic on \( D \), then both \( u \) and \( v \) are Harmonic Functions. \[ \nabla^2 u = 0 \quad \text{and} \quad \nabla^2 v = 0 \]
If \( f \) is analytic on \( D \), then both \( u \) and \( v \) are Harmonic Functions. \[ \nabla^2 u = 0 \quad \text{and} \quad \nabla^2 v = 0 \]
Proof:
Since \( f \) is analytic, \( u \) and \( v \) satisfy the Cauchy-Riemann (C-R) Equations:
\[ u_x = v_y \quad \text{and} \quad u_y = -v_x \]
Differentiate the first equation w.r.t \( x \) and the second w.r.t \( y \):
\[ u_{xx} = v_{yx} \]
\[ u_{yy} = -v_{xy} \]
Summing them up:
\[ u_{xx} + u_{yy} = v_{yx} – v_{xy} \]
Assuming the mixed partial derivatives are continuous, \( v_{yx} = v_{xy} \). Thus:
\[ u_{xx} + u_{yy} = 0 \implies \nabla^2 u = 0 \]
Similarly, it can be shown that \( \nabla^2 v = 0 \).
Note: The existence of these second derivatives is guaranteed because analytic functions are infinitely differentiable (a deeper result in Complex Analysis).
2. Harmonic Conjugates
If \( f(z) = u(x,y) + i v(x,y) \) is analytic, then \( v(x,y) \) is called the Harmonic Conjugate of \( u(x,y) \).
Crucial Note on Order:
If \( \phi_1 \) and \( \phi_2 \) are harmonic functions on \( S \):
We say \( \phi_2 \) is the harmonic conjugate of \( \phi_1 \) \( \iff \phi_1 + i\phi_2 \) is analytic.
However, this does NOT imply that \( \phi_2 + i\phi_1 \) is analytic.
(i.e., The relation is not necessarily symmetric. \( u \) is not necessarily the conjugate of \( v \); actually, \( -u \) is the conjugate of \( v \)).
If \( \phi_1 \) and \( \phi_2 \) are harmonic functions on \( S \):
We say \( \phi_2 \) is the harmonic conjugate of \( \phi_1 \) \( \iff \phi_1 + i\phi_2 \) is analytic.
However, this does NOT imply that \( \phi_2 + i\phi_1 \) is analytic.
(i.e., The relation is not necessarily symmetric. \( u \) is not necessarily the conjugate of \( v \); actually, \( -u \) is the conjugate of \( v \)).
3. Orthogonal Families of Curves
Theorem: If \( w = f(z) = u + iv \) is analytic, then the family of curves:
- \( u(x,y) = \alpha \) (Level curves of Real part)
- \( v(x,y) = \beta \) (Level curves of Imaginary part)
Example: \( f(z) = z^2 = (x^2 – y^2) + i(2xy) \)
u-curves: \( x^2 – y^2 = \alpha \) (Rectangular Hyperbolas opening Left/Right if \( \alpha > 0 \))
v-curves: \( 2xy = \beta \) (Rectangular Hyperbolas in 1st/3rd Quadrants if \( \beta > 0 \))
u-curves: \( x^2 – y^2 = \alpha \) (Rectangular Hyperbolas opening Left/Right if \( \alpha > 0 \))
v-curves: \( 2xy = \beta \) (Rectangular Hyperbolas in 1st/3rd Quadrants if \( \beta > 0 \))
(Visual representation: The red curves \( x^2-y^2=\alpha \) intersect the blue curves \( 2xy=\beta \) at right angles.)
Proof of Orthogonality:
Consider the level curve \( u(x,y) = \alpha \). Differentiate to find the slope \( m_1 \):
\[ du = u_x dx + u_y dy = 0 \implies \frac{dy}{dx} = -\frac{u_x}{u_y} = m_1 \]
Consider the level curve \( v(x,y) = \beta \). Differentiate to find the slope \( m_2 \):
\[ dv = v_x dx + v_y dy = 0 \implies \frac{dy}{dx} = -\frac{v_x}{v_y} = m_2 \]
Calculate the product of slopes \( m_1 m_2 \):
\[ m_1 m_2 = \left( -\frac{u_x}{u_y} \right) \left( -\frac{v_x}{v_y} \right) = \frac{u_x v_x}{u_y v_y} \]
Apply C-R Equations (\( u_x = v_y \) and \( u_y = -v_x \)):
Substitute \( u_x \) and \( u_y \) into the equation:
\[ m_1 m_2 = \frac{(v_y) v_x}{(-v_x) v_y} = -1 \]
Since \( m_1 m_2 = -1 \), the tangents at the point of intersection are perpendicular.
Thus, the curves are orthogonal for all \( \alpha, \beta \in \mathbb{R} \). \(\blacksquare\)
Thus, the curves are orthogonal for all \( \alpha, \beta \in \mathbb{R} \). \(\blacksquare\)