Question

Show that the following functions are harmonic, that is, that they satisfy Laplace's equation, and find for each a function \(f(z)\) of which the given function is the real part. Show that the function \(v(x, y)\) which you find also satisfies Laplace's equation. $$ \cosh y \cos x $$

Short answer

The function is harmonic. The corresponding function is \( f(z) = \cosh(z) \), and \(v(x, y) = \sinh(x) \sin(y) \) also satisfies Laplace's equation.

Step-by-step solution

- Check Harmonic Condition

A function is harmonic if it satisfies Laplace's equation: div(U) = 0 In two dimensions, this means checking if: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \] For the given function, \( u(x, y) = \cosh(y) \cos(x) \). First, find the second partial derivatives with respect to \(x\) and \(y\).

- Compute Second Partial Derivatives

Calculate \( \frac{\partial u}{\partial x} \): \[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (\cosh(y) \cos(x)) = -\cosh(y) \sin(x) \] Next, calculate \( \frac{\partial^2 u}{\partial x^2} \): \[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} (-\cosh(y) \sin(x)) = -\cosh(y) \cos(x) \] Now, compute \( \frac{\partial u}{\partial y} \): \[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (\cosh(y) \cos(x)) = \sinh(y) \cos(x) \] Finally, calculate \( \frac{\partial^2 u}{\partial y^2} \): \[ \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y} (\sinh(y) \cos(x)) = \cosh(y) \cos(x) \]

- Verify Laplace's Equation

Sum the second partial derivatives: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = -\cosh(y) \cos(x) + \cosh(y) \cos(x) = 0 \] Since the sum is zero, \( u(x, y) = \cosh(y) \cos(x) \) is harmonic and satisfies Laplace's equation.

- Identify Corresponding Function \(f(z)\)

Identify a function \( f(z) = f(x + iy) \) whose real part is \( u(x, y) = \cosh(y) \cos(x) \). Consider the exponential definitions: \[ \cos x = \frac{e^{ix} + e^{-ix}}{2} \] \[ \cosh y = \frac{e^y + e^{-y}}{2} \] Then, \( u(x, y) \) can be written as: \[ u(x, y) = \frac{(e^y + e^{-y})(e^{ix} + e^{-ix})}{4} = \frac{e^{y+ix} + e^{y-ix} + e^{-y+ix} + e^{-y-ix}}{4} \] Letting \( z = x + iy \), \( u(x, y) \) is the real part of: \[ f(z) = \frac{e^z + e^{-z}}{2} \] So, \( f(z) = \cosh(z) \)

- Find Corresponding Function \(v(x, y)\)

The imaginary part of \( f(z) = \cosh(z) = \cosh(x + iy) \) is \( v(x, y) \). Using: \[ \cosh(x + iy) = \cosh(x) \cos(y) + i \sinh(x) \sin(y) \] The imaginary part is: \( v(x, y) = \sinh(x) \sin(y) \)

- Verify \(v(x, y)\) Harmonic

Verify if \( v(x, y) = \sinh(x) \sin(y) \) satisfies Laplace's equation. Compute the second partial derivatives as before: \[ \frac{\partial v}{\partial x} = \cosh(x) \sin(y) \] \[ \frac{\partial^2 v}{\partial x^2} = \sinh(x) \sin(y) \] \[ \frac{\partial v}{\partial y} = \sinh(x) \cos(y) \] \[ \frac{\partial^2 v}{\partial y^2} = -\sinh(x) \sin(y) \] Sum the second partial derivatives: \[ \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = \sinh(x) \sin(y) - \sinh(x) \sin(y) = 0 \] Since the sum is zero, \( v(x, y) = \sinh(x) \sin(y) \) also satisfies Laplace's equation.

Key concepts

harmonic functions

Harmonic functions are special solutions to Laplace's equation, which is a second-order partial differential equation. Specifically, a function u(x, y) is said to be harmonic if it satisfies: \[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\] This condition must hold at every point in the function's domain. Harmonic functions represent a wide variety of phenomena in physics and engineering, such as heat distribution and electrostatics. Harmonic functions also have a relation to complex analysis, as the real and imaginary parts of any analytic function are both harmonic.

partial derivatives

Partial derivatives are derivations of a function with respect to one variable while keeping the others constant. For a function u(x, y), the partial derivative with respect to x is notated as \(\frac{\partial u}{\partial x}\). Calculating partial derivatives is a crucial step in evaluating whether a function satisfies Laplace's equation. For example, given u(x, y) = cosh(y) cos(x), its first partial derivatives are: \(\frac{\partial u}{\partial x} = -\text{cosh}(y) \text{sinh}(x)\) \(\frac{\partial u}{\partial y} = \text{sinh}(y) \text{cos}(x)\) To ensure that u(x, y) is harmonic, we then sum the second partial derivatives as shown in the previous steps. This same process applies when checking the harmonicity of the imaginary part function v(x, y) derived from f(z).

complex analysis

Complex analysis studies functions of complex variables. A complex function f(z) is typically expressed where z = x + iy, where x and y are real numbers, and i is the imaginary unit. Many complex functions can be broken down into their real and imaginary parts, each of which might be harmonic. For instance, if f(z) = cosh(z), it can be expressed as: \(\text{cosh}(z) = \text{cosh}(x + iy)\) With \(\text{cos}(x) \text{cosh}(y)\) as its real part and \(\text{sinh}(x) \text{sin}(y)\) as its imaginary part. Complex analysis provides the framework for understanding and deconstructing these relationships, enhancing our understanding of harmonic functions.

hyperbolic functions

Hyperbolic functions are analogs of trigonometric functions but based on hyperbolas rather than circles. Important hyperbolic functions include cosh(x), sinh(x), and tanh(x). For instance, \(\text{cosh}(x) = \frac{e^x + e^{-x}}{2}\) and \(\text{sinh}(x) = \frac{e^x - e^{-x}}{2}\). These functions frequently appear when solving differential equations like Laplace's equation. They also emerge naturally in solutions involving complex variables. In our example, the function u(x, y) = \text{cosh}(y) \text{cos}(x), demonstrates the interplay between trigonometric and hyperbolic components in harmonic functions.