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. $$ e^{-y} \sin x $$

Short answer

The function \( u(x, y) = e^{-y} \sin x \) is harmonic and a function \( f(z) = e^{-iz} \) gives its real part. The imaginary part \( v(x, y) = e^{-y} \cos x \) is also harmonic.

Step-by-step solution

- Define Laplace's equation

Laplace's equation in two dimensions is given by: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \]

- Compute the second partial derivatives

Given the function \( u(x, y) = e^{-y} \sin x \), we need to compute the second partial derivatives: First, compute the first partial derivatives: \[ \frac{\partial u}{\partial x} = e^{-y} \cos x \] \[ \frac{\partial u}{\partial y} = -e^{-y} \sin x \] Now, compute the second partial derivatives: \[ \frac{\partial^2 u}{\partial x^2} = -e^{-y} \sin x \] \[ \frac{\partial^2 u}{\partial y^2} = e^{-y} \sin x \]

- Verify Laplace's equation

To verify that this function satisfies Laplace's equation, we add the second partial derivatives: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = -e^{-y} \sin x + e^{-y} \sin x = 0 \] Since the sum is 0, the function \( u(x, y) = e^{-y} \sin x \) is harmonic.

- Find the complex function \( f(z) \)

Let's find a complex function \( f(z) = u(x, y) + iv(x, y) \) where \( u(x, y) = e^{-y} \sin x \) and \( v(x, y) \) is the imaginary part. To do this, note that \( z = x + iy \), so \( e^{-y} \sin x \) can be the real part of an exponential form: \( f(z) = e^{-iz} \). Express \( f(z) \) in terms of \( x \) and \( y \): \[ f(z) = e^{i(x+iy)} = e^{ix}e^{-y} \] Using Euler's formula, \( e^{ix} = \cos x + i\sin x \), we get: \[ e^{-y}(\cos x + i\sin x) = e^{-y}\cos x + ie^{-y}\sin x \] Thus, \( u(x, y) = e^{-y} \sin x \) and \( v(x, y) = e^{-y} \cos x \).

- Verify that \( v(x, y) \) is harmonic

To verify that \( v(x, y) = e^{-y} \cos x \) is harmonic, compute the second partial derivatives: First, compute the first partial derivatives: \[ \frac{\partial v}{\partial x} = -e^{-y} \sin x \] \[ \frac{\partial v}{\partial y} = -e^{-y} \cos x \] Now, compute the second partial derivatives: \[ \frac{\partial^2 v}{\partial x^2} = -e^{-y} \cos x \] \[ \frac{\partial^2 v}{\partial y^2} = e^{-y} \cos x \] Add the second partial derivatives: \[ \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = -e^{-y} \cos x + e^{-y} \cos x = 0 \] Since the sum is 0, the function \( v(x, y) \) is also harmonic.

Key concepts

Laplace's equation

Laplace's equation is a second-order partial differential equation, which describes the behavior of harmonic functions. In two dimensions, it is given by:

\[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \]

A function that satisfies Laplace's equation is termed 'harmonic.' Harmonic functions are significant in various fields such as physics, engineering, and mathematics due to their properties. For instance, they are solutions to many problems in electrostatics and fluid dynamics.

We often start by computing the second partial derivatives of the given function, adding them to see if the resultant expression becomes zero. If it does, the function is harmonic.

Partial derivatives

Partial derivatives measure how a function changes as one of its variables is varied, while keeping the other variables constant. For a function of two variables, say \( u(x, y) \), the first partial derivatives with respect to \( x \) and \( y \) are denoted as \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \).

Higher-order partial derivatives, such as the second partial derivatives \( \frac{\partial^2 u}{\partial x^2} \) and \( \frac{\partial^2 u}{\partial y^2} \), indicate how the first derivative itself changes. These are crucial in verifying if a function satisfies Laplace's equation.

For the function \( u(x, y) = e^{-y} \sin x \), the computation of its second partial derivatives follows:

• \( \frac{\partial u}{\partial x} = e^{-y} \cos x \)
• \( \frac{\partial u}{\partial y} = -e^{-y} \sin x \)
• \( \frac{\partial^2 u}{\partial x^2} = -e^{-y} \sin x \)
• \( \frac{\partial^2 u}{\partial y^2} = e^{-y} \sin x \)

When these partial derivatives are added together, the resulting sum becomes zero, thus confirming that the function is harmonic.

Complex function

A complex function is a function that involves complex numbers, where a complex number is written as \( z = x + iy \), with \( i \) being the imaginary unit (\( i^2 = -1 \)), and \( x \) and \( y \) are real numbers. Complex functions are of the form \( f(z) = u(x, y) + iv(x, y) \), where \( u(x, y) \) and \( v(x, y) \) are real-valued functions.

In the context of harmonic functions, the real and imaginary parts of a complex function can each be harmonic. That is, if \( f(z) \) is an analytic function, then both \( u(x, y) \) and \( v(x, y) \) satisfy Laplace's equation.

For example, the real part of the complex function \( f(z) = e^{i(x+iy)} = e^{ix}e^{-y} \) can be taken as \( u(x, y) = e^{-y} \sin x \). Using Euler's formula, \( e^{ix} = \cos x + i\sin x \), we find:

\[ f(z) = e^{-y}(\cos x + i\sin x) = e^{-y}\cos x + ie^{-y}\sin x \]

Therefore, the real part \( u(x, y) = e^{-y} \sin x \) and the imaginary part \( v(x, y) = e^{-y} \cos x \) are both harmonic functions.

Euler's formula

Euler's formula is a key relationship in complex analysis that connects the exponential function with trigonometric functions. It is given by:

\[ e^{ix} = \cos x + i\sin x \]

This formula shows how exponentials can be expressed in terms of sines and cosines, and it is particularly useful in converting between different forms of complex numbers.

In the exercise, Euler’s formula helps decompose the complex function \( f(z) \) into its real and imaginary parts. Specifically, if \( f(z) = e^{i(x+iy)} = e^{ix}e^{-y} \), Euler’s formula gives us:

• Real part: \( e^{-y} \cos x \)
• Imaginary part: \( e^{-y} \sin x \)

This decomposition is fundamental in confirming that both the real and imaginary parts of the function are harmonic by verifying Laplace’s equation for each part.