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. $$\cosh y \cos x$$

Harmonic functions are a special class of functions that satisfy Laplace's equation. These functions are vital in many areas of mathematics and physics because they often describe equilibrium states. A function is harmonic if its second partial derivatives with respect to all variables sum to zero: \[ \frac{abla^2 u}{abla x^2} + \frac{abla^2 u}{abla y^2} = 0 \]
Let's break this down. For a given function \(u(x, y) = \text{cosh} y \text{cos} x\), we need to compute its second partial derivatives and show that their sum is zero.

• The second partial derivative with respect to \(x\) is \( \frac{abla^2 u}{abla x^2} = - \text{cosh} y \text{cos} x \).
• The second partial derivative with respect to \( y \) is \( \frac{abla^2 u}{abla y^2} = \text{cosh} y \text{cos} x \).

Adding these together yields zero, confirming that our function is harmonic. A main point to remember is that harmonic functions always combine smoothly, without sudden changes in value.

Partial derivatives are used to see how a function changes as just one of several variables changes while the others are held constant. For instance, if we have a function \(u(x, y)\), we can find its partial derivatives with respect to \(x\) and \(y\).

• First partial derivative with respect to \(x\) for our function \(u(x, y) = \text{cosh} y \text{cos} x\) is \( \frac{abla u}{abla x} = - \text{cosh} y \text{sin} x \).
• Similarly, the first partial derivative with respect to \(y\) is \( \frac{abla u}{abla y} = \text{sinh} y \text{cos} x \).

Second partial derivatives are obtained by applying the partial differentiation process once more. Checking these second derivatives and their sums helps us verify properties like whether a function is harmonic. These smaller steps are crucial for solving more complex problems involving multiple variables.

Complex functions involve a complex variable, which can be written as \( z = x + iy \). In this context, a function \( f(z) \) takes on both real and imaginary parts. For instance, a complex function related to our problem is \( f(z) = \text{cosh} z \text{cos} z \), where \( z \) is composed of real \( x \) and imaginary \( y \) parts.
The idea is to find \( f(z) \) such that its real part corresponds to \( u(x, y) \). Therefore, for our harmonic function \( u(x, y) \), we want to write:
\[ f(z) = u(x, y) + iv(x, y) \]
Given \( u(x, y) = \text{cosh} y \text{cos} x \), we seek \( v(x, y) \) to complete \( f(z) \). This essentially involves using known identities for hyperbolic and trigonometric functions and substituting \( z = x + iy \). Identifying \( v(x, y) \) is crucial as we often solve real-world problems using the complex function's properties.

Hyperbolic functions, such as \( \text{cosh} y \) and \( \text{sinh} y \), are analogs of trigonometric functions but for hyperbolas instead of circles. These functions come up frequently when dealing with Laplace's equation and harmonic functions.
For instance, in our function \( u(x, y) = \text{cosh} y \text{cos} x \), hyperbolic cosine \( \text{cosh} y \) defines how the function behaves along the \( y \)-direction. The identity \( \text{cosh} y = \frac{e^y + e^{-y}}{2} \) shows its relationship to exponential functions. Similarly, \( \text{sinh} y = \frac{e^y - e^{-y}}{2} \) provides the hyperbolic sine function.
Hyperbolic functions satisfy specific differential equations, and discovering their properties helps solve many physical problems involving waves, electricity, and even complex systems in engineering.