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 $\cosh y\cos x$ satisfies Laplace's equation as $\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$. The corresponding complex function is $f(z)=\cosh (x+iy)\cos (x+iy)$.

Step-by-step solution

Define Laplace's Equation

Laplace's equation is defined as: $$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$$ We need to show that the given function satisfies this equation.

Compute the Second Partial Derivative with Respect to x

Given function: $u(x,y)=\cosh y\cos x$First partial derivative with respect to $x$: $$\frac{\partial u}{\partial x}=-\cosh y\sin x$$ Second partial derivative with respect to $x$: $$\frac{{\partial }^{2}u}{\partial x^2}=-\cosh y\cos x$$

Compute the Second Partial Derivative with Respect to y

First partial derivative with respect to $y$: $$\frac{\partial u}{\partial y}=\sinh y\cos x$$ Second partial derivative with respect to $y$: $$\frac{{\partial }^{2}u}{\partial y^2}=\cosh y\cos x$$

Check if Laplace's Equation is Satisfied

Sum of the second partial derivatives:$$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=-\cosh y\cos x+\cosh y\cos x=0$$Thus, the given function $\cosh y\cos x$ satisfies Laplace's equation.

Find a Function f(z) with u(x, y) as its Real Part

A complex function can be written as: $$f(z)=u(x,y)+iv(x,y)$$Given $u(x,y)=\cosh y\cos x$, seek $f(z)$ such that: $$f(z)=\cosh z\cos z$$Since $z=x+iy$: $$f(z)=\cosh (x+iy)\cos (x+iy)$$

Verify that v(x, y) Also Satisfies Laplace's Equation

The imaginary part $v(x,y)$ can be obtained from $f(z)$. For the given real part ($u(x,y)$), verify that the corresponding $v(x,y)$ also satisfies Laplace's equation.

Key concepts

harmonic functions

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: $$\frac{abl{a}^{2}u}{abla{x}^2}+\frac{abl{a}^{2}u}{abla{y}^2}=0$$
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 $\frac{abl{a}^{2}u}{abla{x}^2}=-\text{cosh}y\text{cos}x$.
• The second partial derivative with respect to $y$ is $\frac{abl{a}^{2}u}{abla{y}^2}=\text{cosh}y\text{cos}x$.

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

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 $\frac{ablau}{ablax}=-\text{cosh}y\text{sin}x$.
• Similarly, the first partial derivative with respect to $y$ is $\frac{ablau}{ablay}=\text{sinh}y\text{cos}x$.

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

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

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.