The Laplace Equations

# The Laplace Equations

One application to partial derivatives comes from partial differential equations - that is equations that contains partial derivatives of some function $f$. One particular partial differential equation that arises frequently is known as the Laplacian of a function.

 Definition: Let $z = f(x_1, x_2, ..., x_n)$ be an $n$ variable real-valued function. The Laplacian of $f$ is the equation of the sum of the unmixed second partial derivatives of $f$ denoted $\Delta f = \frac{\partial^2 z}{\partial x_1^2} + \frac{\partial^2 z}{\partial x_2^2} + ... + \frac{\partial^2 z}{\partial x_n^2}$.

Other names for the Laplacian are the "Laplace Operator" and the "Laplace Equation".

For example, suppose that we wanted to determine the Laplacian of the function $f(x, y, z) = \sin x + xy^2 - e^{2z}$. In order to do this, we will need to calculate all three unmixed second partial derivatives of this function. First let's determine the first partial derivatives:

(1)

We can determine the unmixed second partial derivatives:

(2)

Therefore we can construct the Laplacian of $f$ as follows:

(3)
\begin{align} \quad \Delta f = -\sin x + 2x - 4e^{2z} \end{align}
 Definition: Let $z = f(x_1, x_2, ..., x_n)$ be an $n$ variable real-valued function, and suppose that for all $(x_1, x_2, ..., x_n) \in S \subseteq D(f)$ we have the Laplacian of $f$ equals zero, that is $\frac{\partial^2 z}{\partial x_1^2} + \frac{\partial^2 z}{\partial x_2^2} + ... + \frac{\partial^2 z}{\partial x_n^2} = 0$. Then $f$ is said to be Harmonic in $S$.

## Example 1

Suppose that $f(x, y, z) = e^{3x + 4y} \sin (5z)$. Prove or disprove that $f$ is harmonic in $\mathbb{R}^3$.

To show whether or not $f$ is harmonic, we need to see whether the Laplacian of $f$ equals zero for all $(x, y, z) \in \mathbb{R}^3$. We will first determine the first partial derivatives of $f$ as:

(4)
\begin{align} \quad \frac{\partial f}{\partial x} = 3e^{3x + 4y}\sin (5z) \quad , \quad \frac{\partial f}{\partial y} = 4e^{3x + 4y} \sin (5z) \quad , \quad \frac{\partial f}{\partial z} 5e^{3x + 4y} \cos (5z) \end{align}

We will now determine the unmixed second partial derivatives of $f$ as:

(5)
\begin{align} \quad \frac{\partial^2 f}{\partial x^2} = 9e^{3x + 4y} \sin (5z) \quad , \quad \frac{\partial^2 f}{\partial y^2} = 16e^{3x + 4y} \sin (5z) \quad , \quad \frac{\partial^2 f}{\partial z^2} = -25e^{3x + 4y} \sin (5z) \end{align}

Therefore the Laplacian of $f$ is:

(6)
\begin{align} \quad \Delta f = 9e^{3x + 4y} \sin (5z) + 16e^{3x + 4y} \sin (5z) - 25e^{3x + 4y} \sin (5z) = 0 \end{align}

Therefore $f$ is harmonic in $\mathbb{R}^3$.

## Example 2

Let $f$ and $g$ be single variable functions that are twice differentiable and suppose that $w = f(x - at) + g(x + ct)$. Show that $\frac{\partial ^2 w}{\partial t^2} = c^2 \frac{\partial^2 w}{\partial x^2}$.

First we compute $\frac{\partial w}{\partial t}$ as follows.

(7)
\begin{align} \quad \frac{\partial w}{\partial t} = a f'(x - at) + ag'(x + at) \end{align}

Differentiating again with respect to $t$ and we have that:

(8)
\begin{align} \quad \frac{\partial^2 w}{\partial t^2} = a^2 f''(x - at) + a^2 g''(x + at) = a^2 [ f''(x - at) + g''(x + at)] \end{align}

Now we will compute the righthand side of this equation. We first compute $\frac{\partial w}{\partial x}$ as follows:

(9)
\begin{align} \quad \frac{\partial w}{\partial x} = f'(x - at) + g'(x + at) \end{align}

Differentiating again with respect to $x$ and multiplying the result by $a^2$ and we have that:

(10)
\begin{align} \quad \frac{\partial ^2 w}{\partial x^2} = f''(x - at) + g''(x + at) \\ \quad c^2 \frac{\partial ^2 w}{\partial x^2} = c^2 [f''(x - at) + g''(x + at)] \end{align}

Therefore $\frac{\partial ^2 w}{\partial t^2} = c^2 \frac{\partial ^2 w}{\partial x^2}$.