8.1: Laplace Equations
- Page ID
- 35368
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Introduction
The following partial differential equation is called the two-dimensional Laplace equation:
\begin{equation} \displaystyle \frac{\partial^{2}w}{\partial x^{2}}+\frac{\partial^{2}w}{\partial y^{2}}=0 \label{laplace}\end{equation}
where \(w(x,\ y)\) is the unknown function with two variables \(x\) and \(y\). The problem is to find a solution to this equation, namely, find a function \(w(x,\ y)\) which satisfies the Equation \ref{laplace}. This equation is used to model various physical quantities.
Example \(\PageIndex{1}\)
Let \(k\) be a real number. Show that the functions \( w=e^{kx} cos(ky)\) and \( w=e^{kx} sin(ky)\) satisfy the lapalce Equation \ref{laplace} at every point in \(\mathbb {R^2}\).
Solution
Let \( w=e^{kx} cos(ky)\). Then we have,
\(
\dfrac{\partial w}{\partial x}=ke^{kx}\cos(ky),\ \dfrac{\partial w}{\partial y}=-ke^{kx}\sin(ky),\)
which implies \(\displaystyle \frac{\partial^{2}w}{\partial x^{2}}=k^{2}e^{kx}\cos(ky),\ \displaystyle \frac{\partial^{2}w}{\partial y^{2}}=-k^{2}e^{kx}\cos(ky)\).
Consider \(\displaystyle \frac{\partial^{2}w}{\partial x^{2}}+\frac{\partial^{2}w}{\partial y^{2}}=k^{2}e^{kx}\cos(ky)-k^{2}e^{kx}\cos(ky)=0\).
Therefore, that the function \(w=e^{kx}\cos(ky)\) satisfies the Equation \ref{laplace}. Similarly, the function \(w=e^{kx}\sin(ky)\) satisfies the Equation \ref{laplace}.
Definition
A function \(w(x,y)\) of two variables having continuous second partial derivatives in a region of the plane is said to be harmonic if it satisfies the Laplace Equation \ref{laplace}.
Exercise \(\PageIndex{1}\)
Show that \(\ln(y^2+x^2)\) is hamornic everywhere except at the origin.
Converting Laplace's equation to polar co-ordinates
Consider the transformation to polar coordinates, \(x=r \cos(\theta), y=r \sin(\theta),\) imples that \(r^2=x^2+y^2\) and \(\tan(\theta)= y/x.\) We can use these equations to express \(\dfrac{\partial ^2w}{\partial x^2}+\dfrac{\partial ^2w}{\partial y^2} \) in terms of partials of \(w\) with respect to \(r\) and \(\theta.\)