Laplace's equation

If we start from the 1D heat equation and extend to 2D, we have the following PDE:

where $u=u(x, y, t)$ is used to represent temperature. If we now consider 'steady state' as $t \rightarrow \infty$, there is no change of $u$ with time, $t$ and Eq. (2.121) reduces to:

Suppose the PDE is defined in a 'box' with $0 \leq x \leq L$ and $0 \leq y \leq H$. We are looking for solutions that satisfy Laplace's equation given by Eq. (2.122). These are known as harmonic solutions defined as real functions with continuous second derivatives in $x$ and $y$ that satisfy (2.122). We first solve Laplace's equation in a disc followed by solving the PDE in the box defined above.

In a rectangle

We now return to Eq. (2.122) with $0 \leq x \leq L$ and $0 \leq y \leq H$. Again, we need 4 boundary conditions to solve this problem and these are given by:

As indicated in Section 2.4 we expect the solution to the PDE to be separable if the PDE is linear and homogeneous with homogeneous boundary conditions. The problem given by Eq. (2.122) is subject to inhomogeneous boundary conditions as given by (2.123) (see also a picture of the problem we wish to solve in Fig. 2.121 and it is referred to as the Dirichlet problem on a rectangle. We want to use the method of separation of variables and so we need homogeneous boundary conditions. The idea for the solution is to break the problem into simpler problems which have sufficient homogeneous boundary conditions and use superposition of solutions to obtain the final solution to Eq. (2.122) subject to Eqs. (2.123). This decomposition is shown in Fig. 2.7.

Problem 1

We start off with the leftmost box shown in Fig. 2.7 labelled Problem 1. The solution to the first problem (we denote this by $u_{1}$ ), statisfies the following:

The separated solution takes the form:

We use the homogeneous boundary conditions to solve for the eigenvalues, $\lambda$ and eigenfunctions, $\mathcal{X}_{n}, \mathcal{Y}_{n}$ as previously outlined in these notes and finally, we match the solution to the last boundary condition given by $u_{1}(x, 0)=f_{1}(x)$. Upon substitution of the separated solution (2.125) in Eq. (2.122), we have:

Figure 2.6: Inhomogeneous Dirichlet boundary conditions on a rectangular domain as given by PDE (2.122) subject to BCs (2.123)

Figure 2.7: Decomposition of the inhomogeneous Dirichlet boundary value problem into four simpler problems each having 3 boundaries subject to homogeneous conditions (i.e. $u=0$ ) and one boundary that has inhomogeneous conditions.

which leads to:

Note that we could choose $+\lambda$ for the constant in Eq. (2.126) as we still obtain the same solutions from the two second order ODEs. We obtain the eigenvalues and the corresponding eigenfunctions from the $\mathcal{X}$ equation as:

where we have used $\mathcal{X}(0)=\mathcal{X}(L)=0$. Using the eigenvalues, $\lambda_{n}$ and the boundary condition $\mathcal{Y}(H)=0$, we obtain the $\mathcal{Y}$ solutions as follows:

Now, using the $\mathcal{X}$ and $\mathcal{Y}$ solutions in Eq. (2.125), we have:

where $B_{n}$ is an arbitrary constant. The functions $u_{n}$ satisfy all homogeneous boundary conditions of Problem 1 . We match the last condition at $y=0$, i.e. $u(x, 0)=f_{1}(x)$ by superimposing all solutions

evaluating Eq. (2.131) at $y=0$ and equating to $f_{1}(x)$. Finally, the solution to Problem 1 is given as follows:

where $B_{n}$ is obtained from:

We have outlined the solution steps for one of the four cases, namely the solution for $u_{1}$. Repeating the steps for the Problems 2, 3 and 4 shown in Fig. 2.7 allows us to find solutions for $u_{2}, u_{3}$ and $u_{4}$, respectively. As Laplace's equation is linear, we know that it will obey the superposition principle hence we can write the solution to Eq. (2.122) subject to the boundary conditions given by (2.123) as:

In a disc

In what follows, we solve Eq. (2.122) using the method of separation of variables in a disc. We switch to polar coordinates using $x=r \cos \theta$ and $y=r \sin \theta$; upon substitution in Eq. (2.122), we have:

where $u=u(r, \theta)$, defined in the interval $0 \leq r \leq r_{0}$ and $-\pi \leq \theta \leq \pi$. Note that the point $\pm \pi$ identifies in the disc and $r_{0}$ is the radius of the disc.

We need four boundary conditions to solve this problem: two for $r$ and two for $\theta$ (since we have second order derivatives in $r$ and $\theta$ ):

1. Shape of bounded region at $r=r_{0}$

where $f(\theta)$ specifies the temperature at the boundary of the disc.

2. Periodic boundary conditions in $\theta$

since at $\pm \pi$, the two ends meet. These give the 2 boundary conditions we need for $\theta$.

3. Boundedness condition

We need a final condition for $r$. From the PDE (2.135), we see that as $r \rightarrow 0,1 / r \rightarrow \infty$ which leads to Eq. (2.135) blowing up [2.135) is then said to be singular]. Mathematically, this implies that the temperature in the centre of the disc (i.e at $r=0$ ) is infinite. However, physically, we know that the temperature in the centre takes a finite value. We therefore introduce the following boundedness condition:

Equation (2.138) ensures a finite value for the temperature at the centre of the disc.

The PDE (2.135) together with the above boundary conditions constitutes the so-called Dirichlet problem on a disc. We seek solutions of the form,

Following the algorithm given in Subsec. 2.4 we obtain the following two ODEs:

where $R=R(r)$ and $\Theta=\Theta(\theta)$. We first solve Eq. (2.140) to determine the eigenvalues $\lambda$ (note that from Subsec. 2.3.2, for an ODE of the form of (2.140) with periodic boundary conditions, there exist no strictly negative eigenvalues; i.e. in Eqs. (2.140) and (2.141), $\lambda \geq 0$ ). Finally, note that Eq. (2.141) is a second order, linear ODE known as the Euler differential equation which can be solved using the ansatz $R(r)=r^{p}$. The eigenvalues are calculated as $\lambda_{n}=n^{2}$ for $n \geq 0$ which give the eigenfunctions for $\Theta(\theta)$ as:

Next, we need the solutions for $R(r)$ so that we can form the PDE solution [given by (2.139) ] making use of Eq. (2.142). Now, using the ansatz $R=r^{p}$ and the fact that we have found that $\lambda=n^{2}$, we calculate the exponent $p$ as $p= \pm n, n \geq 0$. We need to consider the two cases separately: $n=0$ (in which case $p=n=0$ is a repeated root) and $n>0$. For the case where $n=0$, the ODE (2.141) with $\lambda=n^{2}=0$ becomes $r^{2} R^{\prime \prime}+r R^{\prime}=0$ whose general solution is

For the case where $n>0$,

The boundedness condition (2.138) translates to $|R(0)|<\infty$. To see this, apply (2.138) by setting $r=0$ in Eq. (2.139). Now let us consider the behaviour of solutions (2.143) and (2.144) as $r \rightarrow 0$. In order for $u(r, \theta)$ to be finite, $R(r)$ needs to be finite. So, since as $r \rightarrow 0, \ln r \rightarrow-\infty$ and $r^{-n} \rightarrow \infty$, we need to set the arbitrary constants $c_{1}$ and $c_{4}$ to 0 such that contributions from $\ln r$ and $r^{-n}$, vanish. Taking the above into account, we write down the solution for $R(r)$ as follows:

Finally, substituting Eqs. (2.142) and (2.145) in (2.139), we have:

where $a_{n}=\overline{a_{n} c_{n}}(n \geq 0)$ and $b_{n}=\overline{b_{n}} \overline{c_{n}}(n \geq 1)$.

The solution to Laplace's equation on a disc is given by Eq. (2.146), represented by a full Fourier series where $a_{n}$ and $b_{n}$ are the Fourier coefficients calculated the usual way (see Topic B1). To determine these, we need to use the final boundary condition given by Eq. (2.136). The specific form of the coefficients therefore is determined by the boundary function $f(\theta)$. For the unit disc, $r_{0}=1$, these are:

Poisson integral

While Eq. (2.146) represents the solution to Laplace's equation, it is given in the form of an infinite series. Generally, as we include more modes in the solution (i.e. high $n$ ) then we have a better approximation to the exact solution. It is possible however to transform solution (2.146) into a definite integral independent of a series by making use of the sum to infinity formula for an infinite series. This is done via the Poisson integral.

We start from Eq. (2.146) together with the formulae given by Eqs. (2.147). We express the Fourier coefficient formulae in terms of $\phi$, i.e.:

Upon substitution, we have:

where the terms in blue are the Fourier coefficient formulae. Equation (2.149) simplifies to:

where we have made use of the sum-difference identity to obtain the $\cos (n \alpha)$ term where $\alpha$ is defined as $\alpha=\theta-\phi$.

We continue by defining the complex variable $z$ as:

and $z^{n}$ is:

The real and imaginary parts of (2.152) are:

Using these definitions, Eq. (2.150) becomes:

where the term in the square brackets is known as the Poisson kernel. Now, provided that $|z|<1$, we have the following result for the infinite series:

note that the series starts from $n=0$ while in Eq. (2.154), the series starts from $n=1$. Our assumption $|z|<1$ is obvious from Eq. (2.151), since we know that $|r|<1$ (recall we chose the radius of the disc to be $r_{0}=1$ ). For the more general case, the $r$ term in Eq. (2.151) is replaced by $r / r_{0}$ and the $|z|<1$ assumption still holds. Next, we make use of Eq. (2.155) in Eq. (2.154) to determine the sum given by $\sum_{n=1}^{\infty} \operatorname{Re}\left(z^{n}\right)$. From Eq. (2.155), we expand the first term out of the series to get:

which gives:

We express the term in the square brackets in Eq. (2.154) as:

where $z^{*}$ is the complex conjugate. Note that we only want the real part of the expression given by Eq. (2.158). Let us look at the terms separately:

1. $z-z^{*}=2$ ri $\sin (n \alpha) \Rightarrow$ pure imaginary
2. $z z^{*}=r^{2} \Rightarrow$ real
3. $|1-z|^{2}=(1-z)\left(1-z^{*}\right)=1-2 r \cos (n \alpha)+r^{2} \Rightarrow$ real

The above imply that the term in Eq. (2.158) is:

Finally, replacing the term in the square brackets in Eq. (2.154) with Eq. (2.159) and using $\alpha=\theta-\phi$ yields:

known as the Poisson integral. The advantage of Eq. (2.160) is that the solution to Laplace's equation on a disc is given by a definite integral instead of an infinite series as the solution given by Eq. (2.146).

Gauss' mean value theorem

As previously mentioned, solutions of the Laplace equation, say $u(x, y)$ are called harmonic functions. The latter have the mean-value property which states that the average value of $u$ over any circle is equal to its value at the center. There is a $3 \mathrm{D}$ version of this result - there, the average is over a sphere rather than a circle. For the theorem to hold, $u$ must be harmonic in a region containing the circle (or sphere).

We show this using Eq. (2.160). Setting $r=0$ in (2.160) yields,

Equation (2.161) implies that the temperature at the centre of the disc is equal to the average of the boundary function, $f(\phi)$.

Maximum principle

The maximum principle states that a non-constant harmonic function, $u$ cannot attain a maximum (or minimum) at an interior point of its domain. This result implies that the values of a harmonic function in a bounded domain are bounded by its maximum and minimum values on the boundary. Suppose we have a region $\mathcal{D}$ where $\nabla^{2} u=0$. Then $u$ has no local maximum or minimum inside $\mathcal{D}$. The maximum and minimum values of $u$ are attained on the boundary of $\mathcal{D}$. At a point of maximum in $\mathcal{D}$, say $\left(x_{0}, y_{0}\right)$ we know, from multivariable calculus and the second derivative test, that $u_{x x}\left(\left(x_{0}, y_{0}\right)\right) \leq 0$ and $u_{y y}\left(\left(x_{0}, y_{0}\right)\right) \leq 0$. This would mean that $u_{x x}\left(\left(x_{0}, y_{0}\right)\right)+u_{y y}\left(\left(x_{0}, y_{0}\right)\right) \leq 0$ which is a contradiction if the inequality is strict. The minimum principle states that the minimum value of $u$ in $\mathcal{D}$ is achieved on the boundary.

We can interpret the maximum principle physicaly as follows. Consider a harmonic function $u$ as the equilibrium state (steady state) in the heat equation. If there were a maximum temperature at an interior point of the domain $\mathcal{D}$, there would be heat flux out of this point to points with lower temperature which would consequently decrease the temperature of this point thus rendering it unsteady.

A consequence of the maximum principle is that a solution to the Dirichlet problem (Laplace's equation subject to Dirichlet boundary conditions) in a domain $\mathcal{D}$ is unique. Suppose that $u_{1}$ and $u_{2}$ both solve the Dirichlet problem. This implies that both solutions satisfy the same function at the boundary of the domain (we denote this by $\partial \mathcal{D}$ ), i.e. $u_{1}=u_{2}$ on $\partial \mathcal{D}$. Now, let $u=u_{1}-u_{2}$ and define $\boldsymbol{F}=u \nabla u$. Then, the divergence of $\boldsymbol{F}$ is given by:

which reduces to

since $\nabla^{2} u=0$. By the divergence theorem, we have:

Since $u_{1}=u_{2}$ on the boundary $\partial \mathcal{D}$ then $u$ and $\boldsymbol{F}$ are zero on the boundary. The surface integral therefore is zero. Substituting Eq. (2.163) in (2.164) and using $\oint_{\partial \mathcal{D}} \boldsymbol{F} \cdot \boldsymbol{n} d s=0$, we have:

Since $|\nabla u|^{2}$ is non-negative, Eq. (2.165) holds true only if $\nabla u=0$ which implies that $u=$ constant. But, since $u=0$ on the boundary, then the constant must be zero and therefore $u=u_{1}-u_{2}=0 \Rightarrow u_{1}=u_{2}$ and the solution is unique.