Week 13: Laplace Equation

Week 13: Laplace Equation#

Problem Setup and Initial Conditions#

Laplace equation is nothing new to readers since we have introduced two related special functions in the chapter of Week 9 and 10: Special Functions, Legendre polynomial and Bessel function. While those special functions form a complete set of solution of diffusion equation for the corresponding coordinate, there is an alternative way of solving diffusion problem when certain boundary conditions are satisfied. (i.e., Sturm-Liouville theorem is satisfied).

Dirichlet Problem for a Rectangle#

Let’s begin with a diffusion problem on a Rectangle (figure below),

../_images/Dirichlet_1.png — Fig. 17 The Dirichlet problem on a rectangle.#

the figure above can be written as (213)

(213)#

\begin{array}{r} \begin{aligned} \nabla^{2} u & = 0 for 0 < x < L, 0 < y < K \\ u (x, 0) & = 0 for 0 \leq x \leq L \\ u (0, y) & = u (L, y) = 0 for 0 \leq y \leq K \\ u (x, K) & = f (x) for 0 \leq x \leq L \end{aligned} \end{array}

The rectangle is defined as domain $Ω$ . In such problem, we want a function that is harmonic on $Ω$ , equals $f (x)$ on the upper side and zero on the lower side and two vertical side. Using the same technique of separation of variable, we can assume the structure of $u$ follows

(214)#

\begin{array}{r} u (x, y) = X (x) Y (y) \end{array}

we have

(215)#

\begin{array}{r} X^{^{″}} Y + X Y^{^{″}} = 0 \end{array}

Dividing the entire (215) by $X Y$ , we can get

(216)#

\begin{array}{r} \frac{X^{^{″}}}{X} + \frac{Y^{^{″}}}{Y} = 0 \end{array}

Similar to how we deal with heat and wave equations, the first part is only related to $x$ and the second part is only related to $y$ . This indicates that both are the same constant with opposite sign. i.e.,

(217)#

\begin{array}{r} \begin{aligned} \frac{X^{^{″}}}{X} + λ & = 0 for X (0) = X (L) = 0 \\ \frac{Y^{^{″}}}{Y} - λ & = 0 for Y (0) = 0 \end{aligned} \end{array}

It’s not hard to find that (217) has a solution of (also see solution )

(218)#

\begin{array}{r} λ_{n} = \frac{n^{2} π^{2}}{L^{2}}, X_{n} (x) = \sin (\frac{n π x}{L}) \end{array}

For solution in $y$ direction,

(219)#

\begin{array}{r} Y_{n} (y) = \sinh (\frac{n π y}{L}) \end{array}

The whole solution can therefore be written as

(220)#

\begin{array}{r} u (x, y) = \sum_{n = 1}^{n = \infty} c_{n} \sin (\frac{n π x}{L}) \sinh (\frac{n π y}{L}) \end{array}

For the boundary condition at $y = K$ , we know

(221)#

\begin{array}{r} u (x, K) = \sum_{n = 1}^{n = \infty} c_{n} \sin (\frac{n π x}{L}) \sinh (\frac{n π K}{L}) \end{array}

which infers that

(222)#

\begin{array}{r} c_{n} \sinh (\frac{n π K}{L}) = \frac{2}{L} \int_{0}^{L} f (ξ) \sin (\frac{n π ξ}{L}) d ξ \end{array}

Substitute (222) back to (220), we have

(223)#

\begin{array}{r} u (x, y) = \sum_{n = 1}^{\infty} \frac{2}{L} \int_{0}^{L} (f (ξ) \sin (\frac{n π ξ}{L}) d ξ) \sin (\frac{n π x}{L}) \frac{\sinh (\frac{n π y}{L})}{\sinh (\frac{n π K}{L})} \end{array}

While we only specify boundary condition at the upper end, we can approach the problem the same way if other edges have non-zero boundary conditions. (see figure below)

../_images/Dirichlet_3.png — Fig. 18 The Dirichlet problem on a rectangle with four specified boundaries.#

The solution for the condition above can be decomposed into four cases and each case with only one boundary specified (The lower panel of The Dirichlet problem on a rectangle with four specified boundaries.). The total solution is the linear combination of the four solution above. i.e.,

(224)#

\begin{array}{r} u (x, y) = \sum_{i = 1}^{4} u_{i} (x, y) \end{array}

Dirichlet Problem for a Disk#

If today, we setup a problem on a disk ( $Ω \in x^{2} + y^{2} \leq R^{2}$ ). It’s a rare case when we use Cartesian coordinate but not so rare when we are using polar coordinate i.e., $x = r \cos (θ)$ and $y = r \sin (θ)$ . One can consider Dirichlet’s approach is an alternative way to solve the polar coordinate diffusion problem.

With given boundary condition,

(225)#

\begin{array}{r} u (x, y) = f (x, y) for x^{2} + y^{2} = R^{2} \end{array}

We can first apply chain rule to find the diffusion in polar coordinate. (see Week 9 and 10: Special Functions)

(226)#

\begin{array}{r} \nabla^{2} U (r, θ) = U_{r r} + \frac{1}{r} U_{r} + \frac{1}{r^{2}} U_{θ θ} for - π \leq θ \leq π \end{array}

For (226), we know three basis, $1, r^{n} \cos (n θ), r^{n} \sin (n θ)$ . form a completed solution space. i.e.,

(227)#

\begin{array}{r} U (r, θ) = \frac{1}{2} a_{0} + \sum_{n = 1}^{\infty} (a_{n} r^{n} \cos (n θ) + b_{n} r^{n} \sin (n θ)) \end{array}

Using the boundary condition

(228)#

\begin{array}{r} U (R, θ) = \frac{1}{2} a_{0} + \sum_{n = 1}^{\infty} (a_{n} R^{n} \cos (n θ) + b_{n} R^{n} \sin (n θ)) \end{array}

$a_{0}$ and $a_{n}$ are simply the corresponding Fourier coefficients.

(229)#

\begin{array}{r} a_{0} = \frac{1}{π} \int_{- π}^{π} f (ξ) d ξ \end{array}

and

(230)#

\begin{array}{r} \begin{aligned} a_{n} & = \frac{1}{π R^{n}} \int_{- π}^{π} f (ξ) \cos (n ξ) d ξ \\ b_{n} & = \frac{1}{π R^{n}} \int_{- π}^{π} f (ξ) \sin (n ξ) d ξ \end{aligned} \end{array}

Substituting (229) and (230) back to (227), the complete equation can be written as

(231)#

\begin{array}{r} \begin{aligned} U (r, θ) = & \frac{1}{2 π} \int_{- π}^{π} f (ξ) d ξ + \\ \frac{1}{π} \sum_{n = 1}^{\infty} (\frac{r}{R})^{n} \int_{- π}^{π} (f (ξ) \cos (n ξ) \cos (n θ) + f (ξ) \cos (n ξ) \sin (n θ)) d ξ \end{aligned} \end{array}

Example 1

Find the solution

If the disk has radius $R = 4$ and $U (4, θ) = θ^{2}$ , then the solution by (231) is

\begin{array}{r} \begin{aligned} U (r, θ) = & \frac{1}{2 π} \int_{- π}^{π} ξ^{2} d ξ + \\ \frac{1}{π} \sum_{n = 1}^{\infty} (\frac{r}{4})^{n} \int_{- π}^{π} (ξ^{2} \cos (n ξ) \cos (n θ) + ξ^{2} \cos (n ξ) \sin (n θ)) d ξ \\ = & \frac{1}{3} π^{2} + \sum_{n = 1}^{\infty} \frac{4 (- 1)^{n}}{n^{2}} \end{aligned} \end{array}

The integration used are

\frac{1}{π} \int_{- π}^{π} ξ^{2} \cos (n ξ) d ξ = \frac{4 (- 1)^{n}}{n^{2}}

and

\frac{1}{π} \int_{- π}^{π} ξ^{2} \sin (n ξ) d ξ = 0

Example 2

Solving the Dirichlet problem

\begin{aligned} \nabla^{2} u (x, y) = 0 for x^2+y^2<9 & u (x, y) = x^{2} y^{2} \end{aligned}

We first convert this problem to polar coordinate, letting

\begin{array}{r} U (r, θ) = u (r \cos (θ), r \sin (θ)) \end{array}

and the boundary condition is

\begin{array}{r} x = 3 \cos (θ), y = 3 \sin (θ) \end{array}

so

\begin{array}{r} U (3, θ) = 3^{2} \cos^{2} (θ) \times 3^{2} \sin^{2} (θ) = f (θ) \end{array}

Solve the problem using (231)

\begin{array}{r} \begin{aligned} U (r, θ) = & \frac{1}{2 π} \int_{- π}^{π} 81 \cos^{2} (ξ) \sin^{2} (ξ) d ξ + \\ \frac{1}{π} \sum_{n = 1}^{\infty} (\frac{r}{4})^{n} \int_{- π}^{π} (81 \cos^{2} (ξ) \sin^{2} (ξ) \cos (n ξ) \cos (n θ) + 81 \cos^{2} (ξ) \sin^{2} (ξ) \cos (n ξ) \sin (n θ)) d ξ \end{aligned} \end{array}

All we need is to find the integral of

\begin{array}{r} \int_{- π}^{π} 81 \cos^{2} (ξ) \sin^{2} (ξ) d ξ = \frac{81 π}{4} \end{array}

\begin{array}{r} \int_{- π}^{π} 81 \cos^{2} (ξ) \sin^{2} (ξ) \cos (n ξ) d ξ = {\begin{cases} 0 & if n \neq 4 \\ \frac{- 81 π}{8} & if n = 4 \end{cases} \end{array}

,and

\int_{- π}^{π} 81 \cos^{2} (ξ) \sin^{2} (ξ) \sin (n ξ) d ξ = 0

Poisson Integral Formula#

Poisson integral is a technique of writing a series solution into an integral form.

In a special case, where $R = 1$ and $U (1, θ) = f (θ)$

(232)#

U (r, θ) = \frac{1}{2 π} \int_{- π}^{π} [1 + 2 \sum_{n = 1}^{\infty} r^{n} \cos (n (ξ - θ))] f (ξ) d ξ

Define a Poisson Kernel

(233)#

P (r, ξ) = \frac{1}{2 π} [1 + 2 \sum_{n = 1}^{\infty} r^{n} \cos (n ξ)]

so

(234)#

U (r, θ) = \int_{- π}^{π} P (r, ξ - θ) f (ξ) d ξ

Thinking of a point inside the unit disk as a complex number and also having a polar coordinates $(r, ξ)$ . Use the Euler’s formula to write

(235)#

z = r e^{i ξ} = r [\cos (ξ) + i \sin (ξ)]

then

(236)#

z^{n} = r^{n} e^{i n ξ} = r^{n} [\cos (n ξ) + i \sin (n ξ)]

Substitute (236) back into (234), we can find

(237)#

\begin{aligned} 1 + 2 \sum_{n = 1}^{\infty} \cos (n ξ) & = ℜ (1 + 2 \sum_{n = 1}^{\infty} z^{n}) \\ = ℜ (1 + 2 \frac{z}{1 - z}) \\ = ℜ (\frac{1 + z}{1 - z}) \\ = ℜ (\frac{1 + r e^{i ξ}}{1 - r e^{i ξ}}) \end{aligned}

To find the real part of $\frac{1 + r e^{i ξ}}{1 - r e^{i ξ}}$ , we can apply the following algebraic manipulation,

(238)#

\begin{aligned} \frac{1 + r e^{i ξ}}{1 - r e^{i ξ}} & = \frac{1 + r e^{i ξ}}{1 - r e^{i ξ}} (\frac{1 - r e^{i ξ}}{1 - r e^{i ξ}}) \\ = \frac{1 - r^{2} + r (e^{i ξ} - e^{- i ξ})}{1 + r^{2} - r (e^{i ξ} + e^{- i ξ})} \\ = \frac{1 - r^{2} + r (\cos (ξ) + i \sin (ξ) - \cos (ξ) + i \sin (ξ))}{1 + r^{2} - r (\cos (ξ) + i \sin (ξ) + \cos (ξ) - i \sin (ξ))} \\ = \frac{1 - r^{2} + 2 i r \sin (ξ)}{1 + r^{2} + 2 r \cos (ξ)} \\ = \frac{1 - r^{2}}{1 + r^{2} - 2 r \cos (ξ)} + i \frac{2 r \sin (ξ)}{1 + r^{2} - 2 r \cos (ξ)} \end{aligned}

and the real part is $\frac{1 - r^{2}}{1 + r^{2} - 2 r \cos (ξ)}$

This leads to the final form of Poisson Integral

(239)#

U (r, θ) = \frac{1}{2 π} \int_{- π}^{π} \frac{1 - r^{2}}{1 + r^{2} - 2 r \cos (ξ - θ)} f (ξ) d ξ

For a disk with radius R about the origin, a change of variables yields the Poisson solution.

(240)#

U (r, θ) = \frac{1}{2 π} \int_{- π}^{π} \frac{R^{2} - r^{2}}{R^{2} + r^{2} - 2 r R \cos (ξ - θ)} f (ξ) d ξ

Example 3

Find the Poisson Integral of example 1.

With $R = 4$ and $f (θ) = θ^{2}$

U (r, θ) = \frac{1}{2 π} \int_{- π}^{π} \frac{16 - r^{2}}{16 + r^{2} - 8 r \cos (ξ - θ)} ξ^{2} d ξ = \frac{16 - r^{2}}{2 π} \int_{- π}^{π}

The Neumann Problem#

Different from Dirichlet problem where the boundary temperature (value) is directly given, Neumann problem describes the jump condition. Mathematically, a Dirichlet problem usually has a boundary condition like $u (s, t) = f (s)$ where $s$ is the boundary along a closed region $C$ . While the Neumann usually has a form of $u_{n} (s, t) = g (s)$ , where $u_{n} (s, t)$ is the normal derivative (perpendicular) on the boundary. (see figure below)

../_images/Neumann.png — Fig. 19 The Neumann boundary condition.#

Here, we first define the unit normal vector and unit tangent vector.

(241)#

\frac{d x}{d s} i + \frac{d y}{d s} j

and the unit normal vector

(242)#

n (s) = \frac{d y}{d s} i - \frac{d x}{d s} j

Then the normal derivative of a point on C can be written as

(243)#

\frac{\partial u}{\partial n} = \frac{\partial u}{\partial x} \frac{d y}{d s} - \frac{\partial u}{\partial y} \frac{d x}{d s}

which can be recognized as the dot product of the gradient u with the unit normal vector on C.

Combining with what we learn in previous sector, a complete set of Neumann problem can be written as

(245)#

\begin{array}{r} \nabla^{2} u = 0 for (x, y) interior to D \\ \frac{\partial u}{\partial n} = g (x, y) for (x, y) on C \end{array}

The set of equation above says, if we don’t consider the heat (or some tracers) transport normal to the boundary, the heat flux integrated over the domain should equal 0 (which physically makes sense because the amount of heat loss for a given grid point is equivalent to the amount of heat received by other grid point).

(245)#

\begin{aligned} \nabla^{2} u = 0 for (x, y) interior to D \\ \frac{\partial u}{\partial n} = g (x, y) \end{aligned}

With (245), there is a simple condition must be satisfied.

(246)#

\begin{aligned} \oint_{C} g (x, y) d s & = \oint_{C} \frac{\partial u}{\partial n} d s \\ = \oint_{C} [\frac{\partial u}{\partial x} \frac{d y}{d s} - \frac{\partial u}{\partial y} \frac{d x}{d s}] d s \\ = \oint_{C} - u_{y} d x + u_{x} d y \\ = \iint_{D} (u_{x} x + u_{y} y) d A by Green's theorem \end{aligned}

if $u$ is harmonic on D (i.e., satisfies Sturm-Liouville problem), then $\nabla^{2} u = 0$ . Then we can conclude that $\oint_{C} g (x, y) d s = 0$ .

Example 4

Find if the following set of equation satisfies a Neumann problem

\begin{array}{r} \begin{aligned} \nabla^{2} u & = 0 for 0 < x < 1, 0 < y < 1 \\ u_{n} (x, y) & = {\begin{cases} 0 on the lower, upper, and the left sides \\ y^{2} on the right \end{cases} \end{aligned} \end{array}

This indicates that

\begin{aligned} u_{n} (x, 0) & = u_{n} (x, 1) = u_{n} (0, y) = 0 \\ u_{n} (1, y) & = y^{2} \end{aligned}

To test if the problem above is an Neumann problem, we can take the line integral along the boundary, which leads to

\oint_{C} g (x, y) d s = \int_{0}^{1} y^{2} d y = \frac{1}{3} \neq 0

Immediately, this problem is not a Neumann problem.

Example 5

A very import characteristic of Neumann problem is…there is no unique solution. We will use this example to illustrate the result.

\begin{array}{r} {\begin{cases} \nabla^{2} u (x, y) & = 0 for 0 < x < a, 0 < y < b \\ \frac{\partial u}{\partial y} (x, 0) & = \frac{\partial u}{\partial y} (x, b) = 0 for 0 < x < a \\ \frac{\partial u}{\partial x} (0, y) & = 0 for 0 < y < a \\ \frac{\partial u}{\partial x} (x, y) & = g (y) for 0 < y < a \end{cases} \end{array}

From the very beginning of this chapter, we know the above equations have solution of

u (x, y) = c + \sum_{n = 1}^{\infty} c_{n} \cosh (\frac{n π x}{b}) \cos (\frac{n π y}{b})

also, based on the derivative boundary

u_{n} (0, y) = \sum_{n = 1}^{\infty} \frac{n π}{b} c_{n} \sinh (\frac{n π 0}{b}) \cos (\frac{n π y}{b}) = g (y)

indicating $\frac{n π}{b} c_{n} \sinh (\frac{n π 0}{b})$ is the Fourier $\cos$ coefficient of the solution in y structure. i.e.,

\begin{array}{r} \frac{n π}{b} c_{n} \sinh (\frac{n π a}{b}) = \frac{2}{b} \int_{0}^{b} g (ξ) \cos (\frac{n π ξ}{b}) d ξ or... \\ u (x, y) = c + \sum_{n = 1}^{\infty} c_{n} \cosh (\frac{n π x}{b}) \cos (\frac{n π y}{b}) \end{array}

However, to satisfy Neumann condition, we also know,

c = \frac{1}{b} \int_{0}^{b} g (y) d y = 0

i.e., the line integral of $g (y)$ along the boundary equals 0. One should notice that, $c$ is an arbitrary constant and it’s not a Fourier $\cos$ coefficient for wave number 0 (otherwise, it will violate the Neumann conditions). This also implies that if $u$ is a solution, $u + c$ will be a constant as well (where $c$ is an arbitrary constant).

Here is an application of Neumann problem to a disk. Suppose $D$ is a disk with a radius $R$ about the origins. The boundary of $D$ is the circle $C$ . In polar coordinate, the Neumann problem for $D$ is

(247)#

\begin{array}{r} {\begin{cases} \nabla^{2} u (r, θ) = 0 for \leq r < R, - π \leq θ < π \\ \frac{\partial u}{\partial r} (R, θ) = f (θ) for - π \leq θ \leq π \end{cases} \end{array}

A necessary condition for the existence of a solution is that the line integral of the normal derivative over the boundary is 0.

\int_{- π}^{π} f (θ) d θ = 0

We first attempt a solution of

u (r, θ) = \frac{1}{2} a_{0} + \sum_{n = 1}^{\infty} [a_{n} r^{n} \cos (n θ) + b_{n} r^{n} \sin (n θ)]

Choose a coefficient to satisfy,

(248)#

\begin{array}{r} \begin{aligned} \frac{\partial u}{\partial r} (R, θ) & = f (θ) \\ = \sum_{n = 1}^{\infty} [n a_{n} R^{n - 1} \cos (n θ) + n b_{n} R^{n - 1} \sin (n θ)] \end{aligned} \end{array}

Then using the Fourier transform to find the corresponding coefficients $a_{n}$ and $b_{n}$ .

(249)#

\begin{array}{r} \begin{aligned} a_{0} & = \frac{1}{π} \int_{- π}^{π} f (ξ) d ξ \\ n a_{n} R^{n - 1} & = \frac{1}{π} \int_{- π}^{π} f (ξ) \cos (n ξ) d ξ \\ n b_{n} R^{n - 1} & = \frac{1}{π} \int_{- π}^{π} f (ξ) \sin (n ξ) d ξ \end{aligned} \end{array}

For $n = 1, 2, \dots$ . Then

(250)#

\begin{aligned} a_{n} & = \frac{1}{n π R^{n - 1}} \int_{- π}^{π} f (ξ) \cos (n ξ) d ξ \\ b_{n} & = \frac{1}{n π R^{n - 1}} \int_{- π}^{π} f (ξ) \sin (n ξ) d ξ \end{aligned}

the final solution can therefore be written as

(251)#

\begin{aligned} U (r, θ) & = c + \frac{R}{r} \sum_{n = 1}^{\infty} \frac{1}{n} (\frac{r}{R})^{n} \int_{- π}^{π} [\cos (n ξ) \cos (n θ) + \sin (n ξ) \sin (n θ)] f (ξ) d ξ \\ = c + \frac{R}{π} \sum_{n = 1}^{\infty} \frac{1}{n} (\frac{r}{R})^{n} \int_{- π}^{π} \cos (n (ξ - θ)) d ξ \end{aligned}

Example 6

Solve the following Neumann problem for a unit disk about the origin:

\begin{aligned} \nabla^{2} u (x, y) = 0 for x^{2} + y^{2} < 1 \\ \frac{\partial u}{\partial n} (x, y) = x y^{2} for x^{2} + y^{2} < 1 \end{aligned}

use polar coordinates, letting

\begin{aligned} U (r, θ) & = u (r \cos (θ), r \sin (θ)) \end{aligned}

Now we turn the problem into

\begin{array}{r} \begin{aligned} \nabla^{2} U (r, θ) & = 0 for 0 \leq r < 1, - π \leq θ \leq π \\ \frac{\partial U}{\partial r} (1, θ) & = \cos (θ) \sin^{2} (θ) \end{aligned} \end{array}

First, we observe that

\int_{- π}^{π} \cos (θ) \sin^{2} (θ) d θ = 0

satisfies the Neumann boundary condition. By (251), we know the solution is

U (r, θ) = c + \frac{1}{π} \sum_{n = 1}^{\infty} \frac{1}{n} (r)^{n} \int_{- π}^{π} \cos (n (ξ - θ)) \cos (ξ) \sin^{2} (ξ) d ξ

The integral of the second term is

\begin{array}{r} {\begin{cases} 0 for n = 2, 4, 5, 6, 7 \dots \\ π \cos (θ) / 4 for n = 1 \\ - π \cos^{3} (θ) + 3 π \cos (θ) / 4 for n = 3 \end{cases} \end{array}