Dirichlet Problem for a Rectangle
Let’s begin with a diffusion problem on a Rectangle (figure below),
the figure above can be written as (213)
(213)\[\begin{split}\begin{align}
\nabla ^2 u &= 0 \;\; \textrm{ for } 0<x<L,0<y<K \\
u(x,0) &= 0 \;\; \textrm{ for } 0\leq x\leq L \\
u(0,y) &= u(L,y) = 0 \;\; \textrm{ for } 0\leq y\leq K \\
u(x,K) &= f(x) \;\; \textrm{ for } 0\leq x\leq L
\end{align}\end{split}\]
The rectangle is defined as domain \(\Omega\). In such problem, we want a function that is harmonic on \(\Omega\), 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)\[\begin{align}
u(x,y) = X(x)Y(y)
\end{align}\]
we have
(215)\[\begin{align}
X^{''}Y+XY^{''}=0
\end{align}\]
Dividing the entire (215) by \(XY\), we can get
(216)\[\begin{align}
\frac{X^{''}}{X}+\frac{Y^{''}}{Y}=0
\end{align}\]
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)\[\begin{split}\begin{align}
\frac{X^{''}}{X}+\lambda & =0 \;\; \textrm{ for } X(0)=X(L)=0\\
\frac{Y^{''}}{Y}-\lambda & =0 \;\; \textrm{ for } Y(0)=0 \\
\end{align}\end{split}\]
It’s not hard to find that (217) has a solution of (also see solution )
(218)\[\begin{align}
\lambda_n = \frac{n^2\pi^2}{L^2},X_n(x)=\sin(\frac{n\pi x}{L})
\end{align}\]
For solution in \(y\) direction,
(219)\[\begin{align}
Y_n(y)=\sinh (\frac{n\pi y}{L})
\end{align}\]
The whole solution can therefore be written as
(220)\[\begin{align}
u(x,y) = \sum_{n=1}^{n=\infty} c_n \sin(\frac{n\pi x}{L})\sinh(\frac{n\pi y}{L})
\end{align}\]
For the boundary condition at \(y=K\), we know
(221)\[\begin{align}
u(x,K) = \sum_{n=1}^{n=\infty} c_n \sin(\frac{n\pi x}{L})\sinh(\frac{n\pi K}{L})
\end{align}\]
which infers that
(222)\[\begin{align}
c_n \sinh(\frac{n\pi K}{L}) = \frac{2}{L}\int_{0}^{L} f(\xi)\sin(\frac{n\pi \xi}{L})d\xi
\end{align}\]
Substitute (222) back to (220), we have
(223)\[\begin{align}
u(x,y) = \sum_{n=1}^{\infty} \frac{2}{L}\int_{0}^{L} (f(\xi)\sin(\frac{n\pi \xi}{L})d\xi ) \sin(\frac{n\pi x}{L})\frac{\sinh(\frac{n\pi y }{L})}{\sinh (\frac{n\pi K }{L})}
\end{align}\]
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)
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)\[\begin{align}
u(x,y) = \sum_{i=1}^{4} u_i(x,y)
\end{align}\]
Dirichlet Problem for a Disk
If today, we setup a problem on a disk ( \(\Omega \in x^2+y^2 \leq R^2\)). It’s a rare case when we use Cartesian coordinate but not so rare when we are using polar coordinate i.e., \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\). One can consider Dirichlet’s approach is an alternative way to solve the polar coordinate diffusion problem.
With given boundary condition,
(225)\[\begin{align}
u(x,y) = f(x,y) \; \textrm{ for } x^2+y^2 = R^2
\end{align}\]
We can first apply chain rule to find the diffusion in polar coordinate. (see Week 9 and 10: Special Functions)
(226)\[\begin{align}
\nabla^2 U(r,\theta) = U_{rr}+\frac{1}{r}U_r+\frac{1}{r^2}U_{\theta\theta} \; \textrm{ for } -\pi\leq \theta \leq \pi
\end{align}\]
For (226), we know three basis, \(1,r^n\cos(n\theta),r^n\sin(n\theta)\). form a completed solution space. i.e.,
(227)\[\begin{align}
U(r,\theta) = \frac{1}{2}a_0+\sum_{n=1}^{\infty} (a_n r^{n}\cos(n\theta)+b_n r^{n}\sin(n\theta))
\end{align}\]
Using the boundary condition
(228)\[\begin{align}
U(R,\theta) = \frac{1}{2}a_0+\sum_{n=1}^{\infty} (a_n R^{n}\cos(n\theta)+b_n R^{n}\sin(n\theta))
\end{align}\]
\(a_0\) and \(a_n\) are simply the corresponding Fourier coefficients.
(229)\[\begin{align}
a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi}f(\xi)d\xi
\end{align}\]
and
(230)\[\begin{split}\begin{align}
a_n &= \frac{1}{\pi R^n} \int_{-\pi}^{\pi}f(\xi)\cos(n\xi)d\xi \\
b_n &= \frac{1}{\pi R^n} \int_{-\pi}^{\pi}f(\xi)\sin(n\xi)d\xi \\
\end{align}\end{split}\]
Substituting (229) and (230) back to (227), the complete equation can be written as
(231)\[\begin{split}\begin{align}
U(r,\theta) = & \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\xi)d\xi + \\
& \frac{1}{\pi}\sum_{n=1}^{\infty} (\frac{r}{R})^n \int_{-\pi}^{\pi} (f(\xi)\cos(n\xi)\cos(n\theta)+f(\xi)\cos(n\xi)\sin(n\theta)) d\xi
\end{align}\end{split}\]
Example 1
Find the solution
If the disk has radius \(R=4\) and \(U(4,\theta) = \theta^2\), then the solution by (231) is
\[\begin{split}\begin{align}
U(r,\theta) = & \frac{1}{2\pi}\int_{-\pi}^{\pi}\xi^2 d\xi + \\
& \frac{1}{\pi}\sum_{n=1}^{\infty} (\frac{r}{4})^n \int_{-\pi}^{\pi} (\xi^2\cos(n\xi)\cos(n\theta)+\xi^2\cos(n\xi)\sin(n\theta)) d\xi \\
= & \frac{1}{3}\pi^2 + \sum_{n=1}^{\infty}\frac{4(-1)^n}{n^2}
\end{align}\end{split}\]
The integration used are
\[\frac{1}{\pi} \int_{-\pi}^{\pi} \xi^2 \cos(n\xi)d\xi = \frac{4(-1)^n}{n^2}\]
and
\[\frac{1}{\pi} \int_{-\pi}^{\pi} \xi^2 \sin(n\xi)d\xi = 0\]
Example 2
Solving the Dirichlet problem
\[\begin{align}
& \nabla^2 u(x,y) = 0 \; \; \textrm{for x^2+y^2<9}
& u(x,y) = x^2y^2
\end{align}\]
We first convert this problem to polar coordinate, letting
\[\begin{align}
U(r,\theta) = u(r\cos(\theta),r\sin(\theta))
\end{align}\]
and the boundary condition is
\[\begin{align}
x=3\cos(\theta), y=3\sin(\theta)
\end{align}\]
so
\[\begin{align}
U(3,\theta) = 3^2\cos^2(\theta)\times 3^2\sin^2(\theta) = f(\theta)
\end{align}\]
Solve the problem using (231)
\[\begin{split}\begin{align}
U(r,\theta) = & \frac{1}{2\pi}\int_{-\pi}^{\pi}81\cos^2(\xi)\sin^2(\xi) d\xi + \\
& \frac{1}{\pi}\sum_{n=1}^{\infty} (\frac{r}{4})^n \int_{-\pi}^{\pi} (81\cos^2(\xi)\sin^2(\xi) \cos(n\xi)\cos(n\theta)+81\cos^2(\xi)\sin^2(\xi)\cos(n\xi)\sin(n\theta)) d\xi \\
\end{align}\end{split}\]
All we need is to find the integral of
\[\begin{align}
\int_{-\pi}^{\pi} 81\cos^{2}(\xi)\sin^{2}(\xi)d\xi = \frac{81\pi}{4}
\end{align}\]
\[\begin{split}\int^{\pi}_{-\pi} 81\cos^{2}(\xi)\sin^{2}(\xi)\cos(n\xi)d\xi =
\begin{cases}
0 & \text{if }n\neq 4 \\
\frac{-81\pi}{8} & \text{if }n= 4
\end{cases}\end{split}\]
,and
\[\int^{\pi}_{-\pi} 81\cos^{2}(\xi)\sin^{2}(\xi)\sin(n\xi)d\xi = 0 \]
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 \(U(1,\theta)=f(\theta)\)
(232)\[U(r,\theta) = \frac{1}{2\pi} \int_{-\pi}^{\pi}[1+2\sum_{n=1}^{\infty}r^n\cos(n(\xi-\theta))]f(\xi)d\xi\]
Define a Poisson Kernel
(233)\[P(r,\xi) = \frac{1}{2\pi}[1+2\sum_{n=1}^{\infty} r^n \cos(n\xi)]\]
so
(234)\[U(r,\theta) =\int_{-\pi}^{\pi} P(r,\xi-\theta)f(\xi)d\xi\]
Thinking of a point inside the unit disk as a complex number and also having a polar coordinates \((r,\xi)\). Use the Euler’s formula to write
(235)\[z = re^{i\xi} = r[\cos(\xi)+i\sin(\xi)]\]
then
(236)\[z^n = r^n e^{in\xi} = r^n [\cos(n\xi)+i\sin(n\xi)]\]
Substitute (236) back into (234), we can find
(237)\[\begin{split}1+2\sum_{n=1}^{\infty}\cos(n\xi) & =\Re (1+2\sum_{n=1}^{\infty}z^n) \\
&= \Re(1+2\frac{z}{1-z}) \\
& =\Re(\frac{1+z}{1-z}) \\
& = \Re(\frac{1+re^{i\xi}}{1-re^{i\xi}})\end{split}\]
To find the real part of \(\frac{1+re^{i\xi}}{1-re^{i\xi}}\), we can apply the following algebraic manipulation,
(238)\[\begin{split}\frac{1+re^{i\xi}}{1-re^{i\xi}} & = \frac{1+re^{i\xi}}{1-re^{i\xi}}(\frac{1-re^{i\xi}}{1-re^{i\xi}}) \\
& = \frac{1-r^2+r(e^{i\xi}-e^{-i\xi})}{1+r^2-r(e^{i\xi}+e^{-i\xi})} \\
& = \frac{1-r^2+r(\cos(\xi)+i\sin(\xi)-\cos(\xi)+i\sin(\xi))}{1+r^2-r(\cos(\xi)+i\sin(\xi)+\cos(\xi)-i\sin(\xi))} \\
& = \frac{1-r^2+2ir\sin(\xi)}{1+r^2+2r\cos(\xi)} \\
& = \frac{1-r^2}{1+r^2-2r\cos(\xi)}+i \frac{2r\sin(\xi)}{1+r^2-2r\cos(\xi)}\end{split}\]
and the real part is \(\frac{1-r^2}{1+r^2-2r\cos(\xi)}\)
This leads to the final form of Poisson Integral
(239)\[U(r,\theta) = \frac{1}{2\pi}\int^{\pi}_{-\pi} \frac{1-r^2}{1+r^2-2r\cos(\xi-\theta)}f(\xi)d\xi\]
For a disk with radius R about the origin, a change of variables yields the Poisson solution.
(240)\[U(r,\theta) = \frac{1}{2\pi}\int^{\pi}_{-\pi} \frac{R^2-r^2}{R^2+r^2-2rR\cos(\xi-\theta)}f(\xi)d\xi\]
Example 3
Find the Poisson Integral of example 1.
With \(R=4\) and \(f(\theta) = \theta^2\)
\[U(r,\theta) = \frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{16-r^2}{16+r^2-8r\cos(\xi-\theta)}\xi^2 d\xi = \frac{16-r^2}{2\pi}\int_{-\pi}^{\pi}\]
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)
Here, we first define the unit normal vector and unit tangent vector.
(241)\[\frac{dx}{ds}\mathbf{i} +\frac{dy}{ds}\mathbf{j}\]
and the unit normal vector
(242)\[\mathbf{n}(s) = \frac{dy}{ds}\mathbf{i} -\frac{dx}{ds}\mathbf{j}\]
Then the normal derivative of a point on C can be written as
(243)\[\frac{\partial u}{\partial n} = \frac{\partial u}{\partial x}\frac{dy}{ds}-\frac{\partial u}{\partial y}\frac{dx}{ds}\]
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)\[\begin{split}\nabla^2 u = 0 \; \; \textrm{for $(x,y)$ interior to D} \\
\frac{\partial u}{\partial n} = g(x,y) \; \; \textrm{for $(x,y)$ on C} \\ \end{split}\]
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)\[\begin{split}& \nabla^2 u = 0 \; \; \textrm{for $(x,y)$ interior to D} \\
& \frac{\partial u}{\partial n} = g(x,y)\end{split}\]
With (245), there is a simple condition must be satisfied.
(246)\[\begin{split}\oint_{C} g(x,y) ds &= \oint_{C} \frac{\partial u}{\partial n} ds \\
&= \oint_{C} [\frac{\partial u}{\partial x}\frac{dy}{ds}-\frac{\partial u}{\partial y}\frac{dx}{ds}]ds \\
&= \oint_{C} -u_y dx + u_x dy \\
&= \iint_D(u_xx+u_yy) dA \textrm{by Green's theorem} \end{split}\]
if \(u\) is harmonic on D (i.e., satisfies Sturm-Liouville problem), then \(\nabla^2 u = 0\). Then we can conclude that \(\oint_{C} g(x,y) ds=0\).
Example 4
Find if the following set of equation satisfies a Neumann problem
\[\begin{split}\begin{align}
\nabla^2 u & = 0 \textrm{ for $0<x<1,0<y<1$} \\
u_n(x,y) & =
\begin{cases}
0 \textrm{ on the lower, upper, and the left sides} \\
y^2 \textrm{ on the right}
\end{cases}
\end{align}\end{split}\]
This indicates that
\[\begin{split}u_n(x,0) &= u_n(x,1) =u_n(0,y) = 0 \\
u_n(1,y) &= y^2\\\end{split}\]
To test if the problem above is an Neumann problem, we can take the line integral along the boundary, which leads to
\[\oint_{C} g(x,y) ds = \int_{0}^{1}y^2 dy = \frac{1}{3} \neq 0 \]
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.
\[\begin{split}\begin{cases}
\nabla^2 u(x,y) &= 0 \textrm{for $0<x<a,0<y<b$} \\
\frac{\partial u}{\partial y}(x,0) &= \;\;\; \frac{\partial u}{\partial y} (x,b)=0 \textrm{for $0<x<a$} \\
\frac{\partial u}{\partial x}(0,y) &= 0 \;\;\; \textrm{for $0<y<a$} \\
\frac{\partial u}{\partial x}(x,y) &= g(y) \;\;\; \textrm{for $0<y<a$} \\
\end{cases}\end{split}\]
From the very beginning of this chapter, we know the above equations have solution of
\[u(x,y) = c+\sum_{n=1}^{\infty} c_n \cosh(\frac{n\pi x}{b})\cos(\frac{n\pi y}{b})\]
also, based on the derivative boundary
\[u_n(0,y) = \sum_{n=1}^{\infty} \frac{n\pi}{b} c_n \sinh(\frac{n\pi 0}{b})\cos(\frac{n\pi y}{b}) = g(y)\]
indicating \(\frac{n\pi}{b} c_n \sinh(\frac{n\pi 0}{b})\) is the Fourier \(\cos\) coefficient of the solution in y structure. i.e.,
\[\begin{split}\frac{n\pi}{b}c_n\sinh(\frac{n\pi a}{b}) = \frac{2}{b}\int_{0}^{b} g(\xi) \cos(\frac{n\pi\xi}{b}) d\xi \; \; \; \textrm{or...}\\
u(x,y) = c+\sum_{n=1}^{\infty} c_n\cosh(\frac{n\pi x}{b})\cos(\frac{n\pi y}{b})\end{split}\]
However, to satisfy Neumann condition, we also know,
\[c=\frac{1}{b} \int_{0}^{b} g(y)dy = 0 \]
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)\[\begin{split}\begin{cases}
\nabla^2 u(r,\theta) = 0 \;\;\; \textrm{for $\leq r < R,-\pi\leq\theta<\pi$} \\
\frac{\partial u}{\partial r}(R,\theta)=f(\theta) \;\;\; \textrm{for $-\pi\leq \theta\leq \pi$}
\end{cases}\end{split}\]
A necessary condition for the existence of a solution is that the line integral of the normal derivative over the boundary is 0.
\[\int_{-\pi}^{\pi}f(\theta)d\theta = 0\]
We first attempt a solution of
\[u(r,\theta) = \frac{1}{2}a_0+\sum_{n=1}^{\infty} [a_n r^n \cos(n\theta)+b_n r^n\sin(n\theta)]\]
Choose a coefficient to satisfy,
(248)\[\begin{split}\begin{align}
\frac{\partial u}{\partial r} (R,\theta) & = f(\theta) \\
& = \sum_{n=1}^{\infty} [na_nR^{n-1}\cos(n\theta)+nb_nR^{n-1}\sin(n\theta)]
\end{align}\end{split}\]
Then using the Fourier transform to find the corresponding coefficients \(a_n\) and \(b_n\).
(249)\[\begin{split}\begin{align}
a_0 &= \frac{1}{\pi}\int^{\pi}_{-\pi}f(\xi)d\xi \\
n a_n R^{n-1} &= \frac{1}{\pi}\int^{\pi}_{-\pi}f(\xi)\cos(n\xi)d\xi \\
n b_n R^{n-1} &= \frac{1}{\pi}\int^{\pi}_{-\pi}f(\xi)\sin(n\xi)d\xi \\
\end{align}\end{split}\]
For \(n=1,2,\cdots\). Then
(250)\[\begin{split}a_n & =\frac{1}{n\pi R^{n-1}}\int^{\pi}_{-\pi}f(\xi)\cos(n\xi)d\xi \\
b_n & =\frac{1}{n\pi R^{n-1}}\int^{\pi}_{-\pi}f(\xi)\sin(n\xi)d\xi \\\end{split}\]
the final solution can therefore be written as
(251)\[\begin{split}U(r,\theta) &= c+\frac{R}{r}\sum_{n=1}^{\infty} \frac{1}{n}(\frac{r}{R})^{n}\int^{\pi}_{-\pi}[\cos(n\xi)\cos(n\theta)+\sin(n\xi)\sin(n\theta)]f(\xi)d\xi \\
&= c+\frac{R}{\pi}\sum_{n=1}^{\infty} \frac{1}{n}(\frac{r}{R})^{n}\int^{\pi}_{-\pi} \cos(n(\xi-\theta))d\xi\end{split}\]
Example 6
Solve the following Neumann problem for a unit disk about the origin:
\[\begin{split}& \nabla^2 u(x,y) = 0 \;\;\; \textrm{for $x^2+y^2<1$} \\
& \frac{\partial u}{\partial n} (x,y) = xy^2 \;\;\; \textrm{for $x^2+y^2<1$} \\ \end{split}\]
use polar coordinates, letting
\[\begin{split}U(r,\theta) & = u(r\cos(\theta),r\sin(\theta)) \\\end{split}\]
Now we turn the problem into
\[\begin{split}\begin{align}
\nabla^2 U(r,\theta) &= 0 \;\;\; \textrm{for $0\leq r<1$,$-\pi\leq \theta\leq \pi$} \\
\frac{\partial U}{\partial r}(1,\theta)&=\cos(\theta)\sin^2(\theta)
\end{align}\end{split}\]
First, we observe that
\[\int_{-\pi}^{\pi}\cos(\theta)\sin^2(\theta)d\theta=0\]
satisfies the Neumann boundary condition. By (251), we know the solution is
\[U(r,\theta) = c+\frac{1}{\pi}\sum_{n=1}^{\infty} \frac{1}{n}(r)^{n}\int^{\pi}_{-\pi} \cos(n(\xi-\theta))\cos(\xi)\sin^2(\xi)d\xi\]
The integral of the second term is
\[\begin{split}\begin{cases}
& 0 \;\;\; \textrm{for $n=2,4,5,6,7\cdots$} \\
& \pi\cos(\theta)/4 \;\;\; \textrm{for $n=1$} \\
& -\pi\cos^3(\theta)+3\pi\cos(\theta)/4 \;\;\; \textrm{for $n=3$}\\
\end{cases}\end{split}\]
