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).
(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
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 )
(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)
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.,
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.
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)
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.
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
\[\begin{split}\begin{align}
\nabla^2 u & = 0 \textrm{ for $0<x<1,0<y<1$} \\
u_n(x,y) & =
\begin{cases}
0 \textrm{ on the lower, upper, and the left sides} \\
y^2 \textrm{ on the right}
\end{cases}
\end{align}\end{split}\]
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
