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# 11.1: Modelling the Eye

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# Separation of variables in three dimensions

We have up to now concentrated on 2D problems, but a lot of physics is three dimensional, and often we have spherical symmetry – that means symmetry for rotation over any angle. In these cases we use spherical coordinates, as indicated in figure [fig:spherical].

## Modelling the eye

The temperature on the outside of a simple model of the eye

Let me model the temperature in a simple model of the eye, where the eye is a sphere, and the eyelids are circular. In that case we can put the $$z$$-axis straight through the middle of the eye, and we can assume that the temperature does only depend on $$r,\theta$$ and not on $$\phi$$. We assume that the part of the eye in contact with air is at a temperature of $$20^{\circ}$$ C, and the part in contact with the body is at $$36^{\circ}$$ C. If we look for the steady state temperature it is described by Laplace’s equation,

${ \nabla}^{2} u(r,\theta) =0.$

Expressing the Laplacian $${ \nabla}^{2}$$ in spherical coordinates (see chapter [chap:curvilinear]) we find

$\frac{1}{r^{2}}\dfrac{\partial}{\partial r}{~}\left(r^{2}\dfrac{\partial}{\partial r} u \right) +\frac{1}{r^{2}\sin\theta} \dfrac{\partial}{\partial \theta}{~}\left(\sin\theta\dfrac{\partial}{\partial \theta} u\right) =0.$

Once again we solve the equation by separation of variables,

$u(r,\theta) = R(r) T (\theta).$

After this substitution we realize that

$\frac{[r^{2} R']'}{R} = -\frac{[\sin\theta T']'}{T\sin\theta} = \lambda.$

The equation for $$R$$ will be shown to be easy to solve (later). The one for $$T$$ is of much more interest. Since for 3D problems the angular dependence is more complicated, whereas in 2D the angular functions were just sines and cosines.

The equation for $$T$$ is

$[\sin\theta T']'+\lambda T\sin\theta=0.$

This equation is called Legendre’s equation, or actually it carries that name after changing variables to $$x= \cos\theta$$. Since $$\theta$$ runs from $$0$$ to $$\pi$$, we find $$\sin \theta >0$$, and we have

$\sin\theta = \sqrt{1-x^{2}}.$

After this substitution we are making the change of variables we find the equation ($$y(x)=T(\theta)=T(\arccos x)$$, and we now differentiate w.r.t. $$x$$, $$d/d\theta = - \sqrt{1-x^{2}}d/dx$$)

$\frac{d}{dx}\left[(1-x^{2})\frac{dy}{dx}\right] + \lambda y = 0.$

This equation is easily seen to be self-adjoint. It is not very hard to show that $$x=0$$ is a regular (not singular) point – but the equation is singular at $$x=\pm 1$$. Near $$x=0$$ we can solve it by straightforward substitution of a Taylor series,

$y(x) = \sum_{j=0} a_{j} x^{j}.$

We find the equation

$\sum_{j=0}^{\infty}j(j-1) a_{j} x^{j-2} -\sum_{j=0}^{\infty}j(j-1) a_{j} x^{j} -2\sum_{j=0}^{\infty}j a_{j} x^{j} + \lambda \sum_{j=0}^{\infty} a_{j} x^{j} = 0$

After introducing the new variable $$i=j-2$$, we have

$\sum_{j=0}^{\infty}(i+1)(i+1) a_{i+2} x^{i} -\sum_{j=0}^{\infty}[j(j+1)-\lambda] a_{j} x^{j}=0.$

Collecting the terms of the order $$x^{k}$$, we find the recurrence relation $a_{k+2} = \frac{k(k+1)-\lambda}{(k+1)(k+2)} a_{k}.$

If $$\lambda=n(n+1)$$ this series terminates – actually those are the only acceptable solutions, any one where $$\lambda$$ takes a different value actually diverges at $$x=+1$$ or $$x=-1$$, not acceptable for a physical quantity – it can’t just diverge at the north or south pole ($$x=\cos\theta=\pm 1$$ are the north and south pole of a sphere).

We thus have, for $$n$$ even,

$y_{n} = a_{0}+a_{2}x^{2}+\ldots+a_{n}x^{n}.$

For odd $$n$$ we find odd polynomials,

$y_{n} = a_{1}x+a_{3}x^{3}+\ldots+a_{n}x^{n}.$

One conventionally defines $a_{n} = \frac{(2n)!}{n!^{2} 2^{n}}.$

With this definition we obtain

$\begin{array}{lllllll} P_{0}& =& 1, &~~~&P_{1} & =& x,\\ P_{2}&=& \frac{3}{2}x^{2}-\frac{1}{2},&&P_{3}&=&\dfrac{1}{2}(5x^3-3x),\\ P_4 &=& \frac{1}{8}(35x^{4}-30x^{2}+3) ,&& P_5 &=& \frac{1}{8}(63x^{5}-70x^{3}+15x) . \end{array}$

A graph of these polynomials can be found in figure [fig:Pn].

The first few Legendre polynomials $$P_n(x)$$.

## Properties of Legendre polynomials

### Generating function

Let $$F(x,t)$$ be a function of the two variables $$x$$ and $$t$$ that can be expressed as a Taylor’s series in $$t$$, $$\sum_{n} c_n(x) t^{n}$$. The function $$F$$ is then called a generating function of the functions $$c_{n}$$.

Exercise $$\PageIndex{1}$$

Show that $$F(x,t) = \frac{1}{1-xt}$$ is a generating function of the polynomials $$x^{n}$$.

Answer

Look at

$\frac{1}{1-xt} = \sum_{n=0}^{\infty} x^{n}t^{n}\;\;(|xt|<1).$

Exercise $$\PageIndex{2}$$

Show that $$F(x,t) = \exp\left(\frac{tx-t}{2t}\right)$$ is the generating function for the Bessel functions,

$F(x,t) = \exp(\frac{tx-t}{2t}) = \sum_{n=0}^{\infty} J_{n}(x)t^{n}.$

Answer

TBA

Exercise $$\PageIndex{4}$$

(The case of most interest here) $F(x,t) =\frac{1}{\sqrt{1-2xt+t^{2}}} = \sum_{n=0}^{\infty} P_{n}(x) t^{n}.$

Answer

TBA

### Rodrigues’ Formula

$P_{n}(x) = \frac{1}{2^{n} n!} \frac{d^{n}}{dx^{n}} (x^2-1)^n.$

### A table of properties

1. $$P_{n}(x)$$ is even or odd if $$n$$ is even or odd.
2. $$P_{n}(1)=1$$.
3. $$P_{n}(-1)=(-1)^{n}$$.
4. $$(2n+1) P_{n}(x) = P'_{n+1}(x)-P'_{n-1}(x)$$.
5. $$(2n+1)x P_n(x) = (n+1) P_{n+1}(x) + n P_{n-1}(x)$$.
6. $$\int_{-1}^{x} P_n(x') dx'= \frac{1}{2n+1} \left[P_{n+1}(x)-P_{n-1}(x)\right]$$.

Let us prove some of these relations, first Rodrigues’ formula. We start from the simple formula

$(x^{2}-1) \frac{d}{dx} (x^{2}-1)^{n} - 2 n x (x^{2}-1)^{n}=0,$ which is easily proven by explicit differentiation. This is then differentiated $$n+1$$ times,

\begin{aligned} { \frac{d^{n+1}}{dx^{n+1}}\left[ (x^{2}-1) \frac{d}{dx} (x^{2}-1)^{n} - 2 n x (x^{2}-1)^{n}\right]} \nonumber\\ &=& n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n + 2(n+1) x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n+(x^2-1) \frac{d^{n+2}}{dx^{n+2}} (x^2-1)^n \nonumber\\ &&-2n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n - 2n x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n \nonumber\\ &=&-n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n + 2 x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n+(x^2-1) \frac{d^{n+2}}{dx^{n+2}} (x^2-1)^n \nonumber\\ &=& -\left[\frac{d}{dx}(1-x^2)\frac{d}{dx}\left\{\frac{d^{n}}{dx^{n}}(x^2-1)^n \right\} +n(n+1)\left\{\frac{d^{n}}{dx^{n}}(x^2-1)^n\right\}\right] =0.\end{aligned}

We have thus proven that $$\frac{d^n}{dx^n}(x^2-1)^n$$ satisfies Legendre’s equation. The normalization follows from the evaluation of the highest coefficient, $\frac{d^n}{dx^n} x^{2n} = \frac{2n!}{n!} x^n,$ and thus we need to multiply the derivative with $$\frac{1}{2^n n!}$$ to get the properly normalized $$P_n$$.

Let’s use the generating function to prove some of the other properties: 2.: $F(1,t) = \frac{1}{1-t} = \sum_n t^n$ has all coefficients one, so $$P_n(1)=1$$. Similarly for 3.: $F(-1,t) = \frac{1}{1+t} = \sum_n (-1)^nt^n .$ Property 5. can be found by differentiating the generating function with respect to $$t$$:

\begin{aligned} \frac{d}{dt} \frac{1}{\sqrt{1-2tx +t^2}} &=& \frac{d}{dt} \sum_{n=0}^{\infty} t^n P_n(x)\nonumber\\ \frac{x-t}{(1-2tx+t^{2})^1.5} &=& \sum_{n=0} n t^{n-1} P_{n}(x)\nonumber\\ \frac{x-t}{1-2xt +t^{2}} \sum_{n=0}^{\infty} t^n P_n(x)&=& \sum_{n=0} n t^{n-1} P_{n}(x)\nonumber\\ \sum_{n=0}^{\infty} t^n x P_n(x)- \sum_{n=0}^{\infty} t^{n+1} P_n(x) &=& \sum_{n=0}^{\infty} nt^{n-1} P_n(x) - 2\sum_{n=0}^{\infty} nt^n xP_n(x) +\sum_{n=0}^{\infty} nt^{n+1} P_n(x)\nonumber\\ \sum_{n=0}^{\infty} t^n(2n+1)x P_n(x) &=& \sum_{n=0}^{\infty} (n+1)t^n P_{n+1}(x) + \sum_{n=0}^{\infty} n t^{n} P_{n-1}(x)\end{aligned}

Equating terms with identical powers of $$t$$ we find $(2n+1)x P_n(x) = (n+1) P_{n+1}(x) + n P_{n-1}(x).$

Proofs for the other properties can be found using similar methods.

## Fourier-Legendre series

Since Legendre’s equation is self-adjoint, we can show that $$P_n(x)$$ forms an orthogonal set of functions. To decompose functions as series in Legendre polynomials we shall need the integrals

$\int_{-1}^1 P_n^2(x) dx = \frac{2n+1}{2},$

which can be determined using the relation 5. twice to obtain a recurrence relation

\begin{aligned} \int_{-1}^1 P_n^2(x) dx &=& \int_{-1}^1 P_n(x) \frac{(2n-1)x P_{n-1}(x)-(n-1)P_{n-2}(x)}{n} dx \nonumber\\ &=&\frac{(2n-1)}{n}\int_{-1}^1 x P_n(x) P_{n-1}(x) dx \nonumber\\ &=&\frac{(2n-1)}{n}\int_{-1}^1 \frac{(n+1)P_{n+1}(x) + n P_{n-1}(x)}{2n+1} P_{n-1}(x) dx \nonumber\\ &=&\frac{(2n-1)}{2n+1}\int_{-1}^1 P_{n-1}^2(x) dx,\end{aligned}

and the use of a very simple integral to fix this number for $$n=0$$,

$\int_{-1}^1 P_0^2(x) dx = 2.$

So we can now develop any function on $$[-1,1]$$ in a Fourier-Legendre series

\begin{aligned} f(x) & = & \sum_n A_n P_n(x) \nonumber\\ A_n & = & \frac{2n+1}{2} \int_{-1}^1 f(x) P_n(x) dx\end{aligned}

Find the Fourier-Legendre series for

$f(x) = \left\{ \begin{array}{ll} 0, & -1 < x < 0\\ 1, & 0 < x <1 \end{array} \right. .$

We find \begin{aligned} A_0 &=& \dfrac{1}{2} \int_0^1 P_0(x) dx = \dfrac{1}{2},\\ A_1 &=& \frac{3}{2} \int_0^1 P_1(x) dx= \frac{1}{4}, \\ A_2 &=& \frac{5}{2} \int_0^1 P_2(x) dx= 0, \\ A_3 &=& \frac{7}{2} \int_0^1 P_3(x) dx= -\frac{7}{16} .\end{aligned}

All other coefficients for even $$n$$ are zero, for odd $$n$$ they can be evaluated explicitly.

## Modelling the eye–revisited

Let me return to my model of the eye. With the functions $$P_n(\cos\theta)$$ as the solution to the angular equation, we find that the solutions to the radial equation are

$R=Ar^n + Br^{-n-1}.$

The singular part is not acceptable, so once again we find that the solution takes the form

$u(r,\theta) = \sum_{n=0}^\infty A_n r^n P_n(\cos\theta)$ We now need to impose the boundary condition that the temperature is $$20^\circ$$ C in an opening angle of $$45^\circ$$, and $$36^\circ$$ elsewhere. This leads to the equation

\begin{aligned} \sum_{n=0}^\infty A_n c^n P_n(\cos\theta) = \left\{ \begin{array}{ll} 20 & 0<\theta<\pi/4\\ 36 & \pi/4 < \theta < \pi \end{array}\right.\end{aligned} This leads to the integral, after once again changing to $$x=\cos\theta$$,

$A_n = \frac{2n+1}{2} \left[\int_{-1}^1 36 P_n(x) dx -\int_{\frac{1}{2}\sqrt{2}}^1 16 P_n(x) dx\right].$ These integrals can easily be evaluated, and a sketch for the temperature can be found in figure [fig:eyeT].

A cross-section of the temperature in the eye. We have summed over the first 40 Legendre polynomials.

Notice that we need to integrate over $$x=\cos\theta$$ to obtain the coefficients $$A_n$$. The integration over $$\theta$$ in spherical coordinates is $$\int_0^\pi \sin \theta d\theta = \int_{-1}^1 1 dx$$, and so automatically implies that $$\cos\theta$$ is the right variable to use, as also follows from the orthogonality of $$P_n(x)$$.