Modelling the eye

Figure 11.1: The temperature in 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^o C, and the part in contact with the body is at 36^o C. If we look for the steady state temperature it is described by Laplace's equation,

$$\nabla \theta$$ (11.1)

Expressing the Laplacian $\nabla$² in spherical coordinates (see chapter 7) we find

$${\frac{{1}}{{r^{2}}}} {\frac{{\partial ~}}{{\partial r}}} \left(\vphantom{r^{2}\frac{\partial u}{\partial r} }\right. {\frac{{\partial u}}{{\partial r}}} \left.\vphantom{r^{2}\frac{\partial u}{\partial r} }\right) {\frac{{1}}{{r^{2}\sin\theta}}} {\frac{{\partial ~}}{{\partial \theta}}} \left(\vphantom{\sin\theta\frac{\partial u}{\partial \theta}}\right. \theta {\frac{{\partial u}}{{\partial \theta}}} \left.\vphantom{\sin\theta\frac{\partial u}{\partial \theta}}\right)$$ (11.2)

Once again we solve the equation by separation of variables,

$$\theta \theta$$ (11.3)

After this substitution we realize that

$${\frac{{[r^{2} R']'}}{{R}}} {\frac{{[\sin\theta T']'}}{{T\sin\theta}}} \lambda$$ (11.4)

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

$$\theta \lambda \theta$$ (11.5)

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

$$\theta \sqrt{{1-x^{2}}}$$ (11.6)

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)

$${\frac{{d}}{{dx}}} \left[\vphantom{(1-x^{2})\frac{dy}{dx}}\right. {\frac{{dy}}{{dx}}} \left.\vphantom{(1-x^{2})\frac{dy}{dx}}\right] \lambda$$ (11.7)

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,

$$\sum_{{j=0}}^{}$$ (11.8)

We find the equation

$$\sum_{{j=0}}^{{\infty}} \sum_{{j=0}}^{{\infty}} \sum_{{j=0}}^{{\infty}} \lambda \sum_{{j=0}}^{{\infty}}$$ (11.9)

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

$$\sum_{{j=0}}^{{\infty}} \sum_{{j=0}}^{{\infty}} \lambda$$ (11.10)

Collecting the terms of the order x^k, we find the recurrence relation

$${\frac{{k(k+1)-\lambda}}{{(k+1)(k+2)}}}$$ (11.11)

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 = a₀ + a₂x² +...+ a_nxⁿ. (11.12)

For odd n we find odd polynomials,

y_n = a₁x + a₃x³ +...+ a_nxⁿ. (11.13)

One conventionally defines

$${\frac{{(2n)!}}{{n!^{2} 2^{n}}}}$$ (11.14)

With this definition we obtain

P0	=	1,		P1	=	x,
P2	=	${\frac{{3}}{{2}}}$x2 - ${\frac{{1}}{{2}}}$,		P3	=	${\frac{{1}}{{2}}}$(5x3 - 3x),
P4	=	${\frac{{1}}{{8}}}$(35x4 -30x2 + 3),		P5	=	${\frac{{1}}{{8}}}$(63x5 -70x3 + 15x).

(11.15)

A graph of these polynomials can be found in figure 11.2.

Figure 11.2: The first few Legendre polynomials P_n(x).