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# 11.4: Modelling the eye–revisited

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Let me return to my model of the eye. With the function $$P_n(\cos\theta)$$ 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 $$\PageIndex{1}$$.

Figure $$\PageIndex{1}$$: 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 $$P_n(x)$$.