Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 1 / 14D (Methods)

Substitute $y(x)={\sum }_{j\ge 0}{c}_j{x}^j$ into the Legendre differential equation. Matching the coefficient of $x^j$ givesChoose the parity of $n$: start with nonzero $c_0$ for even $n$, or nonzero $c_1$ for odd $n$, and set the other parity to zero. The numerator is nonzero for every preceding index of that parity below $n$, so $c_n\ne 0$, but at $j=n$ it vanishes. All higher coefficients are zero. This constructs a nonzero polynomial solution of degree exactly $n$.

Its value at $x=1$ cannot be zero. If it had a zero there of multiplicity $k\ge 1$, write $y=(x-1{)}^{k}R$ with $R(1)\ne 0$. The lowest-order term of $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }$ would be $-2k^{2}R(1)(x-1{)}^{k-1}$, whereas $n(n+1)y$ starts at order $k$. The differential equation could not hold. Hence the polynomial can be scaled to satisfy $P_n(1)=1$.

The resulting Legendre polynomials beginEach satisfies both the differential equation and the specified normalization.

Set $U=R(r)\Theta (\theta )$ in the Laplace equation, multiply by $r^2/U$, and separate:Putting $x=\cos \theta$ turns the angular equation into the Legendre differential equation. For a solution smooth on the whole spherical axis, the allowed eigenvalues are $\lambda =n(n+1)$ with angular functions $P_n(\cos \theta )$. The radial equation has the two solutions $r^n,r^{-n-1}$. Superposition yields the regular axisymmetric expansionThe coefficients and convergence region are set by the domain and boundary data. Without regularity at the poles, singular Legendre functions and more general separated degrees would also be possible; they are excluded for this exterior spherical field.

The far-field matching fixes $A_1=v_0$ and all other growing coefficients to zero. The radial Neumann boundary condition at $r_0$ isIndependence of the angular modes gives $B_1=v_0{r}_0^3/2$ and all other decaying coefficients zero. Thus the exterior harmonic potential with no flux through a sphere isIts derivative vanishes on the sphere and its difference from $v_{0}r\cos \theta$ tends to zero at infinity. A spatial constant may be added if the far-field condition is intended only to specify the gradient or the leading growing term; matching the potential difference to zero fixes that constant to zero.