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This subpart is also present in the original PDF but missing from the local TeX. Put $p(x)=1-x^2$ and ${\lambda }_n=n(n+1)$. Multiply the Legendre differential equation for $P_n$ by $P_m$, multiply the equation for $P_m$ by $P_n$, and subtract. The product rule givesIntegrating from $-1$ to $1$, the boundary term vanishes: $p(\pm 1)=0$, and the Legendre polynomials and their derivatives are finite there. For distinct nonnegative integers $m,n$, ${\lambda }_m\ne {\lambda }_n$. Thus the Orthogonality of Legendre polynomials isThe vanishing coefficient at the endpoints causes no difficulty here because the solutions are polynomials.

Differentiate the Legendre differential equation and put $Q_n=P_n^{\prime }$. This givesMultiplication by $1-x^2$ puts it in the self-adjoint formThe boundary term vanishes at $x=\pm 1$. Distinct eigenvalues are therefore orthogonal with weight $1-x^2$:

Direct differentiation of $v=(x^2-1{)}^l$ gives $(x^2-1)v^{\prime }=2lxv$; the identity also holds for $l=0$. Differentiate it $l+1$ times using the Leibniz rule. The quadratic factor has only its first two derivatives nonzero, soWriting $y=v^{(l)}$ and collecting terms gives $(x^2-1)y^{\prime \prime }+2x{y}^{\prime }-l(l+1)y=0$. Multiplication by $-1$ gives the Legendre differential equation:

For a vacuum electric potential, put $\Phi =R(r)Y(\theta )$ in Laplace's equation in spherical coordinates. Multiplication by $r^2/(RY)$ separates the radial and angular terms. With separation constant $\ell (\ell +1)$ and $x=\cos \theta$, the angular Legendre differential equation isRegularity at both polar axes selects nonnegative integers $\ell$ and the Legendre polynomials. The radial equation is $(r^2{R}^{\prime }{)}^{\prime }-\ell (\ell +1)R=0$, whose independent solutions are $r^{\ell }$ and $r^{-\ell -1}$. Hence the angularly regular axisymmetric vacuum solution isThe radial domain and its boundary conditions determine which terms are admissible.

For the given charge density, only $P_1(\cos \theta )=\cos \theta$ is forced. Write $\Phi =f(r)\cos \theta$. The interior radial Poisson equation is $f^{\prime \prime }+2f^{\prime }/r-2f/r^2=-4$. The identity for the Poisson equation with a spherical harmonic source gives the particular solution $f_p=-r^2$, since the radial operator takes $r^2$ to four. Finiteness at zero and decay at infinity therefore giveContinuity of the electric potential and its radial derivative at $a$ requires $Aa-a^2=B/a^2$ and $A-2a=-2B/a^3$. Solving gives $A=4a/3$, $B=a^4/3$, soAll other angular modes have zero coefficients after imposing regularity, matching and decay. At the origin the interior expression is $(4a/3)z-rz$, which extends with a continuous gradient and value zero. The angular charge density has no unique value exactly at that point; Poisson's equation holds classically for $r>0$ and as a weak solution through the origin. There is no additional point charge.

Termwise differentiation of the convergent Frobenius series and collection of powers in the Bessel differential equation gives the lowest coefficient $({\sigma }^2-{\nu }^2)a_0$, the next coefficient $[(\sigma +1{)}^2-{\nu }^2]a_1$, and, for $n\ge 2$, the coefficient $[(\sigma +n{)}^2-{\nu }^2]a_n+a_{n-2}$. Since $a_0\ne 0$, the stated choices $\sigma =\pm \nu$, $a_1=0$ and the recurrence make all coefficients vanish.

For $\nu =1/2$, the two resulting even-coefficient series areandThese are linearly independent on $x>0$ and give the Bessel functions of order one half, up to normalization. The specified condition $a_1=0$ is sufficient; in the negative-exponent branch at this particular order the next coefficient equation alone would allow $a_1$ to be free, adding a multiple of the positive-exponent solution.

For the Legendre differential equation, substitution gives the full conditionsThus $\sigma =0$ permits arbitrary $a_1$, while $\sigma =1$ requires $a_1=0$. This Frobenius recurrence for the Legendre equation separates the even and odd series. The nearest singular points of the differential equation are $\pm 1$, so the series about zero has radius at least one, and the generic nonterminating solutions have radius exactly one. Particular cancellations can give a larger radius.

Termination occurs when the degree reaches a nonnegative integer $N$ with $\lambda (\lambda +1)=N(N+1)$. For $\lambda \ge 0$ this givesChoose the even branch for even $N$ or the odd branch for odd $N$ to obtain a Legendre polynomial. The recurrence then terminates and gives a solution for all real $x$, with infinite radius of convergence. A nonzero entire solution at other $\lambda \ge 0$ is impossible: each nonterminating parity has coefficient ratio tending to one in the variable $x^2$, and so radius one.

Insert the power series and equate the coefficient of $x^n$. In the Legendre differential equation this givesHence the recurrence relation isThe free coefficients $a_0,a_1$ generate the independent even and odd series, respectively. They converge at least for $|x|<1$, up to the nearest singular points of the equation.

Let $W=P_{\ell }{Q}_{\ell }^{\prime }-P_{\ell }^{\prime }{Q}_{\ell }$. Differentiation cancels the cross terms, and the Legendre differential equation givesThus $[(1-x^2)W{]}^{\prime }=0$, sowith $A_{\ell }$ independent of $x$ on each interval avoiding the singular endpoints. This is the Abel identity. Linear independence forces $A_{\ell }\ne 0$, and its value depends on the chosen normalization of $Q_{\ell }$.

Substitute $y=v{P}_n$ into the Legendre differential equation and expand using the product rule. The terms proportional to $v$ cancel because $P_n$ is itself a solution. What remains isMultiplying by $P_n$ identifies an exact derivative:The latter form is convenient for reduction of order on intervals where $P_n\ne 0$ and $x\ne \pm 1$.

Write $P_n(x)\sim c_n{x}^n$ with $c_n\ne 0$. For $n\ge 1$, this branch grows. Choosing the other reduction of order solution by an integral from infinity gives, on $x>1$,Thus for $n\ge 1$, every bounded exterior solution decays as $|x{|}^{-n-1}$, up to a constant factor, on each exterior interval. Equivalently the dominant-balance exponents in the Legendre differential equation are $n$ and $-n-1$.

There is an important exception at degree zero. Since $P_0=1$,These solutions are bounded even when $C_1\ne 0$; boundedness alone does not then imply decay to zero. The constants on $x>1$ and $x<-1$ are independent unless continuation across the singular endpoints is specified.

For a separated axisymmetric harmonic function $\Psi =R(r)\Theta (\theta )$, multiplying the Laplace equation by $r^2/(R\Theta )$ givesThe two sides depend on different variables, hence both equal a constant. Put $u=\cos \theta$ in the angular equation. Since $\sin \theta {\Theta }^{\prime }=-{\sin }^2\theta d\Theta /du$, it becomes the Legendre differential equationRegularity on both parts of the polar axis restricts $\lambda =\ell (\ell +1)$, with $\ell =0,1,\dots$, and selects the Legendre polynomial $P_{\ell }(u)$. The other angular solution is singular at an endpoint.

The radial Euler differential equation isTrying $R=r^s$ gives $s(s+1)=\ell (\ell +1)$, so $s=\ell$ or $s=-\ell -1$. Superposition of these separated harmonic functions givesThe Legendre polynomials form the angular expansion for regular data with rotational symmetry about the polar axis; the radial domain and the boundary conditions determine which coefficients survive and where the expansion converges.

Differentiate the power series term by term inside its radius of convergence, and reindex the second derivative so that every coefficient multiplies $x^m$. The differential equation becomesThe uniqueness of power-series coefficients givesThe even coefficients depend only on $a_0$ and the odd coefficients only on $a_1$. These are the two local solution branches of the Legendre differential equation around its ordinary point $x=0$.

Seek a nonzero separated solution $\psi =T(t)R(r)\Theta (\theta )$. Substitution into the heat equation and division by the product givesThe left side depends only on time, so set it to $-k^2$. Multiplication by $r^2$ then separates the angular dependence, with separation constant $\ell (\ell +1)$. The three ordinary differential equations areThe temporal solution is $T=T_0{e}^{-\kappa k^{2}t}$. With $q=\cos \theta$, the angular equation is the Legendre differential equation; regularity at both angular endpoints selects $\ell =0,1,2,\dots$ and $\Theta \propto P_{\ell }(q)$, the Legendre polynomial. The radial equation has the Spherical Bessel functions $j_{\ell }(kr)$ and $y_{\ell }(kr)$ for $k>0$. Regularity at $r=0$, if imposed, excludes the singular second solution. For $k=0$ the radial powers are $r^{\ell }$ and $r^{-\ell -1}$. Without specified radial boundary conditions, no discrete set of $k$ is selected; $-k^2$ is a convention for the temporal separation constant.

For an axisymmetric harmonic function, put $\Psi =R(r)\Theta (\theta )$ in the Laplacian in spherical coordinates and multiply by $r^2/(R\Theta )$. Separation of variables givesWith $u=\cos \theta$, the angular equation becomes the Legendre differential equation. Regularity at both poles selects the Legendre polynomials $P_{\ell }(u)$ with integer $\ell \ge 0$; the singular second solutions are excluded. The radial Euler-Cauchy equation has solutions $r^{\ell }$ and $r^{-\ell -1}$. Superposition therefore giveswith coefficients chosen to ensure convergence and the required boundary regularity. The axisymmetric harmonic extension from a sphere gives these coefficients directly from boundary data.

The integral representation $y_2=y_1\int e^{-\int p}/y_1^2$ is derived where $y_1\ne 0$. At a zero, its integral factor may have a pole while the product has a removable singularity. For regular coefficients, continue the actual second solution by its finite initial data and initial-value uniqueness; the Abel identity preserves a nonzero Wronskian. In the degree-one Legendre differential equation, $y_1=x$ and $y_2=x\mathrm{arctanh}x-1$ illustrate this cancellation at zero.

This instance of Rodrigues' formula constructs the Legendre polynomial for every nonnegative integer $l$. Put $v=(x^2-1{)}^l$; then $(x^2-1)v^{\prime }=2lxv$. Differentiating $l+1$ times by the Leibniz rule gives $(x^2-1)y^{\prime \prime }+2x{y}^{\prime }-l(l+1)y=0$ for $y=v^{(l)}$, which is the Legendre differential equation after reversing signs. The leading term of the normalized derivative is $(2l)!x^l/[2^l(l!{)}^2]$, proving its polynomial degree is exactly $l$. At $x=1$, the product $(x+1{)}^l(x-1{)}^l$ contributes only the term with all $l$ derivatives on the second factor, giving $v^{(l)}(1)=2^{l}l!$. Thus the normalization is $P_l(1)=1$.

Differentiating the Legendre differential equation gives $((1-x^2{)}^2{R}_n^{\prime }{)}^{\prime }+(n-1)(n+2)(1-x^2)R_n=0$ for $R_n=P_n^{\prime }$. Multiply the equations for two distinct indices by the opposite polynomial and subtract. The endpoint term vanishes because of $(1-x^2{)}^2$, giving the displayed weighted orthogonality. The zero polynomial $R_0$ is not a nonzero eigenfunction.