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A missing explicit dependence produces a first integral of the Euler-Lagrange equation. The two cases follow by differentiating the candidate conserved quantity, as shown in the child solutions.

For a variation $y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$, the first variation of the functional isThe boundary term vanishes by the fixed endpoints. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation

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 normalized Legendre polynomials, $P_n(1)=1$ and $P_n(-1)=(-1{)}^n$. Near the endpoints the integral factor has the asymptoticsandThus an independent solution, for which $C_2\ne 0$, has the logarithmic endpoint singularitiesThe constants are understood for a solution continued consistently across the ordinary interior points. Adding any multiple of $P_n$ cannot cancel these logarithms. This explains why boundedness at an endpoint selects the Legendre polynomial and excludes the independent Legendre function of the second kind.

Two integrations give the general reduction of order formulaThe integral is local on intervals avoiding zeros of $P_n$. A pole of the integral at an interior polynomial zero need not be a singularity of $y$: the product extends by reduction of order across a zero of the known solution. After a standard normalization the independent second solution is the Legendre function of the second kind $Q_n$.

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 $x\ne \pm 1$.

Summing the geometric series givesThere is a particularly short verification using the product rule:Also $y(0)=1$ and $y^{\prime }(0)=0$. The radius of convergence of the series is one, set by the three complex cube roots of unity; the closed expression continues the solution beyond that disk wherever its denominator is nonzero. The real initial-value solution extends on the interval $(-\infty ,1)$.

Set $y={\sum }_{j\ge 0}{a}_j{x}^j$ and equate coefficients in the power series. The constant coefficient gives $a_2=0$. For $m\ge 1$, the coefficient of $x^m$ givesThe square bracket is $(m+1)(m+2)$, so the recurrence relation is $a_{j+3}=a_j$ for $j\ge 0$. The initial conditions give $a_0=1$, $a_1=0$, $a_2=0$. Hence

Use the Wirtinger derivatives ${\partial }_z=({\partial }_x-i{\partial }_y)/2$ and ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$, for which $\Delta =4{\partial }_z{\partial }_{\bar{z}}$. Since $H$ and $G$ are holomorphic,The Laplacian has real coefficients, so taking real and imaginary parts provesThis is the complex analytic representation of biharmonic functions; $F$ itself is generally not holomorphic because of its $\bar{z}$ factor.

On the specified ellipse, the semiaxes are $\sqrt{2}$ and $1$, so its elliptic coordinate is $u_0=\mathrm{arcosh}\sqrt{2}=\log (\sqrt{2}+1)$. Thus in the exterior,is a harmonic function with a homogeneous Dirichlet boundary condition on that ellipse. Choose the square root with $\sqrt{z^2-1}\sim z$ at infinity. Although a global complex logarithm requires a cut, its real part $\log |z+\sqrt{z^2-1}|$ is well defined throughout the exterior and gives the same $\varphi$.

Expanding at large $|z|$ yields $z+\sqrt{z^2-1}=2z[1-1/(4z^2)+O(|z{|}^{-4})]$. ThereforeIn particular the potential grows logarithmically in the distance, in every direction at infinity. It is the exterior solution; no harmonic continuation through the focal segment in the ellipse's interior is claimed.

Inverting $u+iv=\mathrm{arcosh}z$ gives $z=\cosh (u+iv)$, hence the elliptic coordinatesEliminating $v$ at fixed $u>0$ gives the ellipseIts focal distance is $\sqrt{{\cosh }^{2}u-{\sinh }^{2}u}=1$. At fixed $v$, eliminating $u$ instead gives the hyperbolaIts focal distance is also one, so both families have foci $(\pm 1,0)$. A fixed $v$ traces an appropriate half of the hyperbola; complementary coordinate values complete it. Degenerate values give the focal segment and portions of the coordinate axes. Away from these degeneracies the curves are orthogonal by part (i), or by ${\mathbf{r}}_u\cdot {\mathbf{r}}_v=0$ in the displayed parametrization. A branch of the inverse is used locally; its real part with $u\ge 0$ is single-valued outside the focal segment.

Differentiate the Cauchy-Riemann equations and use equality of mixed partial derivatives:Thus the real part is a harmonic function. The same calculation gives $\Delta v=0$. Every holomorphic function has the required smoothness.

For a holomorphic function, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. ConsequentlyA gradient is normal to its level set, so these identities give orthogonal level curves wherever the gradients are nonzero. At a critical point where $f^{\prime }=0$, both gradients vanish and a regular tangent to a level set need not exist; the geometric orthogonality statement is restricted to regular intersections.

Collect the paired orthonormal bases from part (iv), followed by an orthonormal basis of $\ker M^2=\ker M$. The matrix $Q$ of these columns is an orthogonal matrix, andIndeed, each pair obeys $M{e}_j={\mu }_j{f}_j$, $M{f}_j=-{\mu }_j{e}_j$, and all kernel vectors are annihilated. This proves the real block form of a skew-symmetric matrix directly, including repeated block parameters. In particular its rank is $2r$, always even.

Normalize $e_1$ and put $f_1=M{e}_1/{\mu }_1$. The two vectors are an orthonormal basis of a plane $E$ invariant under $M$, since $M{e}_1={\mu }_1{f}_1$ and $M{f}_1=-{\mu }_1{e}_1$. Its orthogonal complement is also invariant: for $w\perp E$ and $z\in E$,Apply the spectral theorem for real symmetric matrices to the restriction of $M^2$ to $E^{\perp }$. Any remaining negative eigenvalue supplies another orthonormal pair by the same construction. Induction leaves only the kernel. Thus every nonzero eigenvalue of $M^2$ has even multiplicity. The pairs can share an eigenvalue, so this does not assert distinctness of the different $-{\mu }_j^2$.

Choose $e_1$ real, as permitted by the spectral theorem for real symmetric matrices. Matrix powers commute, soThe proposed partner $e_2=M{e}_1$ is nonzero becauseAlso the real scalar $e_1^{T}M{e}_1$ equals its own transpose $e_1^T{M}^T{e}_1=-e_1^{T}M{e}_1$, so it vanishes. Therefore $e_1$ and $e_2$ are orthogonal eigenvectors with the same eigenvalue. This is the basic pairing in the paired eigenspaces of the square of a real skew-symmetric matrix.

Since $M$ is real, $(M^2{)}^T=(M^T{)}^2=(-M{)}^2=M^2$. For every real vector $x$,Thus $M^2$ is a real symmetric matrix and is negative semidefinite. When $n$ is odd, part (i) supplies a nonzero $x$ in the kernel of $M$, and then $M^{2}x=0$. Hence zero is an eigenvalue of $M^2$. In fact $\ker M^2=\ker M$, as the displayed norm identity shows.

For a skew-symmetric matrix, $M^T=-M$. Invariance of the determinant under transpose givesFor odd $n$, $2\det M=0$, so $\det M=0$ over the real or complex numbers.