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At $t=0$, the initial conditions implyDifferentiate the first relation and solve for the two derivatives:Fix a reference point $x_{\ast }$ and set $V(x)={\int }_{x_{\ast }}^{x}v(s)ds$. Taking antiderivatives givesThe constants must be opposite to preserve the first initial condition; the arbitrary constant of integration $K$ does not change $\psi$. Substitution yields D'Alembert's formula:The fundamental theorem of calculus verifies that its time derivative at zero is $v(x)$, and its value at zero is $u(x)$. The oriented integral also makes the formula valid for negative $t$.

Expand the radial modified Helmholtz equation as $Ly=-y^{\prime \prime }-2y^{\prime }/x+y$. Setting $y=z/x$ gives $Ly=(-z^{\prime \prime }+z)/x$, soBoundedness at zero requires $A+B=0$, whereas boundedness at infinity requires $A=0$. The respective adapted homogeneous solutions are thereforeThe first has limit one at zero. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-x^{-2}$. Since the highest-derivative coefficient of $L$ is $-1$, the Green function must have first-derivative jump $-1$. The radial modified Helmholtz Green function on the half-line isIt is continuous at the source, and its derivative jump is $-1$, giving $L_{x}G=\delta (x-\xi )$. For fixed source it tends to $\xi e^{-\xi }$ as $x\downarrow 0$ and decays at infinity. This kernel is for the original operator and ordinary measure $d\xi$; multiplying the equation by $x^2$ would change the source normalization.

For the compact source, split the integral at $\xi =x$ when $x<R$. The needed antiderivatives areThuswhile for $x>R$ the result is $e^{-x}{x}^{-1}{\int }_0^R\xi \sinh \xi d\xi$. Simplifying givesThe continuous value at zero is $1-(R+1)e^{-R}$, and $y^{\prime }(0)=0$. Both $y$ and $y^{\prime }$ match at $R$, so no extra Dirac delta appears there. The only homogeneous solution bounded at both ends is zero, establishing uniqueness.

Differentiate the eigenvalue equation along the assumed smooth normalized eigenbranch:Since $A$ is real symmetric, $x^{T}A=\lambda x^T$. Multiplying the differentiated equation by $x^T$ and using $x^{T}x=1$ cancels the terms containing $x^{\prime }$:Thus the finite-dimensional Hellmann–Feynman theorem givesEquivalently, differentiating the normalization gives $2x^T{x}^{\prime }=0$ and eliminates those terms from the derivative of $\lambda =x^{T}Ax$. The proof needs the smooth eigenbranch assumed in the question; the smallest eigenvalue need not be differentiable at an arbitrary crossing without that assumption.

Evaluate the derivative along real and imaginary increments. For a holomorphic function, both give the same value:Equating real and imaginary parts yields the Cauchy-Riemann equations

In monic form the Chebyshev differential equation has $P(x)=-x/(1-x^2)$ and $Q(x)=p^2/(1-x^2)$. Both are analytic near zero, so it is an ordinary point. At $x=1$, $(x-1)P=x/(x+1)$ and $(x-1{)}^{2}Q=-p^2(x-1)/(x+1)$ are analytic; at $x=-1$, $(x+1)P=x/(x-1)$ and $(x+1{)}^{2}Q=-p^2(x+1)/(x-1)$ are analytic. Thus both endpoints are regular singular points.

Insert $y_q={\sum }_{n\ge 0}{a}_n{x}^{n+q}$ and equate coefficients. The recurrence relation isExplicitly the factors are $(n^2-p^2)/[(n+2)(n+1)]$ for $q=0$ and $[(n+1{)}^2-p^2]/[(n+3)(n+2)]$ for $q=1$. The lowest equation for $q=1$ forces $a_1=0$. For $q=0$, one may choose $a_1=0$ to select an even solution, independent of the odd $q=1$ solution. The leading coefficient $a_0$ is free in either chosen parity branch.

For a nonterminating branch the consecutive parity-term ratio tends to $x^2$. Hence the ratio test gives convergence for $|x|<1$, and radius of convergence one. Terminating branches are polynomials with infinite radius. For $p=m\ge 0$ an integer, choose $q$ of the same parity as $m$: the recurrence stops once $n+q=m$, producing a degree-$m$ Chebyshev polynomial. Normalization at $1$ gives all nonzero coefficients through degree three:

Set $Y(\theta )=y(\cos \theta )$. Direct differentiation gives $Y^{\prime \prime }={\sin }^2\theta y^{\prime \prime }-\cos \theta y^{\prime }$, so the equation becomes $Y^{\prime \prime }+m^{2}Y=0$. For $m>0$, $Y=A\cos (m\theta )+B\sin (m\theta )$. A polynomial $y$ has finite derivative at $x=1$, and therefore $Y^{\prime }(0)=-\sin 0y^{\prime }(1)=0$, forcing $B=0$. Also $Y(0)=T_m(1)=1$, so $A=1$. For $m=0$, the transformed solution is affine in $\theta$ and the same derivative condition removes its linear term. ThusThe endpoint values follow by continuity. The finite-derivative condition is what excludes the independent nonpolynomial branch, rather than normalization alone.

The appropriate weighted inner product and Sturm-Liouville operator areFor two functions obeying the endpoint conditions, integration by parts gives the Lagrange identity for a Sturm-Liouville operatorAt $R$, both $u$ and $v$ vanish. At $1$, the factor $x(x-1)$ tends to zero while the functions and their first derivatives remain bounded. Thus the boundary form vanishes at both ends.

This gives the regular self-adjoint differential operator realization: its singular-end condition is ${\lim }_{x\downarrow 1}x(x-1)v^{\prime }=0$, together with $v(R)=0$. The bounded classical eigenfunctions satisfy exactly this no-log choice. More precisely the operator domain is the maximal differential-operator domain in the weighted space with these separated endpoint conditions; the preceding calculation verifies its self-adjoint boundary form, rather than merely the differential expression's formal symmetry.

Multiplying by $x^2$ makes the first two terms a total derivative:The resulting Sturm-Liouville form isEquivalently, an integrating multiplier $\mu$ must satisfy $(\mu (1-1/x){)}^{\prime }=\mu (2/x-1/x^2)$, which reduces to ${\mu }^{\prime }/\mu =2/x$ and gives $\mu =x^2$ up to a constant. The coefficient $p$ is positive for $1<x<R$ and vanishes linearly at the left endpoint; the weight $w$ is positive throughout. The boundedness condition removes the logarithmic local branch, giving the no-log boundary condition at a linearly vanishing Sturm-Liouville endpoint.

For real eigenfunctions $y_i,y_j$, multiply their Sturm-Liouville problems crosswise and subtract. Integration givesAt $R$, both eigenfunctions vanish, while at $1$ their bounded values and bounded first derivatives multiply a factor tending to zero. The right side is zero. Distinct eigenvalues therefore give the orthogonality of Sturm-Liouville eigenfunctionsFor complex eigenfunctions, replace $y_i$ by $\overline{y_i}$, in accordance with the weighted inner product.

For two real eigenfunctions, multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The $q$ terms cancel. The resulting Lagrange identity for a Sturm-Liouville operator isIntegrating first over $[ϵ,R]$ and then taking the endpoint limits givesThe needed assumptions are that the improper integral exists and that the displayed boundary expression tends to zero at both endpoints. Bounded $y_m,y_n$ and their first derivatives are sufficient because $p\to 0$, but are stronger than necessary. Merely knowing that $p\to 0$ does not control arbitrarily growing eigenfunctions. With the vanishing boundary form and distinct eigenvalues, division by ${\lambda }_n-{\lambda }_m$ proves orthogonality of Sturm-Liouville eigenfunctions:For complex eigenfunctions in a real self-adjoint differential operator problem, use $\overline{y_m}$ instead, giving the corresponding weighted inner product.

Multiplying the particular equation by $e^{-x}$ puts it in Sturm-Liouville form:ConsequentlyIts polynomial eigenfunctions are the Laguerre polynomials. Products of these polynomials, with or without one derivative, grow only algebraically at infinity. The exponential in $p$ therefore makes the boundary form vanish there; its factor $x$ makes it vanish at zero. Their orthogonality is thus

For $\lambda =1$, substitution of $Ax+B$ gives $A+B=0$. For $\lambda =2$, substitution of $A{x}^2+Bx+C$ gives $4A+B=0$ and $B+2C=0$. Choosing $y_n(0)=1$, which fixes the otherwise arbitrary nonzero multiplicative constants, yieldsTheir product is $1-3x+\frac{5}{2}{x}^2-\frac{1}{2}{x}^3$. Using the Gamma function identity ${\int }_0^{\infty }{x}^k{e}^{-x}dx=k!$, also obtainable by repeated integration by parts, verifies the required orthogonality directly:

For the insulating wall, the image instead has the same strength $q$. With the distances defined above,It has the required source in the physical domain. Its normal derivative isAt $x=b$ the terms are $b/(b^2+y^2)$ and $-b/(b^2+y^2)$, so they cancel. The Neumann boundary condition leaves the additive constant $C$ free.

The derivative of a holomorphic function can be evaluated using an increment along the real axis:Identifying the components of the gradient $\mathbf{V}=\nabla u$ gives

Use the boundary-adapted homogeneous solutionsFor a source point $\xi \ne 0$, the Green function isIt vanishes at both endpoints and is continuous at $x=\xi$. Its first derivative has the jumpas required for a monic second-order differential operator with a Dirac delta source. Away from that source it is a homogeneous solution and extends smoothly through $x=0$.

The interior singular point requires a qualification. This is a Green kernel with a vanishing Wronskian at an interior singular point, with a $1/\xi$ singularity in its source variable. It is not an ordinary integral inverse for unrestricted forcing. For smooth forcing $f(\xi )=\xi q(\xi )$, however, the product $Gf$ extends continuously at zero and the integral is well defined. This includes part (iii). We use solutions with two continuous derivatives across zero, interpreting the equation there after multiplication by $x$. For such a solution, expansion at zero forces $f(0)=0$; arbitrary nonzero forcing at zero cannot satisfy this regularity. Merely imposing one continuous derivative would leave a piecewise homogeneous ambiguity, so the original singular coefficients alone do not specify an unrestricted boundary-value inverse.

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 $x\ne \pm 1$.

The boundary conditions on the positive half-line, absent from the transcription, select the half-harmonic oscillator. The trial state $x{e}^{-\alpha x}$ with $\alpha >0$ satisfies both endpoints and has finite norm and energy. Integration by parts, with vanishing endpoint term, givesUsing ${\int }_0^{\infty }{x}^n{e}^{-2\alpha x}dx=n!/(2\alpha {)}^{n+1}$, the three required integrals areThus the exponential trial state for a half-harmonic oscillator gives $\Lambda (\alpha )={\alpha }^2+3/{\alpha }^2$. Its derivative vanishes at ${\alpha }^4=3$, and the arithmetic-geometric mean inequality confirms the global minimum:For comparison, the exact lowest half-line state is proportional to $x{e}^{-x^2/2}$ and has eigenvalue three. The exponential trial family gives a valid upper bound rather than an exact eigenvalue.

Use the full field in the PDF, including the exponential term missing from the transcription. Its divergence isThe exponential in the third component has zero $z$ derivative. In cylindrical coordinates, the solid has $0\le r\le 2$ and $0\le z\le 4-r^2$. Therefore the divergence theorem gives the outward fluxThe integration includes the entire closed boundary, both cap and base.

Work in a Cartesian coordinate system, write ${\partial }_j=\partial /\partial x_j$, and use the Einstein summation convention. For a twice continuously differentiable vector field, the contraction of two Levi-Civita symbols givesHere the Kronecker deltas select components, and equality of mixed partial derivatives gives the curl of the curl identityFor the second identity, the same contraction followed by the product rule yieldsThus the curl of a cross product isThe directional derivatives in the last two terms act on the vector field immediately to their right.

Kernel conditions. The clamped-beam Green function must obey $G(0,\xi )=G_x(0,\xi )=G(1,\xi )=G_x(1,\xi )=0$. Away from $x=\xi$, $G_{xxxx}=0$. At the load point, $G,G_x,G_{xx}$ are continuous andA jump in any lower derivative would create a derivative of the Dirac delta, rather than the single point load required by $D_x^{4}G=\delta (x-\xi )$.

Construction. A particularly simple representation isIt already enforces the left-end conditions and the correct source jump. The right-end conditions give $a+b=-(1-\xi {)}^3/6$ and $2a+3b=-(1-\xi {)}^2/2$, henceFor $x<\xi$ this is already factored; for $x>\xi$, expanding and refactoring givesThe formulas agree at the junction. The forced displacement is obtained by integrating this kernel against $f$; for continuous loads it satisfies the equation and all four end conditions. Uniqueness follows because a cubic with zero value and slope at both ends is identically zero.

Reciprocity. In terms of $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$, the same expression isIt is visibly symmetric under interchange of $x$ and $\xi$. Thus the reciprocity of the clamped-beam Green function givesEquivalently, two integrations by parts show self-adjointness of $D_x^4$ under the clamped conditions, with bilinear form $\int u^{\prime \prime }{v}^{\prime \prime }dx$.