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An adjoint Green's function solves the corresponding adjoint point-source problem, with boundary or terminal conditions chosen so that integration by parts gives a representation for a solution of the original operator. For $L={\partial }_t-{\partial }_x^2+c$ on the real line, the formal adjoint is $L^{\ast }=-{\partial }_t-{\partial }_x^2+c$. The backward kernel with terminal Dirac delta at $(\xi ,\tau )$ is the translated heat kernel with constant absorption, defined for $t<\tau$. Its pairing with a homogeneous forward solution is constant in $t$; taking the terminal limit evaluates that solution at $(\xi ,\tau )$.

For $L=d^2/d{x}^2-2/x^2$ on $x>0$, the homogeneous solutions $x^2,x^{-1}$ satisfy respectively the zero and infinity boundary conditions. Their Wronskian is $-3$. The displayed Green's function is continuous at $x=\xi$ and its first derivative has jump one, so $L_{x}G=\delta (x-\xi )$ in the distribution sense. Integration against a compactly supported forcing gives the unique solution decaying at both boundaries.

For $Ly=(x^2{y}^{\prime }{)}^{\prime }-2y$ on $x>0$, choose solutions bounded at zero and infinity. The homogeneous Cauchy-Euler equation has basis $x,x^{-2}$. Joining the left-regular and right-regular solutions continuously and imposing $s^2[G_x{]}_{s^{-}}^{s^{+}}=1$ gives the displayed Green's function with ordinary integration measure $ds$. For bounded forcing,The integrals converge and give the unique bounded solution: no nonzero homogeneous combination is bounded at both ends. If $0\le f\le M$, negativity of the kernel gives $-M/2\le y\le 0$. Differentiation cancels the two moving-endpoint terms, leavingwhich implies $|y^{\prime }(x)|\le M/(3x)$. For constant forcing $f=M$, the bounded solution is $-M/2$; the endpoint conditions are boundedness, rather than zero endpoint values.

Let $0<a<b$ and impose zero Dirichlet boundary conditions at $a,b$. With the convention $L_{x}G(x,s)=\delta (x-s)$ for the Dirac delta distribution, the Green's function is continuous and its first derivative has jump one. For $n\ne 0$ choose $\ell (x)=x^{-n}-a^{-n}$ and $r(x)=x^{-n}-b^{-n}$. Their Wronskian is $n(a^{-n}-b^{-n})s^{-n-1}$, soFor $n=0$, take the continuous limit or use $\ell =\log (x/a)$ and $r=\log (x/b)$ to obtainThe solution is $u(x)={\int }_a^{b}G(x,s)f(s)ds$. These kernels use ordinary measure $ds$. The factor $s^{n+1}$ comes from the nonconstant Wronskian and must not be discarded by confusing the original operator with its multiplied divergence form $(x^{n+1}{u}^{\prime }{)}^{\prime }$.

A continuous forcing gives the solution with zero initial data $z(x)=-{\int }_0^x\sin (x-s)f(s)ds$. Every solution vanishing at zero is $A\sin x+z(x)$. If $\sin L\ne 0$, the remaining endpoint determines $A$ uniquely, with Green's functionIts first derivative has jump $-1$, the correct sign for $-y^{\prime \prime }-y=f$. If $L=m\pi$ with a positive integer $m$, solvability instead requires ${\int }_0^{L}f(s)\sin sds=0$, and the coefficient $A$ is free. This is the Fredholm alternative with a one-dimensional homogeneous kernel.

For $L=a_2(x)D^2+a_1(x)D+a_0(x)$ with $a_2(\xi )\ne 0$, a Green's function solving $L_{x}G=\delta (x-\xi )$ is homogeneous on either side of $\xi$, continuous there, and has first derivative jump $[G_x{]}_{-}^{+}=1/a_2(\xi )$. A nonzero jump in $G$ itself would produce an unwanted distributional derivative of the Dirac delta. Once continuity removes that term, integrating the equation across $\xi$ fixes the derivative jump. Endpoint boundary conditions complete the classical characterization.

Suppose a linear ordinary differential operator satisfies $L(qv)=rMv$ on an interval, where the functions $q,r$ never vanish and homogeneous endpoint boundary conditions for $y=qv$ correspond to those used for $v$. If $G_M$ is a Green's function for $M$, then $Ly=f$ is equivalent to $Mv=f/r$, soThus the displayed kernel is the Green's function for the original, possibly non-normalized operator. The source factor belongs at $s$, while the dependent-variable factor belongs at $x$.

For $Ly=x^2{y}^{\prime \prime }-2x{y}^{\prime }+(x^2+2)y$ on $[\pi /2,\pi ]$ with zero endpoint values, $q=x$ and $r=x^3$ reduce the equation to $Mv=v^{\prime \prime }+v$. The kernel of $M$ is $\cos (x_{<})\sin (x_{>})$, where $x_{<}=\min (x,s)$ and $x_{>}=\max (x,s)$. Therefore $G_L(x,s)=x\cos (x_{<})\sin (x_{>})/s^3$. Its first derivative has jump $1/s^2$, as required by the leading coefficient $s^2$ of $L$.

The homogeneous Cauchy-Euler equation $Ly=0$ has fundamental solutions $x,x^{-1}$ on $x>0$, with Wronskian $-2/x$. Requiring decay at both endpoints gives the unweighted Green's functionRequiring zero initial value and derivative at zero instead gives the forward kernelBoth kernels are continuous and have first-derivative jump one at $x=\xi$, so $L_{x}G=\delta (x-\xi )$. Away from that point they solve the homogeneous equation. Integrating against the forcing with ordinary measure $d\xi$ yields the respective solutions whenever the integral and endpoint limits converge. Their difference is $G_0-G_D=x/2$, independent of $\xi$, so the solutions differ by the homogeneous term $(x/2)\int f(\xi )d\xi$ for integrable forcing. For example, the two choices for an indicator forcing on $(0,1)$ differ by $x/2$. Specifying a differential operator without its endpoint conditions does not specify its Green kernel.

Let ${\delta }_c=C_g{\tau }^2+C_d{\tau }^{-3}$ be the matter-era growing and decaying density modes, with radiation neglected in the background gravity. The radiation density contrast isThe formula follows by the Green's function of the forced harmonic oscillator. For $k\tau \gg 1$, the slowly varying particular response to the decaying mode begins at $16C_d{\tau }^{-5}/{\omega }^2$; lower-endpoint oscillations are absorbed into $C_1,C_2$. The growing mode drives a constant radiation offset, while cold dark matter continues growing as ${\tau }^2$.

Multiplying the equation for $v=y^{\prime }$ by the integrating factor $e^x$ gives $v(x)=-e^{-x}{\int }_x^1{e}^{t}f(t)dt$. Integrating once more and reversing the order of integration gives the Green's functionIt is continuous across $x=t$, its first derivative has jump $+1$, and it satisfies the mixed Dirichlet boundary condition and Neumann boundary condition. Both homogeneous solutions, $1$ and $e^{-x}$, are eliminated by these two conditions, giving uniqueness. For $f=e^{-x}$ the solution is $-x{e}^{-x}$.

For mass-normalized normal modes, an instantaneous step in a point seismic moment tensor excites mode $\alpha$ with $C_{\alpha }=\mathbf{M}:{\mathbf{e}}_{\alpha }^{\ast }({\mathbf{x}}_s)$ and $\tau =t-t_s$. Convolving the causal oscillator Green's function with the step gives $a_{\alpha }=C_{\alpha }[1-\cos ({\omega }_{\alpha }\tau )]H(\tau )/{\omega }_{\alpha }^2$. The expression contains a static response and undamped oscillations; attenuation requires an extension of the conservative model.

With $y=xv$, direct differentiation givesThe homogeneous ordinary differential equation therefore has general solution $y=x(A\cos x+B\sin x)$. Chooseso $y_1(\pi /2)=0$ and $y_2(\pi )=0$. Their Wronskian in the convention $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ is $W(x)=x^2$. There is no nonzero homogeneous solution satisfying both boundary conditions, since the first forces $B=0$ and the second forces $A=0$.

For the operator as printed, whose leading coefficient is $x^2$, the Green's function isIt satisfies the endpoint boundary conditions, is continuous across $x=s$, and its first derivative has jump $1/s^2$. Multiplying that jump by the leading coefficient $s^2$ produces the unit source required by a Green's function. This factor matters because the operator was not normalized to leading coefficient one; it is an instance of a Green function under dependent-variable rescaling.

Thus $y(x)={\int }_{\pi /2}^{\pi }G(x,s)f(s)ds$. For $f(s)=s^3$, split the integral at $s=x$ to obtainThereforeThe substitution above gives $v^{\prime \prime }+v=1$, so the forcing is indeed $x^3$, and both endpoint values vanish.

If $Au+Bv=0$, evaluation at zero gives $A=0$ and differentiation at zero gives $B=0$. Thus the solutions are linearly independent. For $w=u{v}^{\prime }-u^{\prime }v$,so $\boxed{w(x)=\exp (-{\int }_0^{x}p(s)ds)>0}$.

Variation of parameters sets $y=A(x)u+B(x)v$ with $A^{\prime }u+B^{\prime }v=0$, $A^{\prime }{u}^{\prime }+B^{\prime }{v}^{\prime }=f$. Solving gives $A^{\prime }=-vf/w$ and $B^{\prime }=uf/w$. Zero integration constants enforce both initial conditions. ThereforeFor fixed $t$, this kernel is a linear combination of homogeneous solutions, so $K_{xx}+p{K}_x+qK=0$. On the diagonal $K(t,t)=0$ and $K_x(t,t)=(v^{\prime }(t)u(t)-u^{\prime }(t)v(t))/w(t)=1$. Differentiating the integral twice now verifies the forced equation as well as $y(0)=y^{\prime }(0)=0$. Standard uniqueness of the linear initial-value problem makes this the solution, for continuous $f$.

If $u,v$ have period $T>0$, their derivatives do also, hence $w$ is periodic. The kernel is periodic in its first variable. With $f=v$, split the integral at $x$ and use this periodicity to getAt $x=0$, the right side is $-{\int }_0^{T}v(t{)}^2/w(t)dt<0$, because $w>0$ and $v^{\prime }(0)=1$ ensures $v$ is not identically zero. Thus $\boxed{y(T)\ne y(0)}$, so $y$ cannot have period $T$. This direct sign argument also rules out any apparent cancellation between the two terms.

For the linear differential operator $L=D^2+4D+5$, the characteristic equation is $m^2+4m+5=0$, with roots $-2\pm i$. Take the fundamental system of solutionsThe Wronskian is never zero, so variation of parameters is valid on the whole interval. Write a particular solution as $a(x)u(x)+b(x)v(x)$, and impose the auxiliary condition $a^{\prime }u+b^{\prime }v=0$. Substitution then gives $a^{\prime }{u}^{\prime }+b^{\prime }{v}^{\prime }=f$. These two equations giveChoosing both integrals to start at zero makes the resulting particular solution and its first derivative vanish there:A general solution is $Au+Cv+y_p$. The boundary condition at zero sets $A=0$. The condition at $\ell =\pi /2$ givesbecause $\sin \ell =1$ and $\sin (\ell -t)=\cos t$. Combining the two integrals yields the Green's function representationIndeed, for $t<x$ the kernel is $-e^{2(t-x)}\sin t\cos x$, since $\sin (x-t)-\sin x\cos t=-\cos x\sin t$; for $x<t$ it is $-e^{2(t-x)}\sin x\cos t$.

For fixed interior $t$, the Green's function vanishes at both endpoints, is continuous at $x=t$, solves the homogeneous equation away from $t$, and satisfiesThis derivative jump is the normalization required by the leading coefficient one in $L$. Finally, a homogeneous solution satisfying the first boundary condition is $C{e}^{-2x}\sin x$, and the second boundary condition at $\pi /2$ forces $C=0$. Thus the boundary value problem has exactly one solution for every continuous forcing, justifying the unique inverse represented by this kernel.

For $\sin l\ne 0$, use variation of parameters with the homogeneous solutions $\cos x,\sin x$, whose Wronskian is one. Write $y=A(x)\cos x+B(x)\sin x$ and impose $A^{\prime }\cos x+B^{\prime }\sin x=0$. The differential equation gives $-A^{\prime }\sin x+B^{\prime }\cos x=f$, henceAfter integrating from zero and using the left boundary condition,The right boundary condition determinesso the unique solution isCombining the two integrals gives $y(x)={\int }_0^{l}G(x,t)f(t)dt$, with Green's functionIt is symmetric in $x,t$, as expected from self-adjointness, and has zero endpoint values. At $x=t$ it is continuous, while its first derivative obeys the jump conditionThus $({\partial }_x^2+1)G=\delta (x-t)$, with the positive normalization of the Dirac delta function required for $\mathcal{L}y=f$.

The homogeneous solution $h(x)=e^{-x}\cos x$ already has the prescribed endpoint values. Thus $z=y-h$ has zero boundary conditions and satisfies $\mathcal{L}z=f$. The Green's function givesEquivalently, splitting the integral at $\xi =x$ gives the explicit formulaAny difference between two solutions would be $e^{-x}(C\cos x+D\sin x)$ with zero endpoint values, forcing $C=D=0$. The displayed solution is therefore unique.

Set $a=\pi /2$. To construct the Green's function from a Dirac delta function, require ${\mathcal{L}}_{x}g(x,\xi )=\delta (x-\xi )$ and zero endpoint values. Away from $x=\xi$, the homogeneous solutions satisfying the left and right boundary conditions are respectivelyWrite $g=A(\xi )u(x)$ for $x<\xi$ and $g=B(\xi )v(x)$ for $x>\xi$. A discontinuity in $g$ would generate a derivative of the Dirac delta function, which is absent from the desired equation. Thus $g$ is continuous. Integrating that equation through $\xi$ then gives the jump conditionThe resulting two equations are $Au(\xi )=Bv(\xi )$ and $B{v}^{\prime }(\xi )-A{u}^{\prime }(\xi )=1$. Since the Wronskian $u{v}^{\prime }-u^{\prime }v=-e^{-2\xi }$, they giveBoth endpoint values vanish. At $x=\xi$, subtracting the two first derivatives gives ${\sin }^2\xi +{\cos }^2\xi =1$, verifying the sign and normalization of the Dirac delta function.

At $a=\pi$, the nonzero homogeneous solution $e^{-x}\sin x$ satisfies both zero boundary conditions. Thus the boundary value problem has a nontrivial kernel and cannot have an inverse for arbitrary forcing.

More directly, the adjoint operator has the zero-endpoint homogeneous solution $w(x)=e^x\sin x$. If an ordinary Green's function existed, the integration by parts identity would implyBut $w(\xi )\ne 0$ for $0<\xi <\pi$. This contradiction rules out the Green's function for the unmodified operator. It is the Fredholm alternative obstruction.

Choose the positive causal sign convention $\mathcal{L}={\partial }_y-x{\partial }_x^2$, so the forced equation is $\mathcal{L}u=-x{y}^{-2}{e}^{-x/y}$. With source $(\xi ,\tau )$, a forward Green's function satisfiesIt vanishes at $x=0$, decays at spatial infinity in the chosen bounded evolution class, and has the temporal jump $G(x,{\tau }^{+};\xi ,\tau )=\delta (x-\xi )$. Away from the source it solves the homogeneous equation. Equivalently one may reverse both the Green-function operator and its source sign to use the original printed operator.

To select the physical zero-initial-data solution one also excludes extra corner-source terms of the kind found in (b). With this causal bounded convention the solution representation isA mere pointwise statement of the three boundary limits does not by itself provide that uniqueness convention.

For each $0<\xi <2\pi$, the Green's function is a distribution in $x$ satisfyingEquivalently, $g$ is a classical homogeneous solution on each side of $\xi$, satisfies the same endpoint conditions, and obeys the Green function jump conditionsTo see equivalence, let $[g]$ and $[g_x]$ denote its jumps. The singular part of $L_{x}g$ is $[g]{\delta }^{\prime }(x-\xi )+([g_x]+2[g])\delta (x-\xi )$. Matching a unit Dirac delta distribution with no derivative first forces continuity, then the unit derivative jump. Conversely these conditions produce exactly that distribution. The solution for a regular forcing is $y(x)={\int }_0^{2\pi }g(x,\xi )f(\xi )d\xi$. The homogeneous endpoint problem has only the zero solution, which ensures uniqueness of this kernel.