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Let $h_1,\dots ,h_n$ be a basis of homogeneous solutions of a linear ordinary differential operator $L$ of order $n$. Let $B_1,\dots ,B_n$ be independent homogeneous linear boundary conditions, which may be coupled. Suppose a fundamental solution $G_0(x,\xi )$ satisfies $L_x{G}_0=\delta (x-\xi )$. Define the boundary matrix $M_{ij}=B_i{h}_j$ and $b_i(\xi )=B_i{G}_0(\cdot ,\xi )$. If $M$ is invertible, thenThe added terms are homogeneous solutions, so the Dirac delta source is preserved. Applying the boundary operators gives $b-M{M}^{-1}b=0$. Any two such Green functions differ, for fixed source position, by a homogeneous solution with coefficient vector in the null space of $M$; invertibility therefore proves uniqueness. A singular boundary matrix signals a nontrivial homogeneous solution satisfying the endpoint conditions, where a compatibility condition and a nonunique inverse may arise. For $L=D^2$, one convenient fundamental solution is $(x-\xi )H(x-\xi )$.

A causal Green function vanishes before its source acts. For a time differential operator $L_t$, its defining conditions areThe Dirac delta function fixes the appropriate Green function jump conditions; the past-zero condition selects the response from an initial state at rest. A Green-function representation then integrates only earlier source times.

For the harmonic oscillator operator $L_t={\partial }_t^2+{\omega }_0^2$, with ${\omega }_0>0$,where $H$ is the Heaviside step function. This Green function is continuous at $t=s$ and its first derivative jumps from zero to one. Therefore $y(t)={\int }_0^{t}G(t,s)f(s)ds$ solves $y^{\prime \prime }+{\omega }_0^{2}y=f$ for continuous $f$ and zero initial data. It is the oscillator's impulse response.^[ref]

Causality fixes time support, while stability concerns later growth or decay. They are separate conditions, as explained in causality versus stability of a Green function.

For a real linear differential operator $L$, one has $L(\mathrm{Im}u)=\mathrm{Im}(Lu)$. A forcing term written as the imaginary part of a polynomial times $e^{i\omega x}$ can therefore be handled by solving one complex equation, then taking the imaginary part. For $L=d^2/d{x}^2+a$, substituting $u=p(x)e^{i\omega x}$ gives $p^{\prime \prime }+2i\omega p^{\prime }+(a-{\omega }^2)p$. If the forcing frequency is resonant, the polynomial ansatz must include the appropriate additional power of $x$.

For a divergence-form differential operator $-\nabla \cdot K\nabla$, its conormal derivative is the flux factor appearing in integration by parts. It generally differs from the ordinary normal derivative ${\partial }_{n}u=\mathbf{n}\cdot \nabla u$ when $K$ is anisotropic. Decomposing the gradient along a smooth boundary into its normal and tangential components gives ${\partial }_{n,K}u=(\mathbf{n}\cdot K\mathbf{n}){\partial }_{n}u+(\mathbf{n}\cdot K\mathbf{t}){\partial }_{\mathbf{t}}u$ in two dimensions. Thus zero normal derivative does not in general remove the boundary term in the usual symmetric weak formulation. A zero conormal derivative does remove it and produces the natural homogeneous Neumann boundary condition for that energy.

If a differential operator $L$ is a linear map and has nonzero kernel of a linear map, the equation $LA=J$ determines $A$ only up to a homogeneous solution. Lorentz covariance of $L$ and $J$ therefore does not force independently selected solutions in different inertial frames to obey a particular tensor transformation law. A covariant Green function and consistently transformed homogeneous data do provide such a choice. For the four-potential in Lorenz gauge, $\square A^{\mu }=-{\mu }_0{J}^{\mu }$ still permits $A^{\mu }\mapsto A^{\mu }+{\partial }^{\mu }\chi$ with $\square \chi =0$, a residual Lorenz gauge symmetry. Constant spatial potentials for zero sources are a concrete example.

A Dirichlet Green function for a linear differential operator $L$ on a domain $\Omega$ satisfieswith the Dirac delta function normalization and zero Dirichlet boundary conditions understood in the appropriate classical or weak sense. When the corresponding boundary-value operator is invertible, Green-function representation gives the solution of $Lu=f$ with zero boundary values as $u(x)={\int }_{\Omega }{G}_D(x,\xi )f(\xi )d\xi$. Nonzero boundary values require an additional homogeneous contribution.^[ref]

For $L=-d^2/d{x}^2$ on $(0,1)$,It vanishes at both endpoints, is continuous at $x=\xi$, and its first derivative has jump $-1$, exactly the Green function jump conditions for $-d^2/d{x}^2$. Integrating this Green kernel against the constant forcing $f=1$ gives $u(x)=x(1-x)/2$. The operator sign, the boundary conditions and possible zero modes determine which Green's function is meant.

A differential operator is in divergence form when its highest derivatives occur inside a divergence, for example $Lu=\nabla \cdot (A\nabla u)$ or $(p(x)u^{\prime }{)}^{\prime }$. This form makes integration by parts express the operator through a lower-order bilinear form and boundary flux.

An elliptic complex on a smooth manifold is a finite cochain complex of differential operators between sections of vector bundles,with $D_{i+1}{D}_i=0$, whose principal symbol sequence is exact at every nonzero real cotangent vector. Thus for $\xi \in T_x^{\ast }M\setminus \{0\}$, the kernel of each symbol map equals the image of the preceding one, including injectivity and surjectivity at the ends.^[ref]

The de Rham complex is the basic example. Up to an inessential nonzero scalar convention, the symbol of $d$ is $\alpha \mapsto \xi \wedge \alpha$. Choose a tangent vector $v$ with $\xi (v)=1$. The interior product identitygives a contracting homotopy for that symbol sequence and proves exactness.

The condition belongs to the whole sequence: the individual maps need not be elliptic operators between equal-rank bundles. On a compact manifold without boundary, ellipticity gives finite-dimensional cohomology. Boundary problems require additional conditions. This finiteness is what makes an alternating cohomological trace meaningful in the Atiyah-Bott fixed-point theorem.

The generalized Leibniz rule extends composition of differential operators to all integer $n$. Define $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n(n-1)\cdots (n-r+1)/r!$. For nonnegative $n$ the sum stops at $r=n$. For $n=-s<0$, its coefficients are $(-1{)}^r(\genfrac{}{}{0ex}{}{s+r-1}{r})$; in particular ${\partial }^{-1}f=f{\partial }^{-1}-f^{\prime }{\partial }^{-2}+f^{\prime \prime }{\partial }^{-3}-\cdots$. Applying $\partial$ on the left cancels consecutive derivative terms and recovers multiplication by $f$, which explains the inverse expansion.

A Fourier mode is a single complex oscillatory component $e^{i\omega x}$. Its real and imaginary parts are a cosine and a sine; $\omega$ is the spatial angular frequency. Every constant-coefficient differential operator acts on it by multiplication: $P({\partial }_x)e^{i\omega x}=P(i\omega )e^{i\omega x}$. In particular, ${\partial }_x^2{e}^{i\omega x}=-{\omega }^2{e}^{i\omega x}$. These eigenfunctions reduce a partial differential equation such as the heat equation to separate equations for each frequency. On the real line a pure Fourier mode is not in the L2 space; the Fourier transform uses such modes in a continuous superposition.

A global relation for a linear boundary value problem is an identity connecting transformed initial or boundary traces of a solution. Its form depends on the differential operator, the auxiliary adjoint solutions and the transform convention.

For $L=\Delta -k^2$, suppose $Lu=f$ and $Lv=0$ on a smooth bounded domain $\Omega$. Green's second identity givesThus a homogeneous equation $f=0$ gives a boundary identity with right side zero; an inhomogeneous equation contributes the known volume term. Exponential choices of $v$ turn this into a spectral family connecting the Dirichlet boundary data and the outward normal derivative.

A global relation is a necessary identity. Recovering unknown traces requires enough independent relations and an admissible analytic inversion or a rank-checked numerical scheme; the identity alone is not an existence theorem for arbitrary boundary data.

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 $L$.

A Green's function is the response of a specified linear differential operator $L$ to a point source, together with specified boundary conditions or causality conditions. In distributional notation, $L_{x}G(x,y)=\delta (x-y)$, where $\delta$ is the Dirac delta function. Integrating this response against a source gives $u(x)=\int G(x,y)f(y)dy$ when the representation is valid.

For $L=-d^2/d{x}^2$ on $0<x<1$ with zero Dirichlet boundary conditions, $G(x,y)=\min (x,y)(1-\max (x,y))$. It is linear on either side of $y$, continuous there, and its first derivative jumps by $-1$, so $-G^{\prime \prime }={\delta }_y$. Changing either the operator sign or the boundary conditions changes the kernel. The example shows how a difficult source is assembled from elementary impulse responses.

For a fixed formal pseudodifferential operator $P={\sum }_j{p}_j{\partial }^j$ and a monic differential operator $L={\partial }^m+{\sum }_{l=0}^{m-1}{u}_l{\partial }^l$, cyclicity of the Adler trace gives $l_P(L)=\mathrm{Tr}(PL)$ with $c=\int p_{-m-1}dx$ and $a_l=\mathrm{res}({\partial }^{l}P)={\sum }_{k=0}^l\left(\genfrac{}{}{0ex}{}{l}{k}\right)p_{k-l-1}^{(k)}$. The function is affine in the variable coefficients; its variational derivative with respect to $u_l$ is $a_l$.

For the sign convention $\Delta =\mathrm{div}\nabla$, the heat operator is the differential operator $L={\partial }_t-\Delta$. The heat equation is $Lu=0$. If $P=-\Delta$ is the positive Laplace-Beltrami operator, the same operator is ${\partial }_t+P$. A Riemannian heat kernel is a fundamental solution with the initial Dirac delta distribution.

A homogeneous linear differential equation is a linear differential equation with identically zero forcing term, $L[y]=0$. On a common interval, its solutions of a differential equation form the vector space $\ker L$, the kernel of a linear map. They include the zero function and are closed under addition and scalar multiplication. For any two solutions of a differential equation $u,v$ and constant scalars $\alpha ,\beta$, linearity gives $L[\alpha u+\beta v]=\alpha 0+\beta 0=0$: this is the superposition principle. The scalars must be constant; variable multipliers introduce extra derivatives.

For example, $y^{\prime }+2y=0$ has general solution $y=C{e}^{-2t}$, where $C$ is any constant real number. A short verification uses the product rule: $(e^{2t}y{)}^{\prime }=e^{2t}(y^{\prime }+2y)=0$, so $e^{2t}y$ is constant. For $y_1=e^{-2t}$ and $y_2=2e^{-2t}$, the sum is $3e^{-2t}$ and also solves the same homogeneous linear differential equation. At $t=1$, their values are approximately $0.13534$, $0.27067$ and $0.40601$. Every curve is a multiple of one exponential function; the solution space of a homogeneous linear differential equation has dimension one. Both $t$ and $y$ in this example and the following diagram are dimensionless mathematical variables.

Now keep the same differential operator $L[y]=y^{\prime }+2y$ but prescribe the forcing term $1$. The inhomogeneous linear differential equation $y^{\prime }+2y=1$ has particular solution $y_p=1/2$ and general solution $y=1/2+C{e}^{-2t}$. The initial conditions $y(0)=0,1,2$ select, respectively,At $t=1$, their values are approximately $0.43233$, $0.56767$ and $0.70300$. The midpoint $\frac{1}{2}(u_0+u_2)=u_1$ solves the same inhomogeneous linear differential equation, because $L[\frac{1}{2}(u_0+u_2)]=\frac{1}{2}(1+1)=1$. The sum $s=u_0+u_2=1+e^{-2t}$ instead satisfies $s^{\prime }+2s=2$, so it fails the equation with forcing term $1$. More generally, a linear combination of solutions of a differential equation for the same fixed nonzero forcing term retains that forcing term precisely when its scalar weights add to one. This is the affine space structure, rather than unrestricted superposition within that fixed family.

On the left, all three curves solve the same homogeneous linear differential equation. On the right, $u_0,u_1,u_2$ share the particular solution $1/2$, and the shaded difference $u_2-1/2$ is the homogeneous solution $\frac{3}{2}{e}^{-2t}$. The dashed sum has twice the forcing term, even though its starting value coincides with that of $u_2$. Adding the two initial conditions also adds their inputs; taking their midpoint averages both. This distinction makes the superposition principle useful for assembling responses to different inputs without mistaking that assembly for a solution of a differential equation with the original input. See MIT's superposition notes, including the particular solution one half.

A homogeneous solution solves the zero-forcing equation $Ly=0$ associated with a linear ordinary differential equation $Ly=f$. Such solutions form the kernel of the linear differential operator. If $y_1,y_2$ solve the same forced equation, $L(y_1-y_2)=f-f=0$, so their difference is homogeneous.

For $y^{\prime }+2y=3$, the homogeneous family is $C{e}^{-2t}$ and a particular solution is $3/2$. Hence every solution is $y=3/2+C{e}^{-2t}$. The homogeneous term carries the free initial condition; when the coefficient is positive it decays, but other linear equations can have growing or oscillating homogeneous terms. The adjective homogeneous here refers to zero forcing, not to a function's scaling homogeneity.

An integral kernel is a function or distribution specifying an integral operator through $(Tf)(x)=\int K(x,y)f(y)dy$. A Green function is an integral kernel that inverts a differential operator under specified boundary conditions.

Two Riemannian manifolds are isospectral for a specified differential operator if their eigenvalues, counted with multiplicities and with the same boundary conditions, agree. Riemannian isometries preserve the Laplace-Beltrami operator spectrum, but the converse can fail. The transplantation theorem and the Sunada theorem give systematic constructions.

The Jacobi identity is the compatibility condition on a Lie bracketTogether with an Alternating Lie bracket, it defines a Lie algebra. Unlike associativity, it relates three cyclic bracketings rather than equating two bracketings. Equivalently, using antisymmetry of a Lie bracket,Thus the linear map ${\mathrm{ad}}_x:y\mapsto [x,y]$ is a derivation of a Lie algebra: acting on a bracket is the sum of acting on each argument. This makes infinitesimal symmetries act consistently on each other's products. It also gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, the identity that makes the Adjoint representation of a Lie algebra a Lie algebra representation.

For the commutator in an associative algebra, the cancellation is directly visible:Each word occurs once with each sign. This proves the Jacobi identity without a restriction on the characteristic of a field.

For a computed check in the sl2 matrix commutator example,Here the last two terms are individually nonzero over characteristic zero; the identity records their cancellation. In the cross-product Lie algebra, the same cyclic cancellation is the vector triple product identity.

The historical connection is unusually concrete: Lie and Engel's volume I expands nested commutators of differential operators into twelve terms and cancels them cyclically, in Chapter 5, §26, equation (7). See Lie and Engel, §26, in Merker's translation, pp. 105–106.