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For $f\in C([-1,1])$, the equation $((1-x^2)y^{\prime }{)}^{\prime }=f$ has a bounded solution on $(-1,1)$ if and only if ${\int }_{-1}^{1}f=0$. All bounded solutions differ by a constant. To prove necessity, integrate the equation once. The quantity $(1-x^2)y^{\prime }$ has finite endpoint limits, and any nonzero endpoint limit would force logarithmic growth in $y$. Both limits must therefore vanish, yielding the zero-mean condition. For sufficiency set $F(x)={\int }_{-1}^{x}f(t)dt$ and $y(x)=C+{\int }_0^{x}F(s)/(1-s^2)ds$. Zero mean implies $|F(s)|\le \Vert f{\Vert }_{\infty }\min (1+s,1-s)$, so $y^{\prime }$ is bounded and $y$ extends continuously to the endpoints. The homogeneous solutions are a constant and $\frac{1}{2}\log ((1+x)/(1-x))$, proving uniqueness up to a constant. This is the Fredholm alternative for the zero eigenvalue of this singular Sturm-Liouville problem.

The eigenfunction expansion in Legendre polynomials is $y=C-{\sum }_{k\ge 1}{P}_k{\int }_{-1}^{1}f{P}_k/[k(k+1){\int }_{-1}^1{P}_k^2]$. It follows by projecting onto the eigenfunctions with eigenvalues $-k(k+1)$. Completeness follows because the Legendre polynomials span the polynomials, which are dense by the Weierstrass approximation theorem.

Continuity only in the open interval requires more care. Even if $f$ is integrable and has zero mean, the primitive of $F/(1-x^2)$ must still be bounded at both ends. For example, $F(x)=1/\log (e/(1-x^2))$ has zero endpoint limits and absolutely integrable derivative $f=F^{\prime }$, but $F/(1-x^2)$ has a divergent primitive. Thus zero mean alone is insufficient without endpoint control.

For the heat equation with zero Dirichlet boundary conditions on $(0,L)$,It is also obtained by the method of images, taking the difference between Gaussian sources at $y+2jL$ and $-y+2jL$. The eigenfunction expansion is useful at long times; the image expansion is useful at short times.

Multiply the eigenfunction expansion by ${\psi }_m^{\ast }$ and integrate over the common domain. Orthogonality and wavefunction normalization giveThus $\boxed{c_m=⟨{\psi }_m,f⟩}$. For a normalized $f$, the Born rule identifies $|c_m{|}^2$ as the probability of this basis outcome.

Expand in the Fourier sine basis, using unnormalized eigenfunctions:Projection of $\mathcal{L}y=f$ onto each eigenfunction givesWhen $l\notin \pi {\mathbb{N}}_{>0}$, every eigenvalue is nonzero and the eigenfunction expansion is uniquely determined:When $l=m\pi$, the exceptional index is $n=m$, for which the eigenfunction is $\sin x$. If $f_m\ne 0$, its coefficient equation is $0=f_m$, an inconsistency: there is no solution. If $f_m=0$, the coefficient equation is $0=0$, and $c_m$ is arbitrary. The full family isThis is the Fredholm alternative directly in the coefficient equations: nonzero eigenvalues fix their coefficients; a zero eigenvalue requires zero forcing in that direction and leaves the corresponding coefficient free. An inconsistent equation at a zero eigenvalue cannot be repaired by assigning an infinite coefficient.

For convergence, Lipschitz continuity implies absolute continuity of $f$, with bounded derivative almost everywhere. One integration by parts gives $f_n=O(n^{-1})$; since the denominators grow quadratically, $c_n=O(n^{-3})$ apart from the single free coefficient. The solution series and its first derivative therefore converge uniformly. The second derivative converges in $L^2$, and the resulting Sobolev space solution satisfies $y^{\prime \prime }=f-y$. As $f-y$ is a continuous function, this is a classical $C^2$ solution, with both boundary conditions satisfied. The forcing's Fourier sine series is interpreted on the open interval or in $L^2$; its endpoint values need not equal the endpoint values of $f$.

First take the usual endpoint-regular interpretation $f\in C([-1,1])$. Integrating the divergence form shows that $F=(1-x^2)y^{\prime }$ has finite endpoint limits. A nonzero limit would make $y^{\prime }$ proportional to $1/(1\mp x)$ near that endpoint, forcing a logarithmically unbounded $y$. Thus boundedness requires $F(-1)=F(1)=0$, and consequentlyThis is orthogonality to the constant eigenfunction $P_0$, hence the Fredholm alternative condition. It is also sufficient under the stated endpoint regularity: putIf $M=\max |f|$, the zero-mean condition gives $|F(s)|\le M\min (1+s,1-s)$. Hence the last integrand is bounded, so $y$ extends continuously to both endpoints. Any two bounded solutions differ by a constant: their difference solves $((1-x^2)z^{\prime }{)}^{\prime }=0$, whose other solution is $\frac{1}{2}\log ((1+x)/(1-x))$.

For the requested eigenfunction expansion, setThe Legendre polynomials span all polynomials, so their completeness follows from the Weierstrass approximation theorem. Projecting $\mathcal{L}y=f$ onto each eigenfunction gives $-k(k+1)y_k=f_k$ for $k\ge 1$. The zero mode is free, while $f_0=0$. Thus the full family isThe formula is unchanged by rescaling any $P_k$. For the usual normalization $P_k(1)=1$, the facts $|P_k(x)|\le 1$ on $[-1,1]$ and $N_k=2/(2k+1)$, together with the Cauchy-Schwarz inequality, bound each summand by a constant times $k^{-3/2}$. The series therefore converges absolutely and uniformly. Its identification with the preceding integral solution avoids unjustified termwise second differentiation.

There is a source ambiguity: continuity only on $(-1,1)$ does not supply the endpoint control used above. Even for integrable forcing, zero mean alone is insufficient. For example, letThis $f$ is continuous in the interior, absolutely integrable, and has zero mean because $F(\pm 1)=0$. Yet every candidate with zero endpoint values of $(1-x^2)y^{\prime }$ haswhose integral diverges logarithmically in $H$ at each endpoint. Thus no bounded solution exists. For a merely interior-continuous $f$, the exact criterion is that some constant $B$ makesbounded as $x$ approaches both endpoints. The Bounded solvability for the Legendre operator criterion above covers the intended regular forcing and makes this endpoint limitation explicit.

On the torus ${\mathbb{T}}^n=(\mathbb{R}/2\pi \mathbb{Z}{)}^n$, choose normalized Haar measure, so that $e_k(x)=e^{ik\cdot x}$, $k\in {\mathbb{Z}}^n$, is an orthonormal basis of $L^2$. A periodic Sobolev space is defined, for any real $s$, byFor negative $s$, the coefficients describe distributions rather than necessarily functions. For a nonnegative integer $s$, the weight is comparable to ${\sum }_{|\alpha |\le s}|k^{\alpha }{|}^2$, so this norm is equivalent to the sum of the squared $L^2$ norms of all weak derivatives through order $s$. Fourier truncation proves that smooth trigonometric polynomials are dense. The $L^2$ pairing identifies the continuous dual of $H^s$ with $H^{-s}$, giving Sobolev duality.

Two Fourier estimates explain the usefulness of these spaces. If $s>m+n/2$, the Cauchy-Schwarz inequality givesUniform convergence of the differentiated Fourier series proves the Sobolev embedding $H^s\hookrightarrow C^m$. If $\delta >0$, a bounded subset of $H^{s+\delta }$ has uniformly small high-frequency tails in $H^s$:Finite-dimensional Fourier truncation therefore proves the compact embedding $H^{s+\delta }\hookrightarrow H^s$, the periodic version of Rellich-Kondrachov compactness theorem. In addition, for $s>n/2$ multiplication is continuous $H^s\times H^s\to H^s$, making it a Sobolev algebra. Indeed the weighted coefficient of a product is bounded by the sum of the convolutions of $⟨k{⟩}^s|\hat{u}(k)|$ with $|\hat{v}(k)|$ and of $|\hat{u}(k)|$ with $⟨k{⟩}^s|\hat{v}(k)|$; Cauchy-Schwarz puts unweighted coefficients in ${\ell }^1$, and the ${\ell }^1\ast {\ell }^2\to {\ell }^2$ case of Young's convolution inequality finishes the estimate.

For eigenfunction expansions, distinguish ellipticity from self-adjointness. An arbitrary non-self-adjoint elliptic operator need not have an orthonormal basis of eigenfunctions. A useful definite setting iswith smooth real periodic coefficients, $a_{ij}=a_{ji}$, real $V$, and uniform ellipticityTake the self-adjoint realization in $L^2$ with domain $H^2$. Its construction and spectral properties can be seen from the associated form on $H^1$,For a sufficiently large constant $C$, $b_C=b+C(\cdot ,\cdot )$ is a positive coercive form: $b_C(u,u)\ge c_1\Vert u{\Vert }_{H^1}^2$, and $L+C\ge I$. The Lax-Milgram theorem supplies the unique weak solution of $(L+C)u=f$ for $f\in H^{-1}$; the form is Hermitian, which gives symmetry of its inverse on $L^2$.

The regularity input is the periodic elliptic estimateHere is why an elliptic operator gains two derivatives. With coefficients frozen at a point, its principal symbol $\sum a_{ij}{\xi }_i{\xi }_j$ dominates $c|\xi {|}^2$, and the estimate follows by multiplying Fourier coefficients by that symbol. On a sufficiently small coordinate patch, the difference of the leading coefficients from these constants has arbitrarily small supremum norm. In the integer-order differentiated estimate, this absorbs the highest derivatives into the left side; derivatives falling on coefficients contribute only lower-order terms. A partition of unity reduces the variable-coefficient problem to such patches. Its commutators with $L$ have order one, controlled bywhich follows by splitting low and high Fourier modes. Absorbing these terms gives the displayed estimate. For a weak $H^1$ solution, the same localization with difference quotients first bounds its second weak derivatives for $f\in L^2$; differentiating and repeating then gives $f\in H^s\Rightarrow u\in H^{s+2}$ for nonnegative integer $s$. Interpolation and duality extend the scale to real orders. In particular $(L+C{)}^{-1}:L^2\to H^2$ is bounded.

Let $R=(L+C{)}^{-1}$. Compactness of $H^2\hookrightarrow L^2$ makes $R$ a compact operator. Symmetry of the form makes $R$ self-adjoint, and positivity of $L+C$ makes $R$ positive and injective. Its range is dense because $\ker R^{\ast }=\ker R=0$. The spectral theorem for compact self-adjoint operators consequently gives an orthonormal basis ${\varphi }_j$ withElliptic regularity bootstraps every ${\varphi }_j$ to a smooth function. Each eigenvalue has finite multiplicity. Thus a self-adjoint uniformly elliptic second-order operator on the torus has a complete smooth orthonormal eigenfunction expansion with discrete eigenvalues tending to infinity.

Writing $u_j=(u,{\varphi }_j)$, every $u\in L^2$ has $u={\sum }_j{u}_j{\varphi }_j$ in $L^2$, with $\Vert u{\Vert }_2^2={\sum }_j|u_j{|}^2$. Iterating the elliptic estimate shows that the graph norm of $(L+C{)}^m$ is equivalent to the $H^{2m}$ norm. Interpolation and duality give the more general spectral descriptionThe comparison constants depend on the operator and $s$. In particular smooth functions have coefficients decaying faster than every power of ${\lambda }_j+C$, and their expansions converge in every Sobolev norm and hence in every $C^m$ norm.

This gives a practical way to solve elliptic equations. The equation $Lu=f$ is solvable if and only if $f_j=0$ whenever ${\lambda }_j=0$; then one may take $u_j=f_j/{\lambda }_j$ on the other eigenspaces and add an arbitrary vector in $\ker L$. The Sobolev description proves the two-derivative gain and the Fredholm alternative, rather than leaving them as formal series manipulations. Similarly the solution of the heat equation ${\partial }_{t}u+Lu=0$ isFor every $t>0$ the exponential decay at high eigenvalues places this solution in every Sobolev space, hence it is smooth.

For the Laplacian $-\Delta$ the eigenfunctions are precisely $e^{ik\cdot x}$ with eigenvalues $|k{|}^2$. The number with $|k{|}^2\le R$ is the number of lattice points in a radius-$\sqrt{R}$ ball, asymptotic to its volume by comparison with unit cubes. For the general self-adjoint operator above, bounds on its form above and below by positive multiples of $\Vert \nabla u{\Vert }_2^2$ plus constants, together with the Courant–Fischer min-max principle, imply ${\lambda }_j+C\asymp j^{2/n}$. Thus the growth reflects dimension and the two-derivative order. Compactness of the torus is essential: on ${\mathbb{R}}^n$, translated bumps prevent the compact embedding, and the Laplacian instead has continuous spectrum described by the Fourier transform.

Use the positive Laplace-Beltrami operator on a compact Riemannian manifold without boundary. Its heat kernel is the smooth integral kernel $K(t,x,y)$, $t>0$, of the solution operators for the heat equation:The last condition holds, in particular, in $L^2$ for smooth initial functions. The heat trace isThe heat invariants are the coefficients in the small-time heat kernel expansion of this trace:Equivalently, the local diagonal coefficients $u_j(x,x)$ integrate to $a_j$; $R_g$ is the scalar curvature. Some normalizations absorb $(4\pi {)}^{-d/2}$ into the coefficients. These definitions here use the displayed normalization consistently.

To prove uniqueness, let $K_1,K_2$ satisfy the heat kernel conditions and let $w$ be the difference of their solutions with the same smooth initial function. It solves $w_t=-\Delta w$ and tends to zero in $L^2$ as $t\downarrow 0$. The heat equation energy identity, using integration by parts on the closed manifold, gives for $t>0$Integrating from $s>0$ to $t$ gives $\Vert w(t){\Vert }_{L^2}\le \Vert w(s){\Vert }_{L^2}$; sending $s\downarrow 0$ proves $w(t)=0$. Thus for every smooth test function $f$,For fixed positive $t$ and $x$, smoothness makes this difference a smooth function of $y$. Taking its conjugate as test function shows its squared norm is zero. The heat kernel is unique.

The compact elliptic spectral theorem for the self-adjoint elliptic Laplace-Beltrami operator gives a complete orthonormal basis of smooth eigenfunctions ${\varphi }_j$, with discrete eigenvalues $0\le {\lambda }_0\le {\lambda }_1\le \cdots$ counted with multiplicity. Solving the heat equation for initial data ${\varphi }_j$ gives $e^{-t{\lambda }_j}{\varphi }_j$, so the unique heat kernel has its eigenfunction expansion and its heat trace isThe series converges for every $t>0$. This proves that the spectrum with multiplicities determines the heat trace.

Conversely, its large-time behaviour recovers the distinct eigenvalues and their multiplicities successively. The multiplicity of zero is $m_0={\lim }_{t\to \infty }{Z}_M(t)$. After subtracting the already determined terms, letbe the positive remaining sum, and let $\mu$ be its smallest remaining eigenvalue. ThenIndeed, after factoring $e^{-t\mu }$, the next distinct eigenvalue is separated from $\mu$ by a positive gap. The remaining tail tends to zero: for fixed $t_0>0$ its absolute sum is bounded by a constant times $e^{-({\mu }_{\mathrm{next}}-\mu )(t-t_0)}$, using finiteness of $Z_M(t_0)$. Induction recovers every spectral term. Thus the heat trace and the spectrum with multiplicities determine one another.

Finally the leading coefficient among the heat invariants is positive, soThe common heat trace of isospectral manifolds therefore forces them to have the same dimension. Once the dimension is recovered, its leading coefficient also recovers their common volume.

Let $M$ be a closed manifold with a Riemannian metric and use $\Delta =-\mathrm{div}\mathrm{grad}$, the positive Laplace-Beltrami operator. This is the setting in which the final discrete eigenfunction expansion applies without additional boundary or noncompact spectral hypotheses. A Riemannian heat kernel is a smooth $H(t,x,y)$ for $t>0$ such thatfor every smooth initial function $f$. Here $\nu$ solves the heat equation. The limit is the statement that the initial kernel is the Dirac delta distribution on the diagonal. In the closed setting this uniquely determines the heat kernel; it is symmetric, preserves constants, is nonnegative, and satisfies the semigroup property.

A heat parametrix $P(t,x,y)$ is an approximate version with the same delta initial limit and with errorregular enough at $t=0$ to correct by a convergent integral series. One may construct it by the local Gaussian ansatzwhere $\chi =1$ near the diagonal and is supported in a convex normal neighborhood, and the smooth coefficients solve the usual radial transport equations. The leading coefficient accounts for the Riemannian volume form. Taking $q$ sufficiently large makes the residual extend continuously, with any prescribed finite number of derivatives, to $t=0$. Alternatively a smooth asymptotic summation of all transport coefficients gives a residual vanishing to every order. These local construction, extension and differentiability facts are used here as subsidiary results, as permitted.

For the correction proof take a heat parametrix whose residual is smooth up to $t=0$ on a short interval $[0,T]$. We also use its uniform integral bound ${\sup }_{0<t\le T,x}{\int }_M|P(t,x,y)|d{V}_y<\infty$, its approximate-identity limit, and the following standard differentiability property: convolution with a smooth time-dependent kernel gives a smooth kernel for $t>0$, and differentiating it yields the identity below. These properties follow from the Gaussian estimates and the delta initial limit; no conclusion about the final exact heat kernel is assumed.

Define the Volterra convolution of kernels byIt is associative wherever these integrals converge. The delta term at the upper endpoint givesSeek $H=P+P\ast Q$. Then its error vanishes exactly when $Q+R+R\ast Q=0$. The solution is the Volterra parametrix correctionThe signs matter: the first correction is $-P\ast R$.

To prove convergence, let $|R|\le C$ and $V=\mathrm{vol}(M)$. The time variables in $R^{\ast k}$ range over a simplex of volume $t^{k-1}/(k-1)!$, soThe series for $Q$ therefore converges uniformly on $[0,T]\times M\times M$. Derivatives obey analogous bounds, with polynomial factors in $k$, by the quoted residual extension and differentiation properties. Thus the series can be convolved and differentiated as above. Associativity and absolute convergence give $Q+R+R\ast Q=0$, proving $({\partial }_t+{\Delta }_x)H=0$. The correction $P\ast Q$ has integral norm $O(t)$ by the integral bound on $P$ and boundedness of $Q$. It has zero initial limit, so $H$ has precisely the required delta initial data.

For uniqueness, a smooth solution $w$ with zero initial data satisfies the heat equation energy identityIts initial norm is zero, hence $w=0$. Applying uniqueness to consecutive evolutions proves the semigroup property. Choose $s_0<T$ and, for arbitrary $t>0$, compose enough kernels $H(t/r)$ that $t/r<s_0$. The resulting kernel is independent of the subdivision by uniqueness, is smooth for positive time by the quoted differentiability property, and extends the construction to every $t>0$. This proves a parametrix determines the global heat kernel in the closed setting. The heat equation maximum principle gives nonnegativity and applying uniqueness to the constant initial function gives conservation of total mass. Self-adjointness of the positive Laplace-Beltrami operator $\Delta$ gives symmetry.

Finally let $\{{\varphi }_j\}$ be a complete complex orthonormal eigenbasis of $\Delta$, with eigenvalues ${\lambda }_j\ge 0$, repeated by multiplicity. As subsidiary analytic facts we use the compact elliptic compact elliptic spectral theorem, elliptic regularity bounds making each fixed derivative of ${\varphi }_j$ grow at most polynomially in $1+{\lambda }_j$, and a polynomial eigenvalue-counting bound. The spectral expansion of the Riemannian heat kernel converges because exponential decay makes the following series converge with every derivative when $t\ge \epsilon >0$:Indeed, evolving initial data ${\varphi }_j$ gives $e^{-t{\lambda }_j}{\varphi }_j$ by direct substitution into the heat equation and uniqueness. For general $f\in L^2(M)$, completeness and the heat equation energy identity giveThe smoothly convergent kernel series represents this same operator, and equality for all smooth $f$ identifies it pointwise with the constructed heat kernel. The complex conjugate is required by the inner product; for a real eigenbasis it may be omitted. A noncompact Riemannian manifold may instead require a spectral integral, and the unqualified discrete formula should not be asserted there.

For each of the four standard pairs, integration by parts givesConsequently the eigenvalues are nonnegative. For a positive eigenvalue ${\mu }_n=A{k}_n^4$, the eigenvalue equation is $W_n^{\prime \prime \prime \prime }=k_n^4{W}_n$. Its four characteristic roots are $\pm k_n,\pm i{k}_n$, soHere $C_j$ are coefficients, avoiding a collision with the filament bending modulus $A$. The regular finite-interval self-adjoint operator has compact resolvent; applying the spectral theorem for compact self-adjoint operators to a shifted inverse supplies a complete orthonormal basis of eigenfunctions.

For clamped boundary conditions at zero, a clamped--clamped bending mode can be writtenAt $L$, writing $\beta =kL$, the two remaining boundary conditions areThe determinant is $2(1-\cos \beta \cosh \beta )$. Hence the positive wavenumbers obeyEquivalently, intersect $\cos \beta$ with $\mathrm{sech}\beta$. Numerical root bracketing givesThe entire sequence has the useful large-$n$ descriptionIndeed, put $\beta =(n+\frac{1}{2})\pi +\delta$ in $\cos \beta =\mathrm{sech}\beta$ and use $\cos \beta =(-1{)}^{n+1}\delta +O({\delta }^3)$ and $\mathrm{sech}\beta =2e^{-\beta }+O(e^{-3\beta })$.

The apparent root $\beta =0$ in the determinant equation is spurious for the clamped problem: at zero eigenvalue, $W$ is a cubic polynomial, and its four clamped conditions force $W=0$. For other endpoint choices, zero-energy filament modes must be treated separately from the trigonometric formula. The free-free kernel consists of affine functions, the torqued-torqued kernel consists of constants, and the hinged-hinged kernel is trivial. Including those kernels is necessary for a complete eigenfunction expansion.

Choose real eigenfunctions with ${\int }_0^L{W}_n{W}_{m}dx={\delta }_{nm}$. For positive modes, self-adjointness and integration by parts diagonalize the energy:At temperature $T$, the canonical ensemble is a product of centered Gaussian distributions. The equipartition theorem, with Boltzmann constant $k_B$, givesThe resulting thermal covariance of an elastic filament isFor an unnormalized eigenfunction of squared L2 norm $N_n$, divide its summand by $N_n$. This normalization factor cannot be absorbed silently into the modal variance.

For clamped boundary conditions at both ends, the inverse of $d^4/d{x}^4$ has Green function, for $x\le y$,It is cubic on each side of $y$, satisfies the four clamped conditions, has continuous first two derivatives, and has unit jump in its third derivative. Thus ${\partial }_x^{4}G=\delta (x-y)$, and its eigenfunction expansion is the sum above. In particular,These finite variances require removal of every zero-energy filament mode. An unconstrained free-free elastic filament can translate and tilt at no energy cost; a torqued-torqued elastic filament can translate. Their unrestricted Boltzmann distributions are not normalizable, so the full displacement variance is undefined. Fix those rigid degrees of freedom before applying the positive-mode formula.

Use the Bromwich inversion formula in the real variable $\omega$, with $p=\gamma +i\omega$:When the relevant Laplace transforms and spatial integral transforms are explicit, every sample of this integrand can be evaluated directly, without a time-stepping approximation to the partial differential equation. Split the initial condition integral at $y=x$: the two pieces of $R_q$ are linear combinations of $e^{\pm qy}$, so truncated spatial exponential transforms suffice. For real data the negative-frequency half is the complex conjugate of the positive-frequency half.

A practical numerical integration is to truncate to $|\omega |\le \Omega$, apply an adaptive quadrature rule, and independently increase $\Omega$ and refine the quadrature. Choose $\gamma >0$ large enough to remain to the right of all singularities, but avoid an unnecessarily large $e^{\gamma t}$. Evaluate hyperbolic function ratios in scaled form; for example,This prevents overflow when $\mathrm{Re}qL$ is large. The boundary contributions at an interior point are damped by $e^{-x\mathrm{Re}q}$ or $e^{-(L-x)\mathrm{Re}q}$, with $\mathrm{Re}q\sim \sqrt{|\omega |/2}$. Near an endpoint more frequencies are needed; at the endpoint itself use the prescribed Dirichlet boundary condition. The initial condition resolvent has a generally only $O(1/\omega )$ tail. Treat its oscillatory tail accurately, subtract a known transform with the same leading term, or evaluate that contribution separately with the heat kernel.

An equally useful check is the eigenfunction expansion. DefineThe causal modal formula isIt follows either from the heat kernel formula or directly from integration by parts against a sine eigenfunction. Exponential-transform formulas evaluate the time integral explicitly when available. The initial condition contribution has a Gaussian function cutoff in $n$ for each positive time. Nonzero endpoint forcing produces a more slowly convergent sine series tail. A boundary lifting $\ell (x,t)=(1-x/L)g_0(t)+(x/L)e^{aL}h(t)$ gives $w=\ell +z$, where $z$ has zero Dirichlet boundary conditions and forcing $-{\ell }_t-a^2\ell$; expanding $z$ gives better convergence and imposes the endpoint values exactly. At very small times the method of images is often more efficient than retaining many modes.

Check convergence by increasing both the frequency cutoff and quadrature resolution, and compare with a separately truncated causal modal or image-kernel evaluation. Causality in finite-time Laplace contour inversion means a leftward contour deformation must respect both resolvent operator poles and growth of the transformed data. In particular, the transform of data extended by zero after $T$ can grow like $e^{-pT}$ in the left half-plane: for $t<T$ one cannot close that contour indiscriminately and discard its large arc. The modal forcing integral with upper limit $t$ avoids this causality error.

The drift makes the natural measure weighted. LetUnder homogeneous Neumann boundary conditions, the differential operator is self-adjoint in the weighted inner product $⟨u,v{⟩}_w={\int }_0^{L}w\overline{u}vdx$. Indeed integration by parts giveswhen the derivative boundary terms vanish. In particular its eigenvalues are nonpositive and it has the constant stationary eigenfunction. We first construct the weighted Neumann heat kernel with constant drift, then incorporate the two prescribed coordinate derivatives.

Put $b=\beta /2$. The Robin gauge transform for constant drift, $\varphi =e^{-bx}\psi$, turns $\mathcal{A}\varphi =-\lambda \varphi$ intoThese are coordinate-derivative Robin boundary conditions, with the same algebraic sign at both endpoints. The left condition for an oscillatory solution of wavenumber $k$ gives $\psi =\cos (kx)+(b/k)\sin (kx)$. The right condition then reduces to $(k+b^2/k)\sin (kL)=0$, giving $k_n=n\pi /L$. There is also a separate mode $\lambda =0$, $\psi =e^{bx}$, which must not be discarded by restricting to oscillatory $k$. For $0\le \lambda <b^2$, write $k=i\eta$: the endpoint determinant reduces to $\lambda \sinh (\eta L)=0$, leaving only that zero mode. At $\lambda =b^2$, the linear solution $\psi =A+Bx$ satisfies both endpoint conditions only if it vanishes, because $b>0$. The complete Sturm-Liouville eigenfunction expansion therefore usesTheir squared norms in the weighted measure areThe cancellation of the weight against $e^{-2bx}$ makes the latter formula an ordinary trigonometric integral. DefineFor $t>0$ this kernel is smooth in the interior. It integrates initial values against $w(y)dy$, not against unweighted Lebesgue measure. It is symmetric in $x,y$ as a weighted kernel, even though the drift operator does not look symmetric in an unweighted inner product.

To determine the boundary forcing, set $a_n(t)=⟨{\varphi }_n,q(\cdot ,t){⟩}_w/N_n$. Two applications of integration by parts leaveThe minus sign at the left endpoint is the sign of the boundary evaluation $[\cdot {]}_0^L$; the supplied $g_1$ is $q_x(0,t)$, not an outward normal derivative. Solving this linear differential equation by an integrating factor and recombining the modes gives the requested space-time integral representation:This is the Neumann boundary-forcing formula. The zero mode also gives the useful consistency checkOrdinary unweighted mass instead obeys $d{\int }_0^{L}qdx/dt=h_1-g_1+\beta [q(L,t)-q(0,t)]$. Replacing the weighted measure by $dy$ or deleting the constant mode would therefore give an incorrect solution.