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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.

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.

There is a subtle point when checking nonzero derivative data: every homogeneous eigenfunction has zero endpoint derivative, but differentiating the boundary-forcing sum and its time integral term by term at an endpoint is not valid. The kernel is singular as $t-s\downarrow 0$, and the prescribed derivatives are interior limits of the full solution. An explicit Laplace transform form verifies those limits without this interchange and gives an alternative purely contour-integral version.

For $\mathrm{Re}p>0$, set $\kappa =\sqrt{p+b^2}$ using the principal square root, and defineBoth solve $(p-\mathcal{A})f=0$, with $u_p(0)=v_p(L)=1$, $u_p^{\prime }(0)=v_p^{\prime }(L)=0$. Their weighted Wronskian is the nonzero constantHence the resolvent kernel for Neumann advection-diffusion on an interval isThe first derivative in $x$ has jump $-e^{-\beta y}$, giving $(p-{\mathcal{A}}_x){\mathcal{R}}_p=e^{-\beta y}{\delta }_y$ in the sense of distributional derivatives. If $G(p)={\int }_0^T{e}^{-ps}{g}_1(s)ds$ and $H(p)={\int }_0^T{e}^{-ps}{h}_1(s)ds$, an equivalent transformed solution isIn fact $v_p^{\prime }(0)=-D_p$ and $u_p^{\prime }(L)=e^{-\beta L}{D}_p$, so $Q_x(0,p)=G(p)$ and $Q_x(L,p)=H(p)$ exactly. The inverse Bromwich contour integral of $e^{pt}Q(x,p)/(2\pi i)$ is an alternative final answer. Its poles at $p=0$ and $p=-{\lambda }_n$ reproduce the constant and decaying modes of $K_{\beta }$, respectively; apparent square-root singularities in the normalized hyperbolic factors are removable. This also independently checks the kernel normalization and boundary signs.

As in question 1, the printed dot on $q_0$ denotes its spatial derivative: the two corner conditions are $q_0^{\prime }(0)=g_1(0)$ and $q_0^{\prime }(L)=h_1(0)$. The initial eigenfunction expansion converges to $q_0$ in the weighted space and, for the stated smooth compatible data, has the appropriate classical initial and boundary limits. Differences of solutions with zero data satisfy $\frac{1}{2}d\Vert q{\Vert }_w^2/dt=-{\int }_0^{L}w|q_x{|}^{2}dx$, proving uniqueness in the usual regular solution class. In the limit $\beta \to 0$, $N_0\to L$, ${\varphi }_n\to \cos (n\pi x/L)$, and the ordinary Neumann heat kernel on an interval and its boundary signs are recovered.

Reset $t=0$ when gravity is switched off. The subsequent overdamped relaxation of a clamped--free filament obeys ${\zeta }_{\perp }{h}_t=-B{h}_{xxxx}$ with the same homogeneous endpoint conditions. Separate a normal mode as $h=\varphi (x/L)e^{-\Gamma t}$. Writing $s=x/L$ givesThe two clamp conditions reduce the general hyperbolic/trigonometric solution toThe two free-end conditions have matrixIts determinant is $2(1+\cosh \alpha \cos \alpha )$. Hence the clamped--free bending mode values satisfyThere is no zero mode: at zero eigenvalue $\varphi$ is cubic, and the four clamp/free conditions force all coefficients to vanish. With the following normalization a mode isThe fourth-order operator is self-adjoint and positive under these endpoint conditions, since $\int f{g}^{\prime \prime \prime \prime }=\int f^{\prime \prime }{g}^{\prime \prime }$. Its orthogonal eigenfunction expansion givesFor the previously attained steady profile, the first coefficient is nonzero: $h_s$ is positive and the first mode can be chosen positive in the interior. Two applications of integration by parts, using both functions' endpoint conditions, also giveThus the dominant long-time answer, including its shape and amplitude, isThe next mode decays much faster. For other initial shapes with an accidentally zero first-mode projection, the lowest mode with $c_n\ne 0$ determines the asymptotic behaviour. The decay time has the correct units and exhibits the strong $L^4$ dependence characteristic of bending opposed by local viscous drag.

Adding $\frac{1}{2}\int \sigma (x)(h^{\prime }{)}^{2}dx$ to the quadratic bending energy of an elastic filament contributes $-(\sigma h^{\prime }{)}^{\prime }$ to its variational derivative. The endpoint term is $A{h}^{\prime \prime }{\eta }^{\prime }+(\sigma h^{\prime }-A{h}^{\prime \prime \prime })\eta$. If the real filament tension vanishes at both ends, the usual self-adjoint endpoint conditions for filament bending remain valid. A complete eigenfunction expansion diagonalizes the energy even when the eigenfunctions have no explicit formula. Thermal equipartition theorem arguments require positive eigenvalues after fixing any kernel; strong compression can violate this requirement.

For a time-independent strictly positive Dirichlet realization of an elliptic operator and $\psi \in L^2(U)$, the eigenfunction expansionsolves $u_t+Lu=0$ with homogeneous Dirichlet boundary conditions. For $t\ge \epsilon >0$, every power ${\lambda }_m^N$ times the exponential is bounded, so the series and all its time derivatives converge in all the Sobolev domains of powers of an elliptic Dirichlet operator. Elliptic regularity and the Sobolev embedding theorem give smoothness up to the spatial boundary for positive time. Parseval identity and the dominated convergence theorem give $u(t)\to \psi$ in $L^2$ as $t\downarrow 0$.