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For a uniform elastic filament clamped below and free above, its own weight gives filament tension $T(z)=-\lambda g(L-z)$. With filament bending modulus $B=EI$, the static slope $u$ satisfies $B{u}^{\prime \prime }+\lambda g(L-z)u=0$, $u(0)=0$, and $u^{\prime }(L)=0$. Setting $\eta =(2/3)\sqrt{\lambda g/B}(L-z{)}^{3/2}$ gives $u={\eta }^{1/3}{J}_{-1/3}(\eta )$ up to amplitude; the $J_{1/3}$ branch violates the zero-moment condition. The first positive zero $j_{-1/3,1}$ therefore yields $\lambda g{L}_c^3/B\simeq 7.83735$. This is a self-buckling of a vertical rod threshold, not a constant-end-load Euler threshold.

An inextensible filament keeps the length of every material segment fixed. With material arc length $s$, its centerline obeys $|{\mathbf{r}}_s|=1$ and has unit tangent vector $\mathbf{t}={\mathbf{r}}_s$. A filament tension acts as a Lagrange multiplier for this constraint.

Put $B=E{I}_2$, the filament bending modulus in the easier bending direction; $I_2$ is the relevant second moment of area. Take filament tension as positive in extension. A cut at height $z$ supports the weight of the material above it, so its axial force is compressive:The small transverse displacement induces a transverse tension force density $(T{X}_z{)}_z$, while bending produces $-B{X}_{zzzz}$. Their balance givesThe same result follows from elastic energy and gravitational energy. To quadratic order, local inextensibility shortens the height of the material point labelled $z$ by $\frac{1}{2}{\int }_0^z{X}_z(s{)}^{2}ds$. Integrating its weight and interchanging the integrals gives the second-order potential energyTwo integration by parts steps in its first variation giveThe clamped boundary conditions fix displacement and slope at the base, and the unconstrained tip must have zero bending moment and transverse end force. Thus the complete four conditions areSince the filament tension vanishes at the tip, the last condition is $X_{zzz}(L)=0$. Integrate the ordinary differential equation once: $B{X}_{zzz}-T{X}_z$ is constant, and its tip value makes that constant zero. For the slope $u=X_z$, it follows thatThe other tip condition $u_{zz}(L)=0$ already follows from this equation for a regular solution; it is not a third independent condition on the second-order slope problem. Finally reconstruct $X(z)={\int }_0^{z}u(s)ds$ to impose the base position.

Set $x=L-z$, $\alpha =\lambda g/B$, and $\eta =(2/3)\sqrt{\alpha }{x}^{3/2}$. Then $d\eta /dx=3\eta /(2x)$ and $d^2\eta /d{x}^2=3\eta /(4x^2)$, so $u_{xx}+\alpha xu=0$ becomesSubstituting $u={\eta }^{1/3}F(\eta )$ givesThis is the Bessel differential equation of order $1/3$, whose two independent Bessel functions of the first kind giveThe tip is $x=\eta =0$. To decide which branch is admissible, retain both leading behaviours, not just the fact that $u$ stays finite:Since ${\eta }^2\propto x^3$ and ${\eta }^{2/3}\propto x$, the first branch has zero tip derivative, while the second has a nonzero tip derivative. The moment-free condition therefore sets $b=0$, and it does not require the tip slope itself to vanish. The base condition now requiresLet $j_{-1/3,1}$ be the first positive zero. The Bessel threshold for self-buckling of a vertical rod isTo identify this as the first loss of stability, write the quadratic energy in terms of the slope as $\frac{1}{2}[B\int u_z^{2}dz-\lambda g\int (L-z)u^{2}dz]$. On functions with $u(0)=0$, its critical coefficient is the minimum Rayleigh quotient $\int u_z^{2}dz/\int (L-z)u^{2}dz$; variation supplies the moment-free condition at the tip and the same Sturm-Liouville problem. The first zero gives the lowest eigenvalue, so the straight state is stable below this load, neutral at it, and unstable above it, with the easier $I_2$ direction buckling first.

For spatially varying tension in filament bending, keep the derivative of the filament tension as well as the curvature term. The first variation isTherefore the Euler-Lagrange equation and fluctuation operator areFor real, sufficiently smooth $\sigma$, the boundary form is the bending boundary form minus $[\sigma (\overline{u}{v}^{\prime }-\overline{u^{\prime }}v){]}_0^L$. Because $\sigma$ vanishes at both ends, the four self-adjoint endpoint conditions for filament bending still apply. The natural endpoint force also reduces there to the bending shear term. Thus $K_{\sigma }$ is a self-adjoint fourth-order scalar differential operator on the same chosen domain, with compact resolvent.

Choose a real orthonormal basis ${\psi }_n$ of eigenfunctions, $K_{\sigma }{\psi }_n={\lambda }_n{\psi }_n$, and write $h={\sum }_n{b}_n{\psi }_n$. Using the endpoint conditions in integration by parts givesThe equipartition theorem now gives, on the strictly positive subspace,This is a formal modal construction; no explicit eigenfunctions are needed. Nonnegative filament tension makes the energy nonnegative. Any surviving zero-energy filament modes must again be fixed. If signed $\sigma$ permits compression, self-adjointness still holds but does not guarantee a canonical ensemble: sufficiently strong compression can create negative eigenvalues and Euler buckling of an elastic filament. For instance, on $A=L=1$, take $\sigma =-Cx(1-x)$ and the clamped trial function $h=x^2(1-x{)}^2$. Thenso the energy is negative when $C>252$, despite $\sigma (0)=\sigma (1)=0$. The equipartition theorem requires a stable positive quadratic energy, not merely a real modal spectrum.

The first term is the local viscous force per unit length from resistive-force theory. With unit tangent vector $\mathbf{t}={\mathbf{r}}_s$, the drag tensor gives $-c_{\parallel }{\mathbf{v}}_{\parallel }-c_{\perp }{\mathbf{v}}_{\perp }$, where $\mathbf{v}={\mathbf{r}}_t$ is velocity relative to the fluid.

The second term is the bending force of an inextensible filament. Varying its energy $E_b=(B/2)\int |{\mathbf{r}}_{ss}{|}^{2}ds$ gives bulk force density $-B{\mathbf{r}}_{ssss}$. The final term, ${\partial }_s(T{\mathbf{r}}_s)$, is the force density from filament tension; $T$ is the Lagrange multiplier enforcing $|{\mathbf{r}}_s|=1$. At low Reynolds number, these forces balance without a filament acceleration term.

With mass, length, and time dimensions $M,\mathsf{L},\mathsf{T}$, the coefficients haveThus the drag coefficients have the dimensions of dynamic viscosity, the filament bending modulus has dimensions force times length squared, and tension has dimensions force.

In the small-slope Monge representation, $s=x+O(y_x^2)$, $\mathbf{t}=(1,y_x)+O(y_x^2)$, and the leading transverse velocity is $y_t$. With no imposed axial prestress, the induced filament tension is second order in transverse amplitude, so its transverse contribution is higher order. The transverse component of force balance reduces to the small-slope elastohydrodynamic filament equationAn externally imposed zeroth-order tension would instead add $T{y}_{xx}$.

For the transverse motion, the off-diagonal part of the resistive-force theory tensor gives axial propulsive force density $(c_{\perp }-c_{\parallel })y_x{y}_t$ to second order. Its integral isSubstitute the bending equation and integrate by parts:The boundary expression for transverse filament thrust is thereforeOnly endpoint slope, bending moment, and shear enter this expression. Exact inextensibility also generates second-order longitudinal material motion; if that motion is retained in instantaneous total axial drag, it adds $-c_{\parallel }\int v_{x}dx$. For a periodic deformation with fixed axial base position, that additional term has zero cycle average. Thus the displayed formula is the transverse propulsive contribution and also gives the cycle-averaged propulsive force.

For a straight, unloaded inextensible filament with small transverse displacement, resistive-force theory gives viscous force density $-c_{\perp }{y}_t$. Variation of the bending energy gives $-B{y}_{xxxx}$, while the induced filament tension is higher order. Their instantaneous balance yields $c_{\perp }{y}_t=-B{y}_{xxxx}$. It is an overdamped fourth-order diffusion equation: short-wavelength bends relax much faster than long ones. Wiggins and Goldstein's flexive-propulsion paper develops the elastic-wave mechanism for low-Reynolds-number propulsion.

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.