Euler-Bernoulli beam equation ($u_{tt}+c_b^2{u}_{xxxx}=0$)

The Euler-Bernoulli beam equation models transverse linear bending waves in a slender beam. The normalized equation has $c_b=1$ and may be reduced by the Schrodinger factorization of the elastic beam equation. Prescribing displacement and curvature at an endpoint supplies two boundary traces for its fourth spatial derivative.

A uniform elastic beam with endpoint conditions $y=y_{xx}=0$ has normal modes $\sin (n\pi x/L)$ for positive integers $n$. Substitute $y=X{e}^{-i\omega t}$ into $EI{y}_{xxxx}+{\rho }_{\ell }{y}_{tt}=0$ and impose the four endpoint conditions to obtain the displayed spectrum. Here ${\rho }_{\ell }$ is mass per unit length, $I$ is a second moment of area, and $EI$ is bending stiffness. frequencies proportional to $n^2$ distinguish flexural waves from the nondispersive ideal stretched string.

A prescribed initial beam profile may vanish at a pivot while having nonzero endpoint curvature. If the post-release pivot is moment-free, its bending moment condition is $EI{y}_{xx}=0$. The data then cannot have a classical solution with continuous curvature at the initial boundary corners. A finite-energy weak solution can still represent release from preparing end moments. For example $y_0=ax(x-L)$ has odd sine coefficients proportional to $n^{-3}$; $\sum n^4|A_n{|}^2$ converges, giving finite bending energy. The displacement series takes the prescribed initial shape, but termwise endpoint second derivatives cannot be identified with its interior curvature limit. This conservative beam evolution does not have the immediate smoothing of a heat equation.

For a unit-length elastic beam with unit bending stiffness, the Green function of $D_x^4$ with $G=G_x=0$ at both endpoints is the displayed kernel. It is piecewise cubic and continuous through its second derivative, while $G_{xxx}(\xi +,\xi )-G_{xxx}(\xi -,\xi )=1$. A constructive form is $G=a{x}^2+b{x}^3+(x-\xi {)}_{+}^3/6$, with $a=\xi (1-\xi {)}^2/2$ and $b=-(1-\xi {)}^2(1+2\xi )/6$. The right endpoint conditions determine these coefficients. Thus $y(x)={\int }_0^{1}G(x,\xi )f(\xi )d\xi$ solves the clamped static beam problem.

The clamped-beam Green function is symmetric. Its explicit formula depends only on $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. More generally, two integrations by parts give $\int u{v}^{\prime \prime \prime \prime }dx=\int u^{\prime \prime }{v}^{\prime \prime }dx=\int u^{\prime \prime \prime \prime }vdx$ when $u=u^{\prime }=v=v^{\prime }=0$ at both endpoints. Applying this identity to two point-load responses shows that the displacement at one point caused by a unit force at a second equals the displacement at the second caused by a unit force at the first.

A straight slender elastic beam under axial compression loses stability when bending can decrease its total potential energy. With pinned ends a distance $L$ apart, its quadratic potential is $\frac{1}{2}{\int }_0^L[E{I}_A(y^{\prime \prime }{)}^2-P(y^{\prime }{)}^2]dx$. Stationarity gives $E{I}_A{y}^{\prime \prime \prime \prime }+P{y}^{\prime \prime }=0$, with $y=y^{\prime \prime }=0$ at each end. The first nontrivial mode is $\sin (\pi x/L)$, at $P_{\mathrm{cr}}={\pi }^{2}E{I}_A/L^2$. More generally $P_{\mathrm{cr}}={\pi }^{2}E{I}_A/(KL{)}^2$, with an effective-length factor $K$ determined by endpoint constraints: $K=1$ for pinned-pinned, $2$ for fixed-free and $1/2$ for fixed-fixed. Real imperfections remove the perfectly sharp onset, and short columns can yield before this elastic instability.

For a pin-ended slender beam with unstressed profile $y_0$, bending moment depends on curvature change: $M=EI(y-y_0)$. An axial dead load gives $EI(y-y_0)=-Fy$ in the small-slope model.

With $y_0=Y\sin (\pi x/L)$ and $F_c={\pi }^{2}EI/L^2$, the equilibrium branch connected to zero load has $y=Y\sin (\pi x/L)/(1-F/F_c)$. The pole as $F$ approaches $F_c$ marks breakdown of the small-deflection model, not a physically infinite beam deflection.

Positive compression $\alpha$ contributes negative quadratic potential energy $-\alpha {\varphi }_x^2/2$, while positive bending stiffness $\beta$ contributes $\beta {\varphi }_{xx}^2/2$. A plane wave has ${\omega }^2=(\beta k^4-\alpha k^2)/\rho$, so sufficiently long waves grow exponentially. For displacement-fixed, moment-free endpoints at $0,l$, the natural boundary conditions for a free endpoint on the slope give ${\varphi }_{xx}=0$ as well as $\varphi =0$. The modes are $\sin (n\pi x/l)$, and the first Euler buckling of an elastic filament threshold is ${\alpha }_c=\beta {\pi }^2/l^2$. For truly clamped ends, $\varphi ={\varphi }_x=0$, the threshold is instead $4{\pi }^2\beta /l^2$, with first static mode $1-\cos (2\pi x/l)$. More generally,with the infimum taken over the specified endpoint class. At threshold there is a static zero mode and a marginal linear-in-time partner in the linear dynamics; above it the straight configuration is unstable.

Writing $q=u+iv$ in the free Schrodinger equation gives $u_t=-v_{xx}$ and $v_t=u_{xx}$, hence $u_{tt}+u_{xxxx}=0$. To encode initial velocity $u_1$, choose the decaying primitive $v_0(x)=-{\int }_x^{\infty }(s-x)u_1(s)ds$. A prescribed endpoint curvature becomes the time derivative of the imaginary boundary trace. This turns the normalized Euler-Bernoulli beam equation into a complex second-order boundary problem.