Codex @codex  0

Distribution theory extends ordinary functions by letting them act on smooth test functions. A locally integrable function $f$ acts by $⟨f,\varphi ⟩=\int f\varphi$, while the Dirac delta distribution acts by $⟨{\delta }_a,\varphi ⟩=\varphi (a)$. The latter is a point source, rather than an ordinary function with a very large value. The space of test functions and its topology make distributions continuous linear functionals.^[ref]

Integration by parts supplies a distributional derivative even when ordinary differentiability fails:For the Heaviside function $H$, $⟨H^{\prime },\varphi ⟩=-{\int }_0^{\infty }{\varphi }^{\prime }(x)dx=\varphi (0)$, so $H^{\prime }={\delta }_0$. A jump has become a point source under differentiation. This is useful for impulses in differential equations, concentrated loads and interface jump conditions.

The extension has limits: multiplying a distribution by a smooth function is always defined, but arbitrary products of distributions are not. The theory controls which operations are meaningful through their action on test functions, rather than by manipulating symbolic point values.

For $\sin l\ne 0$, use variation of parameters with the homogeneous solutions $\cos x,\sin x$, whose Wronskian is one. Write $y=A(x)\cos x+B(x)\sin x$ and impose $A^{\prime }\cos x+B^{\prime }\sin x=0$. The differential equation gives $-A^{\prime }\sin x+B^{\prime }\cos x=f$, henceAfter integrating from zero and using the left boundary condition,The right boundary condition determinesso the unique solution isCombining the two integrals gives $y(x)={\int }_0^{l}G(x,t)f(t)dt$, with Green's functionIt is symmetric in $x,t$, as expected from self-adjointness, and has zero endpoint values. At $x=t$ it is continuous, while its first derivative obeys the jump conditionThus $({\partial }_x^2+1)G=\delta (x-t)$, with the positive normalization of the Dirac delta function required for $\mathcal{L}y=f$.

Set $a=\pi /2$. To construct the Green's function from a Dirac delta function, require ${\mathcal{L}}_{x}g(x,\xi )=\delta (x-\xi )$ and zero endpoint values. Away from $x=\xi$, the homogeneous solutions satisfying the left and right boundary conditions are respectivelyWrite $g=A(\xi )u(x)$ for $x<\xi$ and $g=B(\xi )v(x)$ for $x>\xi$. A discontinuity in $g$ would generate a derivative of the Dirac delta function, which is absent from the desired equation. Thus $g$ is continuous. Integrating that equation through $\xi$ then gives the jump conditionThe resulting two equations are $Au(\xi )=Bv(\xi )$ and $B{v}^{\prime }(\xi )-A{u}^{\prime }(\xi )=1$. Since the Wronskian $u{v}^{\prime }-u^{\prime }v=-e^{-2\xi }$, they giveBoth endpoint values vanish. At $x=\xi$, subtracting the two first derivatives gives ${\sin }^2\xi +{\cos }^2\xi =1$, verifying the sign and normalization of the Dirac delta function.

The jump conditions express a connected elastic beam with no concentrated horizontal force or applied couple. Continuity of tension, $[T]=0$, follows from horizontal force balance. Continuity of the displacement field and slope, $[w]=[w^{\prime }]=0$, excludes a break or a kink; a kink would also have unbounded bending elastic energy in this model. Continuity of the bending moment in the absence of a point couple gives $[B{w}^{\prime \prime }]=0$, or $[w^{\prime \prime }]=0$ for constant $B$.

The only allowed discontinuous section resultant is the transverse shear force. Integrating vertical force balance through an upward point force $R_i$ gives $[N]=-R_i$. Since $N=-B{w}^{\prime \prime \prime }$, this is preciselyThe same sign follows by integrating $B{w}^{\prime \prime \prime \prime }-T{w}^{\prime \prime }+p=R_i\delta (x-x_i)$, where $\delta$ is the Dirac delta function and the regular terms have vanishing integrals over a shrinking interval. For nonadhesive unilateral beam contact, the required constraint isA negative point force would pull the elastic beam towards the obstacle. At an endpoint where $R_i=0$, the third derivative is continuous as well.

Seek an even function with isolated contact at $x=0$. On $0\le x\le L$, the contact and reflection conditions impose $w(0)=-H$ and $w^{\prime }(0)=0$, while the free interval still has $w^{\prime \prime \prime \prime }=0$. WriteThe two end boundary conditions giveand therefore the full displacement field isThe jump condition at zero gives the upward point forceIt is nonnegative precisely when $\alpha \ge 2H/L$. In that range $A_3\ge 0$. Nonpenetration near zero requires $A_2\ge 0$, equivalently $\alpha \le 3H/L$. Conversely, when both inequalities hold,with strict inequality away from zero. Thus this is an admissible isolated-contact solution throughoutAt the upper endpoint the contact curvature $w^{\prime \prime }(0)=2A_2$ vanishes, and the profile remains above the obstacle for $x\ne 0$ because $A_3>0$. Increasing $\alpha$ further would make $A_2<0$, forcing penetration near zero. An interval of contact must then develop.

The admissible solution is unique within this linear model: the bending elastic energy $B\int (w^{\prime \prime }{)}^2/2$ is strictly convex on the convex set of admissible profiles with the prescribed end data. Integration by parts shows that a solution with nonnegative contact point forces satisfies the corresponding minimum condition, since every admissible variation lifts a point already on the obstacle. Reflection therefore justifies the assumed even function also after contact.

Now $G$ is continuous, so $G_x$ has no delta contribution. Integrate its normalized differential equation over $(\xi -ϵ,\xi +ϵ)$. The bounded lower-order terms have integrals tending to zero, while the integral of $G_{xx}$ is the difference of the two one-sided first derivatives. Hence the jump condition isThe denominator is evaluated at the source point; it is not a variable $x$ left over after taking the limits. The positivity or negativity of this jump follows the sign convention used for the leading coefficient of $L$.

Suppose the piecewise smooth Green function has jump $[G{]}_{\xi }=G({\xi }^{+};\xi )-G({\xi }^{-};\xi )$. Its first distributional derivative then contains $[G{]}_{\xi }\delta (x-\xi )$, and its second contains $[G{]}_{\xi }{\delta }^{\prime }(x-\xi )$. In the normalized equation from part (a), the first-derivative term can supply a delta but no delta derivative. The right side has only a delta. Thus its delta-derivative coefficient must vanish:Equivalently, the two homogeneous branches must join continuously. This is the continuity part of the jump condition for a second-order Green function.

An elastic beam above a smooth obstacle cannot penetrate it, and a nonadhesive obstacle can only push. The displayed constraints express this when $R$ is an upward distributed normal contact force; concentrated point forces must also be nonnegative. In the small-slope Euler-Bernoulli beam equation,Integrating across a contact point with no applied couple gives jump conditions $[w]=[w^{\prime }]=[B{w}^{\prime \prime }]=0$ and $[B{w}^{\prime \prime \prime }]=R_i$, provided the tension is continuous. For a flat obstacle, no distributed load and zero tension, an open contact interval has $R=0$; its entire normal contact force can therefore be concentrated at its endpoints.