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An integral representation expresses a function through an integral of specified data and an integral kernel. Green-function representations solve differential equations this way, while the Bromwich inversion formula reconstructs a function from its Laplace transform.

First use a homogeneous solution to meet the nonzero endpoint values:It satisfies $y_h^{\prime \prime }+k^2{y}_h=0$, $y_h(0)=a$ and $y_h(\pi )=b$. The Green-function representation then givesApplying the differential operator to the integral gives $f(x)$ by the defining Dirac delta property of $G$, and that integral vanishes at the endpoints. For continuous $f$ this is a classical solution; integrable $f$ also gives the corresponding weak solution. The same nonresonant homogeneous argument as in part (a) proves uniqueness.

Interpret the three conditions at zero as an initial-value problem on $x\ge 0$, with interior source $\xi >0$. A causal Green function satisfies ${\mathcal{L}}_{x}G=\delta (x-\xi )$ and vanishes for $x<\xi$. Since the leading coefficient of the third derivative is one, the jump conditions for a third-order Green function require $G$ and $G^{\prime }$ continuous and $G^{\prime \prime }$ to jump by one. For $t=x-\xi >0$, the homogeneous roots are $0,2,-1$. Solving $g(0)=g^{\prime }(0)=0$, $g^{\prime \prime }(0)=1$ yieldsFor $\xi >0$ it vanishes near zero, so all three initial boundary conditions hold. The matching conditions ensure that ${\partial }_x^{3}G$ supplies precisely a Dirac delta, without unwanted delta derivatives.

The homogeneous solution carrying the nonzero initial data is $y_h=\frac{1}{2}-\frac{1}{6}{e}^{2x}+\frac{2}{3}{e}^{-x}$. The Green-function representation therefore givesFor $x<1$ the integral vanishes. For $x\ge 1$, set $t=x-1$ and integrate to obtain $\frac{e}{6}(e^{2t}-3e^t-e^{-t}+3)$. ThusThe bracket and its first two derivatives vanish at $x=1$, so $y,y^{\prime },y^{\prime \prime }$ match continuously there; its third derivative jumps by the stated forcing. The value assigned to $\Theta (0)$ is immaterial. The source at the initial endpoint $\xi =0$ would need a separate endpoint-delta convention, which is not used in this integral.

The Green-function representation givesThe bracket simplifies to $x(x-1)/2$, soIt vanishes at both endpoints, and applying $(D-2{)}^2$ returns $e^{2x}$.

Apply the time Laplace transform to the Airy equation. Extend $g$ past $T$ in any suitable way and writeFor example, setting $g=0$ after $T$ makes $G(p)={\int }_0^T{e}^{-pt}g(t)dt$; the final solution for $t<T$ is independent of this extension. The Laplace transform of a derivative givesTake the principal cube root $\lambda =p^{1/3}$ for $\mathrm{Re}p>0$. The three characteristic roots of the spatial ordinary differential equation areSince $|\arg \lambda |<\pi /6$, only $-\lambda$ has negative real part. Thus spatial decay leaves just one homogeneous exponential, which the single prescribed Neumann boundary condition determines.

The Airy resolvent kernel on the whole real line isIt decays at both ends, is continuous together with its first derivative, and satisfies $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. Consequently $({\partial }_s^3+p)R_p={\delta }_0$ as a distribution. A particular solution is the Green-function representationAdding the decaying homogeneous mode to impose the boundary derivative givesEvery term is known. If ${\tilde{u}}_0(r)={\int }_0^{\infty }{e}^{-ry}{u}_0(y)dy$ denotes the spatial Laplace transform, thenThe Bromwich inversion formula now gives the required integral representation:Here $\gamma >0$ is to the right of any singularities required by the growth of the data. The usual decay or growth hypotheses are understood for this Laplace transform construction; when absolute inversion is unavailable, the vertical integral is interpreted as the limit of truncated Bromwich contours. The transformed ordinary differential equation verifies the partial differential equation and initial condition, while differentiating at $x=0$ gives exactly $G(p)$ and hence $u_x(0,t)=g(t)$. The derivative compatibility $u_0^{\prime }(0)=g(0)$ makes the two data agree at the corner.

Choose the left homogeneous solution $v(x)=\sinh x$ and the right one $w(x)=\sinh (1-x)$. Their Wronskian is $v{w}^{\prime }-v^{\prime }w=-\sinh 1$. The Dirichlet modified Helmholtz Green function on an interval, with operator sign $D_x^2-1$, is thereforeIt vanishes at both endpoints, is continuous at $x=\xi$, and its first derivative jumps by $+1$, giving $(D_x^2-1)G=\delta (x-\xi )$. The Green-function representation then givesThe two integrals are $(e^{2x}-1)/4-x/2$ and $e(1-x)/2-(e-e^{2x-1})/4$, respectively. Substituting them and simplifying givesDirect differentiation verifies $u^{\prime \prime }-u=e^x$ and both zero boundary values. The negative kernel also explains why the solution is negative inside the interval despite the positive right-hand side.

The Green-function representation isThe delta identity gives $L{y}_p=f$ and the endpoint conditions follow from those of $G$. If a nonzero homogeneous solution also satisfies both endpoint conditions, it may be added; uniqueness is not automatic in that case.

Insert the source into the Green-function representation and split the integral at its source discontinuity in derivative:Using ${\sin }^{2}x+{\cos }^{2}x=1$, the answer isBoth endpoint values are zero. As a direct check, conjugating by $e^x$ sends $y^{\prime \prime }-2y^{\prime }+2y$ to $e^x(v^{\prime \prime }+v)$, and $v=1-\cos x-\sin x$ gives $v^{\prime \prime }+v=1$. Thus the differential equation and both boundary conditions are satisfied.

Use the Green-function representation to integrate over the charged segment:For $y\ne 0$, a primitive with respect to $u$ isas direct differentiation verifies using the derivative of inverse tangent. ThereforeThis expression is even in $y$, since $y\arctan (u/y)$ is even. On $y=0$, take the continuous limit $F(u,0)=u\log u^2-2u$, with $u\log u^2=0$ at $u=0$. The finite segment has a logarithmically growing potential at large distance, but its electric field decays; the boundary condition constrains the field rather than the potential.

Differentiate the Green-function representation, valid away from the charged sheet:For $0<x<L$, the two inverse tangent limits at $y\downarrow 0$ are $\pi /2$ and $-\pi /2$; approaching from below reverses both signs. ThusThis agrees with the normal electric field jump obtained by applying Gauss's law to a thin pillbox enclosing unit surface charge density.