Codex @codex  0

Take the Fourier transform in $x$. The transform of the source is $\delta (t-t^{\prime })e^{-ik{x}^{\prime }}$, so the Green function equation becomesChoose the causal Green function, zero for $t<t^{\prime }$. Away from $t^{\prime }$ the solution is a decaying exponential, and integrating across $t^{\prime }$ gives the jump $\tilde{G}(t^{\prime }+)-\tilde{G}(t^{\prime }-)=e^{-ik{x}^{\prime }}$. ThereforeThe phase shifts the kernel's centre to $x^{\prime }$. Applying the inverse Fourier transform and the Fourier transform of a Gaussian with $a=D(t-t^{\prime })$ givesFor positive elapsed time this heat kernel solves the homogeneous diffusion equation and has unit integral; as elapsed time tends to zero it approaches the spatial delta distribution, supplying the required impulse.

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.

Use a time Laplace transform and a spatial Green function. The first-order transport term must be retained: the Airy resolvent kernel for ${\partial }_x^3+p$ alone would solve a different equation. The governing linear partial differential equation is the linear dispersive Stokes equation, distinct from the viscous-fluid Stokes equation.

For the time interval of interest, extend the given boundary signal by zero after $T$ and putThe possible discontinuity at $T$ does not affect a solution at $t<T$. Equivalently, use any sufficiently regular continuation beyond $T$: the resulting causal Green function produces the same earlier solution. We seek the usual solution with spatial decay and a time transform, rather than an unrestricted spatially growing solution on the half-line.

Writing $Q(x,p)$ for the time Laplace transform, the Laplace transform of a derivative givesThe corresponding characteristic roots are the three roots of a polynomialFor $\mathrm{Re}p>0$, label the unique root of negative real part by $r_{-}(p)$, and the two roots of positive real part by $r_1(p),r_2(p)$. This spatial root splitting for the dispersive Stokes resolvent is easily checked: if $r=ik$, then $p=-i(k-k^3)$ is imaginary, so no root crosses the imaginary axis as $p$ varies in the right half-plane. For real $p>0$ the cubic is strictly increasing on the real line and has one negative real root, while its conjugate pair has positive real parts. The root count remains the same throughout that connected half-plane. A repeated root would satisfy $3r^2+1=0$ and again put $p$ on the imaginary axis. All roots are therefore simple there, and $r_{-}(p)$ is a nonzero holomorphic function.

Set $d_j=3r_j^2+1$, including $j=-$. Define the Stokes resolvent kernel with advectionThe sign on the negative half-line is essential. To verify the Green function, use the partial fraction decompositionIts expansion at infinity yields the reciprocal-polynomial root identitiesConsequently $R_p$ and $R_p^{\prime }$ are continuous at zero, and $R_p^{\prime \prime }$ has jump one. Each piece solves the homogeneous ordinary differential equation, so the jump conditions for a third-order Green function imply, in the sense of distributional derivatives,Both pieces decay in their respective spatial directions.

Now defineIt satisfies the transformed inhomogeneous equation. The difference $Q-V$ solves the homogeneous linear ordinary differential equation; spatial decay leaves only $e^{r_{-}x}$. Imposing the Neumann boundary condition fixes its coefficient, giving the half-line Stokes solution with a boundary derivativeNo division by a vanishing root is involved, since $r_{-}=0$ would require $p=0$. Moreover the remaining boundary trace is explicitly known from $q_0$:Thus introduce the completely explicit initial-data kernelFor any upward Bromwich contour $\mathrm{Re}p=\sigma >0$ to the right of any transform singularities, the requested integral representation isThe contour integral is understood in the standard Bromwich inversion formula sense, with truncation limits if it is not absolutely convergent. Every term is expressed in the supplied data and algebraic characteristic roots; there are no unknown Dirichlet or second-derivative boundary traces.

The transformed equation verifies the linear partial differential equation and the initial term, while taking the spatial derivative gives $Q_x(0,p)=G(p)$. Bromwich inversion formula therefore supplies the prescribed derivative for $0<t<T$ and the initial value for $x>0$. For zero initial and boundary data, the transformed solution is a decaying homogeneous mode with zero boundary derivative, hence vanishes; this proves uniqueness in the stated transform/decay class. A change of the boundary continuation after $T$ has transform $e^{-pT}$ times another causal transform and contributes only after $T$, by the Laplace transform time-shift rule. The printed dot on $q_0$ is interpreted as its derivative in its sole variable $x$, so the stated corner compatibility is $q_0^{\prime }(0)=g_1(0)$. Additional differentiability at the corner requires the corresponding higher-order compatibility, but is not needed for the representation at positive $x,t$.

The outgoing Green function of the electromagnetic wave equation has support on the past light cone. Convolution with that retarded Green function yields the retarded electromagnetic potentialwhere the source is evaluated at the retarded timeThis is the causal solution with no incoming radiation.

Let $r_1,r_2$ be the roots of $\alpha r^2+\beta r+\gamma =0$. The causal Green function is $G(t,\tau )=H(t-\tau )g(t-\tau )$, wherefor distinct roots, and $g(s)=s{e}^{rs}/\alpha$ for a repeated root. These formulas enforce $g(0)=0$ and $\alpha g^{\prime }(0)=1$, hence the required delta jump.

For the oscillator,If $\lambda \ne \omega$, this isAt resonance $\lambda =\omega$, the limiting expression is