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A jump condition relates the one-sided limits of a solution or its derivatives across a discontinuity or singular forcing. Integrating a differential equation across a shrinking interval containing a Dirac delta source determines the permitted jump. A jump condition for an impulse and Rankine-Hugoniot condition are specific instances with different governing equations.

Integrate the first equation over $(1-\epsilon ,1+\epsilon )$. The integral of the ordinary cooling term tends to zero as $\epsilon \downarrow 0$, while the Dirac delta function integrates to one. Thus the jump condition for an impulse isAfter the jump there is no further forcing, so Newton's law of cooling givesEquivalently,The last term has jump minus one at time one and then decays at the same exponential rate as the unforced temperature.

The sign of the Dirac delta distribution in the equation is positive after the outer minus sign is expanded. Integrating through zero, the nonsingular potential contributes a term tending to zero, whereas the delta term contributes $3k\psi (0)$. The jump condition for an impulse is thereforeContinuity at zero gives $C_{+}=C_{-}=C$. The two one-sided derivatives are $-3kC/2$ and $+3kC/2$, whose difference is exactly $-3kC$. Hence the matching condition introduces no further restriction on $C$, and the globally bounded solution isThis continuous solution has precisely the derivative jump required for its second distributional derivative to cancel the delta term. Its amplitude remains arbitrary because no normalization has been imposed.

Away from the impulse, the harmonic oscillator equation has a homogeneous solution. Before $x=\pi /2$, the initial conditions select $y=\cos x$. The jump condition for an impulse requires continuity of $y$ and a unit jump in $y^{\prime }$: a jump in $y$ would produce a derivative of the Dirac delta distribution, which is absent from the forcing, while integrating through the impulse gives $[y^{\prime }]=1$.

Just before the impulse, $y=0$ and $y^{\prime }=-1$. Just afterwards, $y=0$ and $y^{\prime }=0$, so uniqueness for the harmonic oscillator equation makes the subsequent solution identically zero. Equivalently the causal response is $H(x-\pi /2)\sin (x-\pi /2)$, with $H$ the Heaviside step function, and it cancels the original oscillation.

The solution isThe impulse cancellation of a harmonic oscillator produces a continuous graph with a corner at $(\pi /2,0)$, followed by a horizontal zero line.

Write $G(x;\xi )=e^{2x}H(x;\xi )$. Since $(D-2{)}^2(e^{2x}H)=e^{2x}{H}^{\prime \prime }$, the Green function must have $H(0;\xi )=H(1;\xi )=0$, be continuous at $x=\xi$, and satisfy the jump condition for an impulse $H_x({\xi }^{+};\xi )-H_x({\xi }^{-};\xi )=e^{-2\xi }$. Solving the two linear pieces gives

Integrating the equation through $t=1$ gives the jump condition for an impulseFor $1<t<2$ the equation is again homogeneous, so propagation of the post-jump temperature givesTherefore

A causal Green function vanishes before the impulse. For $t>\tau$ it solves the homogeneous linear ordinary differential equation, whose characteristic roots are $-q\pm 2i$. Thus it has the formContinuity at $t=\tau$ is necessary to avoid a delta derivative in $G_{tt}$, so $A=0$. Integrating its forced equation across the impulse gives the jump condition for an impulse, $G_t({\tau }^{+},\tau )-G_t({\tau }^{-},\tau )=1$. The derivative from below is zero and the derivative from above is $2B$, givingHere $\Theta$ is the Heaviside step function. The zero initial data exclude any additional homogeneous response. By the superposition principle, the output for a general locally integrable input isThe upper endpoint causes no unwanted term in the first derivative because $G(t,t)=0$; in the second derivative its unit derivative jump reproduces the input.