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Let $D_{ij}=T_{ij}^{+}-T_{ij}^{-}$, so the Lighthill stress tensor is $T_{ij}^{-}+D_{ij}H(S)$. Write $S_i={\partial }_{i}S$ and $S_{ij}={\partial }_i{\partial }_{j}S$. The distributional derivative of the Heaviside step function gives ${\partial }_{i}H(S)=S_i\delta (S)$. Applying the product rule a second time yields the surface sources of a discontinuous Lighthill stress tensor:Repeated indices are summed. The first line gives the two bulk acoustic quadrupole distributions. The second line consists of additional surface delta distributions and their derivatives, all supported on the shock wave $S=0$. For a symmetric Lighthill stress tensor, the first two coefficients of $\delta (S)$ combine to $2D_{ij,i}{S}_j$ after relabelling indices.

A discontinuity therefore adds both a surface-delta source and a surface-delta-derivative source. The level function need not be a signed distance: retaining its derivatives makes the expression invariant under a smooth reparametrization of the same surface. If the two bulk fields are only continuously differentiable, their second derivatives in the first line are read weakly; no classical second derivative is being assumed. The coefficients multiplying ${\delta }^{\prime }(S)$ are kept as extensions, not replaced prematurely by traces.

Choose the shock wave normal to point from $S<0$ to $S>0$:The latter identity follows by differentiating $S({\mathbf{x}}_s(t),t)=0$ along the moving surface. Write $[a]=a^{+}-a^{-}$ and $[\mathbf{b}]={\mathbf{b}}^{+}-{\mathbf{b}}^{-}$. The distributional derivative of the Heaviside step function is ${\partial }_{t}H(S)=S_t\delta (S)$ and $\nabla H(S)=\nabla S\delta (S)$. The regular terms cancel using the conservation equations on each side, leavingThus the explicitly requested vector isOnly its normal component matters; adding tangential velocity to the surface parametrization changes neither result. The invariant surface delta distribution is ${\delta }_s=|\nabla S|\delta (S)$, so the formula is the moving-interface conservation jump identity ${\partial }_{t}a+\nabla \cdot \mathbf{b}=([\mathbf{b}]\cdot \mathbf{n}-v_n[a]){\delta }_s$.

The distributional derivative of the Heaviside step function gives $H^{\prime }={\delta }_0$, while $((x-1)H(x-1){)}^{\prime \prime }={\delta }_1$. A particular solution is therefore $H(x)-2(x-1)H(x-1)$.

The difference from any other solution has second derivative zero. Applying a distribution with zero derivative is constant twice shows that difference is affine. ThusThe ramp function is continuous at $x=1$, and its first derivative jumps there, producing the indicated delta source in its second derivative.

The distributional derivative of the Heaviside step function is $H^{\prime }={\delta }_0$. The Leibniz rule and $x{\delta }_0=0$ therefore give

For a compactly supported test function $f$,Thus this is the distributional derivative of the Heaviside step function: $H^{\prime }=\delta$.

Write a piecewise smooth Lighthill stress tensor as $T_{ij}=T_{ij}^{-}+D_{ij}H(S)$, where $D_{ij}=T_{ij}^{+}-T_{ij}^{-}$ and the regular level surface $S=0$ is a shock wave. The distributional derivative of the Heaviside step function givesHere repeated indices are summed, $S_i={\partial }_{i}S$ and $S_{ij}={\partial }_i{\partial }_{j}S$. Keep the coefficients as smooth extensions before multiplying distributions: prematurely replacing the coefficient of ${\delta }^{\prime }(S)$ by its surface trace can lose a $\delta (S)$ contribution. The extra surface delta distributions and their derivatives are supported entirely on the discontinuity.