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Let $u$ be the actual displacement field, let $u^{\prime }$ be an admissible trial displacement, and set $v=u^{\prime }-u$. On the displacement-prescribed part of the boundary, $v=0$. With actual stress $\sigma =D_{\epsilon }W(\epsilon )$, the supporting-plane inequality for the convex strain energy density givesBoth $\sigma$ and ${\sigma }_0$ satisfy static elastic equilibrium with the same body force, so $\nabla \cdot (\sigma -{\sigma }_0)=0$. On the traction-prescribed boundary their tractions agree. Since ${\epsilon }^{\prime }-\epsilon =\mathrm{sym}\nabla v$, integration by parts givesSubtracting the prescribed external-work representative proves the minimum potential energy principle in elasticity:Strict convexity gives uniqueness of the minimizing strain, although rigid displacements can remain when the boundary conditions allow them.

For the comparison bound, put $F(e,x)=W(e,x)-W_0(e)$ and use the lower conjugate, with its infimum sign convention,Its defining inequality is $F_{\ast }(\tau ,x)\le \tau :e-F(e,x)$, hence $W(e,x)\le W_0(e)+\tau :e-F_{\ast }(\tau ,x)$ for every $e$. Integrate this pointwise majorant for each compatible strain, then take the infimum over the same admissible set. This provesNo convexity of $F$ is used for this inequality. If $F_{\ast }(\tau ,x)=-\infty$, it yields only the extended-real upper bound $+\infty$. For example, $W_0=0$ and $W(e)=|e{|}^2/2$ give $F_{\ast }(\tau )=-\infty$ for every finite $\tau$. A useful finite bound requires a suitable comparison energy; choosing a sufficiently stiff quadratic comparison makes the usual excess energies concave. The derivative used below additionally presupposes finite differentiable duals, or an appropriate supergradient formulation.

Now normalize $|\Omega |=1$, take positive constant $C^0$, and write $F_r(e)=W_r(e)-e:C^{0}e/2$. For phasewise constant elastic polarization $\tau ={\sum }_r{\chi }_r{\tau }^r$, its lower conjugate contribution is constant within each phase. The remaining minimization isWrite the minimizing strain as $\hat{e}=E+\eta$, where $\eta$ comes from a zero-boundary displacement field. The static elastic equilibrium equation is $\nabla \cdot (C^0\eta +\tau )=0$, so the supplied elastic strain Green operator gives $\eta =-\Gamma \tau$. Its zero mean removes the cross term with the constant $E$. Using its reciprocity and projection property $\Gamma C^0\Gamma =\Gamma$, we obtainHere the brackets in the last two terms denote the spatial integral and tensor contraction, not just a phase average.

Let $c_r={\int }_{\Omega }{\chi }_r$. The zero output mean and reciprocity of $\Gamma$ imply that it annihilates constant elastic polarizations on either side. Thus subtracting $c_r$ and $c_s$ from the phase indicators does not change their bilinear integral. Define the fourth-order block tensorsConsequently $A_{rs}=A_{sr}^T$, ${\sum }_s{A}_{rs}=0$, andThese identities follow from reciprocity and ${\sum }_s{\chi }_s={\sum }_s{c}_s=1$; no phase independence assumption is needed. The phase polarization comparison bound, valid for every polarization with finite dual terms, is thereforeIn particular, it can be optimized over the phase elastic polarizations without changing the upper-bound direction.

To derive the final displayed form, impose the stationarity equations in the question. For $c_r>0$, put $e^r=D{F}_{r\ast }({\tau }^r)$; these are dual strains, initially distinct from the comparison solution's phase averages. Reciprocity of $A$ givesThus stationarity is exactlyThe comparison solution has phase meanso the dual strains become its phase means at stationarity. Multiplying the stationarity equations by ${\tau }^r$ and summing gives ${\sum }_{r,s}{\tau }^r:A_{rs}{\tau }^s={\sum }_r{c}_r{\tau }^r:(E-e^r)$.

For clarity, the concave biconjugate here uses another infimum: $F_{r\ast \ast }(e)={\inf }_s\{s:e-F_{r\ast }(s)\}$. The lower conjugate $F_{r\ast }$ is concave, since it is an infimum of affine functions. At a differentiable finite point, its supporting-plane inequality shows that $s={\tau }^r$ minimizes this last expression when $e=e^r$. HenceSubstitute both identities into $\mathcal{B}$ to obtainThis is the requested final bound with the accompanying phase equations. It remains valid at any well-defined stationary choice; asserting existence of such a differentiable choice for arbitrary $W_0$ would require extra hypotheses. Without stationarity, the earlier $\mathcal{B}(E,\tau )$ is the correct general bound. In the concave closed excess-energy case $F_{r\ast \ast }=F_r$; for general excess energies replacing the concave biconjugate by $F_r$ without proof would reverse the needed majorant logic.

Let $\Omega$ be the solid exterior to the cavity, and choose $\mathbf{n}$ outward from the solid, hence into the cavity on its boundary $S$. Subtraction of the incident solution from the total solution gives the scattered displacement field equationsThe traction-free cavity condition for the total field gives ${\sigma }_{ij}^s{n}_j=-{\sigma }_{ij}^0{n}_j$ on $S$. The scattered field has zero initial displacement and initial velocity before the pulse reaches the cavity and satisfies an outgoing radiation condition, excluding an additional incoming elastic wave. Equivalently one seeks the causal response to this prescribed boundary traction. Although the incident plane pulse extends across an infinite plane, the causal scattered field from the bounded cavity has finite propagation distance at each finite time.

To prove energy uniqueness for traction-driven elasticity, take the difference $\mathbf{w}$ of two solutions with the same initial conditions, body force, boundary traction and outgoing condition. It satisfiesFor a large enclosing ball $B_L$, multiply by ${\dot{w}}_i$ and integrate over $\Omega \cap B_L$. Integration by parts givesSymmetry of stress replaces the last derivative by ${\dot{e}}_{ij}(w)$, and major symmetry of the elastic stiffness tensor makes that integral the derivative of the strain energy. ThereforeThe cavity contribution is zero. On any fixed finite time interval choose $L$ beyond the causal support of the scattered difference; the outer contribution is then zero as well. This gives conservation of its nonnegative elastic energy. Positive mass density and positive strain energy imply that zero initial energy forces $\dot{\mathbf{w}}=0$ everywhere; the zero initial displacement field then gives $\mathbf{w}=0$. The same argument works for finite-energy outgoing solutions by taking the limit of enclosing surfaces and excluding incoming energy. Hence the causal scattered field ${\mathbf{u}}^s$ is unique. Specifying the incident field and cavity traction alone without the initial and outgoing conditions would not establish uniqueness.

Use fixed spatial coordinates in a stationary heterogeneous Earth and define the operatorHere the elastic stiffness tensor has its minor and major symmetries, and the physical body force per unit volume is $q_i=\rho f_i$. The retarded elastodynamic Green tensor $G_i^j(\mathbf{x},\mathbf{\xi },t)$ is the displacement in direction $i$ due to a unit physical force impulse in direction $j$ at $\mathbf{\xi }$:It satisfies the actual Earth's homogeneous boundary conditions, including zero traction on the free surface and continuity of displacement and traction at bonded internal interfaces. Integration through $t=0$ gives $\rho (\mathbf{x}){\partial }_t{G}_i^j(0^{+})={\delta }_i^j\delta (\mathbf{x}-\mathbf{\xi })$, with $G(0^{+})=0$. No spherical symmetry is required.

For zero initial conditions, form the convolutionApplying $\mathcal{L}$ to this expression uses the defining Dirac delta function to give $\rho (\mathbf{x})f_i(\mathbf{x},t)$. Causality gives zero initial conditions, and each Green column supplies the required boundary conditions. The positive-energy uniqueness argument then identifies this expression with the physical response. The source factor is $\rho (\mathbf{\xi })$ because $f$ is force per unit mass, whereas the Green tensor was normalized to unit force.

A force dipole is a limiting pair of opposite point forces separated by a small vector. Put $+F_j$ at $\mathbf{\xi }+\epsilon \mathbf{d}/2$ and $-F_j$ at $\mathbf{\xi }-\epsilon \mathbf{d}/2$, and hold $D_{jk}=\epsilon F_j{d}_k$ fixed as $\epsilon \to 0$. Expansion of the two Dirac delta functions givesIts resultant force is zero, and integration by parts gives its radiated displacement as ${\int }_0^t{D}_{jk}(\tau ){\partial }_{{\xi }_k}{G}_i^j(\mathbf{x},\mathbf{\xi },t-\tau )d\tau$. Thus the source derivative of the elastodynamic Green tensor represents a dipole with force direction $j$ and separation direction $k$.

For the surface source choose a normal $\mathbf{n}$ from the minus side to the plus side and defineThe same normal is used in both tractions. If the elastic coefficients are continuous through the surface, the distributional gradient of a piecewise smooth displacement is ${\partial }_j{u}_i=({\partial }_j{u}_i{)}_{\mathrm{reg}}+b_i{n}_j{\delta }_{\Sigma }$. Its stress therefore has a singular part $c_{ijkl}{b}_k{n}_l{\delta }_{\Sigma }$. Taking the divergence includes the regular stress jump, so a field satisfying the source-free equations on both sides has equivalent sourceIn particular, the signs follow from the chosen jump convention, not from an unspecified orientation. The elastodynamic surface-jump representation is consequentlywith the Green tensor evaluated at $(\mathbf{x},\mathbf{\xi },t-\tau )$. The first term is a surface distribution of forces, and the second is a surface distribution of dipoles.

One can derive the same formula without multiplying discontinuous material coefficients by a surface delta. Apply integration by parts on the two sides of $\Sigma$ to the field $u$ and the reciprocal Green field $v$. Major symmetry cancels the volume strain terms; the spatial identity iswhen the temporal terms are handled by the retarded time convolution. More explicitly, take $v_i(\mathbf{\xi },\tau )=G_i^a(\mathbf{\xi },\mathbf{x},t-\tau )$, extend the zero-initial-data solution causally to $\tau <0$, and integrate over all source times. Integration by parts cancels the two temporal terms, with zero endpoint terms; the causal supports subsequently reduce the representation to $0\le \tau \le t$. The Green impulse contributes $-u_a(\mathbf{x},t)$ on the left. The two outward face normals are $\mathbf{n}$ and $-\mathbf{n}$, so their sum on the right is $-[u_i]t_i(v)+v_i[t_i]$. Moving it to the other side gives $[u_i]t_i(v)-v_i[t_i]$. The same identity applied to two force Green fields with interchanged source and receiver proves the reciprocal relation $G_i^j(\mathbf{x},\mathbf{\xi },t)=G_j^i(\mathbf{\xi },\mathbf{x},t)$; hence the displayed surface formula follows. On a bonded material interface the Green displacement and common Green traction are continuous: $[u_i]t_i(v)$ is then the unambiguous version of the dipole term, evaluated with the matching one-sided coefficients. External free-surface terms vanish because both tractions vanish.

For a fault in isotropic material, the Lamé parameters giveTangential slip obeys $\mathbf{b}\cdot \mathbf{n}=0$, and ordinary fault traction is continuous across the two faces. Therefore the source is a double-couple fault source withFor a uniform patch of area $A$ with $\mathbf{b}=b{\mathbf{e}}_1$ and $\mathbf{n}={\mathbf{e}}_3$, its integrated seismic moment tensor isThe $13$ dipole is a force in the slip direction separated in the normal direction; the $31$ dipole is a normal force separated in the slip direction. Their equal and opposite torques cancel. The symmetric trace-free seismic moment tensor has eigenvalues $M_0,-M_0,0$, with zero resultant force and torque. A temporal stress drop on a fault is not a jump of traction between its two simultaneously matching faces; it does not add a single-force layer to this tangential-slip representation.

Let $c=\cos (\alpha {\xi }_3)$, $s=\sin (\alpha {\xi }_3)$ and $r^2=x_1^2+x_2^2=({\xi }_1^2+{\xi }_2^2)/\lambda$. Differentiating the deformation givesFor the neo-Hookean solid, $W_{,A_{i\alpha }}=\mu A_{i\alpha }$. The constrained nominal stress tensor is therefore $N_{\alpha i}=\mu A_{i\alpha }+q(A^{-1}{)}_{\alpha i}$. The multiplier cannot be set arbitrarily to a constant; equilibrium and the traction-free curved surface determine it.

Push the nominal stress to the Cauchy stress tensor: since $J=1$, $\sigma =AN=\mu A{A}^T+qI$. In the current cylindrical orthonormal basis, the left Cauchy-Green deformation tensor has entriesThus ${\sigma }_{rr}=\mu /\lambda +q$, ${\sigma }_{\vartheta \vartheta }=\mu /\lambda +\mu {\alpha }^2{r}^2+q$, and ${\sigma }_{\vartheta 3}=\mu \alpha \lambda r$. Static radial equilibrium without body force givesThe current outer radius is $a=R/\sqrt{\lambda }$. Setting ${\sigma }_{rr}(a)=0$ integrates the reaction field:The other curved-surface traction components vanish because ${\sigma }_{r\vartheta }={\sigma }_{r3}=0$.

On the reference end ${\xi }_3=H$, the outward reference normal is $e_3$, and the nominal traction has components $t_i=N_{3i}$. The third row of $A^{-1}$ is $(0,0,{\lambda }^{-1})$. Hence the nominal end traction of a twisted neo-Hookean cylinder isThe current $x_1,x_2$ here are evaluated at ${\xi }_3=H$. These components are per unit reference area, not per unit current area.

For the moment, use current lever arms and reference-area traction. Its axial component isThe axial traction gives no axial moment. This torque also agrees with the reference-energy derivative at fixed axial stretch: the twist-dependent energy per reference length is $\mu {\alpha }^2\pi R^4/(4\lambda )$. It is important that $\alpha$ is twist per reference length, not twist per current length.

Let $d(\mathbf{y},t)=[u^{[n]}]$ denote the signed opening, positive when the faces separate, and let $S$ be the crack surface. Its displacement jump is $\mathbf{b}=d\mathbf{n}$. The singular strain associated with this jump has moment density $c_{ijkl}{b}_k{n}_l$. For the isotropic elastic stiffness tensor, contraction givesThe equivalent distributional body force is the negative divergence of this prescribed moment density:The resulting forced elastic wave in an isotropic solid equation is $\rho u_{i,tt}={\partial }_j{\sigma }_{ij}+f_i$. This is a tensile-crack seismic source; it need not be represented as an ordinary smooth volume force. For a small nearly planar patch of area $S_0$, uniform normal $\mathbf{n}$ and opening $d(t)$, define its volume change $V(t)=S_{0}d(t)$. Its seismic moment tensor and point-source force areThe sign convention makes a positive opening a positive extensional moment. Reversing the convention for the displacement jump changes both signs together; the force is independent of reversing the orientation of the crack normal.

For a uniform material and an affine static base deformation, the deformation gradient, incremental elastic moduli and reference mass density are constant. With zero incremental body force, the equation reduces to $c_{\alpha i\beta j}{v}_{j,\beta \alpha }={\rho }_0{\ddot{v}}_i$. Let $\nu$ be a unit material propagation direction, $m$ a constant polarization and $s=t-{\nu }_{\alpha }{\xi }_{\alpha }/c$. Direct differentiation of the plane wave givesThus every twice differentiable profile $f$ satisfies the incremental equation ifThis is the acoustic tensor eigenvalue problem. The acoustic tensor is a real symmetric matrix: interchange the index pairs in the incremental elastic moduli and then interchange the dummy material indices to obtain $Q_{ji}=Q_{ij}$. Its eigenvectors give the displacement polarizations. A positive eigenvalue gives a real nonzero speed $c$; arbitrary-profile propagation imposes precisely this algebraic condition. Boundary conditions determine which combinations of these local plane waves are allowed in a finite body.

For a static base state under dead loading, put $v_i=u_i(\xi )e^{i\omega t}$ and take zero incremental body force and traction. The admissible squared frequencies are the eigenvalues of the weighted boundary value problemMultiply by $\overline{u_i}$ and integrate over the reference body. The boundary term in integration by parts vanishes, giving the elastic normal-mode energy identityFor a nonzero mode and positive mass density, the denominator is real and positive. The numerator is real by major symmetry of the incremental elastic moduli. Explicitly, if $u=p+iq$ with real $p,q$, the two imaginary cross terms cancel, leaving the sum of the quadratic energies of $\nabla p$ and $\nabla q$. Consequently every admissible ${\omega }^2$ is real. A negative value, when allowed by indefinite moduli, corresponds to exponential growth or decay rather than an oscillation with real frequency.

If the moduli quadratic form is strictly positive on every nonzero real matrix, that same decomposition shows that the numerator is positive whenever $\nabla u$ is nonzero. Thus ${\omega }^2>0$ for every nontranslation mode. Under the usual assumptions of a bounded connected body, regular coefficients and uniform positivity, the Rayleigh quotient defines a self-adjoint operator with discrete positive squared frequencies on the mass-weighted mean-zero displacement space. Its eigenfunctions give the nontrivial small harmonic oscillations.

There is an unavoidable zero-frequency qualification for the pure traction problem as printed: $u_i=b_i$, for any nonzero constant vector $b$, has zero gradient, zero incremental nominal stress and ${\omega }^2=0$. These three constant translations obey all the homogeneous incremental boundary conditions. Under the strict quadratic-form hypothesis they are the entire zero eigenspace on a connected body. Therefore the unrestricted conclusion isStrict positivity for every nonzero displacement requires excluding these modes, for example by imposing ${\int }_{{\Omega }_0}{\rho }_{0}udV=0$, or by adding a displacement constraint. If harmonic perturbations are defined to have nonzero frequency from the outset, the requested positivity holds for all those perturbations.

For antiplane shear, the infinitesimal displacement is $(0,0,w(x_1,x_2))$. The isotropic linear elasticity constitutive law gives ${\sigma }_{13}=\mu w_{,1}$ and ${\sigma }_{23}=\mu w_{,2}$, together with the symmetric entries ${\sigma }_{31},{\sigma }_{32}$. All other stress tensor components vanish. Static force balance without body force becomes $\mu (w_{,11}+w_{,22})=0$, so $w$ satisfies the Laplace equation.

Locally choose a harmonic conjugate $\varphi$ by ${\varphi }_{,1}=w_{,2}$ and ${\varphi }_{,2}=-w_{,1}$. Compatibility follows from the Laplace equation. These are the Cauchy-Riemann equations for $F=\varphi +iw$, and henceThus the complex stress potential for antiplane shear supplies both stresses and, by integration, the displacement. A global potential must have the displacement periods appropriate to the cracked body; these periods also help fix homogeneous solutions.

For the additional crack-induced field, set $G=\mu F^{\prime }$ and $p(t)={\sigma }_{23}^A(t,0)$. Only $p$ loads the prospective faces, since ${\sigma }_{13}^A$ is tangential to their normal. Denote upper and lower boundary values by $G^{+},G^{-}$. The imposed cancellation of face tractions is $\mathrm{Re}{G}^{\pm }=-p$. The traction data and geometry are invariant under the displacement reflection $w(x_1,x_2)\mapsto -w(x_1,-x_2)$. For the induced correction with no independently added loading, uniqueness up to a rigid translation allows the odd displacement choice. Then $w_{,2}$ is even, $w_{,1}$ is odd, and $G^{-}=\overline{G^{+}}$ on the real axis. Therefore the Hilbert problem for an antiplane crack isThe second condition expresses continuation through uncracked material. There $G$ is real and the additional ${\sigma }_{13}$ vanishes. This scalar Riemann-Hilbert problem is supplemented by at most inverse-square-root stress singularities at the crack tips and the usual correction with no additional remote resultant or imposed dislocation. Without a far-field or period normalization, the face conditions alone permit homogeneous terms. The formulas below select this normalized induced field and explicitly identify the excluded terms.

Choose field $a$ to be $({\hat{\mathbf{u}}}^a{)}^{\ast }$ at frequency $-{\omega }_a$, and field $b$ to be ${\hat{\mathbf{u}}}^b$ at ${\omega }_b$. Zero body forces and zero boundary tractions make all source and surface terms vanish. ThereforeFor distinct squared frequencies this proves weighted orthogonality,Assume positive mass density bounded above and below and finite-energy fields in the ordinary bounded domain. The weighted space $H=L^2(\mathcal{D},\rho dV{)}^3$ is a separable Hilbert space. Select one unit eigenfunction for each distinct squared frequency. They form an orthonormal set. Let $\{d_j\}$ be a countable dense subset of $H$; assign to each eigenfunction a point $d_j$ within distance $1/3$. Two orthogonal unit eigenfunctions have distance $\sqrt{2}$, so the same $d_j$ cannot be assigned to both. This gives an injection into a countable set.

Hence there are at most countably many squared eigenfrequencies, and at most twice that many signed frequencies. Multiplicity at one frequency does not change the argument. A zero frequency associated with rigid translations or rotations is included. This proves the countability of elastic eigenfrequencies directly; proving discreteness or finite multiplicity would require additional spectral regularity arguments.

If two harmonic solutions have the same body force and prescribed boundary traction, their difference has no body force and is traction-free. A nonzero difference would therefore be a free eigenfunction at that frequency. For any frequency outside the countable exceptional set found above, the difference must vanish, so the forced boundary-value solution, when it exists, is unique. At an eigenfrequency, a free eigenfunction can be added to any existing solution, and a resonant forcing may fail to admit a steady harmonic response. This distinguishes uniqueness from existence.

For the full time-dependent problem, let $\mathbf{w}$ be the difference between two fields with identical forces, tractions and initial displacement field and velocity. It obeys $\rho {\mathbf{w}}_{tt}=\nabla \cdot \mathbf{\sigma }(\mathbf{w})$, zero boundary traction, and zero initial data. Its energy uniqueness for traction-driven elasticity follows by direct integration:Multiply the equation by ${\mathbf{w}}_t$ and integrate by parts. Major symmetry turns the strain term into the derivative of elastic energy, giving $E^{\prime }(t)={\int }_{\partial \mathcal{D}}{\mathbf{w}}_t\cdot \mathbf{t}(\mathbf{w})dS=0$. Stable elasticity makes $E\ge 0$, and $E(0)=0$ gives ${\mathbf{w}}_t=0$. Thus $\mathbf{w}$ equals its initial value zero. Even rigid zero-strain motions cannot survive these initial conditions.

The PDF correctly specifies zero initial displacement field and zero initial velocity; the TeX duplicates displacement field in the second condition. The complete causal displacement field is unique, including when its Fourier representation contains resonant modal contributions fixed by the initial data.

The inextensible fibre constraint gives $|{\mathbf{x}}_{,1}|=1$. Write the fibre direction and its transverse unit direction asThen $x_{1,1}=\cos \theta$ and $x_{2,1}=\sin \theta$. In plane strain, incompressibility gives $\det F=1$. Decomposing the second column of the deformation gradient along $e,n$ therefore gives ${\mathbf{x}}_{,2}=\gamma e+n$, where $\gamma ={\mathbf{x}}_{,2}\cdot e$. HenceThis is a local rotation followed by simple shear.

The compatibility of the deformation map requires ${\mathbf{x}}_{,12}={\mathbf{x}}_{,21}$. Since $e_{,j}={\theta }_{,j}n$ and $n_{,j}=-{\theta }_{,j}e$, this isTaking components gives the compatibility relations for incompressible inextensible-fibre plane strain:Along a reference curve,Thus $\theta$ is constant when $d{X}_2/d{X}_1=-1/\gamma$. The equivalent condition $d{X}_1+\gamma d{X}_2=0$ also covers $\gamma =0$. Its current tangent is proportional to $F(-\gamma ,1{)}^T=n$, so these transverse curves are straight wherever the deformation is regular.

To construct the stress with zero body force, specify the local shear stress $h$ as the material's function or functional of $\gamma$ under simple shear. The normal stresses are constraint reactions: incompressibility supplies pressure, and the inextensible fibre constraint supplies fibre tension. WriteThe compatible geometry has $\nabla \theta =\kappa e$ with $\kappa ={\theta }_{,1}$, so $\nabla \cdot e=0$, $\nabla \cdot n=-\kappa$, $(e\cdot \nabla )e=\kappa n$ and $(n\cdot \nabla )e=0$. Resolving $\nabla \cdot \sigma =0$ along the two directions givesThe triangular equilibrium construction for inextensible-fibre plane strain now determines $s$ by integration along fibres and then $t$ along transverse curves. Thus any regular compatible deformation has a local zero-body-force stress construction once the shear law and suitable boundary integration data are supplied. There is no additional constitutive normal-stress law to specify: the two normal components are fixed by force balance and boundary tractions. This does not assert that arbitrary separately prescribed boundary tractions can accompany an arbitrary deformation.

Use the material-first nominal stress tensor convention $P_{Ii}$, with $F_{iI}$ the deformation gradient. For any fixed reference material subvolume $V_0$ with outward unit normal $N$, conservation of energy isHere $b$ is body force per mass, $r$ is heat supply per mass and $q$ is nominal heat flux. Using momentum conservation to remove the kinetic energy balance gives the integral Lagrangian internal energy balance:The integral Lagrangian entropy inequality iswhere $\theta >0$ is temperature and $q/\theta$ is the thermodynamic entropy flux.

Since the subvolume is arbitrary, the divergence theorem givesExpand $\dot{U}=U_{,F_{iI}}{\dot{F}}_{iI}+U_{,\eta }\dot{\eta }+U_{,{\xi }_r}{\dot{\xi }}_r$ and eliminate the heat supply between these relations. The Clausius-Duhem inequality becomesUnder the usual local constitutive independence assumptions, the reversible mechanical and thermal variations can have either sign. The Coleman-Noll procedure therefore imposesDefine the thermodynamic force conjugate to an internal variable by $f_r=-{\rho }_0{U}_{,{\xi }_r}$. The energy balance and the remaining entropy constraint now give, respectively,

For constant density $\rho$, the inviscid Euler equations with conservative body force areThe vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times (\nabla \times \mathbf{u})$ simplifies for potential flow $\mathbf{u}=\nabla \varphi$, since its vorticity vanishes. Also ${\partial }_t\mathbf{u}=\nabla {\varphi }_t$. ThereforeOn each connected fluid region the expression is spatially constant but may depend on time, giving the Unsteady Bernoulli equationThe arbitrariness of $f$ corresponds to the freedom to add a function of time to the velocity potential without changing $\mathbf{u}$. Replacing $\varphi$ by $\varphi -{\int }^{t}f(s)ds$ sets the right-hand side to zero.