Mechanical energy storage function ($E$)

A mechanical energy storage function assigns a nonnegative stored-energy quantity $E$ to the state of a mechanical model so that its increase never exceeds entering work. For a one-port passive mechanical network with entering power $Fv$, the mechanical dissipation inequality isThis state-based accounting includes energy present before the measurement begins. A temporary outward power is compatible with release of stored energy.^[ref]

For an ideal linear spring, $E_k=\frac{1}{2}k{x}^2$ is elastic potential energy. For an ideal inerter, $E_b=\frac{1}{2}b{v}^2$ is kinetic energy associated with terminal relative motion. An ideal dashpot need not store mechanical energy: its entering power $c{v}^2$ is dissipated. If these elements share terminals, their state variables obey $v=\dot{x}$ andThis equality proves the inequality for $c\ge 0$. In a larger network, summing element balances cancels internal transfers, leaving power through the external terminals and total dissipation.

For an original numerical check, let $k=800N/m$, $b=5\mathrm{kg}$ and $c=10\mathrm{N}s/m$. Start at $x=0.050\mathrm{m}$ and $v=0.20m/s$, so $E=1.10\mathrm{J}$. Maintain that relative velocity for $0.10\mathrm{s}$: displacement reaches $0.070\mathrm{m}$ and stored energy becomes $2.06\mathrm{J}$. Dissipation is $c{v}^2\Delta t=0.040\mathrm{J}$, so input work is $0.96+0.04=1.00\mathrm{J}$. Counting all entering work as newly stored energy would overestimate storage.

The mechanical available storage is the maximum net work that could be recovered under allowed future terminal motions; it is a related operational quantity, rather than automatically every form of internal physical energy. A model's storage function is a certificate that must respect its accessible states and port variables. With zero external power, a positive-definite storage function can supply a Lyapunov function, but nonincreasing energy alone does not establish that every motion decays to rest: a lossless spring–inerter oscillation keeps a constant nonzero stored energy.

Mechanical available storage is the supremum of net work that can be extracted from a system's current state through its allowed mechanical terminals. If $Fv$ is entering power and the initial state is $x_0$, defineThe zero-duration choice gives $S_a\ge 0$. Any nonnegative mechanical energy storage function satisfying the mechanical dissipation inequality bounds the extracted work, so $S_a(x_0)\le E(x_0)$. Available storage is thus an operational recoverability quantity; it need not equal every form of internal stored energy. The general supply/storage construction was developed by Jan C. Willems.^[ref]

Consider a spring and dashpot in parallel, with $F=kx+c\dot{x}$ and initially $x=x_0$. Unload to $x=0$ at constant speed over duration $T$, so $\dot{x}=-x_0/T$. The outgoing work isThe first term is released elastic potential energy; the second is lost in the dashpot. For $k=1000N/m$, $c=10\mathrm{N}s/m$ and $x_0=0.10\mathrm{m}$, the initial spring energy is $5\mathrm{J}$. A one-second unloading recovers $4.90\mathrm{J}$ and a ten-second unloading recovers $4.99\mathrm{J}$. With unrestricted duration and suitable terminal control, these paths approach the energy upper bound, proving $S_a=5\mathrm{J}$ for this ideal example. A supremum need not be attained in finite time.

The allowed operations are essential. If the same example must finish at zero extension within one second, the Cauchy-Schwarz inequality gives ${\int }_0^T{\dot{x}}^{2}dt\ge x_0^2/T$, so the best recoverable work under that extra requirement is $4.90\mathrm{J}$. A physically inaccessible internal motion may store energy without allowing its recovery at the chosen terminals. Thus stating the initial state, usable ports, time and actuator limits is part of an available-storage claim, not an optional qualification.

The mechanical dissipation inequality states that a mechanical energy storage function can increase no faster than energy supplied through the mechanical terminals. With one port and entering power $w=Fv$, its differentiable form and interval form areThe second form remains meaningful when a pointwise time derivative is unavailable. The mechanical supply rate $w$ records the external input; the difference $Fv-\dot{E}\ge 0$ records irreversible loss in the model. This is the mechanical specialization of dissipativity, not a demand that entering power be positive at every instant.^[ref]

For a spring, inerter and dashpot sharing a port,and multiplication by $v$ gives the exact balanceA spring may return energy with $Fv<0$ while $E$ decreases; that does not violate the inequality. Conversely, a negative viscous coefficient fails this storage balance by supplying energy to motion.

For an original free-decay example, omit the spring and take grounded inertial coefficient $b=2\mathrm{kg}$, $c=1\mathrm{N}s/m$ and initial velocity $v_0=1m/s$. With zero applied port force, $b\dot{v}+cv=0$, soAt $t=(b/(2c))\ln 2\approx 0.6931\mathrm{s}$, stored energy has fallen from $1\mathrm{J}$ to $0.5\mathrm{J}$; the missing $0.5\mathrm{J}$ equals ${\int }_0^{t}c{v}^{2}dt$. No external work entered during that interval.

For several ports, use ${\sum }_j{F}_j{v}_j$ as the total supply. When passive elements are connected with power-conserving internal constraints, their internal supply terms cancel, so the sum of their storage functions satisfies the same external inequality. This explains why passive interconnections can be checked element by element, provided every power-carrying terminal is accounted for.

A mechanical supply rate specifies the rate at which energy enters a mechanical system through its stated ports. For a translational mechanical power port, entering force $F$ and its conjugate velocity $v$ give $w=Fv$; for rotation, torque $\tau$ and angular velocity $\omega$ give $w=\tau \omega$. The supply rate has units of power, while its time integral is supplied work.^[ref]

A two-terminal element makes the sign convention explicit. If forces on the element are $F$ and $-F$ at terminal velocities $v_1,v_2$, total entering power isOnly when terminal2 is stationary may one replace relative velocity by $v_1$. For an original example, $F=100\mathrm{N}$, $v_1=0.030m/s$ and $v_2=0.010m/s$ give $w=2\mathrm{W}$. Over two seconds the first terminal supplies $6\mathrm{J}$ and the second returns $2\mathrm{J}$, leaving net input $4\mathrm{J}$. Ignoring terminal2 would overstate the net supply.

For several translational and rotational ports, sum all entering power terms:The mechanical dissipation inequality then requires $\dot{E}\le w$ for an appropriate nonnegative mechanical energy storage function. Negative instantaneous $w$ is permitted when stored energy returns through the ports; it does not automatically contradict a passive mechanical network.

The chosen supply rate also defines the question being certified. A mathematical quadratic supply used in dissipativity can encode another input/output property, but it represents physical mechanical power only when its variables, units and sign convention match the actual energy exchange. This is why port bookkeeping must precede interpreting a fitted response model as passive.

A mechanical power port specifies the variables and sign convention through which a mechanical system exchanges energy. For translational motion these variables are an entering force $F$ and the corresponding signed velocity $v$; entering power is $p=Fv$. A rotational port uses entering torque $\tau$ and angular velocity $\omega$, giving $p=\tau \omega$. The pairing matters: multiplying a force by a displacement gives work, not its rate. A mechanical supply rate sums the powers of all specified ports.^[ref]

A two-terminal element makes mechanical power-port orientation explicit. Equal and opposite terminal forces $F,-F$ acting at velocities $v_1,v_2$ giveThus the port's flow variable is relative velocity, unless the second terminal is fixed. Reversing both chosen directions leaves the physical power unchanged. For $F=100\mathrm{N}$, $v_1=0.030m/s$ and $v_2=0.010m/s$, net entering power is $2\mathrm{W}$; the separate terminal contributions are $3\mathrm{W}$ and $-1\mathrm{W}$.

At an ideal rigid interconnection, connected terminals have the same velocity, while their entering forces sum to zero by Newton's third law. Their port powers therefore cancel in the combined system. This cancellation explains why internal connections disappear from an overall mechanical energy balance. A spring at the remaining external port obeys $p=kx\dot{x}=d(k{x}^2/2)/dt$; a dashpot obeys $p=c{v}^2$. One stores energy, while the other dissipates it.

For sinusoidal peak complex amplitudes, mean entering power is $\overline{p}=\frac{1}{2}\mathrm{Re}(F{V}^{\ast })$. A pure mass can have nonzero force and velocity amplitudes yet zero mean power, because their phases differ by a quarter cycle. Measuring magnitudes alone therefore does not determine energy transmission or establish the passivity of a mechanical network synthesis realization.

Mechanical power-port orientation specifies the signs of the conjugate variables used to describe entering power at a mechanical power port. For a translational port with signed force $F$ and velocity $v$, use $p=Fv>0$ for net supply into the chosen system. Reversing the coordinate axis reverses both variables, so $(-F)(-v)=Fv$. Reversing just one variable instead changes the entering-versus-leaving convention and must be stated explicitly. This bookkeeping makes a mechanical supply rate meaningful in a mechanical dissipation inequality.^[ref]

For an element acted on by terminal forces $F,-F$ and terminal velocities $v_1,v_2$,Exchanging terminal labels gives $F^{\prime }=-F$ and $v_1^{\prime }-v_2^{\prime }=-(v_1-v_2)$, preserving power. For $F=100\mathrm{N}$, $v_1=0.030m/s$ and $v_2=0.010m/s$, both descriptions give $2\mathrm{W}$. Taking force from the exchanged convention while retaining the original relative velocity would incorrectly report $-2\mathrm{W}$.

More general coordinate changes require the corresponding conjugate-force change. If physical displacement coordinates satisfy $\mathbf{x}=\mathbf{Tq}$, then $\dot{\mathbf{x}}=\mathbf{T}\dot{\mathbf{q}}$. Equality of virtual work, or instantaneous power, requiresFor the scalar coordinate $q=2x$, the generalized force is $F_q=F_x/2$. Thus $F_x=10\mathrm{N}$ and $\dot{x}=0.10m/s$ correspond to $F_q=5\mathrm{N}$ and $\dot{q}=0.20m/s$, both giving1W. Merely rescaling the motion channel without its conjugate force changes the apparent power. At an ideal connection, equal terminal velocities and opposite entering forces cancel internal port powers, enabling the combined mechanical energy balance.