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Dissipativity expresses an input–output balance for a dynamical system: the increase of a nonnegative storage function cannot exceed the accumulated supply rate. For state $x$, control input $u$, plant output $y$ and chosen supply rate $w(u,y)$, the requirement isfor every admissible trajectory and time interval, with $S\ge 0$. The storage function and supply rate need not be literal physical energy and power; choosing them makes a particular performance statement precise. Jan C. Willems's1972paper develops this general storage–supply formulation.^[ref]

An original mechanical example is a mass with viscous damping, driven by an applied force $u$:Choose the kinetic energy $S(v)=m{v}^2/2$ and entering power $w(u,y)=uy$. ThenIntegration gives the required inequality. The nonnegative integral $\int b{v}^{2}dt$ accounts for the mechanical energy dissipated by the damper. With no applied force, the stored kinetic energy cannot increase. A dashpot driven at constant velocity provides an experimental interpretation: supplied power is dissipated even when the mass's stored kinetic energy remains constant.

A different choice tests an input–output gain. If a storage function establishes the inequality forthen $S(x(0))=0$ and nonnegative final storage implyThe physical power balance and this squared-signal bound use the same logical structure. Neither inequality, by itself, proves that every internal state converges to zero: a hidden lossless motion can retain storage.

Jan C. Willems was an electrical engineer and systems theorist whose work helped make dissipativity a general tool for studying dynamical systems. His university-hosted curriculum records an electrical-engineering doctorate at MIT in1968, a systems-and-control professorship at Groningen from1973, and later work as a guest professor at KU Leuven. It records his death on31August2013.^[ref]

His1972general theory expresses a system's balance using a storage function and supply rate. The formulation compares a change in stored quantity with integrated input supply, allowing physical energy accounting and input–output performance bounds to share one mathematical language.^[ref]

An original example illustrates the scientific idea. A2kilogram mass with viscous coefficient3newton-seconds per metre obeys $2\dot{v}=u-3v$. Its kinetic energy is $S=v^2$ joules when $v$ is expressed in metres per second, andAt constant velocity1metre per second, a3newton force supplies3watts, all dissipated while the stored kinetic energy stays1joule. The dissipativity article develops the general inequality; this simple case shows why its supply and storage terms must be distinguished.

Willems also developed behavioral systems theory, which describes a dynamical system by a family of admissible time histories. His curriculum explains the motivation: an interconnected system need not come with a uniquely privileged division into control inputs and plant outputs. Interconnection can instead impose shared variables and compatibility between trajectories.^[ref] The two contributions address complementary questions: which histories are possible, and what balance law must each history obey?

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 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.

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 storage function assigns a nonnegative quantity to the state of a dynamical system so that its increase is bounded by the integrated supply rate. In dissipativity, it must satisfyfor every allowed history and interval. A storage function may represent physical energy, but it can also certify an abstract input–output bound. Jan C. Willems's theory relates possible storage functions to recoverable supply, rather than selecting an arbitrary positive expression.^[ref]

For a differentiable storage function and state equation $\dot{x}=f(x,u)$, the pointwise inequalityis sufficient: integrating it along a solution proves the interval statement. Positivity alone is insufficient; the derivative must be checked against the specified supply rate.

An original dimensionless example makes this check exact. For $\dot{x}=-x+u$, output $y=x$, and supply rate $w=uy=ux$, try $S_a(x)=a{x}^2/2$ with $a>0$. Its derivative isThe difference from supply is ${\dot{S}}_a-ux=-a{x}^2+(a-1)xu$. Requiring it to be nonpositive for every real $u$ and fixed nonzero $x$ forces $a=1$. That value works because ${\dot{S}}_1=ux-x^2\le ux$. Thus even within this simple positive quadratic family, the supply law fixes the admissible scale.

A storage function also limits extraction. Rearranging the interval inequality givesStarting in a given state, one cannot withdraw more net supply than the initial storage certificate permits. A chosen storage function need not make that upper bound attainable.

When unforced operation makes the supply rate zero, storage cannot increase. This resembles a Lyapunov function, but nonincrease alone does not prove asymptotic stability: the function may fail to distinguish some states, or its derivative may vanish along persistent motion. The storage balance and a convergence proof are separate claims.

A supply rate specifies the instantaneous quantity whose time integral is compared with a change of storage function in dissipativity. With control input $u$, plant output $y$ and state $x$, a chosen function $w(u,y)$ enters the balanceThe supply rate can be negative: it then represents net withdrawal in the chosen accounting. Its physical interpretation depends on the variables and sign convention, rather than on the word supply alone.^[ref]

For a mechanical terminal with applied force $F$ and inward-convention velocity $v$, choose $w=Fv$, the entering power. For an electrical terminal with electric current entering its positive-potential terminal, the corresponding choice is $w=Vi$, with $V$ an electric potential difference. Both have units of watts when $S$ is energy. Reversing one terminal sign without changing the other reverses the apparent power flow.

An original capacitor example illustrates the sign. Let $C=2\mathrm{F}$, terminal electric potential difference $V=3\mathrm{V}$, and charging electric current $i=1\mathrm{A}$. Its stored electrostatic energy is $S=C{V}^2/2=9\mathrm{J}$, and at that instant $\dot{S}=Vi=3\mathrm{W}$. With the same potential and $i=-1\mathrm{A}$, the supply rate is $-3\mathrm{W}$ and storage decreases. This is consistent with nonnegative stored energy; negative supply does not mean negative storage.

A different supply rate can express a signal-performance requirement:Nonnegative final storage and zero initial storage then bound accumulated output-square by ${\gamma }^2$ times accumulated input-square. These squares need not be physical power. Scaling or changing the units of the signals requires corresponding changes in the coefficients. A general real quadratic choice can be written as $w=y^{T}Qy+2y^{T}Lu+u^{T}Ru$, with symmetric $Q,R$; the selected quadratic form states which input–output relation is being tested.