Validity of $\mbox{d}H/\mbox{d}t=\partial H/\partial t$ for dissipative systems

Question

It' well known that in Hamiltonian formalism one has $$ \frac{\mbox{d}H}{\mbox{d}t} = \frac{\partial H}{\partial t}.\tag{*} $$ One proof can be found here. Therefore, the total change of energy during an arbitrary process equals $$ \Delta E = \int \mbox{d} H = \int \frac{\mbox{d}H}{\mbox{d}t} \mbox{d}t = \int\frac{\partial H}{\partial t} \mbox{d} t. $$

To the best of my knowledge, there are many difficulties in describing dissipative systems using Hamiltonian theory. So does Eq.(*) holds in such systems?

Answer 1

In principle you can use Hamiltonian mechanics also for dissipative systems, or fon systems which cannot be completely described by a Lagrangian.

When a Lagrangian is provided, the Legendre transformation connects the Lagrangian space of states -- with local coordinates $(t,q,\dot{q})$ -- with the Hamiltonian one -- with local coordinates $(t,q,p)$.

In local coordinates, the said one-to-one bi-differentiable map reads $$p_k = \frac{\partial L(t,q,\dot{q})}{\partial \dot{q}^k}\:,\tag{0}$$ $$q^k =q^k\:,$$ $$t=t\:.$$

I stress that we can invert these relations as $$\dot{q}^k = \dot{q}^k(t,q,p)\:,\tag{01}$$ $$q^k =q^k\:,$$ $$t=t\:.$$ where $\dot{q}^k(t,q,p)$ is a known function.

If, in a local Hamiltonian chart, we define the Hamiltonian function as usual $$H(t,q,p) := -L(t,q,\dot{q}(t,q,p))+ \sum_{k} p_k\dot{q}^k(t,q,p)\:,$$ the following three identities pop out, where the known function in Eq.(01) pervasively enters, $$\left.\frac{\partial H}{\partial t}\right|_{(t,q,p)}= -\left.\frac{\partial L}{\partial t}\right|_{(t,q, \dot{q}(t,q,p))}\:, \quad \left.\frac{\partial H}{\partial q^k}\right|_{(t,q,p)}= -\left.\frac{\partial L}{\partial q^k}\right|_{(t,q, \dot{q}(t,q,p))},\quad \left.\frac{\partial H}{\partial p_k}\right|_{(t,q,p)}= \dot{q}^k(t,q,p)\tag{1}$$ The second pair directly produces the corresponding of the Euler-Lagrange equations in the Hamiltonian space.

The E-L equations for an unknown curve $\gamma(t) = (t,q(t),\dot{q}(t))$ in Lagrangian variables, in whole generality are written as $$\left.\frac{d}{dt}\right|_{\gamma(t)}\frac{\partial L}{\partial \dot{q}^k}= \left.\frac{\partial L}{\partial q^k}\right|_{\gamma(t)}+ Q_k(\gamma(t))\:, \qquad \dot{q}^k(t)= \left.\frac{dq^k}{dt}\right|_{\gamma(t)}\:.$$ Above $L(t,q,\dot{q}) = T(t,q,\dot{q})+ V(t,q,\dot{q})$ includes all (non-reactive) forces which can be described in terms of a standard or generalized potential $V=V(t,q,\dot{q})$. The functions $Q_k(t,q,\dot{q})$ include all the remaining (non-reactive) forces which are usually dissipative. The typical case is $$Q_k= -\sum_k \gamma_k \dot{q}^k\tag{3}$$ for constants $\gamma_k \geq 0$.

Taking advantage of (0) and (1), if $\gamma$ satisfies the E-L equations in the Lagrangian setting, the associated curve in the Hamiltonian world $$\hat{\gamma}(t) := (t, q(t), p(t))\quad with \quad p_k(t) = \left.\frac{\partial L}{\partial \dot{q}^k}\right|_{\gamma(t)}$$ satisfies the generalized Hamilton equations (and vice versa) $$\frac{dp_k}{dt} = -\left.\frac{\partial H}{\partial q^k}\right|_{\hat{\gamma}(t)} + \left. \hat{Q}_{k}(t,q,p)\right|_{\hat{\gamma}(t)}\:, \quad \frac{dq^k}{dt} = \left.\frac{\partial H}{\partial p_k}\right|_{\hat{\gamma}(t)}\tag{2}$$ where, obviously, the dissipative generalized forces in Hamiltonian language are written as $$\hat{Q}_{k}(t,q,p):= Q_{k}(t,q,\dot{q}(t,q,p))\:.$$ We are now in a position to answer the question on the ground of the equations (2).

Consider a Hamiltonian function $H=H(t,q,p)$, a set of generalized (dissipative) forces $\hat{Q}_{k}(t,q,p)$, and a curve $\hat{\gamma}(t)= (t,q(t),p(t))$ solution of (2). A direct computation immediately leads to $$\left.\frac{d}{dt}\right|_{\hat{\gamma}(t)}H(t,q,p) = \left.\frac{\partial H}{\partial t}\right|_{\hat{\gamma}(t)} + \sum_k \left.\frac{dq^k}{dt} \hat{Q}_{k}(t,q,p)\right|_{\hat{\gamma}(t)}\:.$$

The conclusion is that, in the dissipative case, the known identity valid for the case of absence of dissipativie interactions fails and has to be corrected as written above.

In the specific case of Eq.(3), we have

$$\left.\frac{d}{dt}\right|_{\hat{\gamma}(t)}H(t,q,p) = \left.\frac{\partial H}{\partial t}\right|_{\hat{\gamma}(t)} - \sum_k \gamma_k \left.\left|\frac{dq^k}{dt}\right|^2\right|_{\hat{\gamma}(t)}\:.$$ It is here evident that even if the Hamltonian does not depend explicitily on $t$, the total energy -- when the Hamiltonian represents the energy of the system- is not conserved, but it decreases in view of the dissipative forces.

It is also possible to show that the added term to the standard Hamiltonian equations is responsible for the failure of Liouville theorem: dissipative terms imply that the volume in the phase space is not preserved under the evolution of the system.

(Reference: V. Moretti: Analytical Mechanics Classical, Lagrangian and Hamiltonian Mechanics, Stability Theory, Special Relativity, Springer 2023)

Answer 2

I'll split my answer in some paragraphs: 1. Newton's mechanics in strong and weak forms; 2. Lagrangian mechanics; 3. Hamiltonian mechanics.

1.a Newton's mechanics: strong form. Newton's second principle of mechanics provides the dynamical equation governing the motion of a closed system. Using third principle (action-reaction), it's possible to derive the equations of motion for complex systems, like rigid bodies or multi-body systems. Here I'm avoiding all the complications in the development of the equations for "complex" systems and focusing on point particle only,

$$m \frac{d \mathbf{v}}{d t} = \mathbf{F}$$

I'm doing so to avoid over-complication of the subject: for every kind of systems, Lagrange equations have the same formal expression.

You can find all the details for "complex systems" here: point mass; system of point masses; continuum; rigid bodies.

1.b Newton's mechanics: weak form $-$ principle of virtual works. Given the strong form of the equations of motion, a weak formulation is derived as

$$0 = \mathbf{w} \cdot \left[ m \frac{d \mathbf{v}}{d t} - \mathbf{F} \right] \ ,$$

being $\mathbf{w}$ a test function.

2. From Newton to Lagrange mechanics. Writing the state of the mechanical system (here only position, and its time derivative) as a function of generalized coordinates $q^k(t)$,

$$\mathbf{r}(q^k(t),t) \ ,$$

and

$$\mathbf{v}(\dot{q}^k(t),q^k(t),t) = \frac{d }{dt}\mathbf{r}(q^k(t),t) \ ,$$

and choosing the test function (for the point mass), $\mathbf{w} := \dfrac{\partial \mathbf{r}}{\partial q^k} = \dfrac{\partial \mathbf{v}}{\partial \dot{q}^k}$,

after some manipulations, it's possible to write Lagrange equations as

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}^k} \right) - \frac{\partial L}{\partial q^k} = Q_{q^k} \ ,$$

being $L = T + U$ the Lagrangian function of the system being $T$ the kinetic energy and $U$ the potential and $Q_{q^k} = \dfrac{\partial \mathbf{r}}{\partial q^k} \cdot \mathbf{F}^{n.c.}$ the non-conservative (or the forces that may be included in the potential but they're not).

3. From Lagrange to Hamilton mechanics. Defining the conjugate momenta $p_k = \frac{\partial L}{\partial \dot{q}^k}$ and the Hamiltonian function $H:= p_k \dot{q}^k - L$, and evaluating its differential as

$$\begin{aligned} dH & = dp_k \dot{q}^k + p_k d \dot{q}^k - d \dot{q}^k \partial_{\dot{q}^k} L - d q^k \partial_{q^k} L - \partial_t L dt = \\ & = dp_k \dot{q}^k - d q^k \partial_{q^k} L - \partial_t L dt \end{aligned}$$

so that

$$\begin{cases} \dfrac{\partial H}{\partial p_k} = \dot{q}^k\\ \dfrac{\partial H}{\partial q^k} = - \dfrac{\partial L}{\partial q^k} = - \dot{p}_k + Q_{q^k} \\ \dfrac{\partial H}{\partial t} = - \dfrac{\partial L}{\partial t} \end{cases} \qquad \rightarrow \qquad \begin{cases} \dot{q}^k = \dfrac{\partial H}{\partial p_k}\\ \dot{p}_k = - \dfrac{\partial H}{\partial q^k} + Q_{q^k} \end{cases} $$

The relation between the total and partial time derivative of the Hamiltonian function thus reads

$$\begin{aligned} \frac{d H }{d t} & = \dot{p}_k \dot{q}^k - \dot{q}^k \partial_{q^k} L + \partial_t H = \\ & = \dot{p}_k \dot{q}^k - \dot{q}^k \left( \dot{p}_k - Q_{q^k} \right) + \partial_t H = \\ & = \dot{q}^k Q_{q^k} + \partial_t H \ . \end{aligned}$$

If the Hamiltonian function represents the "mechanical" energy of the system (kinetic + potential), the last equation tells us that the time derivative of the mechanical energy equals the total power of the mechanical actions acting on the system itself, being $\partial_t H$ the "external" contribution of conservative forces and $\dot{q}^k Q_k$ the power of non-conservative forces,

$$\dot{q}^k Q_{q^k} = \dot{q}^k \dfrac{\partial r}{\partial {q}^k} \cdot \mathbf{F}^{n.c.} = \left[ \dfrac{d \mathbf{r}}{d t} - \frac{\partial \mathbf{r}}{\partial t} \right] \cdot \mathbf{F}^{n.c.} \ , $$

that, for $\partial_t \mathbf{r}|_{q^k} = 0$, gives

$$= \mathbf{v} \cdot \mathbf{F}^{n.c.} \ .$$

Extra a. Modified Liouville's theorem. Since we're already here, I'm elaborating the comment of Valter Moretti in his answer about Liouville's theorem. This theorem exploits Hamilton equations in the continuity equation for probability density distribution in the phase space $\rho(q^k, p_k, t)$,

$$\begin{aligned} 0 & = \partial_t \rho + \partial_{q^k} (\rho \dot{q}^k) + \partial_{p_k} (\rho \dot{p}_k) = \\ & = \partial_t \rho + \dot{q}^k \partial_{q^k} \rho + \dot{p}_k \partial_{p_k} \rho + \rho \left( \partial_{q^k} \dot{q}^k + \partial_{p_k} \dot{p}_k \right) = \\ & = \frac{D \rho}{D t} + \rho \left( \partial_{q^k} \partial_{p_k} H + \partial_{p_k} \left( - \partial_{q^k} H + Q_{q^k} \right) \right) = \\ & = \frac{D \rho}{D t} + \rho \partial_{p_k} Q_{q^k} \end{aligned}$$