The Euler-Lagrange Equation and Hamilton's Equations
Let \(L(t, x,v)\) be a continuously differentiable function defined on \(\mathbb{R} \times \mathbb{R}^d \times \mathbb{R}^d\), which is called the Lagrangian.
Fix \(T>0\), an initial point \(x_0\in\mathbb{R}^d\), and an end point \(x_T\in\mathbb{R}^d\).
Consider the following optimization problem:
$$
\min_{ x(\cdot) } J[x],\quad
J[x] := \int_0^T L( t, x(t), \dot{x}(t) )\ \text{d}t,
$$
where the minimization is over all continuously differentiable curves \(x:[0,T]\to\mathbb{R}^d\) satisfying \(x(0)=x_0\) and \(x(T) = x_T\).
Fundamental Lemma of the Calculus of Variations
Lemma 1:
Let \(f:[a,b]\to\mathbb{R}^d\) be a continuous mapping.
If
$$
\int_a^b \langle f(t), h(t) \rangle \ \text{d}t = 0
$$
for all continuously differentiable \(h:[a,b]\to\mathbb{R}^d\) satisfying \(h(a) = h(b) = 0\), then \(f\equiv 0\) on \([a,b]\).
Proof
When \(d=1\), the proof of Lemma 1 can be found at
Wikipedia.
Observe that the \(d>1\) case can be reduced to the \(d=1\) case.
Lemma 2:
Let \(f:[a,b]\to\mathbb{R}^d\) be a continuous mapping.
If
$$
\int_a^b \langle f(t), h(t) \rangle \ \text{d}t = 0
$$
for all continuous \(h:[a,b]\to\mathbb{R}^d\) satisfying \(\int_a^b h(t)\ \text{d}t = 0\), then \(f\) is a constant on \([a,b]\).
Proof
It suffices to prove the \(d=1\) case.
Assume \(f\) is not a constant.
Let \(M = \max_{a\leq x\leq b} f(x)\) and \(m = \min_{a\leq x\leq b} f(x)\).
Since \(f\) is not a constant, we have \(M>m\).
Define \(\tilde{f}(x) = f(x) - (M+m)/2\).
By continuity, there exists an interval \( [M_1,M_2] \) such that \( \tilde{f}(x) > 0 \) for \( x\in[M_1,M_2] \), and another interval \( [m_1,m_2] \) such that \( \tilde{f}(x) < 0 \) for \( x\in[m_1,m_2] \).
Define the normalized
bump function by
$$ \Psi_{[c,d]}(x) = \begin{cases}
Z^{-1}\text{e}^{ \frac{-1}{(x-c)(d-x)} },&\text{if }c < x < d\\
0,&\text{otherwise}
\end{cases},$$
where \(Z\) is the normalizing constant such that \(\int_a^b \Psi_{[c,d]}(x)\ \text{d}x = 1\).
Let \(h(x) = \Psi_{[M_1,M_2]}(x) - \Psi_{[m_1,m_2]}(x)\).
It is clear that \(\int_a^b h(x)\ \text{d}x = 0\).
Moreover, we have
$$
\begin{split}
\int_a^b f(x)h(x)\ \text{d}x
&= \int_a^b \tilde{f}(x)h(x)\ \text{d}x \\
&= \int_{M_1}^{M_2} \tilde{f}(x)\Psi_{[M_1,M_2]}(x)\ \text{d}x
- \int_{m_1}^{m_2} \tilde{f}(x)\Psi_{[m_1,m_2]}(x)\ \text{d}x \\
&> 0,
\end{split}
$$
which contradicts the assumption.
This completes the proof.
The Euler-Lagrange Equation
Let \(x^\star(\cdot)\) be a curve that achieves the minimum of the above optimization problem.
Then, there exists a constant vector \(C\in\mathbb{R}^d\) such that
$$
\nabla_v L( t, x^\star(t), \dot{x}^\star(t) ) = \int_0^t \nabla_x L( s, x^\star(s), \dot{x}^\star(s) )\ \text{d}s + C,\quad\forall t\in[0,T],
$$
which implies that
$$
\frac{\text{d}}{\text{d}t} \nabla_v L( t, x^\star(t), \dot{x}^\star(t) ) = \nabla_x L( t, x^\star(t), \dot{x}^\star(t) ),\quad\forall t\in[0,T].
$$
Proof
For any continuously differentiable curve \(\eta:[0,T] \to \mathbb{R}^d\) satisfying \(\eta(0) = \eta(T) = 0\), the function \(J[x^\star + \alpha \eta]\), as a function of \(\alpha\in\mathbb{R}\), achieves its minimum at \(\alpha=0\).
This implies
$$
0 = \frac{\text{d}}{\text{d}\alpha} J[x^\star + \alpha \eta]\bigg|_{\alpha=0}
= \frac{\text{d}}{\text{d}\alpha}\int_0^T L( t, x^\star(t)+\alpha\eta(t), \dot{x}^\star(t)+\alpha\dot{\eta}(t) )\ \text{d}t\bigg|_{\alpha=0}.
$$
By
the Leibniz integral rule, we obtain
$$
\int_0^T \langle \nabla_x L(t, x^\star(t), \dot{x}^\star(t)), \eta(t) \rangle \ \text{d}t
+
\int_0^T \langle \nabla_v L(t, x^\star(t), \dot{x}^\star(t)), \dot{\eta}(t) \rangle \ \text{d}t = 0.
$$
By integration by parts, the first term equals
$$
\begin{split}
&\bigg\langle \int_0^t \nabla_x L(s, x^\star(s), \dot{x}^\star(s))\ \text{d}s, \eta(t) \bigg\rangle \bigg|_{t=0}^{t=T}
- \int_0^T \bigg\langle \int_0^t \nabla_x L(s, x^\star(s), \dot{x}^\star(s))\ \text{d}s, \dot{\eta}(t) \bigg\rangle \ \text{d}t \\
&\qquad = - \int_0^T \bigg\langle \int_0^t \nabla_x L(s, x^\star(s), \dot{x}^\star(s))\ \text{d}s, \dot{\eta}(t) \bigg\rangle \ \text{d}t,
\end{split}
$$
where we use \(\eta(0) = \eta(T) = 0\) in the equality.
Therefore, we obtain
$$
\int_0^T \bigg\langle \nabla_v L(t, x^\star(t), \dot{x}^\star(t)) - \int_0^t \nabla_x L(s, x^\star(s), \dot{x}^\star(s))\ \text{d}s, \dot{\eta}(t) \bigg\rangle \ \text{d}t = 0,
$$
for all continuously differentiable curve \(\eta(\cdot)\) satisfying \(\eta(0)=\eta(T) = 0\).
Finally, for any continuous function \(h:[0,T] \to \mathbb{R}^d\) satisfying \(\int_0^T h(t)\ \text{d}t = 0\), let \(\eta(t) := \int_0^t h(s)\ \text{d}s\).
Note that \(\eta\) is continuously differentiable, \(\dot{\eta}=h\), and \(\eta(0) = \eta(T) = 0\).
Therefore, we have
$$
\int_0^T \bigg\langle \nabla_v L(t, x^\star(t), \dot{x}^\star(t)) - \int_0^t \nabla_x L(s, x^\star(s), \dot{x}^\star(s))\ \text{d}s, h(t) \bigg\rangle \ \text{d}t = 0,
$$
The theorem then follows from Lemma 2.
Consider the above optimization problem d.
Assume that for any \( t\in\mathbb{R} \) and \( x,p\in\mathbb{R}^d \), we can solve \( p=\nabla_v L(t,x,v) \) for \( v \).
Denote the solution by \( v(t,x,p) \).
Assume that \( v:\mathbb{R}\times\mathbb{R}^d\times \mathbb{R}^d\to\mathbb{R}^d \) is continuously differentiable.
Define the Hamiltonion by
$$
H(t,x,p) := \langle p, v(t,x,p) \rangle - L( t, x, v(t, x,p) ).
$$
Let \( x^\star(\cdot) \) be the curve achieving the minimum, and define \( p^\star(t) := \nabla_v L(t, x^\star(t), \dot{x}^\star(t)) \), which is called the momentum.
Then, \( (x^\star, p^\star) \) satisfies the following differential equations on \( [0,T] \):
$$
\dot{x}^\star(t) = \nabla_p H( t, x^\star(t), p^\star(t) ),
\quad \dot{p}^\star(t) = -\nabla_x H( t, x^\star(t), p^\star(t) ).
$$
Proof
By the Euler-Lagrange equation, \( p^\star \) is continuously differentiable.
By a direct calculation and the definition of \( v(t,x,p) \), we have
$$
\nabla_x H(t,x,p) = -\nabla_x L(t,x, v(t,x,p)),
\quad
\nabla_p H(t,x,p) = v(t,x,p).
$$
Hamilton's equations then follows from the Euler-Lagrange equation.
- Lawrence C. Evan. An Introduction to Mathematical Optimal Control Theory. 2024.
- Daniel Liberzon. Calculus of Variations and Optimal Control Theory: A Concise Introduction. Princeton University Press. 2012.