The Euler-Lagrange Equation and Hamilton's EquationsProblem FormulationLet \(L(t, x,v)\) 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 \(x(T) = x_T\). The Fundamental Lemma of the Calculus of VariationsLemma 1 is called the fundamental lemma of the calculus of variations, and Lemma 2 is a similar statement. We will use Lemma 2 in the proof of the Euler-Lagrange equation. Lemma 1Let \(f:[a,b]\to\mathbb{R}^d\) 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 \(f=0\) on \([a,b]\). Proof of Lemma 1When \(d=1\), 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 2Let \(f:[a,b]\to\mathbb{R}^d\) 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 of Lemma 2It 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 \[ \int_a^b f(x)h(x)\ \text{d}x = \int_a^b \tilde{f}(x)h(x)\ \text{d}x = \int_{M_1}^{M_2} \tilde{f}(x)\Psi_{[M_1,M_2]}(x)\ \text{d}x - \int_{m_1}^{m_2} \tilde{f}(x)\Psi_{[m_1,m_2]}(x)\ \text{d}x > 0, \] which contradicts the assumption. This completes the proof. The Euler-Lagrange EquationStatementLet \(x^\star(\cdot)\) 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 \[ \frac{\text{d}}{\text{d}t} \nabla_v L( t, x^\star(t), \dot{x}^\star(t) ) = \nabla_x L( t, x^\star(t), \dot{x}^\star(t) ),\quad\forall t\in[0,T]. \] ProofFor 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 \[ \bigg\langle \int_0^t \nabla_x L(s, x^\star(s), \dot{x}^\star(s))\ \text{d}s, \eta(t) \bigg\rangle \bigg|_{t=0}^{t=T} - \int_0^T \bigg\langle \int_0^t \nabla_x L(s, x^\star(s), \dot{x}^\star(s))\ \text{d}s, \dot{\eta}(t) \bigg\rangle \ \text{d}t = - \int_0^T \bigg\langle \int_0^t \nabla_x L(s, x^\star(s), \dot{x}^\star(s))\ \text{d}s, \dot{\eta}(t) \bigg\rangle \ \text{d}t, \] 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. Hamilton's EquationsStatementConsider the above optimization problem. 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 \([0,T]\): \[ \dot{x}^\star(t) = \nabla_p H( t, x^\star(t), p^\star(t) ), \quad \dot{p}^\star(t) = -\nabla_x H( t, x^\star(t), p^\star(t) ). \] ProofNote that 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. References
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