Minimization of a Functional Summary
SUMMARY
Notation \color{blue}\dfrac{\mathrm{d}}{\mathrm{d}\alpha}\bigg|_{\alpha=0}=\delta
Problem: minimize \color{blue}\int_a^b L(x,x^\prime,t)\,\mathrm{d}t \;+\; \color{SeaGreen}\Phi(x(a),x(b))
Boundary conditions: \color{OrangeRed} x(a) = x_a, \quad x(b)=\textrm{free}, \quad b_k(x(a),x(b)) = 0, \cdots
Integral contraints: \color{Brown} \int_a^b g_k(x,x^\prime,t)\,\mathrm{d}t \;+\; \Psi_k( x(a),x(b)) = 0
Higher order problem: minimize
\color{blue}\int_a^b L(x,x^\prime,x^{\prime\prime},\ldots,x^{(n)},t)\,\mathrm{d}t \;+\;\color{SeaGreen}\Phi(x(a),x(b))
Spaces
- \Bbb{V} = \Big\{x\in C^m([a,b]),\ldots\Big\}
- \Bbb{D} = \Big\{\delta x\in C^\infty([a,b]),\delta x(a) = \delta x(b) = 0, \delta x^\prime(a) = \delta x^\prime(b) = 0, \ldots \Big\}
Variation of problem 2.
\begin{aligned} \mathcal{F}(x) &= \int_a^b L(x,x^\prime,t)\,\mathrm{d}t + \Phi(x(a),x(b),x^\prime(a),x^{\prime}(b)) \\ \delta\mathcal{F}(x) &= \delta\int_a^b L(x,x^\prime,t)\,\mathrm{d}t + \delta \Phi(x(a),x(b),x^\prime(a),x^{\prime}(b)) \\ &= \int_a^b \delta L(x,x^\prime,t)\,\mathrm{d}t + \delta \Phi(x(a),x(b)) \\ &= \int_a^b L_x(x,x^\prime,t)\delta x+ L_{x^\prime}(x,x^\prime,t)\delta x^\prime\,\mathrm{d}t \\ & \quad + \Phi_{x_a}(x(a),x(b)){\color{OrangeRed}\delta x(a)} + \Phi_{x_b}(x(a),x(b)){\color{OrangeRed}\delta x(b)} \\[1em] &\color{red}\textrm{integration by part} \\[1em] &= \int_a^b \left[L_x(x,x^\prime,t)- \dfrac{\mathrm{d}}{\mathrm{d}t}L_{x^\prime}(x,x^\prime,t)\right]{\color{OrangeRed}\delta x}\,\mathrm{d}t \\[0.5em] & \quad+ \left(\Phi_{x_a}(x(a),x(b))- L_{x^\prime}(x(a),x^\prime(a),a)\right){\color{OrangeRed}\delta x(a)} \\[0.5em] & \quad+ \left(\Phi_{x_b}(x(a),x(b))+L_{x^\prime}(x(b),x^\prime(b),b)\right){\color{OrangeRed}\delta x(b)} \\[0.5em] \end{aligned}
Variation of problem 3.
Define \begin{aligned} \mathcal{F}(x) & = \int_a^b L(x,x^\prime,t)\,\mathrm{d}t- \sum_{k+1}^p \lambda_k b_k(x(a),x(b)) \\ \delta\mathcal{F}(x) &= \delta\int_a^b L(x,x^\prime,t)\,\mathrm{d}t + \delta\sum_{k+1}^p \lambda_k b_k(x(a),x(b)) \\ &= \int_a^b L_x(x,x^\prime,t)\delta x+ L_{x^\prime}(x,x^\prime,t)\delta x^\prime\,\mathrm{d}t - \sum_{k=1}^p b_k(x(a),x(b)){\color{OrangeRed}\delta\lambda_k} \\ & \quad - \sum_{k=1}^p \lambda_k \left( \dfrac{\partial b_k}{\partial x_a}(x(a),x(b)){\color{OrangeRed}\delta x(a)}+ \dfrac{\partial b_k}{\partial x_b}(x(a),x(b)){\color{OrangeRed}\delta x(b)} \right) \\[2em] &\color{red}\textrm{integration by part} \\[1em] &= \int_a^b \left[L_x(x,x^\prime,t) - \dfrac{\mathrm{d}}{\mathrm{d}t} L_{x^\prime}(x,x^\prime,t)\right]{\color{OrangeRed}\delta x}\,\mathrm{d}t - \sum_{k=1}^p b_k(x(a),x(b)){\color{OrangeRed}\delta\lambda_k} \\ & \quad - \left(\sum_{k=1}^p \lambda_k \dfrac{\partial b_k}{\partial x_a}(x(a),x(b)) +L_{x^\prime}(x(a),x^\prime(a),a)\right){\color{OrangeRed}\delta x(a)} \\ & \quad - \left( \sum_{k=1}^p \lambda_k \dfrac{\partial b_k}{\partial x_b}(x(a),x(b)) -L_{x^\prime}(x(b),x^\prime(b),b) \right) {\color{OrangeRed}\delta x(b)} \end{aligned}
Variation of problem 4.
Define \begin{aligned} M(x,x^\prime,\bm{\lambda},t) &= L(x,x^\prime,t)-\sum_{k=1}^m \lambda_k g_k(x,x^\prime,t) \\ N(x_a,x_b,\bm{\lambda}) &= \Phi(x_a,x_b)-\sum_{k=1}^m \lambda_k \Psi_k(x_a,x_b) \end{aligned}
\begin{aligned} \mathcal{F}(x,\bm{\lambda}) &= \int_a^b M(x,x^\prime,\bm{\lambda},t)\,\mathrm{d}t +N(x(a),x(b),\bm{\lambda})\\ \delta\mathcal{F}(x,\bm{\lambda}) &= \delta\int_a^b M(x,x^\prime,\bm{\lambda},t)\,\mathrm{d}t +\delta N(x(a),x(b),\bm{\lambda}) \\ &= \int_a^b M_x(x,x^\prime,\bm{\lambda},t){\color{OrangeRed}\delta x}+ M_{x^\prime}(x,x^\prime,\bm{\lambda},t){\color{OrangeRed}\delta x^\prime}+ M_{\bm{\lambda}}(x,x^\prime,\bm{\lambda},t)\cdot {\color{OrangeRed}\delta\bm{\lambda}} \,\mathrm{d}t\\ & \quad +N_{x_a}(x(a),x(b),\bm{\lambda}) {\color{OrangeRed}\delta x(a)} +N_{x_b}(x(a),x(b),\bm{\lambda}) {\color{OrangeRed}\delta x(b)} +N_{\bm{\lambda}}\cdot {\color{OrangeRed}\delta\bm{\lambda}} \\[1em] &\color{red}\textrm{integration by part} \\[1em] &= \int_a^b \left[M_x(x,x^\prime,\bm{\lambda},t)- \dfrac{\mathrm{d}}{\mathrm{d}t}M_{x^\prime}(x,x^\prime,t)\right]{\color{OrangeRed}\delta x}\,\mathrm{d}t\\ & \quad - \sum_{k=1}^m {\color{OrangeRed}\delta\lambda_k} \left[\int_a^b g_k(x,x^\prime,t)\,\mathrm{d}t+ \Psi_k(x(a),x(b))\right]\\ & \quad+\Big(N_{x_a}(x(a),x(b),\bm{\lambda})- M_{x^\prime}(x(a),x^\prime(a),\bm{\lambda},a)\Big){\color{OrangeRed}\delta x(a)} \\ & \quad+\Big(N_{x_b}(x(a),x(b),\bm{\lambda})+M_{x^\prime}(x(b),x^\prime(b),\bm{\lambda},b)\Big){\color{OrangeRed}\delta x(b)} \end{aligned}
Problem 5
Transform to optimal control problem
\begin{aligned} \textrm{minimize:}\qquad & \color{blue}\int_a^b L(x_1,x_2,x_3,\ldots,x_n,u,t)\,\mathrm{d}t\\ \textrm{subject to:}\qquad & \left\{ \color{red} \begin{aligned} x_1^\prime & = x_2\\ x_2^\prime & = x_3\\ \quad & \vdots \\ x_{n-1}^\prime & = x_n \\ x_n^\prime & = u \end{aligned} \right. \end{aligned}
General Problem
\begin{aligned} \textrm{minimize:}\quad & \color{blue}\int_a^b L(x,x^\prime,t)\,\mathrm{d}t \;+\; \color{SeaGreen}\Phi(x(a),x(b)) \\[2em] \textrm{subjedt to:}\quad & \color{brown}b_k(x(a),x(b)) = 0, \quad & k&=1,2,\ldots,p \\[1em] & \color{purple}\int_a^b g_k(x,x^\prime,t)\,\mathrm{d}t \;+\; \color{RoyalBlue}\Psi_k( x(a),x(b)) = 0, \quad & k& =1,2,\ldots,m \end{aligned}
defining
\begin{aligned} M(x,x^\prime,\bm{\lambda},t) &= L(x,x^\prime,t)-\sum_{k=1}^m \lambda_k g_k(x,x^\prime,t) \\ N(x_a,x_b,\bm{\lambda},\bm{\mu}) &= \Phi(x_a,x_b) -\sum_{k=1}^m\lambda_k \Psi_k(x_a,x_b) -\sum_{k=1}^p\mu_k b_k(x_a,x_b) \end{aligned}
the stationary points satisfy the BVP
\left\{\begin{aligned} M_x(x,x^\prime,\bm{\lambda},t)-\dfrac{\mathrm{d}}{\mathrm{d}t}M_{x^\prime}(x,x^\prime,\bm{\lambda},t) &= 0\quad && \textrm{\color{red} Euler Lagrange} \\[1em] N_{x_a}(x(a),x(b),\bm{\lambda},\bm{\mu})- M_{x^\prime}(x(a),x^\prime(a),\bm{\lambda},a)&=0 \quad && \textrm{\color{red} Adjoint BC} \\[1em] N_{x_b}(x(a),x(b),\bm{\lambda},\bm{\mu})+M_{x^\prime}(x(b),x^\prime(b),\bm{\lambda},b) &=0\quad && \textrm{\color{red} Adjoint BC} \\[1em] b_k(x(a),x(b)) &= 0, \quad & k&=1,2,\ldots,p \\[1em] \int_a^b g_k(x,x^\prime,t)\,\mathrm{d}t + \Psi_k( x(a),x(b)) &= 0, \quad & k& =1,2,\ldots,m \end{aligned}\right.