Lagrangians and Euler-Lagrange Equations in Partial Differential Equations 📂Partial Differential Equations

Lagrangians and Euler-Lagrange Equations in Partial Differential Equations

Definition¹

Lagrangian
Let’s assume a smooth function $L : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ is given. This is called the Lagrangian and is denoted as follows.
$L = L(v,x)=L(v_{1}, \dots, v_{n}, x_{1}, \dots, x_{n}) \quad v,x\in \mathbb{R}^{n} \\ D_{v}L = (L_{v_{1}}, \dots, L_{v_{n}}), \quad D_{x}L = (L_{x_{1}}, \dots, L_{x_{n}})$
The reason for using variables $v, x$ is because, in physics, each variable actually signifies velocity and position.
Action, Admissible class
For two fixed points $x,y \in \mathbb{R}^{n}$ and time $t>0$ , the following defined functional $I$ is called action.
$I[ \mathbf{w}(\cdot)] := \int_{0}^tL(\dot{\mathbf{w}}(s), \mathbf{w}(s) ) ds \quad \left( \dot{　}=\dfrac{d}{ds}\right)$
In this case, the function $\mathbf{w}(\cdot)=\big( w^1(\cdot), \cdots, w^n(\cdot) \big)$ is an element of the admissible class defined as follows $\mathcal{A}$ .
$\mathcal{A} := \left\{ \mathbf{w}(\cdot) = \in C^2 \big([0,t];\mathbb{R}^n \big) \ \big| \ \mathbf{w}(0)=y, \mathbf{w}(t)=x\right\}$
In other words, $\mathcal{A}$ means the set of all paths that are twice continuously differentiable, starting from position $y$ to $x$ as time flows from $0$ to $t$ .

Description

The goal of the calculus of variations is to find $\mathbf{x} \in \mathcal{A}$ that makes the integral value of action $I$ minimum. This $\mathbf{x}$ is called the minimizer of $I$ .

$I[ \mathbf{x} (\cdot) ] = \inf_{\mathbf{w}(\cdot)\in \mathcal{A}} I[\mathbf{w}(\cdot)]$

Such $\mathbf{x}$ is sought because the path that minimizes the action of the Lagrangian is actually the path along which an object moves. In essence, this is for understanding the motion of an object, which fundamentally equates to solving $F=ma$ . In classical mechanics, the Lagrangian specifically refers to the difference between kinetic energy and potential energy.

Regarding the determination of minimizers, there’s the following theorem.

Theorem

Assume $\mathbf{x}(\cdot) \in \mathcal{A}$ is the minimizer of action $I$ . Then, $\mathbf{x}(\cdot)$ satisfies the following equation.

$-\dfrac{d}{ds} \Big[ D_{v}L\big( \dot{\mathbf{x}}(s), \mathbf{x}(s) \big) \Big] + D_{x}L\big( \dot{\mathbf{x}}(s), \mathbf{x}(s)\big)=0 \quad (0 \le s \le t)$

This equation is called the Euler-Lagrange equations.

It’s important to note that although a minimizer satisfies the Euler-Lagrange equations, satisfying the Euler-Lagrange equations does not necessarily mean it’s a minimizer. Similar to how differentiating at the minimum value gives $0$ , but not every point where differentiation gives $0$ is a minimum. In this sense, $\mathbf{x}(\cdot) \in \mathcal{A}$ satisfying the Euler-Lagrange equations is called a critical point of $I$ . Therefore, while a minimizer is a critical point, not all critical points are minimizers.

Proof

Assume $\mathbf{x} \in \mathcal{A}$ is the minimizer of action $I$ .

Step 1.
Let’s presume the function $\mathbf{y} : [0,t] \to \mathbb{R}^{n}, \mathbf{y}(\cdot) = (y^1(\cdot), \cdots, y^n(\cdot) )$ satisfies the following equation as a smooth function.
$\begin{equation} \mathbf{y}(0)=\mathbf{y}(t)=\mathbf{0} \label{eq1} \end{equation}$
And for any $\tau \in \mathbb{R}$ , define $\mathbf{w}(\cdot)$ as follows.
$\mathbf{w}(\cdot) : = \mathbf{x}(\cdot) + \tau \mathbf{y}(\cdot) \in \mathcal{A}$
Then $\mathbf{w}$ represents a path that has the same start and end points as $\mathbf{x}$ but differs by $\tau \mathbf{y}(\cdot)$ in between. Furthermore, since $\mathbf{x}(\cdot)$ is the minimizer of $I$ , the following equation holds.
$I[\mathbf{x}(\cdot)] \le I[\mathbf{w}(\cdot)]=I[\mathbf{x}(\cdot) + \tau \mathbf{y}(\cdot)] =: i(\tau)$
Also, the function $i$ has a minimum value at $\tau=0$ by definition of a minimizer. Therefore, if the derivative of $i$ exists, it is $i^{\prime}(0)=0$ .
Step 2.
As defined above, $i$ is as follows.
$i(\tau) = \int_{0} ^t L\big( \dot{\mathbf{x}}(s) + \tau \dot{\mathbf{y}}(s), \mathbf{x}(s)+ \tau \mathbf{y}(s) \big)ds$
Since $L$ , $\mathbf{y}$ are smooth functions, by differentiating $i$ , we get the following.
$i^{\prime}(\tau) = \int_{0}^t \sum_{i=1}^{n} \left[ L_{v_{i}} ( \dot{\mathbf{x}} + \tau \dot{\mathbf{y}}, \mathbf{x}+ \tau \mathbf{y} )\dot{y}^i + L_{x_{i}} ( \dot{\mathbf{x}} + \tau \dot{\mathbf{y}}, \mathbf{x}+ \tau \mathbf{y} ) y^i \right] ds$
Substituting $\tau=0$ , we obtain the following result from Step 1.
$0=i^{\prime}(0) = \int_{0}^t \sum_{i=1}^{n} \left[ L_{v_{i}} ( \dot{\mathbf{x}} , \mathbf{x})\dot{y}^i + L_{x_{i}} ( \dot{\mathbf{x}} , \mathbf{x} )y^i \right] ds$
Applying partial integration to each term of $L_{v_{i}}\dot{y}^i$ , and by the assumption $\mathbf{y}(0)=\mathbf{y}(t)=\mathbf{0}$ , we get the following.
$\int_{0}^t L_{v_{i}}\dot{y}^i ds = \left. L_{v_{i}}y^i \right]_{0}^t- \int_{0}^t \dfrac{d}{ds}L_{v_{i}}y^i=\int_{0}^t-\dfrac{d}{ds} L_{v_{i}}y^i$
Therefore, the following equation holds.
$0=i^{\prime}(0) = \sum_{i=1}^{n} \int_{0}^t \left[ -\dfrac{d}{ds} L_{v_{i}} ( \dot{\mathbf{x}} , \mathbf{x})\ + L_{x_{i}} ( \dot{\mathbf{x}} , \mathbf{x} ) \right]y^i ds$

The result of Step 2. satisfies for all smooth functions $\mathbf{y} : [0,t] \to \mathbb{R}^n$ that fulfill $\eqref{eq1}$ . Therefore, the value inside the brackets must be $0$ . Thus, the following is true.

$-\dfrac{d}{ds} L_{v_{i}}( \dot{\mathbf{x}}, \mathbf{x} ) +L_{x_{i}}( \dot{\mathbf{x}}, \mathbf{x}) =0$

■

This result leads to the Hamiltonian equations.