변분법은 $t$를 인수로 하는 함수 $q$와 그 도함수 $q'$에 대해서 어떤 값 $I$가 $t,q,q'$를 인수로 하는 함수 $L$에 대해서 다음과 같이 주어질 때,
$$I(q)=\int_{a}^{b} L(t,q,q')dt$$
값 $I$가 최대 또는 최소가 될 때의 함수 $q$를 찾는 학문이다.
이때 다음과 같은 조건이 만족되는 것에 대해 오일러-라그랑주 방정식이라는 함수 $q$대한 정보를 얻을 수 있다.
Let $\mathbf{q}$ be a function to be found: $\mathbf{q}=\mathbf{q}(t)$ such that q is differentiable, $\mathbf{q}(a)=q_a$, and $\mathbf{q}(b)=q_b$.
Let $\mathbf{q}'$ be the derivative of $\mathbf{q}$.
Let $L$ be a function with continous first partial derivatives with respect to $t, \mathbf{q},$ and $\mathbf{q}'$.
Then Euler-Largrange equation is given by
$$\frac{\partial}{\partial \mathbf{q}}L-\frac{d}{dt}\frac{\partial}{\partial \mathbf{q}'}L=0$$
Suppose that
$\mathbf{q}$ is a function to be found: $\mathbf{q}=\mathbf{q}(t)$ such that q is differentiable, $\mathbf{q}(a)=q_a$, and $\mathbf{q}(b)=q_b$,
$\mathbf{q}'$ is the derivative of $\mathbf{q}$, and
$L$ is a function with continous first partial derivatives with respect to $t, \mathbf{q},$ and $\mathbf{q}'$.
Define a function $I=I(\mathbf{q})$ by
$$I=\int_{a}^{b} L(t,\mathbf{q},\mathbf{q}')dt.$$
We are interested in the condition of $\mathbf{q}$ where $I(\mathbf{q})$ is extremum of $I$.
So, let $\mathbf{q}$ be a function such that $I(\mathbf{q})$ is extremum.
Consider a function
$$\mathbf{q}+\epsilon\eta$$
where $\epsilon$ is a small value and $\eta=\eta(t)$ is a differentiable arbitrary function with $\eta(a)=\eta(b)=0$.
Then $\mathbf{q}+\epsilon\eta$ satisfies the boundary condtion such that $\mathbf{q}(a)=\mathbf{q}(a)+\epsilon\eta(a)=q_a$ and $\mathbf{q}(b)=\mathbf{q}(b)+\epsilon\eta(b)=q_b$.
Define a function $\Phi$ by $\Phi(\epsilon)=I(\mathbf{q}+\epsilon\eta)$.
Then $\Phi(0)=I(\mathbf{q})$ and $\left. \frac{d}{d \epsilon}\Phi \right|_{\epsilon=0}=0$.
We now differentiate $\Phi(\epsilon)$ with respect to $\epsilon$ to find $\frac{d}{d \epsilon}\Phi$.
\begin{align*} \frac{d}{d \epsilon}\Phi &= \frac{\partial}{\partial \epsilon} I(\mathbf{q}+\epsilon\eta) = \frac{\partial}{\partial \epsilon}\int_{a}^{b}L(t,\mathbf{q}+\epsilon\eta,\mathbf{q}'+\epsilon\eta')dt \\ \\ &= \int_{a}^{b} \frac{\partial }{\partial \epsilon} L(t,\mathbf{q}+\epsilon\eta,\mathbf{q}'+\epsilon\eta')dt \end{align*}
Since, by the chanin rule,
\begin{align*}\frac{\partial }{\partial \epsilon} L(t,\mathbf{q}+\epsilon\eta,\mathbf{q}'+\epsilon\eta') =\frac{\partial }{\partial \epsilon} L(t,\mathbf{y},\mathbf{y}')=\frac{\partial L}{\partial \mathbf{y}}\eta+ \frac{\partial L}{\partial \mathbf{y}'}\eta' \\ \text{where} \qquad \mathbf{y}=\mathbf{q}+\epsilon\eta \qquad \text{and} \qquad \mathbf{y}'=\mathbf{q}'+\epsilon\eta', \end{align*}
\begin{align*} \frac{d}{d \epsilon}\Phi &= \int_{a}^{b}\left( \frac{\partial L}{\partial \mathbf{y}}\eta+ \frac{\partial L}{\partial \mathbf{y}'}\eta' \right)dt \end{align*}
By the partial integral,
\begin{align*} \frac{d}{d \epsilon}\Phi &= \left[ \frac{\partial L}{\partial \mathbf{y}'}\eta\right]_a^b + \int_{a}^{b}\left( \frac{\partial L}{\partial \mathbf{y}}\eta-\frac{d}{dt}\frac{\partial L}{\partial \mathbf{y}'}\eta \right)dt\\ \\ &= 0 + \int_{a}^{b}\left( \frac{\partial L}{\partial \mathbf{y}}-\frac{d}{dt}\frac{\partial L}{\partial \mathbf{y}'} \right)\eta dt\end{align*}
Now, we take $\epsilon=0$, then
$$ \left.\frac{d}{d \epsilon}\Phi\right|_{\epsilon=0} = \int_{a}^{b}\left( \frac{\partial L}{\partial \mathbf{q}}-\frac{d}{dt}\frac{\partial L}{\partial \mathbf{q}'} \right)\eta dt=0.$$
Lemma. <Fundamental lemma of calculus of variations>
Let f be of class $C^k$ (i.e. $k$ times continuously differentiable) on the interval $[a,b]$.
If $$\int_a^bf(x)h(x)dx=0$$ for every function $h$ that is of class $C^k$ on $[a,b]$ with $h(a)=h(b)=0$, then $$f(x)=0 \quad \text{on} \quad [a,b].$$
<Proof>
Since $h$ is arbitray which satisfy the hypothesis, we choose $h$ such that $h(x)=r(x)f(x)$ where $r(x)=-(x-a)(x-b)$.
Then $h$ is of $C^k$ and $h(a)=h(b)=0$.
Now, we have
$$\int_a^b f(x)h(x) dx = \int_a^b r(x)f(x)^2 dx=0.$$
Since $f(x)^2\ge 0$ and $r(x)\ge 0$ on $[a,b]$, $r(x)f^2(x) \ge 0$ on $[a,b]$.
Since $f,r$ are continuous on $[a,b]$, $r(x)f(x)^2$ is continuous on $[a,b]$.
Then the integrand $r(x)f(x)^2=0$ on $(a,b)$ since the integral $\int_a^b r(x)f^2(x)dx=0$ for the non-negative integrand $r(x)f(x)^2=0$ on $(a,b)$.
Thus $f(x)=0$ on $(a,b)$ since $r(x)>0$ on $(a,b)$.
Since $f$ is continuous, $f(x)=0$ on $[a,b]$.
Since $\eta$ is a differentiable arbitrary function, by the Fundamental Lemma of Calculus of Variation,
$$ \frac{\partial L}{\partial \mathbf{q}}-\frac{d}{dt}\frac{\partial L}{\partial \mathbf{q}'}=0.$$
따라서 어떤 값 $I$를 극값으로 할 때의 $\mathbf{q}$를 찾기 위해서는 오일러-라그랑주 방정식을 풀면 된다.
