[proofplan]
We rewrite the weak Euler-Lagrange identity in the standard DuBois-Reymond form by naming the two coefficient functions appearing against $\varphi$ and $\varphi'$. The DuBois-Reymond lemma then gives an integral representation of the momentum term $x \mapsto \partial_v L(x,y(x),y'(x))$. Since the momentum term is assumed $C^1$ and the force term $x \mapsto \partial_y L(x,y(x),y'(x))$ is continuous, we differentiate the representation on the open interval and rearrange to obtain the classical Euler-Lagrange equation pointwise.
[/proofplan]
[step:Name the force and momentum maps in the weak identity]
Define the force map
\begin{align*}
F:(a,b)\to \mathbb{R}^n,\quad F(x)=\partial_y L(x,y(x),y'(x)).
\end{align*}
Define the momentum map
\begin{align*}
G:(a,b)\to \mathbb{R}^n,\quad G(x)=\partial_v L(x,y(x),y'(x)).
\end{align*}
Because $L \in C^1([a,b]\times \mathbb{R}^n\times \mathbb{R}^n;\mathbb{R})$ and $y \in C^1([a,b];\mathbb{R}^n)$, the map $F$ is continuous on $(a,b)$. By hypothesis, $G \in C^1((a,b);\mathbb{R}^n)$.
With this notation, the weak Euler-Lagrange equation says that for every $\varphi \in C_c^1((a,b);\mathbb{R}^n)$,
\begin{align*}
\int_a^b \left(F(x)\cdot \varphi(x)+G(x)\cdot \varphi'(x)\right)\, d\mathcal{L}^1(x)=0.
\end{align*}
[/step]
[step:Apply the DuBois-Reymond lemma to obtain an integral representation]
By the DuBois-Reymond lemma (citing a result not yet in the wiki: DuBois-Reymond Lemma), applied componentwise to the continuous map $F:(a,b)\to \mathbb{R}^n$ and the continuous map $G:(a,b)\to \mathbb{R}^n$, there exists a vector $c \in \mathbb{R}^n$ such that for every $x \in (a,b)$,
\begin{align*}
G(x)=c+\int_a^x F(s)\, d\mathcal{L}^1(s).
\end{align*}
The hypotheses of the lemma are satisfied because $F$ and $G$ are continuous and the displayed weak identity holds for every compactly supported $C^1$ variation $\varphi:(a,b)\to \mathbb{R}^n$.
[guided]
The weak identity contains the two terms that appear in the DuBois-Reymond lemma: a coefficient $F$ paired with the variation $\varphi$, and a coefficient $G$ paired with the derivative $\varphi'$. We have already defined
\begin{align*}
F(x)=\partial_y L(x,y(x),y'(x))
\end{align*}
and
\begin{align*}
G(x)=\partial_v L(x,y(x),y'(x)).
\end{align*}
The map $F$ is continuous because it is the composition of the continuous map $\partial_y L$ with the continuous curve $x\mapsto (x,y(x),y'(x))$. The map $G$ is continuous because it is assumed to be $C^1$ on $(a,b)$.
Now the weak Euler-Lagrange equation becomes
\begin{align*}
\int_a^b \left(F(x)\cdot \varphi(x)+G(x)\cdot \varphi'(x)\right)\, d\mathcal{L}^1(x)=0
\end{align*}
for every $\varphi \in C_c^1((a,b);\mathbb{R}^n)$. This is exactly the vector-valued DuBois-Reymond hypothesis; equivalently, applying the scalar lemma to each coordinate function gives the same conclusion. Therefore there exists $c \in \mathbb{R}^n$ such that
\begin{align*}
G(x)=c+\int_a^x F(s)\, d\mathcal{L}^1(s)
\end{align*}
for every $x \in (a,b)$. This integral representation is the bridge from the weak formulation to a pointwise differential equation: it expresses the momentum $G$ as an antiderivative of the force $F$ up to an additive constant.
[/guided]
[/step]
[step:Differentiate the representation on the open interval]
Since $F:(a,b)\to \mathbb{R}^n$ is continuous, the [fundamental theorem of calculus](/theorems/632) gives
\begin{align*}
\frac{d}{dx}\int_a^x F(s)\, d\mathcal{L}^1(s)=F(x)
\end{align*}
for every $x \in (a,b)$. Differentiating the identity
\begin{align*}
G(x)=c+\int_a^x F(s)\, d\mathcal{L}^1(s)
\end{align*}
therefore yields
\begin{align*}
G'(x)=F(x)
\end{align*}
for every $x \in (a,b)$.
[/step]
[step:Rewrite the derivative identity as the classical Euler-Lagrange equation]
Substituting the definitions of $F$ and $G$ into $G'(x)=F(x)$ gives, for every $x \in (a,b)$,
\begin{align*}
\frac{d}{dx}\partial_v L(x,y(x),y'(x))=\partial_y L(x,y(x),y'(x)).
\end{align*}
Rearranging the two sides gives
\begin{align*}
\partial_y L(x,y(x),y'(x))-\frac{d}{dx}\partial_v L(x,y(x),y'(x))=0.
\end{align*}
This is exactly the classical Euler-Lagrange equation on the open interval $(a,b)$. No endpoint conclusion is asserted, because the weak formulation was tested only against compactly supported variations in $(a,b)$.
[/step]
