Lagrange Multiplier Proof

I have only found proof of Lagrange multiplier with one constraint only. I could not foound proof of multiple constraints. So, I generalized the proof. If you find any mistakes in my proof (as if there would be anyone reading my posts XD), please let me know.

One Constraint

Let $P = (q_1(t_0),...,q_n(t_0),\dot{q}_0(t_0),...,\dot{q}_n(t_0))$ be the point where the Lagragian $L$ is extremized at the constraint $f_1$. Let $\vec{r}(t) = (q_1(t),...,q_n(t),\dot{q}_0(t),...,\dot{q}_n(t))$ be a parametrized function on $f_1$. Define $$\nabla L = (\frac{\partial L}{\partial q},...,\frac{\partial L}{\partial \dot{q}})$$ Let $h(t) = L(q_1(t),...\dot{q}(t))$. Then, $$\begin{align} \frac{d}{dt}h(t) &= \sum_k \frac{\partial L}{\partial q_k}\dot{q}_k + \frac{\partial L}{\partial \dot{q}_k}\frac{d\dot{q}_k}{dt}\\ &= \nabla L \cdot \frac{d\vec{r}}{dt}. \end{align}$$ At extremum $P$, $$\frac{dh}{dt} = \nabla L \cdot \frac{d\vec{r}}{dt} = 0.$$ As $\vec{r}$ is an arbitrary parametrized function of $f_1$, $L$ is orthogonal to any parametrized function of $f_1$. Then $L$ is parallel to $f_1$ $$\nabla L = \lambda_1 \nabla f_1,$$ for some constant $\lambda$.

More Constraints

As we have seen $L$ is extremized at $f_1$ when $$\nabla L = \lambda_1 \nabla f_1,$$ let $$h(t) = L - \lambda_1 f_1.$$ Then, using argument from above, we know that extremum occurs when $$\nabla L - \lambda_1\nabla f_1 = \lambda_2 \nabla f_2.$$ Similarly, for $m$ constraints, $$\nabla L = \sum^m_{\alpha=1}\lambda_\alpha \nabla f_\alpha.$$

Aftermath

As we know $\delta S=0$ can be written as $$\begin{align} \int^{t_2}_{t_1}\left(\delta L - \sum^m_{\alpha=1}\lambda_\alpha\delta f_\alpha\right)dt &= 0 \\ \int^{t_2}_{t_1}\sum_k\left( \frac{\partial L}{\partial q_k}\delta q_k + \left(\frac{\partial L}{\partial \dot{q_k}}\right)\delta\dot{q}_k - \sum^m_{\alpha=1}\lambda_\alpha\left(\frac{\partial f_\alpha}{\partial q_k}\delta q_k + \left(\frac{\partial f_\alpha}{\partial \dot{q}_k}\right)\delta\dot{q}_k \right)\right)dt &=0\\ \end{align}$$ Integrating by part, we have $$\int\left(\frac{\partial L}{\partial \dot{q}}\right)\frac{d\delta q}{dt}dt = \frac{\partial L}{\partial \dot{q}}\delta q - \int\left(\delta q\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right)dt$$ and $$\int\left(\frac{\partial f_\alpha}{\partial \dot{q}}\right)\frac{d\delta q}{dt}dt = \frac{\partial f_\alpha}{\partial \dot{q}}\delta q - \int\left(\delta q\frac{d}{dt}\frac{\partial f_\alpha}{\partial \dot{q}}\right)dt$$ So, $$\begin{align} \sum_k\left[\delta q_k\frac{\partial L}{\partial \dot{q_k}}\right]^{t_2}_{t_1} + \int^{t_2}_{t_1}\sum_k\delta q_k\left(\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}\right) & \\ +\sum^m_{\alpha=1}\left[\lambda_\alpha\left(\frac{\partial f_\alpha}{\partial q_k}-\frac{d}{dt}\frac{\partial f_\alpha}{\partial \dot{q}_k}\right)\delta q_k + \lambda_\alpha\frac{d}{dt}\left(\frac{\partial f_\alpha}{\partial \dot{q}_\alpha}\delta q_k\right)\right]dt &= 0. \end{align}$$ As $\delta q_k(t_2) = \delta q_k(t_1)$, $$\int^{t_2}_{t_1}\sum_k\delta q_k\left(\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}\right) +\sum^m_{\alpha=1}\lambda_\alpha\left[\left(\frac{\partial f_\alpha}{\partial q_k}-\frac{d}{dt}\frac{\partial f_\alpha}{\partial \dot{q}_k}\right)\delta q_k + \frac{d}{dt}\left(\frac{\partial f_\alpha}{\partial \dot{q}_\alpha}\delta q_k\right)\right]dt = 0.$$ Similarly, integrating by part again, we have $$\lambda_\alpha\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{q}}\delta q\right) = \frac{d}{dt}\left(\lambda_\alpha\frac{\partial f_\alpha}{\partial \dot{q}}\delta q\right) - \frac{d\lambda_\alpha}{dt}\left(\frac{\partial f}{\partial \dot{q}}\delta q\right).$$ Then, $$\left[\lambda_\alpha\frac{\partial f_\alpha}{\partial \dot{q}_k}\delta q_k\right]^{t_2}_{t_1}+\int^{t_2}_{t_1}\left[\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}+\sum^m_{\alpha=1}\lambda_\alpha\left(\frac{\partial f_\alpha}{\partial q_k}-\frac{d}{dt}\frac{\partial f_\alpha}{\partial \dot{q}_k}\right)-\sum^m_{\alpha=1}\left(\frac{d\lambda_\alpha}{dt}\right)\left(\frac{\partial f}{\partial \dot{q}_k}\right)\right]\delta q_kdt = 0.$$ Define $$Q_k=\sum^m_{\alpha=1}\left[\lambda_{\alpha}\left(\frac{\partial f_{\alpha}}{\partial q_k}-\frac{d}{dt}\frac{\partial f_\alpha}{\partial \dot{q}_k}\right)-\frac{d\lambda_{\alpha}}{dt}\frac{\partial f_{\alpha}}{\partial \dot{q}_k}\right].$$ As $\delta q_k(t_2) = \delta q_k(t_1)$, $$\int^{t_2}_{t_1}\sum_k\left[\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}-Q_k\right]\delta q_kdt = 0.$$ From above section, we know that $$\nabla L = \sum^m_{\alpha=1}\lambda_\alpha \nabla f_\alpha.$$ So, this time we know that for each coordinate $$\bbox[5px,border:2px solid #666] { \frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}=Q_k }$$