We conclude this chapter by consider the stationary value of a functional with integrand, \(L(q_1,q_2,\dots,\dot{q_1}, \dot{q_2},\dots,t)\), subject to \(k\) constraints on the, \(q_i(t)\), functions. Here, in analogy with mechanics, we chose \(q(t)\) and \(\dot q(t)\) as functions that depend on the independent variable, \(t\) (instead of \(y(x)\) as we had in the previous section).
\begin{align}
f_1(q_1, q_2, \dots, q_n, t)&=0\\
f_2(q_1, q_2, \dots, q_n, t)&=0\\
\dots\\
f_k(q_1, q_2, \dots, q_n, t)&=0
\tag{2.58}\end{align}
where the constraints depend on time. As before, it would be possible to eliminate \(k\) of the \(q_i(t)\) functions, but this may not be the most mathematically convenient and may arbitrarily make some functions “dependent” and others “independent”. Instead, we can use the method of the Lagrangian multipliers. The variation of the constraints are zero for all times:
\begin{align}
\delta f_1&=\frac{\partial f_1}{\partial q_1}\delta q_1 + \frac{\partial f_1}{\partial q_2}\delta q_2 +\dots+\frac{\partial f_1}{\partial q_n}\delta q_n=0\\
\delta f_2&=\frac{\partial f_2}{\partial q_1}\delta q_1 + \frac{\partial f_2}{\partial q_2}\delta q_2 +\dots+\frac{\partial f_2}{\partial q_n}\delta q_n=0\\
\dots\\
\delta f_k&=\frac{\partial f_k}{\partial q_1}\delta q_1 + \frac{\partial f_k}{\partial q_2}\delta q_2 +\dots+\frac{\partial f_k}{\partial q_n}\delta q_n=0
\tag{2.59}\end{align}
We can then construct a new functional of which we want to find a stationary value:
\begin{align}
\delta I = \int_{t_a}^{t_b} \delta L(q_1,q_2,\dots,\dot{q_1}, \dot{q_2},\dots,t)dt+\int_a^b (\lambda_1 \delta f_1+\lambda_2 \delta f_2 +\dots +\lambda_k \delta f_k)dt=0
\tag{2.60}\end{align}
where we have effectively added zero to our original functional. Note that the \(\lambda_i\) are in principle functions of \(t\), since the \(f_i\) are functions of t. The first term of the variation, after integration by parts will have a term containing \(\delta q_i\) of the form:
\begin{align}
\int_{t_a}^{t_b} \left(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)-\frac{\partial L}{\partial q_i}\right) \delta q_i dt
\tag{2.61}\end{align}
and there will be a matching term in the part with the Lagrange multipliers:
\begin{align}
\int_{t_a}^{t_b} \left(\lambda_1 \frac{\partial f_1}{\partial q_i}+\lambda_2\frac{\partial f_2}{\partial q_i} +\dots +\lambda_k \frac{\partial f_k}{\partial q_i} \right) \delta q_i dt
\tag{2.62}\end{align}
This results in the functional being
\begin{align}
\delta I=\sum_{i=1}^n \int_a^b\left[ \left(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)-\frac{\partial L}{\partial q_i}\right) +\left(\lambda_1 \frac{\partial f_1}{\partial q_i}+\lambda_2\frac{\partial f_2}{\partial q_i} +\dots +\lambda_k \frac{\partial f_k}{\partial q_i} \right)\right] \delta q_i dt
\tag{2.63}\end{align}
In principle, the term in square brackets is not zero, because the \(\delta q_i\) cannot all be varied independently because of the constraint equations 2.59. We can however choose the \(\lambda_i\) such that the last \(k\) terms are zero:
\begin{align}
\left[ \left(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)-\frac{\partial L}{\partial q_i}\right) +\left(\lambda_1 \frac{\partial f_1}{\partial q_i}+\lambda_2\frac{\partial f_2}{\partial q_i} +\dots +\lambda_k \frac{\partial f_k}{\partial q_i} \right)\right]=0 \text{ (i=n-k+1…n)}
\tag{2.64}\end{align}
Now, this leaves us with the remaining \(n-k\) terms for which the \(\delta q_i\), can be varied independently. Of course, the whole point of the Lagrange multiplier method is that this gives equations that are identical for all \(q_i\), removing the distinction between dependent and independent variables. The above equation is thus true for all values of \(i=1\dots n\)!
We can thus re-express the problem of varying the integrand \(L\) subject to auxiliary conditions as the equivalent problem of varying a modified integrand, \(L'\), without worrying about auxiliary conditions:
\begin{align}
L'=L-(\lambda_1f_1+\lambda_2f_2+\dots +\lambda_kf_k)
\tag{2.65}\end{align}
where the auxiliary conditions can be used in addition to the Euler-Lagrange equations to determine all of the \(q_i(t)\) and the \(\lambda_i(t)\). This leads to the Euler-Lagrange equations which we can re-write as:
\begin{align}
\left(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)-\frac{\partial L}{\partial q_i}\right) &=-\lambda_1 \frac{\partial f_1}{\partial q_i}-\lambda_2\frac{\partial f_2}{\partial q_i} -\dots -\lambda_k \frac{\partial f_k}{\partial q_i}\\
\left(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)-\frac{\partial L}{\partial q_i}\right) &=-\sum_{j=1}^k\lambda_j \frac{\partial f_j}{\partial q_i}\\
\left(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)-\frac{\partial L}{\partial q_i}\right) &=Q_i
\tag{2.66}\end{align}
where we have introduced:
\begin{align}
Q_i\equiv -\sum_{j=1}^k\lambda_j \frac{\partial f_j}{\partial q_i}
\tag{2.67}\end{align}
and the overall negative sign is not important, since the Lagrange multipliers can be chosen.
Example 2-5
Show that the Euler-Lagrange equation for \(L'=L-\lambda_1f_1-\lambda_2f_2-\dots -\lambda_kf_k\) is equivalent to equation
2.64. That is, show that the sign in front of the Lagrange multipliers is arbitrary.
\begin{align*}
&\frac{d}{dt}\left(\frac{\partial L'}{\partial \dot{q}_i}\right)-\frac{\partial L'}{\partial q_i}\nonumber\\
&=\frac{d}{dt}\left(\frac{\partial}{\partial \dot{q}_i}(L-\lambda_1f_1-\dots -\lambda_kf_k)\right)-\frac{\partial}{\partial q_i}(L-\lambda_1f_1-\dots -\lambda_kf_k)\nonumber\\
\end{align*}
The \(\lambda_if_i\) terms do not explicitly depend on \(\dot{q}_i\), and the \(\lambda_i\) do not depend on the \(q_i\):
\begin{align*}
&=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)-\frac{\partial}{\partial q_i}(L-\lambda_1f_1-\dots -\lambda_kf_k)\nonumber\\
&=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)-\frac{\partial L}{\partial q_i}+\lambda_1\frac{\partial f_1}{\partial q_i}+\dots +\lambda_k\frac{\partial f_k}{\partial q_i}
\end{align*}
where we have a minus sign on all the Lagrange multipliers compared to equation 2.64, which is ok, since we are free to choose them.
Case when auxiliary condition is in the form of an integral
In certain cases, the auxiliary condition may be in the form of an integral:
\begin{align}
\int_{t_a}^{t_b} f(q_1,q_2,\dots,t)dt=C
\tag{2.68}\end{align}
where \(C\) is a constant, and there may be any number of such constraint equations. Again, we can take the variation of this integral:
\begin{align}
\delta\int_{t_a}^{t_b} f(q_1,q_2,\dots,t)dt&=\int_{t_a}^{t_b} \sum \die{f}{q_i}\delta q_i dt=0
\tag{2.69}\end{align}
which must be zero. We can thus use a Lagrange multiplier and add the variation of the constraint to our original integral:
\begin{align}
&\delta\int_{t_a}^{t_b} L(q_1,q_2,\dots,\dot{q_1}, \dot{q_2},\dots,t)dt+\lambda\delta\int_{t_a}^{t_b} f(q_1,q_2,\dots,t)dt\\
&=\delta \int_{t_a}^{t_b}\left[L(q_1,q_2,\dots,\dot{q_1}, \dot{q_2},\dots,t)+\lambda f(q_1,q_2,\dots,t)\right] dt
\tag{2.70}\end{align}
We can solve this problem by considering the stationary value of a new function, \(\bar L\):
\begin{align}
\bar L(q_1,q_2,\dots,\dot{q_1}, \dot{q_2},\dots,t)\equiv L(q_1,q_2,\dots,\dot{q_1}, \dot{q_2},\dots,t)+\lambda f(q_1,q_2,\dots,t)
\tag{2.71}\end{align}
and apply the Euler-Lagrange equation to \(\bar L\).