I have been given a Functional:
$$
I=\int^1_0 \left[(x_1')^2+(x'_2)^2-4tx_2-4x_2\right]dt
$$
where $x_1=x_1(t)$ and $x_2=x_2(t)$
And auxiliary constraint:
$$
\int^1_0 \left[(x'_1)^2-tx_1'-(x'_2)^2\right] dt = 2
$$
and boundary conditions:
$$
x_1(0)=x_2(0)=0\\
x_1(1)=x_2(1)=1.
$$
That gives me
$$
\begin{split}
L(t,x_1,x_2,x'_1,x'_2) &= (x_1')^2+(x'_2)^2-4tx_2-4x_2 \\
L_\mu &= (x'_1)^2-tx_1'-(x'_2)^2 \\
i &= 2 \\
\mu &= 1.
\end{split}
$$
I am given the Euler-Lagrange Equation:
$$
\frac{d}{dt}\left(\frac{\partial L}{\partial x'_i}\right)
- \frac{\partial L}{\partial x_i}
- \sum^p_{\mu=1} \lambda_\mu
\left[\frac{d}{dt} \left(\frac{\partial L_\mu}{\partial x'_i}\right)
- \frac{\partial L_\mu}{\partial x_i}
\right]
= 0
$$
Where $\lambda_\mu$ are Lagrange multipliers, which due to the multiplication rule, is constant
Working through this, I am given 2 Equations for $x''_1$ and $x''_2$:
$$x''_1=\frac{\lambda}{2-2\lambda}$$
and
$$x''_2=\frac{2t-2}{1-\lambda}.$$
These are then simply integrated to find $x_1(t)$ and $x_2(t)$ respectively: $$ \begin{split} x_1(t) &= \frac{1}{2}\frac{\lambda}{2-2\lambda}t^2+at+b \\ x_2(t) &= \frac{1}{6}\frac{2}{1-\lambda}t^3 - \frac{1}{2}\frac{2}{1-\lambda}t^2 + \left(1 - \frac{1}{6}\frac{2}{1-\lambda} + \frac{1}{2}\frac{2}{1-\lambda} \right) t \end{split} $$
in both $x_1$ and $x_2$, $\lambda \ne 1$. In order to get find $\lambda$, I'd need to plug bot $x_1$ and $x_2$ into the auxiliary constraint: $$\int^1_0 \left[ (x'_1)^2-tx_1'-(x'_2)^2 \right]dt = 2.$$ But this I know will end up giving me a situation where $\lambda$ has more than 1 value.
Can anyone help me please?
In this answer let us outline a strategy:
The Euler-Lagrange (EL) equations for $x_1$ and $x_2$ yield families of solutions that depend on the Lagrange multiplier $\lambda$ and 4 integration constants.
Use the 4 boundary conditions to determine the 4 integration constants.
At this stage the pair $(x_{1,\lambda},x_{2,\lambda})$ is a 1-parameter solution with parameter $\lambda$.
Use the constraint to get an equation for $\lambda$. This turns out to be a polynomial equation in $\lambda$ with a finite number of real solutions $\{\lambda_1,\ldots, \lambda_n\}$.
Determine which pair $(x_{1,\lambda_1},x_{2,\lambda_1}), \ldots, (x_{1,\lambda_n},x_{2,\lambda_n})$ minimizes OP's functional $I[x_1,x_2]$.