Using Lagrange multiplier in Euler-Lagrange Equation

Question

I think I am doing something wrong when combining Lagrange multiplier and Euler-Lagrange equation.

I need to maximize a functional of the form: $$ \int\!dx~{L(x, G, \dot{G})}~~~~~\text{where } L(a, b, c) = bc - ac, $$ provided that: $$ C = \int\!dx~{K(x, G, \dot{G})}~~~~~\text{where } K(a, b, c) = v(a) \times c, $$ where $v$ itself is a fixed function. Boundary conditions $G(-b)=0$ and $G(b)=1$ are assumed. Applying the Lagrange multiplier method reduces the problem to maximizing $\int\!dx~({L - \lambda K})$, where $\lambda$ is constant.

Euler-Lagrange equation states that the functional's maximizer must satisfies: $$ \frac{\partial}{\partial b}(L-\lambda K) = \frac{d}{dx}[\frac{\partial}{\partial c}(L-\lambda K)]. $$ This gives: $$ \dot{G} = \frac{d}{dx}[G - x - \lambda v(x)] = \dot{G}-1-\lambda \dot{v}\\ \Longrightarrow \lambda = -\frac{1}{\dot{v}} $$ Which suffers from two problems:

1. All the information about $G$ is lost! I can deduce literally nothing about it.
2. The final result contradicts the assumption of constant $\lambda$.

What am I doing wrong? Thanks for your help!

Answer 1

Due to Dirichlet boundary conditions $G(-b)=0$ and $G(b)=1$ we can remove total derivative terms. The new functional simplifies to $$ \int_{-b}^b\! dx~G(x)$$ and the new constraint simplifies to $$ -\int_{-b}^b\! dx~\dot{v}(x)G(x)~=~\tilde{C}~=~\text{some const}.$$ OP correctly finds that the Euler-Lagrange (EL) equation becomes $$ 1+\lambda\dot{v}~=~0. $$ There are several cases:

1. Case $\dot{v}~=~0$ and $\tilde{C}~=~0$: Constraint is automatically satisfied. Then the functional is unbounded from below and from above.

2. Case $\dot{v}~=~0$ and $\tilde{C}~\neq~0$: Constraint is never satisfied.

3. Case $\dot{v}~=~\text{const}~\neq~0$: Functional takes same value $-\tilde{C}/\dot{v}$ for all configurations $G$ that satisfy constraint.

4. Case $\dot{v}$ non-constant: EL equation is impossible to satisfy. Assume for simplicity that $\dot{v}$ is continuous. Then the functional is unbounded from below and from above.