# Using Lagrange multiplier in Euler-Lagrange Equation

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!

• @Qmechanic Yes it does! I am not sure if it makes any differences but it assumes G(-b)=0 and G(b) = 1. And about maximum/minimum the solution is still the same as above; the functional gradient must be zero at the extremum (i.e Euler-Lagrange equation) Commented Mar 15, 2020 at 20:22

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.