# Euler-Lagrange Equation, Lagrange Multipliers and optimisation

I'm just reading through a section of notes about Lagrange multipliers and the Euler lagrange equation and I could use a bit of clarification to make sure that i'm not missing something:

We're looking to find the extrema of $$J(\textbf{u}) = \int_{0}^{\pi} \frac{|u'|^{2}}{2} dx$$ for $$u \in U = \{u \in C^{1}[0,\pi]: u(0) = u(\pi) = 0\}$$ subject to the constraint $$\int_{0}^{1} u^{2}(x)~dx = 1$$

now i understand that the procedure is to find solutions of euler-lagrange equation when applied to the augmented functional $$\Lambda_{\lambda} = \Lambda + \lambda \Gamma$$ where $$\Lambda$$ is the lagrangian of the function we wish to find the extrema of (in this case J), $$\Gamma$$ is the Lagrangian of the constraints, and $$\lambda$$ is the Lagrange multiplier.

Since we're seeking the constraints to also vanish, ie for $$K(\mathbf{u}) = \int_{a}^{b} \Gamma(x,\mathbf{u},\mathbf{u'})~dx = 0$$ the notes have thus defined K to be $$K(\mathbf{u}) = \int_{0}^{\pi}\left[ \frac{u^2}{2}-\frac{1}{2 \pi}\right] dx$$

This doesn't seem obvious to me as it stands. If it's simply because we require the constraint to vanish and so far we have $$\int_{0}^{1} u^{2}(x)~dx = 1$$ then it seems obvious to set $$K(\mathbf{u}) = \int_{0}^{\pi} u^{2}(x)~dx - 1 \implies \int_{0}^{\pi} u^{2}(x)~dx - \int_{0}^{\pi}\frac{1}{\pi} dx \implies \int_{0}^{\pi} u^{2}(x) - \frac{1}{\pi}~dx$$ has the factor of $$\frac{1}{2}$$ been introduced simply because of J? i mean since $$K(\mathbf{u}) = 0$$ this seems like a legitimate operation. and does give a nice-ish augmented functional of $$J_{\lambda} = \frac{1}{2} \int_{0}^{\pi} \left[ |u'|^2 + \lambda \left( u^{2}-\frac{1}{\pi}\right)\right] dx$$ and so this all seems fine and worth while. but since there's been no explanation I want to make sure there's not another reason for this choice of K

Thanks in advanced, I appreciate it.

As a cheeky side note: as an English man I maintain my right to spell it with an s!!! :P

• As a cheeky English man, do you maintain your "right" to spell "I" with an "i"? Apr 11 '20 at 1:25
• no I default to dyslexia for that one :P
– Vaas
Apr 11 '20 at 14:50

FWIW, a scaling of the Lagrange undetermined multiplier $$\lambda$$ by a non-zero constant factor, e.g. a half, is irrelevant to the variational problem.
$$( u^{'2}+ \lambda u^2 )- u' \cdot 2 u' = c; \quad \lambda u^2 - u^{'2} =c ;$$
$$\frac{du}{dx}=\sqrt{ \lambda u^2 -c } \quad ; \int \frac{du}{\sqrt{ \lambda u^2 -c }} = x +d$$