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

  • 1
    $\begingroup$ As a cheeky English man, do you maintain your "right" to spell "I" with an "i"? $\endgroup$ Commented Apr 11, 2020 at 1:25
  • $\begingroup$ no I default to dyslexia for that one :P $\endgroup$
    – Vaas
    Commented Apr 11, 2020 at 14:50
  • $\begingroup$ The constant 1 appears to be ignored in standard treatments with variational calculus. See this video here: youtube.com/watch?v=L6AvC0YZ9pQ $\endgroup$ Commented Dec 5, 2023 at 15:05

2 Answers 2


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.


Constants vanish, the bare essentials are :

$$( u^{'2}+ \lambda u^2 )- u' \cdot 2 u' = c; \quad \lambda u^2 - u^{'2} =c ;$$

$$ \frac{du}{dx}=\sqrt{ \lambda u^2 -c } \quad ; \int \frac{du}{\sqrt{ \lambda u^2 -c }} = x +d $$

&c. log/hyperbolic function solutions.


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