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"? $\endgroup$ Apr 11 '20 at 1:25
  • $\begingroup$ no I default to dyslexia for that one :P $\endgroup$
    – 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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.