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Given a functional $$J(y)=\int_a^b F(x,y,y')dx, \tag{1}$$
where $y$ is a function of $x$, and a constraint $$\int_a^b K(x,y,y')dx=l, \tag{2}$$ if $y=y(x)$ is an extreme of (1) under the constraint (2), then there exists a constant $\lambda$ such that $y=y(x)$ is also an extreme of the functional $$\int_a^b [F(x,y,y')+\lambda K(x,y,y')]dx. \tag{3}$$ Similarly, if the constraint is $$g(x,y,y')=0, \tag{4}$$ then there exists a function $\lambda(x)$ such that the extreme also holds for the functional $$\int_a^b [F(x,y,y')+\lambda(x)g(x,y,y')]dx. \tag{5}$$

This is known as the Lagrange multiplier rule for calculus of variations. However, I have two questions about this statement.

  1. If the functional (1) has two constraints (2) and (4), does the extreme also hold for the functional $$\int_a^b [F(x,y,y')+\lambda K(x,y,y') +\lambda(x) g(x,y,y')]dx ? \tag{6}$$
  2. Is this statement also valid for multiple variable case? For example, if $J=\iint F(x_1,x_2,y(x_1,x_2))dx_1 dx_2$, and $\iint K(x_1,x_2,y(x_1,x_2))dx_1 dx_2=l$, is this equivalent to $J=\iint F(x_1,x_2,y(x_1,x_2))+\lambda K(x_1,x_2,y(x_1,x_2))dx_1 dx_2$?

Thanks in advance and any suggestion will be appreciated. It is better if you have any reference.

MIMIGA
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1 Answers1

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The short answer is yes and yes. As a reference, I would recommend (highly) an inexpensive general math reference that I have found helpful: Mathematical Handbook for Scientists and Engineers. It has practically every applied math theorem and technique you'd want. Note that it is a math reference, not a computational/engineering reference, so it discusses the theoretical aspects. I found answers to your questions on pages 11.6-3 and 11.6-9, respectively. The general Lagrange multiplier $\lambda(x)$ has, as a specific case the constant function $\lambda_{i}$ for an isoparametric constraint of the type you indicated in (2).

Hope that helps.

  • Yes, that helps. Actually, I am working on a functional $$J(y_1,y_2)=\iint F(x_1,x_2,y_1(x_1),y_2(x_2))dx_1 dx_2$$ and its Lagrange multipliers w.r.t constraint such as $$\iint K(x_1,x_2,y(x_1),y(x_2))dx_1 dx_2=l.$$ Note that in my case, each fucntion only depends on one variable. Unfortunately, I have not seen any textbook regarding this. Do you have any reference suggestions for this kind of problem. Many thanks. – MIMIGA Oct 17 '13 at 23:52
  • That shouldn't be an issue, as you can represent each single-variable function as a degenerate two-variable function: $y_1(x_1) = y^{'}_{1}(x_1,x_2) = y_1(x_1)+0x_2$. A similar arugment can be made for $y_2(x_2)$. Just do that modification and carry though the euler-lagrange equations...you will see that making each y a multi-variable function will have no effect, as the derivatives wrt the ancillary variable will be zero. –  Oct 18 '13 at 12:55
  • You mean the euler-lagrange still holds like $$\frac{\partial{K}}{\partial{y_1}}=0, ~\frac{\partial{K}}{\partial{y_2}}=0.$$ Actually, that is not true. You can refer to my previous post. Thanks. http://math.stackexchange.com/questions/520761/calculus-of-variations-mul-variable-mul-function – MIMIGA Oct 18 '13 at 14:42
  • No, I mean that when you set up the system of euler-lagrange equations for each y, the contribution from the x that is missing from the argument disappears. –  Oct 18 '13 at 14:51
  • Example: for $y_1: \frac{d}{dx_1}(\frac{\partial K}{\partial y_{1|x_2}})+\frac{d}{dx_2}(\frac{\partial K}{\partial y_{1|x_1}}) - \frac{\partial K}{\partial y_1} = 0$. The second term will be zero, as $y_1$ is not sensitive to changes in $x_2$ –  Oct 18 '13 at 14:56
  • Sorry for the confusion. If you do not mind, can you have a look at my problem directly? $$\min \iint F(x_1,x_2,y_1,y_2)dx_1 dx_2,\tag{1}$$ $$s.t. ~\iint K(x_1,x_2,y_1,y_2)dx_1 dx_2=l,\tag{2}$$ $$\int H(x_1,x_2,y_1)dx_1 = y_2, \tag{3}$$ $$g(x_1,x_2,y_1,y_2)=0, \tag{4}$$ where $y_1\equiv y_1(x_1)$ and $y_2\equiv y_2(x_2).$ Perhaps we can add Lagrange multiplier $\lambda$ to (2) and $\lambda(x_1,x_2)$ to (4) such that to minimize $\iint [F+\lambda K+\lambda(x_1,x_2) g] dx_1 dx_2$, but how to do with constraint (3). Thanks a lot. – MIMIGA Oct 18 '13 at 16:01
  • (3) looks like $h(x_2,y_1)-y_2=0$ which we could view, like (4), as $h^*(x_2,y_1,y_2)=0$, so this is theoretically similar to (4), and you would need a $\lambda(x_2)$ i think. –  Oct 18 '13 at 16:13
  • Can I ask for a reference for that '' Lagrange multiplier rule for calculus of variations''? How can we prove that statement? – I0_0I Jul 24 '21 at 18:20