# Constrained variational problems intuition

Problem: minimise $$F(x,y,y')$$ over $$x$$, constrained by $$G(x,y,y')=0$$.

$$J_1(x,y,y')=\large \int_{x_0}^{x_1}F(x,y,y')+ \lambda (x) G(x,y,y')dx$$

I understand the Euler-Lagrange equation and Lagrange multipliers in multivariable (i.e. not variational) calculus, but am having a hard time putting them together, I don't understand the logic behind this equation.

1. Is this the same $$\lambda$$ that appears in the multivariable $$\nabla f (x,y)= \lambda \nabla g (x,y)$$? If so, why is it not also a function of $$y$$?

2. By the fundamental lemma of the calculus of variations, doesn't $$\int_{x_0}^{x_1}\lambda (x) G(x,y,y')dx=\int_{x_0}^{x_1}\lambda (x) 0dx=0$$, thus $$J=J_1$$, and so the constraint has had no effect on the integral?

3. More importantly, what is the proof that if the minimum of $$J_1$$ is found (i.e. the Euler-Lagrange equations are satisfied with this new integrand) $$G$$ will be constrained properly?

In short: what is the logic behind that integrand?

Side-note: Without constraints, the problem is simply to minimise $$J(x,y,y')=\large \int_{x_0}^{x_1}F(x,y,y')dx$$ i.e. solve the Euler-Lagrange equation for the functional $$F$$ and the variable $$y$$. (Side-question: why is $$G$$ usually given as a function of only $$x$$ and $$y$$, and $$J$$ of $$y$$? Surely in $$J$$'s case this severely limits the number of possible functionals $$F$$ [as the integral must not be a function of $$x$$ or $$y'$$], and in $$G$$'s case there are many possible constraints involving $$y'$$ that are overlooked?).

• On 3.: That's not the case; you need to solve the resulting Euler-Lagrange equations together with the constraint equation; the constraint equation isn't automatically satisfied. You've introduced one more unknown, $\lambda$, and you have one more equation to determine it, just as in the multivariable case. Jan 15, 2013 at 20:31
• @joriki Thank you. Why not just solve the 'normal' (nonconstrained) Euler-Lagrange equation and the constraint together (what is special about the 'new' Euler- Lagrange equation?)? I suppose that may be one of the cruces of my bafflement.
– Meow
Jan 15, 2013 at 20:37
• You can't -- without $\lambda$, you don't have the additional freedom required to satisfy them both. If you just solve the normal Euler-Lagrange equation, the solution will generally not fulfil the constraint. It's exactly analogous to the multivariable case. Jan 15, 2013 at 20:40

I think you have some problems, because you use an incorrect notation. Let me rewrite your original problem: \begin{align*} \text{minimize}\quad & J(y) = \int_{x_0}^{x_1} F(x, y(x), y'(x)) \, \mathrm{d} x \\ \text{subject to}\quad & G(x, y(x), y'(x)) = 0 \quad\text{for all } x \in [x_0,x_1]. \end{align*} Here, $$F : \mathbb{R} \times \mathbb R \times \mathbb R \to \mathbb R$$ and $$G : \mathbb R \times \mathbb R \times \mathbb R \to \mathbb R^n$$. Do you see the differences? $$J$$ only depends on the function $$y$$, whereas the integrand $$F$$ and the constraint $$G$$ depend on real numbers.

Now (if a constraint qualification is satisfied), you get a multiplier $$\lambda : [x_0, x_1] \to \mathbb R^n$$ (compare with section 6.2 in your link: you get a multplier for each constraint, that is, for each $$x$$), such that the derivative of the Lagrangian $$J(y) + \int_{t_0}^{t_1} G(x, y(x), y'(x)) \, \lambda(x) \, \mathrm{d} x$$ with respect to $$y$$ is zero (that is, the derivative of your lagrangian w.r.t. the optimization variable). Now, you can continue like for the derivation of the euler-lagrange equation.

@2.: Yes, this is correct, but the derivative of $$J$$ w.r.t. $$y$$ does not equals the derivative of $$J_1$$ w.r.t. $$y$$.

@3.: As joriki already said, you have to solve the resulting Euler-Lagrange equation together with the constraint. In other words: The Euler-Lagrange equation depends on $$\lambda$$. Once you have fixed $$\lambda$$, they have a unique solution $$y$$ (depending on $$\lambda$$). It remains to choose $$\lambda$$, such that the corresponding $$y$$ satisfies the constraints (this is reasonable, since you have as many constraints as degrees of freedom in $$\lambda$$).

• Thanks for the thorough answer, but I have one further question: if we know $G(x,y(x),y'(x))=0$, why don't we just use this knowledge: insert it into $J_1$ and make out lives easier? This doesn't seem incorrect, but implies that $J=J_1$, which is obviously untrue as the unconstrained problem is very different from the constrained one.
– Meow
Jan 22, 2013 at 18:54
• Your $J_1$ is the Lagrangian and $J$ is the objective. Note that $J_1$ depends on your optimization variable and additionally on your multiplier $\lambda$. You stated correctly $J_1(y, \lambda) = J(y)$ iff $y$ satisfies the constraints. However, the multiplier rule tells you to compute the zeros of the derivative of $J_1$. But this differs from the derivative of $J$.
– gerw
Jan 23, 2013 at 11:33