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I am new to calculus of variations, till now I know how to get the extremal functions for a given functional using Euler-Lagrange equation.

What if I have a functional but I am not looking for minimizing/maximizing it, but instead solving equations involving functionals, say: $$I = \int_{x_{1}}^{x_2}{F(x,y,y') \,\mathrm{d}x}=\alpha\quad \,,\text{for }\alpha\in \mathbb{R}$$ How to solve for $y(x)$ that satisfy this equation? can I transform it to a classical problem then solve it using Euler-Lagrange equation?

EDIT: for example say, we have the following problem : $$I = \int_{0}^{1}{\left(f(x)+2f'(x)\right) \, \mathrm{d}x}= 1/2$$

I appreciate any ideas,

Thank you

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    $\begingroup$ It's possible this could be "transform[ed].... to a classical problem", but not enough information about $F$ is given to suggest ways that might succeed. It appears that you are trying to determine a function $y$ (on what interval?) from a single scalar equation, which generically is not enough information to make a full determination of $y$. $\endgroup$
    – hardmath
    Jun 23, 2016 at 0:11
  • $\begingroup$ y(x) for $x \in \, [x_1,\, x_2]$ $\endgroup$
    – Freshman42
    Jun 23, 2016 at 0:17
  • $\begingroup$ I hope you take my point, that a single equation is not enough to determine the infinite degrees of freedom that define $y(x)$ on $[x_1,x_2]$. There are variational formulations (see Galerkin methods) that use a family of integral equations to determine a function $y(x)$ (and for which the variational formulation may be more tractable than the classical formulation of the problem). $\endgroup$
    – hardmath
    Jun 23, 2016 at 0:30
  • $\begingroup$ @Hardmath I really do not understand what kind of additional info to solve the problem, can you give me an example please ( for me it seems logical because I replaced the min/max requirement by $\alpha$)? $\endgroup$
    – Freshman42
    Jun 23, 2016 at 0:36
  • $\begingroup$ @Freshman42 To your example: choose some arbitrary function $g$ which is finite with finite derivative on $[0,1]$. Now, compute $I_g=\int_0^1g(t)+2g'(t)\ dt$. Finally, set $f\equiv \frac{1}{2I_g}g$ (this assumes that $I_g$ isn't zero, of course, but it will only be zero for a small number of functions). Now this $f$ - which is essentially arbitrary - satisfies your equation. $\endgroup$ Jun 23, 2016 at 0:49

2 Answers 2

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You can rig this to have unique solutions for very particular $\alpha$ (e.g. $F = (y-f)^2$, $\alpha = 0$ has unique solution $y=f$) but in general you should expect a large family of solutions.

Intuitively this is because the space of functions is much bigger than the space of possible values of the functional $J(y)=\int F[y]$, so $J$ can't be anything close to injective. Making this rigorous isn't quite as simple as a cardinality argument (since e.g. $C^1([0,1])$ has the same cardinality as $\mathbb R$), but reasonable requirements on $F$ will make $J$ a differentiable map when restricted to a finite-dimensional space of functions. Sard's theorem then tells you that $J(y)=\alpha$ has multiple solutions (or no solution) for almost every $\alpha$, even amongst a 2-parameter family of functions $y$.

This shouldn't be too surprising - prescribing the value of the functional (a single real number) is much less information than prescribing the derivative of the functional (an element of some infinite-dimensional function space).

Another way of putting it: You're prescribing some kind of average of $y$, but not the local behaviour.

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Suppose we have a point particle of mass $m$ moving along the $y$ axis. The Lagrangian is the difference between the kinetic ($K$) and potential ($U$) energies

$$\mathcal{L} (y, \dot y) := K - U = \frac{1}{2} m \dot y^2 - m g y$$

and the action is the functional

$$S (y) := \int_0^T \mathcal{L} (y, \dot y) \, \mathrm{d}t$$

The Euler-Lagrange equation gives us the 2nd order differential equation $\ddot y = -g$. However, suppose we have the equality constraint

$$\int_0^T \mathcal{L} (y, \dot y) \, \mathrm{d}t = S_0$$

which can be rewritten in the form

$$\frac{S_0}{T} = \left(\frac{1}{T} \int_0^T \frac{1}{2} m \dot y^2 \, \mathrm{d}t \right)- \left(\frac{1}{T} \int_0^T m g y \, \mathrm{d}t \right) = \langle K \rangle - \langle U \rangle$$

where $\langle K \rangle$ and $\langle U \rangle$ are the average kinetic and potential energies, respectively. Thus, the equality constraint merely imposes a constraint on the average kinetic and potential energies.

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    $\begingroup$ thank you but $K = \dfrac{1}{2}m\dot y^2$, can you apply this for the given example above? $\endgroup$
    – Freshman42
    Jun 23, 2016 at 0:53
  • $\begingroup$ @Freshman42 Thanks. $\endgroup$ Jun 23, 2016 at 0:55
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    $\begingroup$ @Freshman42 $$\int_0^1 f (x) \, \mathrm{d}x + 2 f (1) - 2 f (0) = \frac{1}{2}$$ $\endgroup$ Jun 23, 2016 at 1:03

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