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In several (good) textbooks in calculus of variations one important step when dealing with the isoperimetric problem seems not to be properly addressed. The problem is as follows: let $G$ be a smooth enough function in three variables and let $$ K[y]=\int_a^b G(x,y,y')\,dx $$ be the corresponding functional on $C^1([a,b]).$ Suppose that $z \in C^1([a,b])$ is not a solution to Euler's equation for $K.$

Given distinct points $x_1,x_2 \in (a,b)$ and any positive $\varepsilon_1,\varepsilon_2 >0,$ we wish to have a pair of functions $h_1,h_2 \in C^1([a,b])$ such that

1) each $h_k$ is of constant sign on an open subinterval $(a_k,b_k)$ of $[a,b]$ such that $x_k \in (a_k,b_k),$ and vanishes outside $(a_k,b_k);$ the intervals $(a_1,b_1)$ and $(a_2,b_2)$ are to be disjoint;

2) $b_1-a_1 < \varepsilon_1$ and $|h_1(x)| < \varepsilon_2$ on $(a_1,b_1);$

3) $K[z]=K[z+h_1+h_2].$

Intuitively, such $h_1,h_2$ must exist (continuity, etc.), but at the moment I fail to see how to prove this simply and neatly (the books I've mentioned either take existence of $h_1,h_2$ in question for granted, or contain some vague remarks on continuity).

Olod
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  • Why do you need two functions $h_1$ and $h_2$ ? And why are no conditions imposed on $h_2$? In other words, could you explain in words what their role is? I presumed they both were small variations. – Raskolnikov Jun 13 '13 at 19:53
  • Since small smooth bumps are needed at two points. Such variations are used in the standard proof of the isoperimetric problem (when one wishes to find an extremal of a functional of the above type, fixing values at endpoints, and leaving the value of some other functional fixed). – Olod Jun 13 '13 at 20:22
  • $h=h_1+h_2$ is a varition of $z$ which must leave the value of $K$ fixed. $h_2$ appers on the scene once, but in a very important role, to be assoiciated with a point $x_2 \in [a,b]$ where the functional variation of $K$ is nonzero. Thereafter, using the condition $K[z]=K[z+h_1+h_2],$ we express the increment $J[z+h_1+h_2]-J[z]$ of another functional $J$ essentially in terms of $h_1$ which is associated with an arbitrary point of $[a,b].$ Sorry, can't write all details here, for I would need to reproduce the whole of the proof of the isoperimetric problem. – Olod Jun 13 '13 at 20:44

1 Answers1

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The integral form of functional $K$ implies that for any smooth functions $h_1,h_2$ with disjoint supports we have $$K[z+h_1+h_2]-K[z] = (K[z+h_1]-K[z])+(K[z+h_2]-K[z]) \tag1$$

The function $t\mapsto K[z+th]$ is smooth (with respect to real variable $t$), because $G$ is smooth.

We pick the intervals $(a_i,b_i)$ where the Euler-Lagrange equation fails. This makes it possible to choose $h_1$ and $h_2$ so that $\frac{d}{dt} K[z+th_i]\ne 0$ at $t=0$. Indeed, if we couldn't, that would imply that the Euler-Lagrange equation holds.

The nonvanishing of derivatives makes it possible to pick small values of $t_1$ and $t_2$ so that $K[z+t_ih_i]-K[z]$ have opposite signs for $i=1,2$. If $K[z+t_ih_i]-K[z]$ have the same absolute value for $i=1,2$, we are done by (1). Otherwise, focus on the larger, say $$|K[z+t_1h_1]-K[z]|>|K[z+t_2h_2]-K[z]|$$ By the intermediate value theorem, there is $\tilde t_1$ between $0$ and $t_1$ such that $$K[z+\tilde t_1h_1]-K[z] = -(K[z+t_2h_2]-K[z])$$

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