Fundamental lemma of variational calculus.?

Question

Look i know this question is very very stupid. However i would like a detailed explanation of the fundamental lemma of the calculus of variations(please I'm only 14) and a proof if possible. It's the one that states, that given the definite integral of M(x)G(x)dx on an interval where G(b) and G(a) are zero and where the integraL IS ZERO it states that M(x) is also zero. I want somone to explain why M(x) has to be zero.

Answer 1

The actual lemma is "if f(x) is continuous on [a,b], and the integral of f(x)h(x)dx from a to b = 0 for all continuous h(x) such that h(a) = h(b) = 0, then f(x) = 0 on the interval [a,b]."

Prove by contradiction. Suppose f(x) in nonzero, say at some x in the interval [y,z]. Then choose h(x) = (x-y)(z-x) in the interval [y,z], 0 otherwise. h(x) does in fact satisfy the conditions of the lemma. Yet the integral from a to b of f(x)h(x)dx = the integral from y to z of f(x)(x-y)(z-x)dx, which is >0. We have a contradiction with the assumption in the lemma, and therefore f(x) must be zero.

Steve

Answer 2

Fundamental Lemma

Answer 3

Are you thinking of a different lemma? Coz I am have a hard time believing the one you wrote. Basically take an integral which is zero (like sin(x) over a period) and then factor the integrand into something positive* original function/that something positive. There ya go, it ain't true: int(sqrt(x) * sin(x)/sqrt(x), x= 2pi to 4pi) Integral =0 , sin(x)/sqrt(x) =0 at the end points, yet sqrt(x) <> 0 in the interval

However, the website http://mathworld.wolfram.com/FundamentalLemmaofCalculusofVariations.html says something different, if the integral is zero for ALL infinitely differentiable functions h(x) then M(x)=0 for all x. I would guess a proof of that uses functions h(x) which approximates zero. You can come up with a sequence of analytic functions hn(x) which approximate Dirac delta function del(x) = 0 if x<>c and del(c) = 1. Then integral (M(x) * hn(x) ) is going to approximate M(c). Idea in the background: you are putting all the weight of the function on the point c. But the integral is 0 so therefore M(c) approximates to 0. Let n-> infinity and M(c) = 0. Finally, c was arbitrary. You need the appropriate hn(x) formulas (probably something like {1-exp(-1/(x-c)^(2n))} or something like that) and theorems about integrals of limits and limits of integrals to tie up the loose ends. But that would be my first guess at how to prove it.

Hey Yashi try paste URL again it isnt a hot link...... OR put spaces in it so we can paste all into our browser then edit

Answer 4

The fundamental theorem of calculus states that ∫(1..9)ln x dx = F(9)-F(1) = F(9)-2. So F(9) = 2 + ∫(1..9) ln x dx. Here the 1..9 means to evaluate the integral from 1 to 9. The problem actually doesn't require you to calculate the integral, just to find F(9) in terms of it.

Answer 5

Definite integral gives the area of the curve/region being considered within the given interval. So, definite integral of M(x)G(x) within bounds a and b is the area of the curve betwwen a and b. This implies area of curve above the x-axis and below it is equal in magnitude, thus nullifying the overall area.
Now there are two possibilities:
1) G(x)=0 for all a,b in the closed interval[a,b]. In this case M(x) need not be zero because integral will be 0. However this is a trivial case.
2) G(x) is not equal to 0 at some point in open interval(a,b). In this case integral of G(x) will have some finite value unless G(x) is symmetric. Whatever the case may be, for the value of integral(G(x)M(x)) to be 0, G(x) has to be multiplied by a zero-valued function throughout the interval
(a,b). However value of M(x) need not be 0 at end-points a and b. Hope things become clear to you.

Answer 6

go to this site it is a pdf file.
www.worldscibooks.com/phy_etextbook/3557/3557_chap1.pdf

it might help u.