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Calculus of Variations

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Section 2.3 A fundamental lemma

In the integrand of the integral (2.2.3) the coefficient of \(\eta'\) is really a function of \(x\text{,}\) since the derivative \(f_{y'}\) contains as its argument the slope \(y'(x)\) of the arc \(E_{12}\text{,}\) and we may denote this coefficient by \(M(x)\text{.}\) It should be noted that the function \(M(x)\) is continuous except possibly at the values of \(x\) defining the corners of the arc \(E_{12}\) where the slope \(y'(x)\) changes abruptly. At those points of the curve it has two values, one corresponding to the backward and one to the forward slope. The lemma which we wish to prove is then as follows:
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Remark 2.3.2.

I want to restate this lemma just slightly so it’s clearer what it’s saying. Suppose that:
  • \(M(x)\) is a piecewise-differentiable function
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  • \(\eta(x)\) is a variation
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  • \(\displaystyle \int_{x_1}^{x_2} M(x)\eta'(x)\,dx = 0\text{,}\) no matter which variation \(\eta(x)\) you choose!
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Then we get to conclude that \(M(x)\) is in fact a constant function.
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Just for one more layer of emphasis here: IF!!!!! the three things in bullets, THEN!!!! \(M(x)\) is constant.
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Activity 2.3.1.

  1. Suppose \(C\) is just some constant, and remember that \(\eta(x)\) is supposed to be a variation in the sense of Remark 2.2.1. Compute \(\int_{x_1}^{x_2} C\cdot\eta'(x)\,dx.\)
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    Solution.
    \begin{equation*} \int_{x_1}^{x_2} C\cdot\eta'\,dx = C \cdot \eta(x) \large\vert_{x_1}^{x_2} = C \cdot (\eta(x_2) - \eta(x_1)) = C\cdot(0-0) = 0. \end{equation*}
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  2. Consider the function \(\displaystyle \eta^*(x) = \left(\int_{x_1}^x M(t)\,dt\right) - C\cdot(x-x_1)\text{.}\) As you may suspect since I used the letter \(\eta\text{,}\) I’d really like it if \(\eta^*(x)\) was a variation. Compute \(\eta^*(x_1)\text{.}\)
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  3. Is there a value of \(C\) (which is supposed to be a constant) such that \(\eta^*(x_2)=0\text{?}\) (Hint: \(\int_{x_1}^{x_2} M(t)\,dt\) is just some area or other.)
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  4. Compute the derivative of this particular variation \(\eta^*(x)\text{.}\) (You’ll need the fundamental theorem of calculus.)
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  5. Finally, and changing gears a little, think of some function \(f(x)\text{.}\) What can you tell me about \(\int_a^b \left[f(x)\right]^2\,dx\text{?}\) Is it positive or negative? Can it ever be zero?
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To see that this is so we note first that the vanishing of the integral of the lemma implies also the equation
\begin{equation} \int_{x_1}^{x_2} \left[M(x)-C\right]\eta'(x)\,dx = 0\tag{2.3.1} \end{equation}
for every constant \(C\text{,}\) since all the functions \(\eta(x)\) to be considered have \(\eta(x_1)=\eta(x_2)=0\text{.}\) The particular function \(\eta(x)\) defined by the equation
\begin{equation} \eta(x) = \int_{x_1}^x M(x)\,dx - C\cdot(x-x_1)\tag{2.3.2} \end{equation}
evidently has the value zero at \(x=x_1\) and it will vanish again at \(x=x_2\) if, as we shall suppose, \(C\) is the constant value satisfying the condition
\begin{equation*} 0=\int_{x_1}^{x_2} M(x)\,dx - C\cdot(x_2-x_1). \end{equation*}
The function \(\eta(x)\) defined by (2.3.2) with this value of \(C\) inserted is now one of those which must satisfy (2.3.1). Its derivative is \(\eta'(x)=M(x)-C\) except at points where \(M(x)\) is discontinuous, since the derivative of an integral with respect to its upper limit is the value of the integrand at that limit whenever the integrand is continuous at the limit. For the special function \(\eta(x)\text{,}\) therefore, (2.3.1) takes the form
\begin{equation*} \int_{x_1}^{x_2} \left[M(x)-C\right]^2\,dx =0 \end{equation*}
and our lemma is an immediate consequence since this equation can be true only if \(M(x) \equiv C\text{.}\)
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Activity 2.3.2.

We’re going to deduce another form of the fundamental lemma that will come in handy. It’s weird to me that in the original version of the fundamental lemma (Lemma 2.3.1), there’s an \(\eta'(x)\) instead of an \(\eta(x)\text{.}\) So, consider the very similar integral
\begin{equation*} \int_{x_1}^{x_2} m(x) \eta(x)\, dx, \end{equation*}
and suppose the same hypotheses as in the original version: \(m(x)\) is piecewise continuous, \(\eta(x)\) is a variation, and the value of this integral is 0 no matter which \(\eta(x)\) you choose.
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  1. Whenever I see something that looks like the integral of a product, I immediately start thinking about integration by parts (which is, after all, the integral version of the product rule). Choose a \(u\) and a \(dv\) that will get you something that looks like the integral in the original version. Why are your choices good?
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    Solution.
    \(u = \eta(x)\) will produce \(du = \eta'(x)\, dx\text{,}\) which definitely appears in the original version, and \(dv = m(x)\,dx\) will produce... well, I’m not sure, because I don’t exactly know what \(m(x)\) is, but let’s just say that the antiderivative of \(m(x)\) is \(\int m(x)\, dx = M(x)\text{.}\)
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  2. Now go ahead and use integration by parts. Careful: this is a definite integral. (Hint: What’s \(\eta(x_1)\) and \(\eta(x_2)\text{?}\))
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    Solution.
    \begin{align*} \int_{x_1}^{x_2} m(x) \eta(x)\, dx \amp= \eta(x) M(x) \bigg\vert_{x_1}^{x_2} - \int_{x_1}^{x_2} M(x) \eta'(x)\,dx \\ \amp = (\eta(x_2) M(x_2) - \eta(x_1) M(x_1)) - \int_{x_1}^{x_2} M(x) \eta'(x)\,dx \\ \amp = - \int_{x_1}^{x_2} M(x) \eta'(x)\,dx. \end{align*}
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  3. The value of your integral is supposed to be 0, by supposition. If you’ve chosen your parts wisely, you’re in a situation where you can invoke Lemma 2.3.1. What can you conclude about \(m(x)\text{?}\)
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    Solution.
    Since \(\displaystyle 0 = - \int_{x_1}^{x_2} M(x) \eta'(x)\,dx\text{,}\) Lemma 2.3.1 implies that \(M(x)\) is a constant. Since \(M(x)\) was the antiderivative of \(m(x)\text{,}\) we can conclude that \(m(x) \equiv 0\) -- that is, that \(m(x)\) is the constant function 0.
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