5
$\begingroup$

Second Lemma of variation Calculus

If $u(x)$ is a differentiable function for $a\leq x \leq b$ and $$ \int_{a}^{b} u(x) \cdot \phi'(x)\: dx = 0$$ for all infinitely often differentiable functions $\phi(x)$ then $$u'(x) = 0 \quad \text{for all} \quad a\leq x \leq b \quad \text{and} \quad u(a) \cdot \phi(a) = u(b) \cdot \phi(b) = 0$$

$\textbf{Proof:}$ we use the integration by parts $$ 0 = \int_{a}^{b}u(x)\cdot\phi'(x)\:dx = u(b) \cdot \phi(b) - u(a) \cdot \phi(a) - \int_{a}^{b}u'(x)\cdot\phi(x)\:dx $$ Considering all test function $\phi(x)$ with $\phi(a) = \phi(b) = 0$ leads to the condition $u'(x) = 0$. We are free to choose test functions with arbitrary values at the end points $a$ and $b$, thus we arrive at $u(a)\cdot\phi(a) = u(b))\cdot\phi(b) = 0$

For a given function $f(x,u,u')$ we try to find $u(x)$ such that the functional $$ F(u) = \int_{a}^{b} f(x,u(x),u'(x))\:dx $$ has a critical value for the function $u$. If the functional $F$ attains its minimal value at the function $u(x)$ we conclude that $$ g(\varepsilon) = F(u + \varepsilon\phi) \geq F(u) \quad \text{for all $\varepsilon \in \mathbb{R}$ and arbitrary functions $\phi(x)$} $$ Thus the scalar function $g(\varepsilon)$ has a minimum at $\varepsilon = 0$ and thus the derivative should vanish. We require that $$ \frac{d\:g(0)}{d\varepsilon} = \frac{d}{d\varepsilon}F(u + \varepsilon\phi)\big|_{\varepsilon = 0} = 0\quad \text{for all functions} \quad \phi $$ To find the equations to be satisfied by the solution $u(x)$ we use linear approximations. For small values of $\Delta u$ and $\Delta u'$ we use a Taylor approximation to conclude \begin{align*} f(x, u + \Delta u, u' + \Delta u') &\approx f(x,u,u') + \frac{\partial f(x,u,u')}{\partial u}\Delta u + \frac{\partial f(x,u,u')}{\partial u'}\Delta u' \\ &= f(x,u,u') + f_u (x,u,u')\Delta u + f_{u'} (x,u,u')\Delta u' \\ f(x, u(x) + \varepsilon\phi(x), u'(x) + \varepsilon\phi'(x)) &= f(x,u(x),u'(x)) + \varepsilon f_u(x,u(x),u'(x))\phi(x) \\ &+ \varepsilon f_{u'}(x,u(x),u'(x))\phi'(x) + O(\varepsilon^2) \\ \end{align*} Now we examine the functional in question \begin{align*} g(0) &= F(u) = \int_{a}^{b}f(x,u(x),u'(x))\:dx \\ g(\varepsilon) &= F(u + \varepsilon\phi) = \int_{a}^{b}f(x,u(x) + \varepsilon\phi(x),u'(x) + \varepsilon\phi'(x))\:dx \\ &= \int_{a}^{b}f(x,u(x),u'(x)) + \varepsilon f_u(x,u(x),u'(x))\phi(x) + \varepsilon f_{u'}(x,u(x),u'(x))\phi'(x)\:dx \\ &= F(u) + \varepsilon \int_{a}^{b} f_u(x,u(x),u'(x))\phi(x) + f_{u'}(x,u(x),u'(x))\phi'(x)\:dx\tag{1} \end{align*} or $$ \frac{d}{d\varepsilon}F(u + \varepsilon\phi) \big|_{\epsilon = 0} = \int_{a}^{b} f_u(x,u(x),u'(x))\phi(x) + f_{u'}(x,u(x),u'(x))\phi'(x)\:dx\tag{2} $$ This integral has to vanish for all function $\phi(x)$ and we may use the Fundamental Lemma, leading to a necessary condition. An integration by parts leads to \begin{align*} 0 &= \int_{a}^{b} f_u(x,u(x),u'(x))\phi(x) + f_{u'}(x,u(x),u'(x))\phi'(x)\:dx\tag{3}\\ &= f_{u'}(x,u(x),u'(x))\phi'(x)\big|_{x = a}^b + \int_{a}^{b}\Big( f_u(x,u(x),u'(x)) - \frac{d}{dx}f_{u'}(x,u(x),u'(x))\:\Big)\: \phi(x)\:dx\tag{4} \end{align*} Since this expression has to vanish for all function $\phi(x)$ we need $$ \int_{a}^{b}f(x,u(x),u'(x))\:dx \: \text{extremal} = \left\{ \begin{array}{ll} \frac{d}{dx}f_{u'}(x,u(x),u'(x)) &= f_u(x,u(x),u'(x))\quad \quad \quad \quad \text{(5)}\\ f_{u'}(a,u(a),u'(a))\cdot\phi(a) &= 0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{(6)}\\ f_{u'}(b,u(b),u'(b))\cdot\phi(b) &= 0 \\ \end{array} \right. $$

Here are my questions

  • how can you go from (1) to (2), or what does the expression $\frac{d}{d\varepsilon}F(u + \varepsilon\phi) \big|_{\epsilon = 0}$ exactly means and how does it connect with the rest?
  • how can you go from (3) to (4)
  • how can you assert $\frac{d}{dx}f_{u'}(x,u(x),u'(x)) = f_u(x,u(x))$ at (5) from that proof

for the answer of each of those points, please detail your answer as much possible with step by step explanation

$\endgroup$
1
$\begingroup$

Partial answer:

First things first: High praise for the beautifully formatted question!

To go from 1 to 2, note that equation 1 looks like (as a function of $\epsilon$ only)

$$ g(\epsilon) = C + \epsilon D $$ where $C$ and $D$ are expressions not involving $\epsilon$. The derivative of $g(\epsilon)$ with respect to $\epsilon$ is evidently $D$.

As for "what does $\frac{d}{d\varepsilon}F(u + \varepsilon\phi) \big|_{\epsilon = 0}$ mean?", it means exactly the computation I just did above.

In general $\frac{d}{dz} H(z) \big|_{z = c}$ means "take the derivative of $H$ with respect to $z$, at the point $z = c$," which could also be written $H'(z)$. When $H$ is given as a more complex expression, rather than having a single name, i.e., when instead of $H$ we have $F(k + z p)$, for instance, there's no easy "prime" notation to express this (especially when $k$ and $p$ are not real numbers, but something more complicated like functions).

One other answer to that question is an explicit formula: it means $$ \lim_{h \to 0} \frac{F(u + (0+h)\cdot\phi) - F(u + 0 \cdot\phi)}{h}, $$ pretty much the same as any other derivative you've ever seen.

To go from 3 to 4, use integration by parts (with upper and lower limits), namely $$ \int_a^b F(x)G'(x) dx = F(x)G(x)\big|_a^b - \int_a^b F'(x) G(x) dx $$ in this case applying the integration-by-parts only to the SECOND integral in (4), and picking $$ F(x) = f_{u'}(x,u(x),u'(x))\\ G(x) = \phi(x) $$ The right-hand side then becomes a sum of the first integral, an evaluation from $a$ to $b$, and another integral, and the authors have combined the two integrals.

To answer the third bullet, I had to go back and look at the text. In the paragraph just after the word "Proof" , I believe the sentence starting "For a given function ..." should begin a new paragraph, or indeed, a new section, in which the lemma just proved will be applied.

When you get to equation (2), you have exactly the situation listed in the Lemma: some function (in this case a complicated mess) times $\phi'$ integrates to $0$ for all possible $\phi$. From that, you can conclude directly the the complicated mess must be everywhere zero (which is exactly what the top line of claim (5) says).

(The remaining integration by parts tells you some other stuff, I guess, I suppose the other two lines of claim 5, but I didn't check this.)

$\endgroup$
  • $\begingroup$ many thanks @JohnHughes. That helped already a lot. I upvoted your answer. Hope you can clarify the third bullet as well $\endgroup$ – ecjb Jan 6 at 17:28
  • $\begingroup$ I've revised to provide an explanation of the first part of claim 5. $\endgroup$ – John Hughes Jan 6 at 18:17
  • $\begingroup$ I edited the paragraph, i'm reading and trying to understand the rest of your explanation $\endgroup$ – ecjb Jan 6 at 18:35
  • $\begingroup$ Many thanks for your time @JohnHughhes. I'm sorry that the (5) didn't appear on the line I wanted to. I used \tabular{} I didn't know how to fix it and that's why I wrote in the question that it was about the first (5) and not the second line (6). The second line (6) was also clear to me for the reason you gave. The first line (5) is the one that I still don't understand. Sorry again for not being clear enough $\endgroup$ – ecjb Jan 6 at 18:47
  • $\begingroup$ The first line, 5, is exactly the one proved by the first conclusion ($u'(x) = 0$) of the lemma! From line 2, you have $\int (mess_1 - mess_2) \phi' =0$. The lemma tells you that this means $mess_1 - mess_2$ is zero; that in turn means that $mess_1 = mess_2$. $\endgroup$ – John Hughes Jan 6 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.