Direct method for Calculus of variation for problem specifying slope and value of the function

Question

I am looking for a direct method proof of the infimum of a calculus of variation problem. $$\inf_{u\in\scr{C}[0,1],u (0)=0}\int_0^1u^2+(u'-b)^2dx \tag 1$$ The first term wants to set $u=0$ locally, while the second term wants to set $u'=b$. There is a competition between the two. The Euler Lagrange equations would give me something like $u(x)=b\dfrac{\sinh (x)}{\cosh 1}$ and the corresponding energy is $\dfrac{b^2}{e\cosh 1}<b^2$. However, this assumes $u\in\scr{C}^2[0,1]$. I think there might be a minimizing sequence that can lower the energy even more. If a minimum exists, then we should be able to prove that. References to any papers would be helpful.

Some More Analysis That I did $$\int_0^1u^2+(u'-b)^2dx=\int_0^1u^2+u'^2dx+b^2-2u(1)b\geq b^2-2u(1)b \tag 2$$ where to get the inequality, we used the fact that the term in the integral is positive so that the smallest possible value would be controlled by u(1). However, to integral only achieves its minimum value if the function locally is zero. This would impose restrictions on $u(1)$.

Answer 1

In this part of the answer we again show that $$\inf_{u\in\scr{C^1}[0,1],u (0)=0}\int_0^1u^2+(u'-b)^2dx \tag{1*}$$ is attained, and we show that the minimizer is $$ v(x)=b\dfrac{\sinh (x)}{\cosh 1}. $$ That means that $F(v) \le F(u)$ for every admissible $u$. We say that $u$ is admissible when $u\in\scr{C^1}[0,1]$ and $u(0)=0$

This time we use the convexity -- maybe. And compared to the previous, I change the order from "how I got that" to "how one thing follows from another".

Let $F(u)=\int_0^1u^2+(u'-b)^2dx$ be the functional.

Obviously $v$ is $C^\infty$ smooth, hence $C^2$ smooth and also $C^1$ smooth. Obviously, $v$ is admissible. By direct calculation we can check that $v''=v$ and $v'(1)=b$.

If $u$ is $C^2$ and $h$ is $C^1$ and $u(0)=h(0)=0$, I calculate $$ F(u+h)=F(u)+DF(u)(h)+\int_0^1 h^2+\int_0^1 h'{}^2 \tag 3 $$ where (in certain anticipations) I wrote $DF(u)(h)$ instead of $$ \int_0^1 2u h + 2(u'-b) h' = \int_0^1 2u h - 2u'' h + \bigl[ (u' - b) h\bigr]_0^1 . \tag 4 $$

As $v$ is $C^2$, I can do the above with $v=u$. This time we use also that $v''=v$ and $v'(1)=b$. If $h$ is $C^1$ and $u(0)=h(0)=0$, we have $$ F(v+h)=F(v)+DF(v)(h)+\int_0^1 h^2+\int_0^1 h'{}^2 \tag {3v} $$ where as before I wrote $DF(v)(h)$ instead of $$ \int_0^1 2v h + 2(v'-b) h' = \int_0^1 2v h - 2v'' h + \bigl[ (v' - b) h\bigr]_0^1 . \tag {4v} = 0. $$

VARIANT 1:

We will now prove that $F$ attains minimum at $v$. This means that $F(v) \le F(u)$ for every admissible $u$.

So let $h=u-v$. Looking at (3v)&(4v), we see that $F(u)=F(v+h)=F(v)+0+positive\ge F(u)$ which is what we had to prove. QED.

VARIANT 2:

(4v) means (**) that the directional derivative (sometimes called the variation) of $F$ at $v$ in the direction of $h$ is $0$. This can be written as $DF(v)(h)=0$. (Forgive me I used the notation to early, in eager anticipation.) There is a (simple) theorem stating that if $F$ is convex and $DF(v)(h)=0$ for every admissible $h$, then $F$ attains the minimum at $v$. QED. Actually, at (**), the argument is broken. I wonder if any one notices that, and I do not want to rewrite the text again. The correct argument means to use the definition of derivative, together with (3v)&(4v) with $h$ substituted by $t h_0$, and to calculate the corresponding limit where we will observe $t/t$ cancel at the proper place and $t^2/t=t$ make diminishes something that we want to disappear. (There is also alternative to use a ready-made statement relating zero derivative to the Euler-Lagrange equation but this is only weakly related in what I found in Gelfand Fomin, p.28, with end-point condition missing.) This works but VARIANT 1 is easier, after all.

Answer 2

I cannot comment yet, so forgive me putting this here.

I think that in (1), you do not want to allow all continuous functions.

If you allow just almost everywhere differentiable functions, the answer actually is: the infimum is zero. For any $\varepsilon>0$, let $u(0)=0$, and from zero follow the equation $u'=b$ until $u$ reaches height $\varepsilon$. Then use (reversed) devil's staircase to go back down to zero on an interval of length $\varepsilon^2$. Repeat this construction periodically to define $u$ on the full length of interval $[0,1]$. The functional evaluates at $u$ to something $\le \varepsilon^2 + b^2 \varepsilon$ (or similar), which is arbitrarily close to $0$.

So, unless this is the desired result, you want to put $C^1$ or $AC$ as the class for $u$ in the definition of the functional.

BTW what is your basic reference for the calculus of variations and direct method?

Answer 3

This is an older version of another answer. The difference is that here we explain that we found the minimizer by solving the Euler-Lagrange equation, which however makes the ordering confusing and the relation between the proposition less graspable.

In this part of the answer we show that $$\inf_{u\in\scr{C^1}[0,1],u (0)=0}\int_0^1u^2+(u'-b)^2dx \tag{1*}$$ is attained, and we show that the minimizer is $$ u(x)=b\dfrac{\sinh (x)}{\cosh 1} $$

I repeat the reasoning behind the following part of the question but provide some new details.

The Euler Lagrange equations would give me something like $u(x)=b\dfrac{\sinh (x)}{\cosh 1}$ and the corresponding energy is $\dfrac{b^2}{e\cosh 1}<b^2$.

Let $F(u)=\int_0^1u^2+(u'-b)^2dx$ be the functional. If $u$ is $C^2$ and $h$ is $C^1$ and $u(0)=h(0)=0$, I calculate $$ F(u+h)=F(u)+DF(u)(h)+\int_0^1 h^2+\int_0^1 h'{}^2 \tag 3 $$ where (in certain anticipations) I wrote $DF(u)(h)$ instead of $$ \int_0^1 2u h + 2(u'-b) h' = \int_0^1 2u h - 2u'' h + \bigl[ (u' - b) h\bigr]_0^1 . \tag 4 $$ Now, if $u$ is a minimizer, and we use $u+h$ for comparison, with all smooth $h$ with $h(1)=0$, we see that $u''=u$. I think this is called E-L. Obviously you were solving $u''=u$ with the initial condition $u(0)=0$ and found $$ ()= k \sinh() \qquad (k\in R) \tag 5 $$ as a solution and I have the same result. The minimizer (if it exists and if it is $C^2$) must be of this form. At this moment I did not know why did you claim $k$ must be $k=b/\cosh 1$, but I will arrive at the same as you. (Addendum: I see, you used condition (29$_{+1}$) for mixed problems on page 26, Sec. 6 of Gelfand Fomin, 1963, 2000)

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Now the main part: if $u$ is as in (5), and if $u'(1) = b$, then we have that (4) is zero, hence in (3) we have $F(u+h) = F(u) + 0 + positive \ge F(u)$ - for every $h$ $C^1$ smooth with $h(0)=0$. That MEANS that $u$ is minimizer of $F$.

Remark: Since I assume $u'(1)=b$, I did not need to assume that $h(1)=0$ to get that $(4)$ is zero in the last paragraph. So we compare $F(u)$ with $F(u+h)$ for all admissible $u+h$.

It only remains to find $k$ for which we have $u'(1)=b$, and this is $k=b/\cosh h$.