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basic_tools:variational_calculus [2018/03/08 22:49]
iiqof [Student] Formating
basic_tools:variational_calculus [2021/04/17 19:03] (current)
cleonis Added assertion that different approaches will converge to the same differential equation
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 ====== Variational Calculus ====== ====== Variational Calculus ======
  
-<tabbox Why is it interesting?> ​+<tabbox Why it is interesting?>​
  
 +Variational calculus is the alternative to the usual calculus methods when we want to find functions that minimize something. As an analogy, usually when we search for the extrema of a function, we differentiate the function, set the derivative of the function to zero and find the point that yields the extrema. Similar results can be computed by using variational calculus.In variational calculus we find extrema of functionals which are functions of functions with respect some function (instead of variable). This is extremely important for the [[formalisms:​lagrangian_formalism|Lagrangian formalism]]. ​
  
-Variational calculus is the alternative to the usual calculus methods when we want to find functions that minimize something. As an analogy, usually when we search for the extrema of a function, we differentiate the function, set the derivative of the function to zero and find the point that yields the extrema. Similar results can be computed by using variational calculus.In variational calculus we find extrema of functionals which are functions of functions with respect some function (instead of variable). This is extremely important for the [[frameworks:​lagrangian_formalism|Lagrangian formalism]]. ​ 
  
 +**Variational calculus of the catenary**
  
-<tabbox Layman> ​+'​Catenary'​ is the name of the curve that represents the shape of a hanging chain. The catenary problem is a problem in statics. Each point of the hanging chain is motionless, as if //all// points of the chain are anchored.
  
-<​blockquote>​Another way of saying a thing is least is to say that if you move the path a little bit at first it +At the anchor point the tension ​in the chain can be calculated as follows: ​the weight tugging at that point is the total weight of the chain. That total weight ​is tugging ​in vertical ​direction. ​Then the horizontal component of the chain tension ​is given by the local angle of the chain.
-does not make any difference. Suppose you were walking around on hills – but smooth hills, since +
-the mathematical things involved correspond to smooth things – and you come to a place where you +
-are lowest, then I say that if you take a small step forward you will not change your height. When +
-you are at the lowest or at the highest point, a step does not make any difference ​in the altitude in +
-first approximation,​ whereas if you are on a slope you can walk down the slope with a step and then +
-if you take the step in the opposite direction you walk up. That is the key to the reason why, when +
-you are at the lowest place, taking a step does not make much difference, because if it did make any +
-difference then if you took a step in the opposite ​direction ​you would go downSince this is the +
-lowest point and you cannot go down, your first approximation ​is that the step does not make any +
-difference. We therefore know that if we move a path a little bit it does not make any difference to the +
-action on a first approximation.<​cite>"​The Character of Physical Law" by RFeynman</​cite></​blockquote>​ +
-<tabbox Student> ​+
  
-Variational calculus ​is a powerful mathetmatical tool to find the extremums (maximaminima and saddle points) ​of functions and even functionals ( we can call them functions ​of functions).+Given that the chain is motionless this evaluation can be repeated everywhere along the length of the chain, from the top anchor point to the lowest pointwhich is in the middle. This means that the state of static equilibrium of the catenary ​can be expressed in the form of a [[https://​en.wikipedia.org/​wiki/​Catenary#​Analysis|differential equation]]. I will refer to this way of obtaining a differential equation as 'the direct approach'​.
  
-==What we know from calculus==+This raises the question: is it fortuitous that the problem of finding the shape of a catenary can be stated as a //​differential equation//?
  
-On calculuswhen we want to find the extremum ​of a function, we use the derivative: ​+In fact this is a general propertyand the underlying reason for that was first recognized by the mathematician Jacob Bernoulli. This was in the context ​of the Brachistochrone problem.
  
-We differentiate the function $f(x)$, then demand that the resulting derivative vanishes: ​ 
-$$\frac{d f(x)}{dx} \stackrel{!}{=} 0 .$$ 
  
-While this mathematical tool is great for functions it does not help us if we want to calculate thee extrema of different objects, like functionals. A **functional** is a function of a function. This means, a functional $S[f(x)]$ has as an argument a **function** $f(x)$ and spits out a number for each function that we put into it. This is to be contrasted with what a function is: A function $f(x)$ eats a //number// $x$ and spits our a number. ​+**Brachistochrone problem**
  
-(Functionals are especially important for the Lagrangian framework.)+When Johann Bernoulli had presented ​the Brachistochrone problem to the mathematicians of the time Jacob Bernoulli was among the few who was able to find the solution independently. The treatment by Jacob Bernoulli is in the Acta Eruditorum, May 1697, pp211-217
  
-We will see in a moment ​that the variational calculus enables us to calculate the extrema of functions **and** functionals+Jacob opens his treatment with an observation concerning the fact that the curve that is sought is a minimum curve.
  
-To "​invent"​ this new theory that is capable of finding the minima of functionals,​ we need to take a step back and think about what characterizes a mathematical minimum.+{{:​basic_tools:​jacob_bernoulli_lemma.png|Jacob'​s Lemma}}
  
-The answer of variational calculus is that minimum is characterised by the neighbourhood of the minimum. For example, let's find the minimum $x_{\mathrm{min }}$ of an ordinary function $f(x)=3x^2+x$. We start by looking at one specific $x=a$ and take close look at its neighborhoodMathematically this means $a+ \epsilon$where $\epsilon$ denotes an infinitesimal (positive or negative) variation. We put this variation ​of $a$ into our function $f(x)$:+<​blockquote>//​Lemma//​. Let ACEDB be the desired curve along which heavy point falls from A to B in the shortest timeand let C and D be two points on it as close together as we like. Then the segment ​of arc CED is among all segments of arc with C and D as end points the segment that heavy point falling from A traverses in the shortest timeIndeedif another segment ​of arc CFD were traversed in shorter time, then the point would move along ACFDB in a shorter time than along ACEDB, which is contrary to our supposition.
  
-$$f(a+\epsilon)=3(a+\epsilon)^2+(a+\epsilon)=3(a^2+2a \epsilon + \epsilon^2)+a+\epsilon. $$+</​blockquote>​
  
-**If $a$ is a minimum, first order variations in $\epsilon$ must vanish**, because otherwise we can choose $\epsilon$ to be negative $\epsilon<​0$ and then $f(a+\epsilon)$ is smaller than  $f(a)$Therefore, we collect all terms linear in $\epsilon$ and demand this to be zero+The historical statement of Jacob'​s Lemma was specifically for the Brachistochrone problem. Generalization of it is straightforward. It makes no difference whether the extremum condition ​is a minimum ​condition or maximum condition.
  
-$$3\cdot 2a \epsilon+ \epsilon \stackrel{!}{=}0 \rightarrow 6a+1 \stackrel{!}{=} 0.$$+If the solution is an extremum for the entire curve then it is also an extremum for any sub-section of the curve, down to infinitisimally short subsections
  
-So we find the minimum+It follows that if a problem can be stated in variational form, then if a solution exists there exists a way to restate ​the condition in the form of a differential equation.
  
-$x_{\mathrm{min }}= a=\frac{-1}{6},​$$ 
  
-which is of course exactly ​the same result we get if we take the derivative $f(x)=3x^2+x \rightarrow f'(x)= 6x+1$ and demand this to be zero. In terms of ordinary functions this is just another way of doing the same thingbut varitional calculus ​is in addition able to find the extrema ​of functionals.+In the case of the catenary ​the static equilibrium can be cast in terms of minimizing potential energy. The middle ​of the chain tends to pull the sides inward. When the sides are pulled inward they are also raisedwhich increases the potential energy of the sides. So the middle can pull the sides inward only so much. The catenary ​is an extremum of the global potential energy ​of the chain
  
-== Mathematics ==+The same reasoning, mirrored, applies in the case of a catenary //arch//. In the case of a catenary arch the extremum is a maximum of potential energy.
  
 +For the catenary the variational problem then is: find the curve such that the derivative of the total potential energy with respect to variation is zero. Notice that since the evaluation looks exclusively at the //​derivative//​ it is not known whether the extremum is a minimum or a maximum. This does not present any problem; the catenary and the catenary arch have the same shape; the shape is the solution to the problem. Whether the extremum condition is a minimum condition or a maximum condition is immaterial.
  
-** Fundamental Lemma of Variational Calculus **                                   +The two approachesdirect approach ​and variational approachconverge; the two approaches arrive ​at the same differential equation
-Assume $f\in\mathcal{C}[a,b]$ and that for all $h\in\mathcal{C}[a,b]$ wich is zero at the endpoints  +
-it holds that $$\int f(x) h(x) d x =0$$Then $f(x)=0$ for all $x \in [a,b]$ +
-//Proof// +
-<​hidden>​ +
-TODO +
-</​hidden>​+
  
-This is analogous to the vector algebra proposition,​ let $v,w\in V$, where $V$ is a vector space. If $v\cdot w = 0$ for all $w\in V$ then $v=0$. In fact, if you dig deeper, it is the same result: the space of continuous functions form $a$ to $b$ is a vector space, and we can define the integral of the multiplication as the inner product!+**Euler-Lagrange equation**
  
-Now let's say that $h$ is in fact a derivative ​of another function, or to simplify notation, we have $\dot h$ instead of $h$ inside ​the integral?+The strategy ​of finding a solution ​to a problem in variational calculus is to find the way to restate the problem in terms of differential calculus.
  
-** Corollary for the Fundamental Lemma ** +The Euler-Lagrange equation is the implementation of that strategy. The derivation of the Euler-Lagrange equation does not make any assumption about the nature of the variational problem. If the solution to the variational problem satisfies an extremum condition then that solution will satisfy the corresponding Euler-Lagrange equation. Hence to find the solution to the variational form it suffices to find the solution to the corresponding Euler-Lagrange equation.
-TODO+
  
 +The existence of the Euler-Lagrange equation is in itself mathematical proof that any problem in variatioal calculus can be restated as a problem in differential calculus. As stated in the previous paragraph: the derivation of the Euler-Lagrange equation is completely general since it requires a single assumption only: that the solution to the variational problem satisfies an extremum condition. ​
  
-** Definition [Functional]** +<tabbox Concrete>​
-Let $\Omega(\mathcal{Q})$ be the set of functions $q:​\mathbb{R} \to \mathcal{Q}$,​ then a //​functional//​ F is a map  +
-$$ +
-F:\Omega \to \mathbb{R}; F[q] \mapsto \alpha \in\mathbb{R} +
-$$ +
  
-So we can see how a functional is a //function of functions// as we said before. ​ 
  
 +**What we know from Calculus**
  
-** References**+On [[basic_tools:​calculus|calculus]],​ when we want to find the extremum of a function, we use the derivative: ​
  
-Gelfand and FominCalculus of Variations+We differentiate the function $f(x)$then demand that the resulting derivative vanishes:  
 +$$\frac{d f(x)}{dx} \stackrel{!}{=} 0 , 
 +$$ 
 +if we solve for $x$, we find an critical point, for this function $f$
  
-<tabbox Researcher> ​ 
  
-<note tip> +**Concept ​of Variational Calculus**
-The motto in this section is: //the higher the level of abstraction,​ the better//. +
-</​note>​+
  
---> Common Question 1#+Instead of a function $f$ that takes numbers to numbers, the object of interest is a [[basic_tools:​variational_calculus:​functional|functional]],​ a function of functions of sorts. To find the stationary functions of the functionals,​ we need to change a bit the differentiation process, and we use the //​[[basic_tools:​variational_calculus:​functional_derivatives|functional derivative]]//​ or variational derivative, and we equate the result to zero.
  
-  +In the same way that the to find the extrema of a function one needs to solve a system of algebraic equation, the result of a variational derivative is a system of differential equations, these being ordinary or partial differential equations, depending on the function space being search.
-<--+
  
---> Common Question 2# 
  
-  +This is the recipe of the [[basic_tools:​variational_calculus:​the_variational_problem|variational problem]]
-<-- +
-   +
-<tabbox Examples> ​+
  
---> Example1# 
  
-  +**References**
-<--+
  
---> Example2:#+  * //Calculus of Variations//,​ Gelfand and Fomin  
 +  * Calculus of Variations by MacCluer.
  
-  +<​tabbox ​Abstract>
-<-- +
-   +
-<​tabbox ​History+
  
-</tabbox>+See [[https://​www.ams.org/​journals/​notices/​201903/​rnoti-p303.pdf|Karen Uhlenbeck and the Calculus of Variations]] by Simon Donaldson
  
 +<tabbox Quotes> ​
  
 +<​blockquote>​
 +Another way of saying a thing is least is to say that if you move the path a little bit at first it does not make any difference. Suppose you were walking around on hills – but smooth hills, since the mathematical things involved correspond to smooth things – and you come to a place where you are lowest, then I say that if you take a small step forward you will not change your height. When you are at the lowest or at the highest point, a step does not make any difference in the altitude in first approximation,​ whereas if you are on a slope you can walk down the slope with a step and then if you take the step in the opposite direction you walk up. That is the key to the reason why, when you are at the lowest place, taking a step does not make much difference, because if it did make any difference then if you took a step in the opposite direction you would go down. Since this is the lowest point and you cannot go down, your first approximation is that the step does not make any difference. We therefore know that if we move a path a little bit it does not make any difference to the action on a first approximation.
 +<​cite>"​The Character of Physical Law" by R. Feynman</​cite>​
 +</​blockquote>​
 +
 +</​tabbox>​
basic_tools/variational_calculus.1520545741.txt.gz · Last modified: 2018/03/08 21:49 (external edit)