Newtonian Mechanics 04

— 1. Variational Calculus —

Definition 1 A functional is a mapping from vector spaces to into real numbers.

Let $ {\displaystyle J = \int _{x_1}^{x^2} f\{y(x),y\prime (x),x\}dx}$. Suppose that $ {x_1}$ and $ {x_2}$ are constants, the functional form of $ {f}$ is known.

According to definition 1 $ {J}$ is a functional and the goal of the Calculus of Variations is to determine $ {y(x)}$ such that the value of $ {J}$ is an extremum.

Let $ {y=y(\alpha, x)}$ be a parametric representation of $ {y}$ such that $ {y(0,x)=y(x)}$ is the function that makes $ {J}$ an extremum.

We can write $ {y(\alpha, x)=y(0,x)+ \alpha\eta(x}$, where $ {\eta (x)}$ is a function of $ {x}$ of the class $ {C^1}$ (that means that $ {\eta}$ is a continuous function whose derivative is also continuous) with $ {\eta (x_1)=\eta (x_2)=0}$.

Now $ {J}$ is of the form $ {\displaystyle J(\alpha) = \int _{x_1}^{x^2} f\{y(\alpha, x),y\prime (\alpha, x),x\}dx}$.  Therefore the condition for $ {J}$ to be an extremum is

$ \displaystyle \frac{dJ}{d\alpha}(\alpha=0)=0$

Example 1 Let $ {y(x)=x}$. Take $ {y(\alpha, x)= x+ \alpha\sin x}$ as a parametric representation of $ {y}$. Let $ {f=\left(dy/dx\right)^2}$, $ {x_1=0}$ and $ {x_2=2\pi}$. Given the previous parametric equation find $ {\alpha}$ such that $ {J}$ is a minimum. Now $ {\eta (0)=\eta (2\pi)=0}$ and $ {dy/dx=1+\alpha\cos x}$. Hence $ {\displaystyle J(\alpha)= \int_0^{2\pi}(1+2\alpha\cos x +\alpha^2\cos ^2x)dx=2\pi+\alpha^2\pi}$. By the previous expression $ {J(\alpha)}$ it is trivial to see that the minimum value is reached when $ {\alpha=0}$

Exercise 1 Given the points $ {(x_1,y_1)=(0,0)}$ and $ {(x_2,y_2)=(1,0)}$, calculate the equation of the curve that minimizes the distance between the points. Now $ {y(\alpha, x)=y(0,x)+\alpha \eta (x) = 0+\alpha(x^2-x)}$. It is $ {\eta (x) = x^2-x}$, $ {ds=\displaystyle \sqrt{dx^2+dy^2}=\sqrt{1+(dy/dx)^2}dx}$ And it is $ {s= \displaystyle \int _0^1 \sqrt{1+(dy/dx)^2}dx}$ with $ {dy/dx=\alpha (2x-1)}$. The rest is left as an exercise for the reader.

— 2. Euler Equations —

In the following section we'll analyze the condition for $ {J}$ to be an extremum:

$ {\begin{aligned} \frac{\partial J}{\partial \alpha} &= \frac{\partial}{\partial \alpha} \int _{x_1}^{x_2}f(y,y\prime,x)dx \\ &= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+ \frac{\partial f}{\partial y\prime}\frac{\partial y\prime}{\partial \alpha}\right) dx \end{aligned}}$

Since it is $ {\partial y /\partial \alpha = \eta (x)}$ and $ {\partial y\prime /\partial \alpha = d\eta/dx}$ it follows

$ \displaystyle \frac{\partial J}{\partial \alpha}= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\eta (x)+ \frac{\partial f}{\partial y\prime}\frac{d \eta}{dx}\right) dx$

Now $ {\displaystyle \int _{x_1}^{x_2}\frac{\partial f}{\partial y\prime}\frac{d \eta}{dx}dx=\frac{\partial f}{\partial y\prime}\eta (x)|_{x_1}^{x_2}- \int _{x_1}^{x_2}\frac{d}{dx}\left( \frac{\partial f}{\partial y\prime} \right)\eta (x) dx}$.

For the first term it is $ {\frac{\partial f}{\partial y\prime}\eta (x)|_{x_1}^{x_2}=0}$ since $ {\eta (x_1)=\eta (x_2)=0}$ by hypothesis.

Hence

$ {\begin{aligned} \frac{\partial J}{\partial \alpha} &= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}- \frac{d}{dx}\left( \frac{\partial f}{\partial y\prime} \right) \frac{\partial y}{\partial \alpha}\right)dx \\ &= \int _{x_1}^{x_2}\left( \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y\prime} \right)\eta (x) dx \end{aligned}}$

Remembering that $ {\partial J / \partial\alpha(\alpha=0)=0}$ and taking into account the fact that $ {\eta (x)}$ is an arbitrary function one can conclude that

$ \displaystyle \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y\prime}=0 $

The previous equation is known as the Euler's Equation

Example 2 As an example let us try to derive the equations of motion of a particle that moves in a constant force field starting its movement from the state of rest. The particles starts from point $ {x_1, y_1}$ and goes to point $ {x_2, y_2}$.From the enunciate it follows $ {K+U=c}$. Let us take our original point as being our reference point for the potential. Then it is $ {k+U=0}$. As always it is $ {k=1/2mv^2}$. For the potential it is $ {U=-Fx=-mgx}$. From the previous equations it follows that $ {v=\sqrt{2gx}}$. From the definition of velocity it follows that $ \displaystyle t=\int _{x_1,y_1}^{x_2,y_2} \frac{ds}{v}=\int _{x_1,y_1}^{x_2,y_2}\frac{\sqrt{dx^2+dy^2}}{\sqrt{2gx}}=\int _{x_1,y_1}^{x_2,y_2}\frac{\sqrt{1+y\prime^2}}{\sqrt{2gx}}dx$ Let $ {f=\sqrt{\frac{1+y\prime^2}{x}}}$ since $ {(2g)^{-1/2}}$ is only a constant factor and can be omitted from our analysis. Given the functional form of $ {f}$ it is $ {df/dy=0}$ and Euler's Equation just is: $ \displaystyle \frac{d}{dx}\frac{\partial f}{\partial y\prime}=0 $ From the previous relationship it is $ \displaystyle \frac{\partial f}{\partial y\prime}=(2a)^{-1/2}=\mathrm{const} $ Hence it is $ {\begin{aligned} \frac{y\prime^2}{x(1+y\prime^2)} &= \frac{1}{2a} \Rightarrow\\ y &= \int \frac{x}{\sqrt{2ax-x^2}}dx \end{aligned}}$ Making the change of variables $ {x=a(1-\cos \theta)}$ it follows $ {dx=a\sin \theta d\theta}$. Hence the expression for $ {y}$ is $ {y=\int a(1-\cos \theta)d\theta\Rightarrow y=a(\theta-\sin \theta)+A}$. Since our particle starts from the origin it is $ {A=0}$. Thus the solution to our initial problem is $ {\begin{aligned} x &= a(1-\cos \theta) \\ y &= a(\theta-\sin \theta) \end{aligned}}$ Which are the parametric equations of a cycloid. Cycloid[/caption]

To close our thoughts on the Euler equation let us say that there also is a second form for the Euler equation. The second form is

$ \displaystyle f-y\prime\frac{\partial f}{\partial y\prime}= \mathrm{const} $

and is used in the cases where $ {f}$ doesn't depend explicitly on $ {x}$.

— 3. Euler Equation for $ {n}$ variables —

Let $ {f}$ be of the form $ {f=f\{ y_1(x),y\prime _1(x),y_2(x),y\prime _2(x),\cdots,y_n(x),y\prime _n(x), x \}}$.

Now we have $ {y_i(\alpha, x)= y_i(0,x)+\alpha \eta (x)}$ and $ {\displaystyle \int _{x_1}^{x_2}\left( \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y _i\prime} \right)\eta _i (x) dx}$ for each of the values of $ {i}$. Since $ {\eta _i(x)}$ are independent functions it follows that for $ {\alpha=0}$

$ \displaystyle \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y _i\prime}=0 $

That is to say we have $ {n}$ independent Euler equations.