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

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.

10 comments on “Newtonian Mechanics 04”

1. […] function can be identified with the function that we saw on Newtonian Mechanics 04 if one makes the obvious […]

2. joeschmo26 says:

“Hence {\displaystyle J(\alpha)= \int_0^{2\pi}(1+\alpha\cos x +\alpha^2\cos ^2x)dx=2\pi+\alpha^2\pi}.”

Should the second term be ${\2\alpha\cos x}$ ?

• joeschmo26 says:

“Hence ${\displaystyle J(\alpha)= \int_0^{2\pi}(1+\alpha\cos x +\alpha^2\cos ^2x)dx=2\pi+\alpha^2\pi}$.”

Should the second term be ${ 2\alpha\cos x }$ ?

• ateixeira says:

Yes it should. Sorry for the typo and thanks for the help.

3. joeschmo26 says:

In Example 2 where you derive the equations of motion for the cycloid… is ${frac{\partial{f}}{\partial{y}} = 0}$ because ${f}$ is not dependent on ${y}$, or is there some other reason?

• joeschmo26 says:

Edit: ${ frac{\partial f}{\partial y} = 0}$

• joeschmo26 says:

Attempt 3: ${ \frac{\partial f}{\partial y} = 0}$

• ateixeira says:

That’s exactly the reason.

4. joeschmo26 says:

And I assume Exercise 1 ( without actually completing the entire integral) that the solution comes from trigonometric substitution which its second to last step is

${\int \sec\theta d\theta} = ln( \sec\theta + \tan\theta )$ After evaluation with the limits in the theta domain the right side becomes 0, hence y = 0 is the equation of the line?

• ateixeira says:

You have to make a trigonometric change of variable to calculate this integral and then you need to make an approximation but it is slightly more complicated than what you present.