Newtonian Mechanics 04

by ateixeira

— 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. Newtonian Mechanics 05 « The Quantum Gang says:
    February 9, 2014 at 12:21 pm

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

    Reply
  2. joeschmo26 says:
    February 9, 2014 at 3:44 pm

    “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} ?

    Reply
  3. joeschmo26 says:
    February 20, 2014 at 12:55 am

    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?

    Reply
  4. joeschmo26 says:
    February 20, 2014 at 2:08 am

    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?

    Reply
    • ateixeira says:
      February 21, 2014 at 6:46 am

      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.

      Reply

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