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Calculas of variation /euler,s equation

  1. Mar 3, 2012 #1
    i m little confusd in (finding the shortest distance b/w two points is a straight line in three dimensions)i have solved it but nothing found any accurate result
    [ds][/2]=[dx][/2]+[dy][/2]+[dz][/2] is the distance in 3-dimension b/w 2 points then
    how we can start and how we can take functional by this information?
     
  2. jcsd
  3. Mar 3, 2012 #2
    The length of a curve is [itex]\int_{s_0}^{s_1}ds=\int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)}\sqrt{(\frac{\partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}ds[/itex]. Setting the variation of this integral to zero one obtains [itex]\delta\int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)}\sqrt{( \frac{\partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}ds=0\Rightarrow \int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)}\frac{\frac{ \partial{x}}{\partial{s}}\delta( \frac{\partial{x}}{\partial{s}})+\frac{\partial{y}}{\partial{s}}\delta( \frac{\partial{y}}{\partial{s}})+\frac{\partial{z}}{\partial{s}}\delta( \frac{\partial{z}}{\partial{s}})}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}ds=0[/itex]. Integrating by parts (requiring the boundry terms to vanish) and using the fundamental theorem of variational calculus will yield the Euler equations (which turn out to be rather simple in this case). Solving the Euler equations will yield a straight line.
     
  4. Mar 7, 2012 #3
    Thanx i have tried but i didn,t get straight line plz if u will provide a comprehensive detail than ,,,,,,,,,,,,,,,,
     
  5. Mar 7, 2012 #4
    If you post what you have done so far I will better know how to help.
     
  6. Mar 9, 2012 #5
    i couldn,t found the method to write the proper derivative in square root (in newton method , dot put on variable) but i tell u that i have taken their dervative with respect to time and using the limits from t1 to t2 and used the variational principle of minima but further solved it but .............as usual:cry:
     
  7. Mar 9, 2012 #6
    Thats why i have not attached my solution
     
  8. Mar 9, 2012 #7
    [itex]\int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)}\frac{\frac{ \partial{x}}{\partial{s}}\delta( \frac{\partial{x}}{\partial{s}})+\frac{\partial{y}}{\partial{s}}\delta( \frac{\partial{y}}{\partial{s}})+\frac{\partial{z}}{\partial{s}}\delta( \frac{\partial{z}}{\partial{s}})}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}ds=0\Rightarrow\int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)}\frac{\frac{ \partial{x}}{\partial{s}}\delta( \frac{\partial{x}}{\partial{s}})}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}ds+\int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)} \frac{\frac{ \partial{y}}{\partial{s}}\delta( \frac{\partial{x}}{\partial{s}})}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}ds+\int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)} \frac{\frac{ \partial{z}}{\partial{s}}\delta( \frac{\partial{x}}{\partial{s}})}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}ds=0[/itex] taking [itex]U=\frac{\frac{ \partial{q}}{\partial{s}}}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}[/itex] and [itex]dV=\delta( \frac{\partial{q}}{\partial{s}})ds=\frac{\partial{(\delta{}q)}}{\partial{s}}ds=d(\delta{q})[/itex], where q is the appropriate coordinate in each integral, one obtains [itex]\left[\frac{\frac{ \partial{x}}{\partial{s}}\delta{x}+\frac{\partial{y}}{\partial{s}}\delta{y}+\frac{\partial{z}}{ \partial{s}}\delta{z}}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}\right]_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)}-\int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)}\frac{\partial}{\partial{s}}\left[\frac{\frac{ \partial{x}}{\partial{s}}}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}\right]\delta{x}+ \frac{\partial}{\partial{s}}\left[ \frac{\frac{ \partial{y}}{\partial{s}}}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}\right]\delta{y}+\frac{\partial}{\partial{s}}\left[ \frac{\frac{ \partial{z}}{\partial{s}}}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}\right]\delta{z}\,\,ds=0[/itex]. By the fundamental principle of variational calculus the coefficients of the variations must independently vanish. This yields three coupled partial differential equations: [itex]\frac{\partial}{\partial{s}}\left[\frac{\frac{ \partial{x}}{\partial{s}}}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}\right]=0\,\,\,\,\frac{\partial}{\partial{s}}\left[ \frac{\frac{ \partial{y}}{\partial{s}}}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}\right]=0\,\,\,\, \frac{\partial}{\partial{s}}\left[ \frac{\frac{ \partial{z}}{\partial{s}}}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}\right]=0[/itex], which are the Euler equations for this problem. Are these steps similar to yours?

    Edit: P.S. If you hit the quote button under a post you can see what was typed to produce the post.
     
    Last edited: Mar 9, 2012
  9. Mar 12, 2012 #8
    Thanx i understood that this partial derivative = zero so entire term inside the bracket is equal to some constant ad by the elimination process of constants we get a stright line in 3-dimension right or,,,,,,,,,,,,,,,,,:smile:
     
  10. Mar 12, 2012 #9
    i think i didn't get the calculas of variation or the euler's method(form).if u vl refer me some books which deals with abc....to xyz...........plz recomend and also plz told me that wht actually we are doing while dealing with this method whts our main purpose is?
     
  11. Mar 12, 2012 #10
    I first learned variational calculus in Intermediate Mechanics (Mechanics for Physics majors). We used John Taylor's Classical Mechanics textbook, in which variational calculus is covered in chapter 6. For a more advanced treatment (at the graduate level), Goldstein's Classical Mechanics is popular. In it variational calculus is covered in chapter 2. I think most popular books on Mechanics for physicists cover variational calculus at some point. There are also books on variational calculus with more mathematical applications (and deriviations), but I am not very familiar with them (they are probably what one would find if one looked it up at a university library in the math section).

    The main purpose of this method is to find paths which extremize the value of the integral. In your problem we wanted to minimize the length of a line. In geometries that are more complex than Euclidean space even this exercise could produce non-intuitive results. Anything that you want to maximize or minimize (or obtain a stationary point for) that you can describe using an integral would be something you could analyze using this technique (assuming the solution is a contiuous function). In Physics variational calculus is used to derive equations of motion (classically, applications of Newton's second law) from equations giving the energy of a system (wikipedia- Hamiltonian Mechanics). These methods apply in all areas of Physics; Newtonian, quantum, relativistic, and just about any other part of Physics I can think of. This formulation of Physics is useful because it usually makes it easy to ensure that new theories/models obey principles such as conservation of energy, conservation of momentum, etc. due to Noether's Theorem.
     
  12. Mar 13, 2012 #11
    I am very thankful to you:smile:
     
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