Arc length paramatrizations satisfy original system of differential equations?Say we have a system of differential equations { x ‴ ( t ) + f ( t ) x ′ ( t ) = 0 y ‴ ( t ) + f ( t ) y ′ ( t ) = 0 on an interval [ a , b ], along with the restriction that x ′ ( t ) 2 + y ′ ( t ) 2 = 1on [ a , b ].Now, Euler proved that as long as x ( t ) , y ( t ) are continuous and not both zero at a point, then we can re-parametrize x , y to some functions x ~ , y ~ which have the same image as x , y but are arc-length parametrized.Now, say we have a solution ( x ∗ ( t ) , y ∗ ( t ) ) to { x ‴ ( t ) + f ( t ) x ′ ( t ) = 0 y ‴ ( t ) + f ( t ) y ′ ( t ) = 0 that satisfies the hypotheses needed to apply the previously mentioned re-parametrization. Say ( x ~ ∗ ( t ) , y ~ ∗ ( t ) ) are this re-parametrization, so they have the same image of ( x ( t ) , y ( t ) ) on [ a , b ] and satisfy x ′ ( t ) 2 + y ′ ( t ) 2 = 1.Is it still the case that { ( x ~ ∗ ) ‴ ( t ) + f ( t ) ( x ~ ∗ ) ′ ( t ) = 0 ( y ~ ∗ ) ‴ ( t ) + f ( t ) ( y ~ ∗ ) ′ ( t ) = 0 ?

Question

Arc length paramatrizations satisfy original system of differential equations?Say we have a system of differential equations      {                                        x            ‴                    (          t          )          +          f          (          t          )                      x            ′                    (          t          )          =          0                                                  y            ‴                    (          t          )          +          f          (          t          )                      y            ′                    (          t          )          =          0                        on an interval   [  a  ,  b  ], along with the restriction that      x    ′    (  t      )    2    +      y    ′    (  t      )    2    =  1on   [  a  ,  b  ].Now, Euler proved that as long as   x  (  t  )  ,  y  (  t  ) are continuous and not both zero at a point, then we can re-parametrize   x  ,  y to some functions       x    ~    ,      y    ~   which have the same image as   x  ,  y but are arc-length parametrized.Now, say we have a solution         (        x    ∗    (  t  )  ,      y    ∗    (  t  )        )   to      {                                        x            ‴                    (          t          )          +          f          (          t          )                      x            ′                    (          t          )          =          0                                                  y            ‴                    (          t          )          +          f          (          t          )                      y            ′                    (          t          )          =          0                        that satisfies the hypotheses needed to apply the previously mentioned re-parametrization. Say         (                x      ~        ∗    (  t  )  ,              y      ~        ∗    (  t  )        )   are this re-parametrization, so they have the same image of         (    x  (  t  )  ,  y  (  t  )        )   on   [  a  ,  b  ] and satisfy      x    ′    (  t      )    2    +      y    ′    (  t      )    2    =  1.Is it still the case that      {                            (                                              x              ~                        ∗                                )            ‴                    (          t          )          +          f          (          t          )          (                                              x              ~                        ∗                                )            ′                    (          t          )          =          0                                      (                                              y              ~                        ∗                                )            ‴                    (          t          )          +          f          (          t          )          (                                              y              ~                        ∗                                )            ′                    (          t          )          =          0          ?

Kharroubip9ej0 · Accepted Answer

Solutions of that system cannot be reparametrized to have   (      x    ′        )    2    +  (      y    ′        )    2   constant for nonconstant   f  (  t  ). Reparametrization alters the velocity vector but here       x    ′    ,      y    ′  v are the &quot;spatial&quot; coordinates and       x    ″    ,      y    ″   the velocities. For variable positive   f  (  t  ) the solution curves cut across the solutions with   f  =  c for different   c, and all of the latter are circles (for   c  &amp;gt;  0).If you did reparametrize to have               x      ″        2    +              y      ″        2   constant, there would be no reason for it to satisfy the same system of differential equations, but   (      x    ′    (  t  )  ,      y    ′    (  t  )  ) would follow the same curves.

Arc length paramatrizations satisfy original system of differential equations? Say we have a system

Answered question

Answer & Explanation

New Questions in Algebra I