I've been told numerical methods in solving ODEs, such as Euler's method and Runge-Kutta, are all in some way approximations to Picard's iteration, and I'm trying to understand how.Suppose we have a differential equation on an interval [ x 0 , x L ]: d y d x = f ( x , y )with initial condition y ( x 0 ) = y 0 I would like to numerically solve the equation on a set of points { x 0 &lt; x 1 &lt; ⋯ &lt; x n }, i.e. obtain approximations y i to the true solution y ( x i ) for each x i .Picard's iteration works as follows: y 0 , 0 = y 0 y 0 , k ( x 1 ) = y 0 + ∫ x 0 x 1 f ( x , y 0 , k − 1 ( x ) ) d x f o r k ≥ 1suppose we stop for k = m, then take y 1 = y 0 , m ( x 1 )We then repeat the process for i ≥ 1. y i , 0 = y i y i , k ( x i + 1 ) = y i + ∫ x i x i + 1 f ( x , y i , k − 1 ( x ) ) d x f o r k ≥ 1 y i + 1 = y i , m ( x i + 1 )So is the idea of a numerical method (e.g. Euler's method) to replace the integral ∫ x i x i + 1 f ( x , y i , k − 1 ( x ) ) d x with an approximation such as ( x i + 1 − x i ) f ( x i , y i ) (Euler's method)? What I don't understand is why numerical methods only iterate once for each point x i (in other words, m = 1) but Picard's iteration suggests you should iterate multiple (potentially many) times for each x i ?

Question

I&#039;ve been told numerical methods in solving ODEs, such as Euler&#039;s method and Runge-Kutta, are all in some way approximations to Picard&#039;s iteration, and I&#039;m trying to understand how.Suppose we have a differential equation on an interval   [      x    0    ,      x    L    ]:            d      y              d      x        =  f  (  x  ,  y  )with initial condition   y  (      x    0    )  =      y    0  I would like to numerically solve the equation on a set of points   {      x    0    &amp;lt;      x    1    &amp;lt;  ⋯  &amp;lt;      x    n    }, i.e. obtain approximations       y    i   to the true solution   y  (      x    i    ) for each       x    i  .Picard&#039;s iteration works as follows:      y          0      ,      0        =      y    0        y          0      ,      k        (      x    1    )  =      y    0    +      ∫                  x        0                            x        1              f  (  x  ,      y          0      ,      k      −      1        (  x  )  )  d  x          f    o    r        k  ≥  1suppose we stop for   k  =  m, then take       y    1    =      y          0      ,      m        (      x    1    )We then repeat the process for   i  ≥  1.      y          i      ,      0        =      y    i        y          i      ,      k        (      x          i      +      1        )  =      y    i    +      ∫                  x        i                            x                  i          +          1                      f  (  x  ,      y          i      ,      k      −      1        (  x  )  )  d  x          f    o    r        k  ≥  1      y          i      +      1        =      y          i      ,      m        (      x          i      +      1        )So is the idea of a numerical method (e.g. Euler&#039;s method) to replace the integral       ∫                  x        i                            x                  i          +          1                      f  (  x  ,      y          i      ,      k      −      1        (  x  )  )  d  x with an approximation such as   (      x          i      +      1        −      x          i        )  f  (      x    i    ,      y    i    ) (Euler&#039;s method)? What I don&#039;t understand is why numerical methods only iterate once for each point       x    i   (in other words,   m  =  1) but Picard&#039;s iteration suggests you should iterate multiple (potentially many) times for each       x    i  ?

lutzantsca885 · Accepted Answer

All these methods do approximate the solution of the ODE. Picard&#039;s method is more a theoretical tool. It can be used to obtain approximate analytical solutions, or to prove theorems. As far as I know, it is never used with a numerical integration method, as that would be quite inefficient.The basic numerical method is indeed Euler&#039;s, and Runge-Kutta are improvements that gives better convergence order for smooth functions. (The same way that the rectangle rule is the basic numerical integration method for explicit functions while Newton-Cotes are improvements.)

hisyhauttaq84w · Accepted Answer

Euler method stops at just one iteration, but Runge-Kutta goes for higher iterates. For example, the RK2 (midpoint method):      y          n      +      1        =      y    n    +  h  f  (      x          n      +              1        2              ,      y    n    +      1    2    h  f  (      x    n    ,      y    n    )  )is the result of the second Picard iteration: the first use the constant       y    0    (  x  )  =      y    n   for       ∫    0          t      h        f  (      x    n    +  ξ  ,      y          0        (      x    n    +  ξ  )  )        d    ξ with Euler and get       y    1    (      x    n    +  t  h  )  =      y    n    +  t  h  f  (      x    n    ,      y    n    ) for   t  ∈  [  0  ,  1  ] and then the second iteration      ∫    0          h        f  (      x    n    +  ξ  ,      y          1        (      x    n    +  ξ  )  )        d    ξwe use the mid-point rule.Similarly, you have predictor-corrector methods which you can view the corrector method as approximating the repeated Picard iterations as many times as you want.

I've been told numerical methods in solving ODEs, such as Euler's method and Runge-Kutta, are all in

Answered question

Answer & Explanation

New Questions in Integral Calculus