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Picard's Method


Date: 08/10/97 at 15:23:23
From: Michael Lam
Subject: Picard's Iteration method of approximating solutions to 
Differential Equations

Hello,

Can you tell me about Picard's Iteration method of solving 
Differential Equations?  I've looked for books in libraries but with 
no luck. Thanks in advance!


Date: 08/10/97 at 16:23:29
From: Doctor Anthony
Subject: Re: Picard's Iteration method of approximating solutions to 
Differential Equations

Suppose you have the differential equation   dy/dx = f(x,y)  given 
that y = b when x = a.

We can write   y = b + INT(a to x)[f(x,y).dx]

For a first approximation we replace the y in f(x,y) by b; for the 
second approximation we replace it by the first approximation, for a 
third by the second, and so on.

Example   dy/dx = x+y^2 and y = 0 when x = 0

Then   y = 0 + INT(0 to x)[x+y^2].dx

First approximation: put y = 0 in x+y^2  giving

       y = INT(0 to x)[x.dx]

         = 1/2 x^2

Second approximation: put y = 1/2 x^2 in  x+y^2

       y = INT(0 to x)[x + 1/4 x^4].dx

         =   x^2/2 + 1/20 x^5

Third approximation: put  y = x^2/2 + 1/20 x^5 for y in x+y^2

       y = INT(0 to x)[x + 1/4 x^4 + 1/20 x^7 + 1/400 x^10]dx

         =  x^2/2 + x^5/20 + x^8/160 + x^11/4400

and so on indefinitely.  Picard's method converts the differential 
equation into an equation involving integrals, which is called an 
integral equation.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   




Date: 08/10/97 at 21:16:28
From: Michael Lam
Subject: Re: Picard's Iteration method of approximating solutions to 
Differential Equations

>         =  x^2/2 + x^5/20 + x^8/160 + x^11/4400

Thanks for the method!!!  Now I think I can do them...however, if I
proceed the iterations indefintely, am I gauranteed to find a 
solution?
Is this solution unique?
And, if I find something which looks like a Taylor's series of some
function, can I write that function as the solution instead?

Thank you very much!!

MIchael



Date: 08/11/97 at 07:42:38
From: Doctor Anthony
Subject: Re: Picard's Iteration method of approximating solutions to 
Differential Equations


As Michael Lam wrote to Dr. Math
On 08/10/97 at 21:16:28 (Eastern Time),
>>         =  x^2/2 + x^5/20 + x^8/160 + x^11/4400
>
>Thanks for the method!!!  Now I think I can do them...however, if I
>proceed the iterations indefintely, am I gauranteed to find a solution?
>Is this solution unique?
>And, if I find something which looks like a Taylor's series of some
>function, can I write that function as the solution instead?
>
>Thank you very much!!
>
>MIchael
>

Remember we are finding a series solution about some point x=a, y=b 
(in 
above example, about the point (0,0))  and so the number of terms you 
require will depend how far from x=a you need the solution to apply.  
It can 
be shown that the series is convergent, and provides a solution to the 
differential equation, but the number of terms you take will depend 
how 
rapidly the series converges, and how far from x=a you want to 
operate.

Your second point, concerning a Taylor series, is perfectly acceptable 
if 
the  series is available.  e.g. dy/dx = x^(1/2).sin(x).  Here a Taylor 
series for sin(x) is available, and could be used.


-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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