Picard's MethodDate: 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/ |
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