24 sept 2009 · PICARD ITERATION DAVID SEAL The differential equation we're interested in studying is (1) y′ = f(t, y), y(t0) = y0 Many first order
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[PDF] Solution of Ordinary Differential Equations using the Picard method
2 Picard Iteration By thinking of the right hand side of this equation as an operator, the problem now becomes one of finding a fixed point for the integral
[PDF] PICARD ITERATION The differential equation were interested in
24 sept 2009 · PICARD ITERATION DAVID SEAL The differential equation we're interested in studying is (1) y′ = f(t, y), y(t0) = y0 Many first order
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Definition 1 (Picard Iterates) Let β : I → Rn be a continuous curve, let X : B ⊆ R As we continue to apply T, the iterates become more complicated functions of t
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PICARD ITERATION
DAVID SEAL
The differential equation we"re interested in studying is (1)y?=f(t,y), y(t0) =y0. Many first order differential equations fall under this category and the following method is a new method for solving this differential equation. The first idea is to transform the DE into anintegral equation, and then apply a new method to the integral equation. We first do a change of variables to transform the initial conditions to the origin. Explicitly, you can definew=y-y0andx=t-t0. With a newf, the differential equation we"ll study is given by (2)y?=f(t,y), y(0) = 0. Note: it"s not necessary to do this substitution, but it makes life a lot easier if we do. Now, we integrate equation (2) froms= 0 tos=tto obtain t s=0y?(s)ds=? t s=0f(s,y(s))ds. Applying the fundamental theorem of calculus, we have t s=0y?(s)ds=y(t)-y(0) =y(t). Hence we reduced thedifferential equationto an equivalentintegral equationgiven by (3)y(t) =? t s=0f(s,y(s))ds. Even though this looks like it"s 'solved", it really isn"t because the functionyis buried inside the integrand. To solve this, we attempt to usethe following algo- rithm, known as Picard Iteration: (1) Choose an initial guess,y0(t) to equation (3). (2) Forn= 1,2,3,..., setyn+1(t) =?t s=0f(s,yn(s))ds Why does this make sense? If you take limits of both sides, andnote thaty(t) = lim nyn+1= limnyn, theny(t) is a solution to the integral equation, and hence a solution to the differential equation. The next question youshould ask is under what hypotheses onfdoes this limit exist? It turns out that sufficient hypotheses are thefandfybe continuous at (0,0). These are exactly the hypotheses given in your existence/uniqueness theorem 2.Date: September 24, 2009.
1 Note:If we stop this algorithm at a finite value ofn, we expectyn(t) to be a very good approximate solution to the differential equation. This makes this method of iteration an extremely powerful tool for solving differential equations! For a concrete example, I"ll show you how to solve problem #3 from section 2-8. Use the method of picard iteration with an initial guessy0(t) = 0 to solve: y ?= 2(y+ 1), y(0) = 0. Note that the initial condition is at the origin, so we just apply the iteration to this differential equation. y1(t) =?
t s=0f(s,y0(s))ds=? t s=02(y0(s) + 1)ds=? t s=02ds= 2t. Hence, we have the first guess isy1(t) = 2t. Next, we iterate once more to gety2: y2(t) =?
t s=0f(s,y1(s))ds=? t s=02(y1(s) + 1)ds=? t s=02(2s+ 1)ds=222!t2+ 2t.
Hence, we have the second guessy2(t) =22
2!t2+ 2t. Iterate again to gety3:
y3(t) =?
t s=02(y2(s) + 1)ds=? t s=02?222!s2+ 2s+ 1?
ds=(2t)33!+(2t)22!+ 2t.It looks like the pattern is
y n(t) =n?k=1(2t)k k! and hence theexactsolution is given by y(t) = limn→∞yn(t) =∞?k=1(2t)k k!=n?k=0(2t)kk!-1 =e2t-1. If you plug this into the differential equation, you"ll see wehit this one on the money. To demonstrate this solution actually works, below is a graph ofy5(t), y15(t) andy(t), the exact solution.
-2-1.5-1-0.500.511.52-10 0 10 20 3040
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