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I want to find a series of functions converging to the solution of $\displaystyle \frac{dx}{dt}=2x$ with $x(0)=x_0$

But, I was to do this using the Picard-Lindolf iteration and am completely stuck on how to use it!! Please could someone please explain?

I know that you do several iterations and then use induction, but how do you iterate?

Arrghh! Been stuck on this for 2 Hours now, I have the iteration x_5 (t) = x_0(1 + 2t + 2t^2 + 4/3t^3 + 3/4^t^4 ..+ but i dont see a pattern with this series at all! I want to obviously prove by induction this series is of the solutions but i have no idea how to write it in series form, not sure if its impossible or just too late in the night! If someone could help me/guide me towards the the series form i would be very grateful! Many thanks

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up vote 4 down vote accepted

For your problem, we are given:

$\tag 1 \displaystyle \frac{dx}{dt}=2x, ~~x(0)=x_0$

The solution to $(1)$ is: $$x(t) = x_0 e^{2t}.$$

The Picard-Lindelof Iteration is given by:

$$\tag 2 \displaystyle x_0(t) = x_0, ~~x_{n+1}(t) = x_0 + \int^t_{t_0} f(s, x_n(s))ds$$

For $(1)$, we have: $f(s, x_n(s)) = 2s$ and using $(2)$, yields:

  • $\displaystyle x(0) = x_0$

  • $\displaystyle x_1(t) = x_0 + \int^t_{t_0} f(s, x_0(s))ds = x_0 + \int^t_{0} 2(x_0) ds = x_0 + 2x_0t$

  • $\displaystyle x_2(t) = x_0 + \int^t_{t_0} f(s, x_1(s))ds = x_0 + \int^t_{0} 2(x_0 + 2x_0 s) ds = x_0 + 2x_0 t (t+1)$

  • $\displaystyle x_3(t) = x_0 + \int^t_{t_0} f(s, x_2(s))ds = x_0 + \int^t_{0} 2(x_0 + 2x_0 s (s+1)) ds = x_0 + \frac{2}{3} x_0t(2t^2+3t+3)$

  • $\displaystyle \ldots$

  • See a pattern for $x_n(t)$?

Now, lets experiment by choosing $x_0 = 1$, and plot the four solutions $\{x(t), x_0, x_1(t), x_2(t), x_3(t)\}$ using Wolfram Alpha.

Does it converge in a suitable interval and diverge for other values of $t$?

I'll leave that up to you because your question implies you understand how to do induction to see what is going on.


If you expand the series for $e^{2t}$, we get:

$$e^{2t} = 1 + 2t + 2t^2 + \frac{4}{3}t^3 + \frac{2}{3}t^4 + \ldots$$

Now, look at my successive $x_n(t)$, and what do you notice?

Got it? (Note, you made an error in your calc in comments.)

Update 2

$$x(t) = x_0e^{2t} = x_0(1 + 2t + 2t^2 + \frac{4}{3}t^3 + \frac{2}{3}t^4 + \ldots) = x_0 \sum_{k=0}^\infty \frac{(2t)^{k}}{t!}$$

  • $\displaystyle x(0) = x_0 = x_0(1)$

  • $\displaystyle x_1(t) = x_0 + 2x_0t = x_0(1+2t)$

  • $\displaystyle x_2(t) = x_0 + 2x_0 t (t+1) = x_0(1 + 2t + 2t^2)$

  • $\displaystyle x_3(t) = x_0 + \frac{2}{3} x_0t(2t^2+3t+3) = x_0(1 + 2t + 2t^2 + \frac{4t^3}{3})$

  • $\ldots$

Now compare the $x(t)$ solution to the $x_n(t)$ and what do you see.


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Thank you for your answer! Cant believe i didnt understand that, just got to the induction part now which im guessing will look something like x(t)= x_0(1 + 2t +...) but we'll see!! Thanks – edna1974 Feb 26 '13 at 2:07
Ahhh why didnt i just expand the solution?! ha yeah so very close, and looking to hard at it! Now to work out the pattern to write as a series...Hopefully it will be easier this time! Thank you so much, i must be a very frustrating person to help! :) – edna1974 Feb 26 '13 at 4:44
yeah i saw the error, i made it while integrating (badly!) im just trying to show this as a sum of a series - would it be x_0 multipied by (the sum of, i = 0) 2t^i+1?? Bit confused how to do this, is there a particular method i can look up or use? I may have to leave this for a little later in the morning and have a sleep - I will be back in around 6 hours to finish this!! Thanks so much for your help :) – edna1974 Feb 26 '13 at 6:02
sorry must have been so tired hah! so i get xk(t)=x0(sum of, i =0 t^2i/i!) – edna1974 Feb 26 '13 at 13:39
+10 for your solid way. Always. Nice to have you here to learn much about ODE's. – Babak S. Feb 27 '13 at 17:37

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