Picard's method does not solve first order differential equation?

Question

I have the following first order differential equation $$x^\prime(t)=-(x(t))^2+2x(t),\quad t\geq 0,\quad x(0)=1$$ Now I want to obtain an approximation of $x(t)$ by using Picard's method. Then the first Picard's iterates of this problem are given by: $$x_1 = x_0 + \int_0^t F(s,x_0(s))ds=1+\int_0^t -(1)^2+2ds=t+1 $$ $$x_2 = 1+ \int_0^t-(s+1)^2+2(s+1)ds=-\frac{1}{3}t^3+t+1 $$ $$x_3 = 1+ \int_0^t -(-\frac{1}{3}s^3+s+1)^2+2(-\frac{1}{3}s^3+s+1)ds=-\frac{1}{63}t^7 + \frac{2}{15}t^5 + -\frac{1}{3}t^3 + t+1 $$ This should converge to the true solution of $x(t)$ but when I derive the solution analytically i get the following: $$x^\prime(t)=-(x(t))^2+2x(t)\Longleftrightarrow \frac{x^\prime(t)}{-(x(t))^2+2x(t)}=1$$ Using the fact that: $$\frac{1}{x(2-x)}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{2-x}\right) $$ differentiating both sides then yields: $$\int_0^t\frac{x^\prime(s)}{x(s)}ds + \int_0^t \frac{x^\prime(s)}{2-x}ds = \int_0^t2ds $$ $$\Longleftrightarrow \ln|x(t)| -\ln|2-x(t)|=2t+c_1,\quad c_1\in\mathbb{R} $$ $$\Longleftrightarrow \frac{x(t)}{2-x(t)} = c_2e^{2t}, \quad c_2 = \pm e^{c_1}$$ $$\Longleftrightarrow x(t) = \frac{2e^{2t}}{c_3 + e^{2t}}, \quad c_3 = \frac{1}{c_2}$$ Combining this with the fact that $x(0)=1$ yields that we have $x(t)$ given by: $$x(t)=\frac{2e^{2t}}{1+e^{2t}}$$ When I plot the functions obtained by the Picard method I get a totally different result then when I plot the above given function of $x(t)$. Why is this? Shouldn't the Picard method converge to its true solution? I also used more Iterations but both methods keep giving a different answer. Does anyone know why this is or am I doing something wrong?

Answer 1

I don't know how you think both results do not agree, as you don't give more details. I calculated the degree 7 Taylor polynomial of $f(t)=\frac{2e^{2t}}{1+e^{2t}}$, and it is $$ p(t)=1+t-\frac{t^3}3+\frac{2t^5}{15}-\frac{17t^7}{315}. $$ So the 7th term does not agree, with the Picard polynomial, but no one said it has to. In any case, the plots look very similar to me.

Answer 2

You should have gotten a plot like this

which gives a visible idea of convergence for $|t|\le 0.5$.

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Let's consider the interval $|x-1|<1$ in state space in the proof construction of the Picard theorem. Then for the Picard iteration operator $P(x)(t)=x_0+\int_0^tf(s,x(s))ds$ we get here $$|P(x)(t)-1|\le \int_0^t|x(s)^2-2x(s)|ds\le \int_0^t(1+|x(s)-1|^2)ds\le 2t.$$ To stay inside the chosen bound one needs $|t|\le \bar t\le\frac 12$.

Next, the Lipschitz constant is the maximum of $|-2x+2|=2|x-1|$ over the given region, which gives $L=2$. For convergence in the Picard-Lindelöf proof, that is, contractivity of the Picard map, one needs also $L\bar t<1$, which is given for any $\bar t<\frac12$, for example for $\bar t=\frac13$.

This quantitative reasoning confirms that you get reasonably fast visible convergence only for $|t|\le\frac12$ or smaller intervals.

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Let's explore obstacles for the convergence of the Picard iteration from the point of view of the result, that is, the convergence of the resulting power series. While the denominator $1+e^{2t}$ does not have real roots, the radius of convergence of the power series that you compute is determined also by the complex roots of the denominator which are poles of the function itself. $$ e^{2t}=-1\iff 2t=i(2k+1)\pi $$ has the smallest solutions at $t=\pm i\frac\pi2$, which gives a more optimistic radius of convergence for the power series expansion, and thus possibly also the Picard iteration, of $\frac\pi2\approx1.5$.

But convergence is much slower the closer you get to the boundary of the region of convergence. Outside of the region of convergence it is to be expected that the partial sums of the power series wildly diverge.