# Contraction operator

In a proof of Picard's theorem using the contraction mapping theorem, we define an operator $T$ which is applied to a function $y$. I don't really see below how $Ty$ is any different from $y$ as the RHS for both are the same. Could someone please explain how they are different?

• But it seems to me like $T$ will always produce the same thing as $y$? – goodcow Nov 3 '15 at 21:21
• OK, now I see it. I must admit I do not know what the difference is. – Peter Nov 3 '15 at 21:26
• The point is to formulate the ODE $y'=f(t,y)$ as a fixed point problem $Ty=y$. So while $Ty$ will indeed give back $y$ itself if $y$ is a solution, the point of the construction is not really to feed the solution into $T$, but to feed an approximate solution into $T$ and (hopefully) get a better approximate solution. – Ian Nov 4 '15 at 0:09
• So the approximate solutions converge uniformly toward $y$, the true solution. I see now that the operator is literally just the Picard approximation. Thanks! – goodcow Nov 4 '15 at 10:20

Perhaps the author wrote it in a slightly confusing way. The first equation

$$y(x) = b + \int_{a}^{x}f(s,y(s))ds$$ is true of the solution $y(x)$ to the differential equation $$\frac{d}{dx}y(x) = f(x,y(x))$$ with initial condition $y(a) = b$. I would think of 1.21 as an equation that is true when the variable $y$ takes on the value "solution to the differential equation with initial condition y(a) = b".

Then the author defines the Picard operator $T$, which takes functions to functions. It might be a little more clear to use a variable other than $y$, for example

$$(Tg)(x) := b + \int_{a}^{x}f(s,g(s))ds$$

I used the $:=$ notation to indicate that here the function $T$ is being defined. The solution $y(x)$ to the differential equation will satisfy $Ty = y$ (in the sense that $(Ty)(x) = y(x)$ for all $x$ in the given time interval), but other functions will not.

For example you can work out for yourself what is the value of $Tg$ when $a=0$, $b=1$ and $f(s,z) = -z$ (This corresponds to the differential equation $\dot{y} = -y$ with initial condition $y(0)=1$). Try the function $g(x) = 0$ or $g(x) = x$. You will see that $Tg \neq g$ in both cases. (the true solution is $y(x) = e^{-x}$)

• I see, the $y(x)$ that is a solution with the initial conditions is the fixed point in the mapping. If I understand correctly, the other non-solutions converge uniformly towards this $y$, so they can be considered 'approximations' of the true solution. – goodcow Nov 4 '15 at 10:18