7

So the uniqueness theorem is: Consider the following IVP. $$y'=f(t, y)$$ $$y(t_o)=y_o$$ If $f(t,y)$ and $df\over dy$ are continuous functions in some rectangle $a < t < b$, $d < y < z$ containing the point $(t_o, y_o)$ then there is a unique solution to the IVP in some interval to $h < t < t_o + h$ that is contained in $a < t < b$.

I cannot seem to wrap my head around why this must be true (the intuition), I'm sure it must be pretty obvious since there are not any other questions concerning this on stackexchange.

Jack M
  • 28,518
  • 7
  • 71
  • 136
TQM
  • 85
  • I really found it instructive to take a few examples, figure out the regions and rectangles and why this is a very important theorem to get your head around. There are a few examples that bring this theorem to life and also some items that show limitations. Highly recommend working this! Regards – Amzoti Jul 13 '13 at 22:28

2 Answers2

5

Imagine you want to draw the graph of the solution $y(t)$. The first equation tells you for each point in which direction your graph has to continue if it happens to go through that point. The second equation tells you one point of the graph. At least to me it is intuitively clear that if you give a start point and for each point a direction (that is, whereever you are, you are told in which direction to continue), then the whole curve is already fixed. That is, what needs explanation is more that this is not always the case, but you need to fulfil those conditions.

The first condition basically demands that the direction changes smoothly when going in the $y$-$t$ plane. Now if that isn't the case, it is not hard to imagine a case where you can have several solutions. For example imagine $f(t,y)=sign(y)$. Then you could have a line that starts at $y(t)=0$ but at some arbitrary point deviates to go either up or down. Since at that point it would leave the $y=0$ line, it would immediately be subject to the other condition, and everything would be fine. By demanding continuity, you basically say that the direction cannot change abruptly, and therefore if you do an arbitrary small displacement, you cannot get a completely different curve.

The second condition restricts the possible direction changes even more, by demanding that when going in $y$ direction, even the change of direction has to be smooth. Maybe the intuitive picture here is that a small deviation in $y$ direction can quickly increase through "self-acceleration" (the small deviation in $y$ gives a small deviation in $y'$, which then causes an even bigger deviation in $y$, see also: chaos theory), therefore you want also the change of the direction under control.

celtschk
  • 44,527
0

What is the equation telling you? It says that, if you find yourself at a given time $t$ at a given point $y$, then the slope of the $y(t)$ should be $f$. This tells you where to go next, at the 'next' $t$. Of course, this only works if you step forward in discrete steps $\delta t$. However, if $f, f_y$ are nice smooth functions it turns out that you can smooth out the function $y(t)$ (roughly you can take $\delta t\to 0$) in a way resulting in a unique, smooth graph, at least locally. (Why locally? Because it's possible that the steps speed off to infinity given enough time - see $f=y^2$.)

not all wrong
  • 16,568
  • 2
  • 37
  • 58