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The existence-uniqueness theorem is an abstract result,
and it probably doesn't mean a whole lot
until you see it applied to an example.
So we're going to apply it to this simple example:
x.(dy/dx) = y.
Now this is kind of a cool example because, in this example,
things go 'right' in some places and go 'wrong' in others.
First let's note that we can rewrite this equation as:
dy/dx = y/x.
Now this is the form we need it in
in order to apply the existence-uniqueness [theorem]
because you remember, in that theorem we have:
dy/dx = f (x,y)
(on the right-hand side) so our 'f (x,y)' is 'y/x'.
So the first thing to do is to check the continuity of that function 'f':
and it's easy to see that the function 'y/x' is continuous
so long as the denominator is not '0'.
So y/x is continuous near any point (a,b)
so long as 'a' is not equal to '0'.
Therefore by the existence theorem (the 'existence' part of thte theorem)
a solution satisfying y (a) = b
'exists' so long as 'a' is not equal to '0'.
Now if 'a = 0' we don't know anything from the theorem
the theorem says nothing
because the hypotheses are not satisfied.
Now what about uniqueness?
Well we have to take the partial derivative of 'y/x'
and that is the same thing as taking the derivative
with respect to y holding x fixed (holding x constant)
That turns out to be: '1/x'.
And it's easy to see again that '1/x' is continuous
so long as that denominator is not '0':
So it's continuous near any point (a,b)
so long as 'a' not equal to '0'.
Therefore, by the uniqueness part of the theorem
a solution with y (a) = b
exists by this first part and is unique
so long as 'a' is not equal to '0'.
Again, here, if 'a' equals '0'
the hypotheses of the theorem are not satisfied
and so I know nothing about 'uniqueness'
(in addition to 'existence', from before.)
It's interesting to try to get a concrete understanding of what's going on here
by actually solving this equation and looking at the solutions.
So it's easy to check that y (x) = c.x solves this equation x.(dy/dx) = y.
(for any value of c)
Now I've sketched a few of these solutions up here:
These are all lines of slope c which pass through the origin
(and there's just a few of them that I've sketched.)
So if I go to a point (a,b) with 'a' not equal to '0',
so I'm 'away' from the y-axis here,
I go to appoint a point (a,b) and say: what solution passes through that point?
Well, choose: y (x) = (b/a)*x and that will pass through that point.
and it will be the unique solution passing through that point.
So you can see, concretely,
that so long as 'a' is not equal to '0'
I'm able to uniquely solve the initial-value problem
'x.(dy/dx) = y' and 'y (a) = b'.
It's interesting to think about what happens on the y-axis:
these are the points (a,b) where 'a = 0'.
The hypotheses of the existence-uniqueness theorem are not satisfied, so anything can happen.
In fact, the two things that could go wrong do go wrong!
First of all
if i look at points (0,b) where 'b' is not '0'
so I'm looking at a point, say, up here on the y-axis..
if i wanted to pass a line through that point, 'y (x) = c.x'
that line would need to be vertical
so I can't have a solution passing through that point.
So there are 'no' solutions
which pass through points up here, on the y-axis.
On the other hand, if I go down to the origin
- the point (0,0) -
then lots of solutions (in fact 'all' of the solutions) to the equation pass through that point.
So if I wanted to satisfy 'y (0) = 0'
I could do it in infinitely many different ways.
So existence fails (up here) and..
existence is 'ok' (down here) but uniqueness fails.
So the the point of all this, the take-home message:
if the hypotheses of the existence-uniqueness theorem are not satisfied,
then anything can happen.