Uniqueness theorem for ODEs
Intuitively, the uniqueness theorem says:
For an -th order ODE, once you specify the value of the function and its first derivatives at one point, you have completely fixed the future trajectory of the solution.
That is why the textbook can say that those two functions must actually be the same function everywhere.
For example, consider a second-order ODE
To know what happens next near , you need to know
Why are those enough? Because the differential equation itself tells you the second derivative:
So if two solutions have the same and , they automatically have the same .
Then, roughly speaking, the ODE keeps telling both functions how they must evolve from that state. They cannot suddenly choose different paths, because at every instant the differential equation prescribes the same โnext motionโ for both.
A good physical analogy is Newtonian mechanics. Suppose
This could describe a spring. If I tell you at time :
then the law
determines the acceleration. From position, velocity, and acceleration you move a tiny amount forward in time; at the new instant the same law determines the new acceleration, and so on.
So two spring motions that start at exactly the same position and velocity cannot branch apart while both obeying the same deterministic law.
For an -th order equation,
the โstateโ is the -dimensional vector
and the ODE tells you how this entire state vector changes:
This is probably the most useful intuition in the context of your Wronskian discussion.
In equation (20), the book has shown that
and
have exactly the same state vector at
They also obey the same ODE.
So the uniqueness theorem says there is only one trajectory through that particular state. Hence the two supposed solutions cannot really be different:
throughout the interval.
The deeper geometric picture is: an
And this also explains something you were asking earlier: why do we care about derivatives up to
are precisely the complete state required to uniquely identify a solution of an