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Uniqueness theorem for ODEs

Intuitively, the uniqueness theorem says:

For an ๐‘›-th order ODE, once you specify the value of the function and its first ๐‘›โˆ’1 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

๐‘ฆโ€ณ=๐น(๐‘ฅ,๐‘ฆ,๐‘ฆโ€ฒ).

To know what happens next near ๐‘ฅ0, you need to know

๐‘ฆ(๐‘ฅ0),๐‘ฆโ€ฒ(๐‘ฅ0).

Why are those enough? Because the differential equation itself tells you the second derivative:

๐‘ฆโ€ณ(๐‘ฅ0)=๐น(๐‘ฅ0,๐‘ฆ(๐‘ฅ0),๐‘ฆโ€ฒ(๐‘ฅ0)).

So if two solutions have the same ๐‘ฆ(๐‘ฅ0) and ๐‘ฆโ€ฒ(๐‘ฅ0), they automatically have the same ๐‘ฆโ€ณ(๐‘ฅ0).

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 โ€œnext motionโ€ for both.

A good physical analogy is Newtonian mechanics. Suppose

๐‘ฆโ€ณ=โˆ’๐‘˜๐‘ฆ.

This could describe a spring. If I tell you at time ๐‘ก0:

position=๐‘ฆ(๐‘ก0),velocity=๐‘ฆโ€ฒ(๐‘ก0),

then the law

๐‘ฆโ€ณ=โˆ’๐‘˜๐‘ฆ

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 ๐‘›-th order equation,

๐‘ฆ(๐‘›)=๐น(๐‘ฅ,๐‘ฆ,๐‘ฆโ€ฒ,โ€ฆ,๐‘ฆ(๐‘›โˆ’1)),

the โ€œstateโ€ is the ๐‘›-dimensional vector

[๐‘ฆ๐‘ฆโ€ฒโ‹ฎ๐‘ฆ(๐‘›โˆ’1)]

and the ODE tells you how this entire state vector changes:

๐‘‘๐‘‘๐‘ฅ[๐‘ฆ๐‘ฆโ€ฒโ‹ฎ๐‘ฆ(๐‘›โˆ’1)]=[๐‘ฆโ€ฒ๐‘ฆโ€ณโ‹ฎ๐น(๐‘ฅ,๐‘ฆ,โ€ฆ,๐‘ฆ(๐‘›โˆ’1))].

This is probably the most useful intuition in the context of your Wronskian discussion.

In equation (20), the book has shown that

๐‘ฆ๐‘›

and

๐‘‘1๐‘ฆ1+โ‹ฏ+๐‘‘๐‘›โˆ’1๐‘ฆ๐‘›โˆ’1

have exactly the same state vector at ๐‘ฅ0:

[๐‘ฆ๐‘ฆโ€ฒโ‹ฎ๐‘ฆ(๐‘›โˆ’1)]|๐‘ฅ=๐‘ฅ0.

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:

๐‘ฆ๐‘›(๐‘ฅ)=๐‘‘1๐‘ฆ1(๐‘ฅ)+โ‹ฏ+๐‘‘๐‘›โˆ’1๐‘ฆ๐‘›โˆ’1(๐‘ฅ)

throughout the interval.

The deeper geometric picture is: an ๐‘›-th order ODE turns the problem into motion through an ๐‘›-dimensional โ€œstate space.โ€ The uniqueness theorem says that through each admissible point in that state space, there is only one solution curve. Two solution curves cannot pass through the same state and then split apart.

And this also explains something you were asking earlier: why do we care about derivatives up to ๐‘›โˆ’1? It is not merely because we want a square matrix for the determinant. Those ๐‘› quantities,

๐‘ฆ,๐‘ฆโ€ฒ,โ€ฆ,๐‘ฆ(๐‘›โˆ’1),

are precisely the complete state required to uniquely identify a solution of an ๐‘›-th order ODE. The Wronskian packages those states into columns.

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