# Uniqueness Theorem for homogenous linear ODEs

Consider the system of linear differential equations:

X' = AX where X is a column vector (of functions) and A is coefficient matrix. We could just as well consider a first order specific case: y'(x) = C(x)y

We know that the soltuion will be a subset of the vector space of continuous functions. We know the function f(x) = 0 (the additive identity) is contained in the set of solutions S. We also know that any scalar multiple of an element in S is also in S, as is any linear combination of elements (all do due the properties of differential operator) . Therefore, because S is a subset of C, and the operations of addition and scalar multiplication are closed in S, S itself is a vector space.

What is the dimension of S? The dimension of S is the number of elements in the column vectors X' = AX, so a first order equation has a solution space of dimension two, etc. Therefore, the solution space of an nth order ODE can be spanned by a basis of n linearly independent vectors.

Then if we find two solutions to a second order equation F(x) and G(x), and we can show they pass the Wronskian test for linear independence, is this sufficient to show that:

S = {aF(x) + bG(x): a,b contained in R}

And thereby show uniqueness?

## Answers and Replies

is this sufficient

Yeap.

HallsofIvy
Science Advisor
Homework Helper
Consider the system of linear differential equations:

X' = AX where X is a column vector (of functions) and A is coefficient matrix. We could just as well consider a first order specific case: y'(x) = C(x)y

We know that the soltuion will be a subset of the vector space of continuous functions. We know the function f(x) = 0 (the additive identity) is contained in the set of solutions S. We also know that any scalar multiple of an element in S is also in S, as is any linear combination of elements (all do due the properties of differential operator) . Therefore, because S is a subset of C, and the operations of addition and scalar multiplication are closed in S, S itself is a vector space.

What is the dimension of S? The dimension of S is the number of elements in the column vectors X' = AX, so a first order equation has a solution space of dimension two, etc. Therefore, the solution space of an nth order ODE can be spanned by a basis of n linearly independent vectors.
But the proof of that requires "existence and uniqueness". Once you have the d.e. written X'= AX you can use the standard existence and uniqueness proof for first order differential equations. (After proving that it extends to "vectors", of course.)

Then if we find two solutions to a second order equation F(x) and G(x), and we can show they pass the Wronskian test for linear independence, is this sufficient to show that:

S = {aF(x) + bG(x): a,b contained in R}

And thereby show uniqueness?

Ι don't think he can answer... he's banned! lol

robphy
Science Advisor
Homework Helper
Gold Member
Ι don't think he can answer... he's banned! lol

...and this question is over 4 years old!

...and this question is over 4 years old!

I kind of like replying to old but interesting questions.

As I answered to an earlier accusation, it's like going on a date with a middle-aged virgin. :P