Existence and Uniqueness Theorem

**JohnDoe** · October 17th 2011, 12:32 PM

Greetings I have a question about the basic existence and uniqueness theorem for nth order homogeneous linear differential equations on a certain interval.If $f(x_0)=0,f'(x_0)=0,...,f^{(n-1)}(x_0)$ where x, then $f(x)=0$ on the whole interval. I did not understand the logic behind this, can someone explain it to me? I would be grateful. I think this is also valid for first order differential equations if $f(x_0)=0$ then $f(x)=0$ if I am not mistaken?

Thanks

**Ackbeet** · October 17th 2011, 04:36 PM

For the linear nth order homogeneous differential equation on an interval, you know that if the coefficients are continuous functions, and the coefficient of the highest-order derivative is nonzero on the interval, then there exists a unique solution to the initial value problem. Does f(x) = 0 solve the initial value problem? What could you say then?

**JohnDoe** · October 18th 2011, 12:46 AM

Thanks for your reply.The attached pdf is from Shepley, Ross Differential Equations 3rd Edition and the theorem I did not understand is theorem 4.1 and especially its corollary. Can you shed some light on that ?

**HallsofIvy** · October 18th 2011, 10:47 AM

What exactly is your confusion? The theorem and it corollary state that the nth order linear differential equation, with continuous coefficients in some neighborhood of $x_0$ , has unique solution to the "initial value" problem- that is finding a function, f, that satisfies the given differential equation and has specified values for the function and its derivatives, up to the n-1 derivative, at $x_0$ . Specifically, the "theorem" says there is a solution and the corollary says there is only one solution.

**JohnDoe** · October 18th 2011, 01:03 PM

I did not understand the corollory in the special case $f(x_0)=0 ... f^{n-1}(x_0)=0$ . Why does it imply that $f(x)=0$ on the whole interval?

**Ackbeet** · October 18th 2011, 04:18 PM

First answer this question: is f(x) = 0 a solution of the initial value problem?

**JohnDoe** · October 19th 2011, 11:28 AM

F is assumed to be the solution of the nth order homogeneous differential equation. The corollary states that if $f(x_0)=0$ and $x_0$ belongs to aforementioned certain interval I, then f(x)=0 for every x that is a member of this interval. How is this possible ?

**Ackbeet** · October 19th 2011, 11:39 AM

Originally Posted by JohnDoe

F is assumed to be the solution of the nth order homogeneous differential equation. The corollary states that if $f(x_0)=0$ and $x_0$ belongs to aforementioned certain interval I, then f(x)=0 for every x that is a member of this interval. How is this possible ?

You don't have to assume anything. Tell me this:

1. Does f(x) = 0 satisfy the differential equation?
2. Does it satisfy the initial conditions?

**JohnDoe** · October 19th 2011, 11:43 AM

Originally Posted by Ackbeet

You don't have to assume anything. Tell me this:

1. Does f(x) = 0 satisfy the differential equation?
2. Does it satisfy the initial conditions?

It satisfies the differential equation and the initial conditions.

**Ackbeet** · October 19th 2011, 11:46 AM

Originally Posted by JohnDoe

It satisfies the differential equation and the initial conditions.

Right. So that tells you that f(x) = 0 is a solution. The question is, is it the only solution? Now, you tell me: does the existence and uniqueness theorem apply to this problem?

**JohnDoe** · October 19th 2011, 11:47 AM

Yes it does apply then I think I answered my own question. I think it is the trivial solution that is unique if I am not mistaken ?

**Ackbeet** · October 19th 2011, 11:48 AM

Originally Posted by JohnDoe

Yes it does apply then I think I answered my own question.

**HallsofIvy** · October 19th 2011, 01:04 PM

Originally Posted by JohnDoe

Yes it does apply then I think I answered my own question. I think it is the trivial solution that is unique if I am not mistaken ?

Your wording implies that you think this solution is unique because it is the trivial solution. In fact, for the conditions given on the coefficients, there exist a unique solution with $f(x_0)$ , $f'(x_0)$ , ..., $f^{n-1}(x_0)$ any given numbers. Since the trival function, f(x)= 0, satisfies both the differential equation and the conditions $f(x_0)= 0$ , $f'(x_0)= 0$ , ..., $f^{n-1}(x_0)= 0$ , it is the only solution to that problem.

By the way, this is not true for "boundary value problems". The functions f(x)= C sin(x) satisfies the differential equation $\frac{d^2f}{dt^2}= -f$ and the conditions $y(0)= 0$ , $y(\pi)= 0$ for any value of C.

**JohnDoe** · October 20th 2011, 05:49 AM

I concluded that this is only valid for an initial value problem and understood what is told in the book many thanks again for your help.

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