Existence and Uniqueness Theorem

Greetings I have a question about the basic existence and uniqueness theorem for nth order homogeneous linear differential equations on a certain interval.If $\displaystyle f(x_0)=0,f'(x_0)=0,...,f^{(n-1)}(x_0) $ where x, then$\displaystyle f(x)=0$ 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 $\displaystyle f(x_0)=0$ then $\displaystyle f(x)=0$ if I am not mistaken?

Thanks

Re: Existence and Uniqueness Theorem

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?

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Re: Existence and Uniqueness Theorem

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 ?

Re: Existence and Uniqueness Theorem

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 $\displaystyle x_0$, 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 $\displaystyle x_0$. Specifically, the "theorem" says there **is** a solution and the corollary says there is only one solution.

Re: Existence and Uniqueness Theorem

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

Re: Existence and Uniqueness Theorem

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

Re: Existence and Uniqueness Theorem

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

Re: Existence and Uniqueness Theorem

Quote:

Originally Posted by

**JohnDoe** F is assumed to be the solution of the nth order homogeneous differential equation. The corollary states that if $\displaystyle f(x_0)=0$ and $\displaystyle x_0$ 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?

Re: Existence and Uniqueness Theorem

Quote:

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.

Re: Existence and Uniqueness Theorem

Quote:

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?

Re: Existence and Uniqueness Theorem

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 ?

Re: Existence and Uniqueness Theorem

Quote:

Originally Posted by

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

(Clapping)

Re: Existence and Uniqueness Theorem

Quote:

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 $\displaystyle f(x_0)$, $\displaystyle f'(x_0)$, ..., $\displaystyle f^{n-1}(x_0)$ **any** given numbers. Since the trival function, f(x)= 0, satisfies both the differential equation and the conditions $\displaystyle f(x_0)= 0$, $\displaystyle f'(x_0)= 0$, ..., $\displaystyle f^{n-1}(x_0)= 0$, 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 $\displaystyle \frac{d^2f}{dt^2}= -f$ and the conditions $\displaystyle y(0)= 0$, $\displaystyle y(\pi)= 0$ for **any** value of C.

Re: Existence and Uniqueness Theorem

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.