1. ## Existence and Uniqueness

Determine the existence and uniqueness of solutions for each of the following. Indicate the interval (for t) where there exists a unique solution (if possible). DO NOT SOLVE THE DIFF EQ, JUST DETERMINE WHETHER A UNIQUE SOLUTION EXISTS AND EXPLAIN HOW YOU KNOW. State the results that you use to get to your conclusions on each one.

1.) $\displaystyle \displaystyle \frac{dy}{dt}=3y^{2/3}, y(0)=1$

2.) $\displaystyle \displaystyle (t-1)y^{(4)}+(t+1)y''+\tan t y=0, y(0)=y'(0)=y''(0)=y'''(0)=0$

3.) $\displaystyle \displaystyle y''+25y=0, y(0)=0, y(\pi)=0$

*** Well, I'm not sure, but I think the 2nd is unique.. not sure how to determine without solving.

2. Originally Posted by Ideasman
Determine the existence and uniqueness of solutions for each of the following. Indicate the interval (for t) where there exists a unique solution (if possible). DO NOT SOLVE THE DIFF EQ, JUST DETERMINE WHETHER A UNIQUE SOLUTION EXISTS AND EXPLAIN HOW YOU KNOW. State the results that you use to get to your conclusions on each one.

1.) $\displaystyle \displaystyle \frac{dy}{dt}=3y^{2/3}, y(0)=1$

2.) $\displaystyle \displaystyle (t-1)y^{(4)}+(t+1)y''+\tan t y=0, y(0)=y'(0)=y''(0)=y'''(0)=0$

3.) $\displaystyle \displaystyle y''+25y=0, y(0)=0, y(\pi)=0$

.
You see the existence/uniqueness theorem only applies to linear differencial equations. If you look at #2 it is linear. Well, not exactly we need to divide by $\displaystyle (t-1)$ but when we do that we need to be sure that this is not zero, i.e. $\displaystyle t\not = 1$. And we require that $\displaystyle \tan t$ to be defined. In that case we are are on the interval $\displaystyle (-\pi/2,1)$ then we are guarenntted to have a unique solution.

If you look at equation #3 this is a linear equation BUT it is not an initival value problem. So the theorem does not apply.*

*)This is actually a Sturm-Louiville boundary value problem which in fact has no solutions.

3. Originally Posted by ThePerfectHacker
You see the existence/uniqueness theorem only applies to linear differencial equations. If you look at #2 it is linear. Well, not exactly we need to divide by $\displaystyle (t-1)$ but when we do that we need to be sure that this is not zero, i.e. $\displaystyle t\not = 1$. And we require that $\displaystyle \tan t$ to be defined. In that case we are are on the interval $\displaystyle (-\pi/2,1)$ then we are guarenntted to have a unique solution.

If you look at equation #3 this is a linear equation BUT it is not an initival value problem. So the theorem does not apply.*

*)This is actually a Sturm-Louiville boundary value problem which in fact has no solutions.
Okay, I'm confused.

For #1, this is non-linear, and therefore it has no unique solution.

For #2, first off I am assuming its tan(t)*y instead of tan(ty) (my teacher should really learn to use parentheses!). Now, I don't see why we divide by (t - 1). It's linear without dividing by (t - 1). And how can you be so sure that t does not equal 1? Also, we know tan(t) is continuous between -Pi/2 and Pi/2. So, according to the unique/existence theorem, if the interval contains 0, it has to have a unique solution..but on this interval? I don't know what you mean by (-Pi/2,1).

I understand your #3, and I'm impressed you knew WHAT type of boundary problem it was!

4. Originally Posted by Ideasman
For #1, this is non-linear, and therefore it has no unique solution.
We cannot say that. My point is that the existence/uniqueness theorem was only proven for linear differencial equations. When we have non-linear differencial equations we have no idea whether it works or not without solving it.
For #2, first off I am assuming its tan(t)*y instead of tan(ty) (my teacher should really learn to use parentheses!). Now, I don't see why we divide by (t - 1). It's linear without dividing by (t - 1). And how can you be so sure that t does not equal 1? Also, we know tan(t) is continuous between -Pi/2 and Pi/2. So, according to the unique/existence theorem, if the interval contains 0, it has to have a unique solution..but on this interval? I don't know what you mean by (-Pi/2,1).
The theorem says that given,
$\displaystyle y^{(n)}+f_{n-1}(x)y^{(n-1)}+...+f_1(x)y'+f_0(x)y = g(x)$
With $\displaystyle y(x_0)=y_0 , y'(x_0)=y'_0 , ... , y^{(n)} (x_0)=y^{(n)} _0$.
Where $\displaystyle f_{n-1}(x),...,f_0(x),g(x)$ are continous functions on some open interval $\displaystyle (a,b)$ and where $\displaystyle x_0\in (a,b)$ then there exists exactly one solution to that differencial equation.

So given (after division),
$\displaystyle y^{(4)} + \frac{t+1}{t-1} y'' + \tan t \ y = 0 \mbox{ and }0=y(0)=y'(0)=y''(0)=y'''(0)=y''''(0)$.
We must work on the interval containing $\displaystyle 0$ because if you look at the theorem the initial point for the equation is contained in the interval. But that is not it. We also require that $\displaystyle \frac{t+1}{t-1}$ and $\displaystyle \tan t$ be continous. Now, $\displaystyle \tan t$ is not continous at certain points (graph it). But if it needs to contain zero so the only open interval (largest interval) where $\displaystyle \tan t$ is continous is on $\displaystyle (-\pi/2,\pi/2)$. But we also require that $\displaystyle (t+1)/(t-1)$ be continous, so we need to exclude the point $\displaystyle t=1$. This point $\displaystyle t=1$ is contained in $\displaystyle (-\pi/2, \pi/2)$ we we need to omit it. This gives us that on $\displaystyle (-pi/2,1)$ these coefficient functions are continous. So there is a unique solution.

5. Originally Posted by ThePerfectHacker
We cannot say that. My point is that the existence/uniqueness theorem was only proven for linear differencial equations. When we have non-linear differencial equations we have no idea whether it works or not without solving it.

The theorem says that given,
$\displaystyle y^{(n)}+f_{n-1}(x)y^{(n-1)}+...+f_1(x)y'+f_0(x)y = g(x)$
With $\displaystyle y(x_0)=y_0 , y'(x_0)=y'_0 , ... , y^{(n)} (x_0)=y^{(n)} _0$.
Where $\displaystyle f_{n-1}(x),...,f_0(x),g(x)$ are continous functions on some open interval $\displaystyle (a,b)$ and where $\displaystyle x_0\in (a,b)$ then there exists exactly one solution to that differencial equation.

So given (after division),
$\displaystyle y^{(4)} + \frac{t+1}{t-1} y'' + \tan t \ y = 0 \mbox{ and }0=y(0)=y'(0)=y''(0)=y'''(0)=y''''(0)$.
We must work on the interval containing $\displaystyle 0$ because if you look at the theorem the initial point for the equation is contained in the interval. But that is not it. We also require that $\displaystyle \frac{t+1}{t-1}$ and $\displaystyle \tan t$ be continous. Now, $\displaystyle \tan t$ is not continous at certain points (graph it). But if it needs to contain zero so the only open interval (largest interval) where $\displaystyle \tan t$ is continous is on $\displaystyle (-\pi/2,\pi/2)$. But we also require that $\displaystyle (t+1)/(t-1)$ be continous, so we need to exclude the point $\displaystyle t=1$. This point $\displaystyle t=1$ is contained in $\displaystyle (-\pi/2, \pi/2)$ we we need to omit it. This gives us that on $\displaystyle (-pi/2,1)$ these coefficient functions are continous. So there is a unique solution.
$\displaystyle y^{(4)} + \frac{t+1}{t-1} y'' + \tan t \ y = 0 \mbox{ and }0=y(0)=y'(0)=y''(0)=y'''(0)=y''''(0)$

Um, why didn't you divide tan(t)y by (t-1) and how did you decide y''''(0) = 0