# Differential Equations - Existence and Uniqueness

1. Feb 4, 2012

### prosteve037

I'm having trouble understanding what uniqueness is/means. When given a slope/direction field I don't know what I should be looking for if asked to determine if a given initial condition has a unique solution.

Example:
$\textit{y' = }\frac{(x - 1)}{y}$

With this equation I can see that as long as $\textit{y ≠ 0}$ a solution exists.

But now if I'm asked to find a given interval where a solution exists and is unique, I'm confused :P What should I be looking for in the equation/direction field?

Here's the direction field for that differential equation:

2. Feb 4, 2012

### LCKurtz

Your differential equation is of the form $y' = f(x,y)$ with $f(x,y) = \frac{(x - 1)}{y}$. What does the existence and uniqueness theorem require about $f(x,y)$? Does your $f(x,y)$ satisfy the requirements? Look at the theorem.

3. Feb 5, 2012

### prosteve037

Is it that there must only be one function $y(x)$ that passes through the given point?

But how do you prove that this is the only solution that passes through that point? Couldn't there be other functions that pass through that point?

From the example above, the point $(0,1)$ has a solution with slope -1. But how can I justify that there is only one function that passes through this point?

4. Feb 5, 2012

### LCKurtz

I presume your text has a theorem called the existence and uniqueness theorem. Theorems have hypotheses and conclusions. The conclusion of the theorem is that there is a unique solution through the point, provided that the hypotheses on $f(x,y)$ are satisfied. That is how you prove there is only one solution through the point. So I will ask you again: what are the hypotheses and does this f(x,y) satisfy them?

5. Feb 6, 2012

### prosteve037

Oh wow, I totally glossed over that in the text :grumpy: It says if $f(x,y)$ and $D_{y}f(x,y)$ are continuous, $(x,y)$ has a unique solution.

So then does this mean if I take the partial derivative of $f(x,y)$ with respect to y and plug in the values for x and y, yielding a real solution, then the point $(x,y)$ has a unique solution?

In other words if $f(x,y)$ is defined, a solution exists. If $D_{y}f(x,y)$ is defined as well, the solution is unique. Is this correct?

6. Feb 6, 2012

### HallsofIvy

Staff Emeritus
"Yielding a real value" is in general NOT sufficient to be continuous. If it happens that your function is a "rational function", a polynomial divided by another, then the function is continuous as long as the denominator is not 0, but other kinds of functions do not satisfy that.

Also, by the way, the partial derivative being continuous is "sufficient" but not "necessary". The necessary condition is that f(x,y) be "Lipschitz" in y- that is, that for some y1 and y2 in a neighborhood of the y value for the initial point, there exist a constant, c, such that |f(x,y1)- f(x,y2)|< c|y1- y2|. It can be shown that being "Lipschitz" is 'stronger' than "continuous" but 'weaker' than "continous differential". That is, any function that has a continuous derivative is Lipschitz but there exist Lipschitz functions that are do not have a continuous differivative and every Lipschitz function is continuous while there exist continuous functions that are not Lipschitz.

7. Feb 6, 2012

### LCKurtz

It is not "the point $(x,y)$" that has a unique solution. It is the differential equation that has a unique solution passing through that point. All you have to figure out is for what $(x,y)$ are both $f(x,y)$ and $f_y(x,y)$ continuous. The theorem guarantees a unique solution through those points.

8. Feb 6, 2012

### prosteve037

Ahh okay thank you so much!

Now how about the example from the above post? How would I explicitly find solutions for these intervals where the solution exists and is unique?

For that D.E. $f(x,y) = \frac{x-1}{y}$ I took the partial derivative:

$f_{y}(x,y) = -\frac{x - 1}{y^{2}}$

Which means that there exists a unique solution for the differential equation wherever $y ≠ 0$. But through which values of x does y ≠ 0?

Would it depend on the function $y(x)$ chosen? And so then could it be any function $y(x)$ where $y(x)\propto{ x}$?

9. Feb 6, 2012

### LCKurtz

x and y are independent variables in the xy plane. Whether $y\ne 0$ has nothing to do with x, which can be anything.
That last statement makes no sense to me. If you actually want to find a solution through, for example, (0,1), then solve the equation by separation of variables. You haven't given an exact statement of the problem you are trying to solve, but solving for a solution and knowing that there is a unique solution through that point are two different things. The theorem tells you that without solving the DE.

10. Feb 6, 2012

### prosteve037

Alright yeah, I'm sorry. I'm just not making sense anymore

Here's the problem statement:

Consider the following differential equation for $y = y(x)$

$yy' = (x - 1)$

a.) Sketch the direction fields and draw possible solution curves.

b.) If the initial condition is given by (x,y) = (0,1) does the solution exist near this point and is it unique if it exists?

The solution does exist near this point and is unique because both the differential equation and its partial derivative with respect to y are continuous at that point.

c.) If the initial condition is given by (x,y) = (1,0) does the solution exist near this point and is it unique if it exists?

The solution does not exist near this point because the differential equation is discontinuous at that point.

d.) Give several intervals where the solution exists and is unique.

... This is where I'm stuck :/

Sorry again for the confusion, I should've chosen the right sentences to express my thoughts more clearly before actually posting them

11. Feb 6, 2012

### LCKurtz

OK, you have drawn what looks like a solution. Have you written down its equation and verified that it is a solution? Is it a solution on [-1/2,1/2]? [-2,2]? Think about that.

12. Feb 6, 2012

### prosteve037

Its equation is $y(x) = -x + 1$ :

$y = -x + 1$

$y' = -1$

$y' = \frac{x - 1}{y}$

$y' = \frac{x - 1}{-x + 1}$

$y' = -1$

And yes, it's a solution in those intervals:

$y(x) = -x +1$

$y(-\frac{1}{2}) = \frac{1}{2} + 1 = \frac{3}{2}$

$y'(-\frac{1}{2}) = -1$

$y(\frac{1}{2}) = -\frac{1}{2} + 1 = \frac{1}{2}$

$y'(\frac{1}{2}) = -1$

$y(-2) = 2 + 1 = 3$

$y'(-2) = -1$

$y(2) = -2 + 1 = -1$

$y'(2) = -1$

13. Feb 6, 2012

### LCKurtz

Really? Does it satisfy your DE at (1,0), which is in the interval [-2,2]?

14. Feb 6, 2012

### prosteve037

Oh oops! No it doesn't, it's discontinuous at that point :P

So then are the intervals $[\frac{-1}{2}≤ x ≤ \frac{1}{2}], [x ≥ 2], [x ≤ -2]$ for $y(x) = -x + 1$?

15. Feb 6, 2012

### LCKurtz

The solution through (0,1) is not valid on two of those intervals.