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Differential Equations  Existence and Uniqueness 
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#1
Feb412, 09:18 PM

P: 105

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: [itex]\textit{y' = }\frac{(x  1)}{y}[/itex] With this equation I can see that as long as [itex]\textit{y ≠ 0}[/itex] 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
Feb412, 09:52 PM

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#3
Feb512, 12:06 PM

P: 105

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
Feb512, 02:06 PM

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Differential Equations  Existence and Uniqueness



#5
Feb612, 04:41 PM

P: 105

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
Feb612, 05:08 PM

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"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)< cy1 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
Feb612, 06:11 PM

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#8
Feb612, 07:17 PM

P: 105

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{x1}{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
Feb612, 07:36 PM

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#10
Feb612, 08:23 PM

P: 105

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
Feb612, 09:09 PM

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#12
Feb612, 09:49 PM

P: 105

##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
Feb612, 10:16 PM

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#14
Feb612, 10:34 PM

P: 105

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


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