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Existence and uniqueness of differential solution, help?

by Lengalicious
Tags: differential, existence, solution, uniqueness
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Lengalicious
#1
Feb28-12, 12:57 PM
P: 164
Ok so ill give an example, x'(t) = log(3t(x(t)-2)) is differential equation where t0 = 3 and x0 = 5
The initial value problem is x(t0) = x0.

So what i'de do is plug into initial value problem to get x(3) = 5, so on a graph this plot would be at (5,3)? Then plop conditions into differential equation so: x'(3) = log(3*3(5 - 2)) = 1.43 which would be at a plot (1.43,3)? So x'(3) < x(3), does this mean that the solution is unique for all t? If so, why is this? Just want to understand 100%.
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HallsofIvy
#2
Feb28-12, 03:09 PM
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I am confused as to what you are saying.
"Ok so ill give an example, x'(t) = log(3t(x(t)-2)) is solution to equation where t0 = 3 and x0 = 5".

x'(t)= ... is NOT a "solution" to a differential equation, it is a differential equation.


Or do you mean to say that x(t)= log(4t(x(t)- 2)) is a solution to some unknown differential equation? Then whether or not that equation is unique depends upon exactly what the differential equation is.
Lengalicious
#3
Feb28-12, 03:22 PM
P: 164
Ok, yes my mistake its a differential equation i guess? This is where I myself am confused =/. If its a differential equation then how do you find out whether it has a unique solution or not?

alan2
#4
Feb28-12, 10:39 PM
P: 199
Existence and uniqueness of differential solution, help?

If dx/dt=f(x,t), and both f and the partial of f w.r.t. x are continuous in some region about your initial point then then exists a unique solution of the initial value problem in some region containing that initial point. This is a local property. The proof is dependent upon finding a sequence of solutions via Picard iteration which converges to the unique solution. You can find this proof in any introductory ode text.
A. Neumaier
#5
Feb29-12, 10:22 AM
Sci Advisor
P: 1,943
Quote Quote by Lengalicious View Post
Ok so ill give an example, x'(t) = log(3t(x(t)-2)) is differential equation where t0 = 3 and x0 = 5
The initial value problem is x(t0) = x0.

So what i'de do is plug into initial value problem to get x(3) = 5, so on a graph this plot would be at (5,3)? Then plop conditions into differential equation so: x'(3) = log(3*3(5 - 2)) = 1.43 which would be at a plot (1.43,3)? So x'(3) < x(3), does this mean that the solution is unique for all t? If so, why is this? Just want to understand 100%.
You must interpret x'(3) as the slope of the tangent at your first point, so you'd draw
x(3+s) approx x(3)+x'(3)s = 5+1.43 s for small s. Then you continue from that point. This gives you an idea of the solution, not very accurate though. It is called Euler's method, but it can be refined to give an existence and uniqueness proof, if you make appropriate assumption about the differential equation.
Lengalicious
#6
Feb29-12, 12:21 PM
P: 164
Ok, I sort of get it but still slightly confused, an exact question I have is: In the following case:

x'(t) = log(3t(x(t)-2)), where t0 = 3 and x0 = 5

Does the theorem of existence and unicity guarantee an existence of a solution for the initial value problem x(t0) = x0? Justify your answer.

What would your answer be to this? Would help me understand if I got a model answer.
A. Neumaier
#7
Feb29-12, 12:54 PM
Sci Advisor
P: 1,943
Quote Quote by Lengalicious View Post
Ok, I sort of get it but still slightly confused, an exact question I have is: In the following case:

x'(t) = log(3t(x(t)-2)), where t0 = 3 and x0 = 5

Does the theorem of existence and unicity guarantee an existence of a solution for the initial value problem x(t0) = x0? Justify your answer.

What would your answer be to this? Would help me understand if I got a model answer.
This belongs in the homework section:
http://physicsforums.com/showthread.php?t=88061
alan2
#8
Feb29-12, 01:01 PM
P: 199
I gave you the conditions and the general approach above. Look in any intro ode book. Here's a link if you don't have one:

http://www.math.uconn.edu/~troby/Mat.../existence.pdf
Lengalicious
#9
Feb29-12, 01:19 PM
P: 164
Ok thanks

EDIT: Wow yeh that link you gave is incredibly useful, thanks alot =)
Lengalicious
#10
Feb29-12, 03:18 PM
P: 164
In this case where the differential equation is dx/dt = √x , I understand that to get the intervals is easy cuz x is independant from t so t would belong to all real numbers and x would be ≥ 0, but in the example I have given I don't understand how to get the interval since x(t) is within the function?

EDIT: basically how do i find the domain of the function in the original differential equation posted.

And just to straighten my understanding out, I find the domain of the functions variables, then the domain of the partial derivates , then consider maximum interval that fullfills both requirements and compare to the initial condition to consider whether a unique solution exists or not.

EDIT: Ok I think i got it, so i use intial condition to figure out x(t) in this case i think x(t) = t+2, so would the domain of the function be x belongs to ℝ and t ≠ 0, then for the partial derivative = 2/t where domain is t ≠ 0. so initial condition has to satisfy t ≠ 0 for there to be unique solution? Due to common subset of t ≠ 0, initial value condition shows that t0 = 3 and therefore belongs to common interval so a unique solution exists?
Lengalicious
#11
Mar1-12, 01:44 PM
P: 164
Can anyone let me know if what i said was correct? I am confused as how to find the domain of f(x,t) cuz the function contains x(t) and t, so whats the domain for x and whats the domain for t? I was thinking t ≠ 0 and x ≠ 0 cuz their both inside the log funct.


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