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 Differential Equations Ordinary and Partial Differential Equations Math Forum

February 24th, 2016, 08:28 PM   #11
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Quote:
 Originally Posted by Endymion The plots I show are x position versus velocity.
That would be ${\mathrm d v \over \mathrm x}=0$. February 25th, 2016, 03:29 PM   #12
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Quote:
 Originally Posted by v8archie That would be ${\mathrm d v \over \mathrm x}=0$.
Right, but there's no way for me to know at which point x that is going to happen. It could be anywhere between x0 and x1.

Here's what I do know:
I HAVE the function y = f(x) which gives me the full path.
I also have $\displaystyle d^2x/dt^2$ as explained in my first post. From that, I used numerical integration to try to find $\displaystyle dx/dt$.

What I don't know:
x(t), y(t) and $\displaystyle dx/dt, dy/dt$

Is there any way to make use of my knowledge of y=f(x)? which I do have

Last edited by skipjack; February 25th, 2016 at 05:07 PM. February 25th, 2016, 05:05 PM #13 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2202 Is that a different function f from the one in your first post (where y didn't appear)? February 25th, 2016, 07:46 PM #14 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Why don't you post the problem you have? It might be easier to get a handle on it if we can see it. February 25th, 2016, 10:59 PM   #15
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Quote:
 Originally Posted by skipjack Is that a different function f from the one in your first post (where y didn't appear)?
Yes, it is different. Sorry for the typo.

Quote:
 Originally Posted by v8archie Why don't you post the problem you have? It might be easier to get a handle on it if we can see it.
I have:
$\displaystyle y = (2 - x^2)^{0.5}$

Differentiating I get:

$\ddot y$ =$- (2 - x^2)^{-1.5} x^2 \dot x^{2} - (2 - x^2)^{-0.5} \dot x^2 - (2 - x^2)^{-0.5} x \ddot x$

I also know that:

$\ddot x$ = ${(x - 1.7)(\ddot y + 4)} \over (y - 0.5)$

And now plugging in for $\ddot y$ and $y$ and rearranging so as to solve for $\ddot x$ gives me an equation for $\ddot x$ in terms of $x$ and $\dot x$. However, the equation is non-linear and has no closed form solution.

I know the starting condition at $x_0$, $\dot x_0$. So to approximate the behavior I used numerical integration:

$\dot x_{k+1}$ = $\dot x_k$ + $\ddot x_k \epsilon$

$x_{k+1}$ = $x_k$ + $\dot x_k \epsilon$ + $0.5 \ddot x_k \epsilon^2$

I also know the starting conditions at another point. Using the same numerical integration approach I get the blue and red lines I showed earlier. Ideally I would like to somehow get the green line since the solution when we measured the data separately looks like the green curve. Any ideas?

btw, much thanks for the help you guys have already provided. February 26th, 2016, 02:41 AM #16 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Byinspection you have $y=\sqrt2 \sin f(t)$ and $x=\sqrt2 \cos f(t)$ for $0 \le f(t) \le \pi$. You don't have any information on $f(t)$ other than being able to determine $f(0)$. Such a parameterisation constrains $(x,y)$ to the given curve, and any $f(t)$ (that satisfies the initial conditions and the requirements of the range) will also satisfy the given equation for all $t$. Where did this come from? Edit: your second equation for $\ddot x$ can specify $f(x)$. Last edited by v8archie; February 26th, 2016 at 03:35 AM. Tags ode, question, solving Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ahsanalisony New Users 0 May 21st, 2014 07:27 PM dthomas86 Algebra 1 January 29th, 2013 02:21 PM xnarutox Algebra 5 September 10th, 2010 06:04 PM Striker2 Algebra 7 December 3rd, 2009 08:39 PM Houshmandz Algebra 2 June 11th, 2009 10:22 PM

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