My Math Forum Question about solving ODE

 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?

 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.

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