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I have a system described an equation, and I want to find an (DE) equation for z(x,t), in the limit as l->0.

First some definitions to simplify it some:
$Z1=z(x,t)-z(x+l,t)$
$Z2=z(x,t)-z(x-l,t)$

These are not constants, just to simplify expr.

c however, is a constant.

EQ:

$c*l^2*d^2z(x,t)/dt^2=(1-l/sqrt(Z1^2+l^2))*Z1+(1-l/sqrt(Z2^2+l^2))*Z2$

So, I want to find the limit of this expression as l->0, from the positive side. (l is a physical distance, and this is a physical realizable system, hopefully in the limit too).

So z(x,t) initially only makes physical sense for x=0,l,2l,3l.
And I want to make it into a continuous function for all x.
I dont know if its relevant but z(x,t) and x is also a distance, and t is time.

Please tell me if I should clarify something, and whether the problem as stated is well defined mathematically.

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    Make a Taylor expansion up to first order in the derivatives to x of $z(x+l,t)$ around the point $x$. Same for $z(x-l,t)$. I think there is a problem with your formula though. There should be an extra factor $1/l$ on the right hand side for the formula to be non trivial.2011-01-11
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    Can you give some background as to how you arrive at this formula?2011-01-11
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    @Rask Yes I have2011-01-11
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    I think you choose your spring constant incorrectly. You should choose them so that they become infinitely stiff in the limit of $l \to 0$. Otherwise you'll just get $0$ for the right hand side.2011-01-11
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    @Rask You are right I had the proportionality reversed, fixed now2011-01-11

1 Answers 1

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OK, I found out how it works. So, Newton's second law and Hooke's law tell you two things. First, for the displacement in the $z$-direction you have:

$$m \frac{d^2z}{dt^2} = -k \Delta l_1 \sin\theta_1 -k \Delta l_2 \sin\theta_2$$

Second, since we assume there is no substantial displacement in the $x$-direction:

$$-k \Delta l_1 \cos\theta_1 -k \Delta l_2 \cos\theta_2 \approx 0$$.

Using this second equation we can simplify the first as

$$m \frac{d^2z}{dt^2} = -k \Delta l_1 \sin\theta_1 +k \Delta l_1 \tan\theta_2 \cos\theta_1$$

or

$$m \frac{d^2z}{dt^2} = k \Delta l_1 \cos\theta_1 (\tan\theta_2 - \tan\theta_1)$$

Now, $\Delta l_1 \cos\theta_1=l$ and

$$\tan\theta_1 = \frac{z(x,t)-z(x-l,t)}{l}$$

while

$$\tan\theta_2 = \frac{z(x+l,t)-z(x,t)}{l}$$

so that we get

$$m \frac{d^2z}{dt^2} = k (z(x+l,t)+z(x-l,t)-2z(x,t))$$

Now, rearranging and introducing a factor $l^2$

$$\frac{d^2z}{dt^2} = \frac{k l^2}{m} \frac{z(x+l,t)+z(x-l,t)-2z(x,t)}{l^2}$$

So, the limit will give the second derivative on the right hand side times a constant factor only if $\lim_{l\to 0}\frac{k l^2}{m}=c$. This can be achieved by letting the parameters scale as:

$$\lim_{l \to 0}\frac{m}{l} = M$$

and

$$\lim_{l \to 0} \; k l = K$$

Hope this clears things up. (I'm sure I made a load of sign mistakes, but I went pretty fast over the calculations, so I just made it so there was an even number of them. ;P )

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    Made a slight correction at the end.2011-01-11
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    It looks right, is it consistent with my equation btw? And do you happend to know how to find the equation for a 2D square grid of springs? Or what it is? I cant find any reference.2011-01-11
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    I think it should work for a 2D square grid the same way. Add an equation for the $y$-direction like I did for the $x$-direction and add 2 terms to the $z$-direction. My very first equation is exactly the same as yours, if you write out the $\sin$ as the proportion of side to hypothenusa, you'll get your equation. It's just more difficult to find out how the second derivative gets into the picture.2011-01-11