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.
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 )