How do you nondimensionalize an equation with substituted variables?

[jessicamorgan]

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For this question I have substituted feta and t into the equation, but am unsure what to do after that. Please can anyone point me in the right direction. Thanks :)

 

[Ackbach]

After doing the chain rule and substituting in, I get
$$\frac{1}{\beta^2} \, \frac{d^2 \tilde{\theta}_i}{d\tilde{t}^2}=\frac{k}{I}\left( \tilde{\theta}_{i-1}-2\tilde{\theta}_i+\tilde{\theta}_{i+1} \right).$$
Is that what you get?

 

[jessicamorgan]

Hi, yes but still am unsure what alpha and beta can be. As k=I=1. so can alpha and beta just =1 also without loss of generality.

 

[jessicamorgan]

After doing the chain rule and substituting in, I get
$$\frac{1}{\beta^2} \, \frac{d^2 \tilde{\theta}_i}{d\tilde{t}^2}=\frac{k}{I}\left( \tilde{\theta}_{i-1}-2\tilde{\theta}_i+\tilde{\theta}_{i+1} \right).$$
Is that what you get?

Actually, I'm not getting that .. please can you explain to me how you've used the chain rule here :)

 

[Ackbach]

Actually, I'm not getting that .. please can you explain to me how you've used the chain rule here :)

Sure! So you've got $\theta_i = \alpha \tilde{\theta}_i$ and $t=\beta \tilde{t}$. In order to substitute into the original DE, we need to assemble the ingredients listed. That is, we need $\dfrac{d^2\theta_i}{dt^2}, \; \theta_{i-1},\; \theta_i,\;$ and $\theta_{i+1}$. The $\theta_j$'s are all fairly straight-forward according to $\theta_i = \alpha \tilde{\theta}_i$ formula:
\begin{align*}
\theta_i&=\alpha\tilde{\theta}_i \\
\theta_{i-1}&=\alpha\tilde{\theta}_{i-1} \\
\theta_{i+1}&=\alpha\tilde{\theta}_{i+1}.
\end{align*}
Now then, we need the second derivative. Whenever I need to do a second derivative change-of-variables like this, I always do it one step at a time - I make fewer mistakes that way. Note that, since $\beta>0$, we have that $\tilde{t}=\dfrac{t}{\beta}$. So get the first derivative first:
$$\frac{d\theta_i}{dt}=\frac{d\theta_i}{d\tilde{t}} \, \frac{d\tilde{t}}{dt}=\frac{d\theta_i}{d\tilde{t}} \, \frac{1}{\beta}.$$
To get the second derivative, we differentiate this equation on both sides with respect to $t$, and get:
$$\frac{d^2\theta_i}{dt^2}=\frac{d}{dt} \, \frac{d\theta_i}{dt}=\frac{d}{dt} \left[ \frac{d\theta_i}{d\tilde{t}} \, \frac{1}{\beta} \right] =\frac{1}{\beta} \left[ \frac{d}{d\tilde{t}} \, \frac{d\theta_i}{d\tilde{t}} \right] \frac{d\tilde{t}}{dt}
=\frac{1}{\beta^2} \, \frac{d^2\theta_i}{d\tilde{t}^2}.$$
So that handles the $t$ to $\tilde{t}$ conversion. To get the $\theta_i$ to $\tilde{\theta}_i$ conversion is very similar, actually:
$$\frac{d\tilde{\theta}_i}{d\tilde{t}}=\frac{d\tilde{\theta}_i}{d\theta_i} \, \frac{d\theta_i}{d\tilde{t}}=\frac{1}{\alpha}\, \frac{d\theta_i}{d\tilde{t}}.$$
Doing the next derivative will get you (you can fill in the details)
$$\frac{d^2\tilde{\theta}_i}{d\tilde{t}^2}=\frac{1}{\alpha^2} \, \frac{d^2\theta_i}{d\tilde{t}^2}.$$

Does that help with the conversion?

Now, at the end, you need to compare what you get with the conversion, which is
$$\frac{1}{\beta^2} \, \frac{d^2 \tilde{\theta}_i}{d\tilde{t}^2}=\frac{k}{I}\left( \tilde{\theta}_{i-1}-2\tilde{\theta}_i+\tilde{\theta}_{i+1} \right),$$
to what you want, which is
$$\frac{d^2 \tilde{\theta}_i}{d\tilde{t}^2}=\tilde{\theta}_{i-1}-2\tilde{\theta}_i+\tilde{\theta}_{i+1}.$$
Can you make those fractions go away? If so, how?