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I have the equation: $$m*x''(t)=a*x(t)+b*x'(t)$$ And the starting values:

$$x(t_0)=x_0 \text{ and } x'(t_0)=v_0$$

I nondemensionalize this with: $$\tau=\frac{t}{\bar{t}} \text{ and } y(\tau)=\frac{x(t)}{\bar{x}}$$

For the first starting value i use $$y(0)=\frac{x_0}{\bar{x}}$$, but what do i have to do with the secound starting value? Can i just introduce $\bar{v}$ and use $$y'(0)=\frac{v_0}{\bar{v}}$$ for my nondimensionalized problem?

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1 Answers 1

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Obviously, you have $$ y(τ)=\frac{x(\bar t τ)}{\bar x} $$ so that in the derivative $$ y'(τ)=\frac{\bar t}{\bar x}x'(\bar t τ) $$ and at the initial value $$ y'(0)=\frac{\bar t}{\bar x}v_0 $$ or if you want, $\bar v=\frac{\bar x}{\bar t}$.