MHB Non-Dimensionalisation of Two Differential Equations: Working Out and Struggles

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The discussion focuses on the non-dimensionalisation of two differential equations, specifically \(\frac{dc}{dT} = \alpha - \mu c\) and \(\frac{dN}{dt} = \frac{rN}{h + rN} - bNC\). The user successfully non-dimensionalised the first equation using substitutions \(N = N_0 n\), \(C = C_0 c\), and \(t = t_0 T\). However, they encountered difficulties with the left-hand side of the second equation. A proposed solution involves dividing the numerator and denominator of \(\frac{rN}{h + rN}\) by \(h\) and substituting \(n = \frac{rN}{h}\), leading to the transformed equation \(\frac{h}{r}\frac{dn}{dt} = \frac{n}{1 + n} - \left(\frac{br}{h}\right)nC\).

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mt91
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I've got two differential equations I need to non-dimensionalise

I've managed to do the \[ dc/dT=α- μc \] with the following working out:

By letting:
\[ N=N0n \]
\[ C=C0c \]
\[ t=t0T \]

1596461095620.png


However I'm struggling with the first equation.

1596461128387.png


I'm up to here, it's just the left hand side of the equation I'm struggling to work out. Any help would be great, cheers
 
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Looking at $\frac{dN}{dt}= \frac{rN}{h+ rN}- bNC$ and comparing that to the desired $\frac{dr}{dt}= \frac{n}{n+1}- \beta nc$ my first thought would be to divide the numerator and denominator of $\frac{rN}{h+ rN}$ by h. That gives $\frac{\frac{rN}{h}}{1+ \frac{rN}{h}}$. So we should let $n= \frac{rN}{h}$. With that substitution, the first equation becomes $\frac{h}{r}\frac{dn}{dt}= \frac{n}{1+ n}- \left(\frac{br}{h}\right)nC$.
 

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