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

  • Thread starter Thread starter mt91
  • Start date Start date
mt91
Messages
13
Reaction score
0
1596460878703.png


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
 
Physics news on Phys.org
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 $\frac{h}{r}\frac{dn}{dt}= \frac{n}{1+ n}- \left(\frac{br}{h}\right)nC$.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Back
Top