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LouiseLøcke
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I don't have any experince in normalizing when the equations depend on each other, so was hoping someone could tell me if what I have done is correct, and if not what the problem is :)
I need to non-dimensional the following differential equations.
L1:= dn/dt = (a-b-k*y)*n
L2:= dy/dt = l*n-d*y
to
dN/dT= (D-Y)*N
dY/dT = N-Y
knowing that D = (a-b)/d.
I start with inserting
n = k[n]*N,
y = k[y]*Y,
t = k[t]*T
L1:= d(k[n]*N))/d(k[t]*T) = (a -b-k*(k[y]*Y)*k[n]*N
L2:= d(k[y]*Y)/d(k[t]*T)= l*(k[n]*N) - d*k[y]*Y
I look at L2 first, and divide with k[y]/k[t] on both sides.
Then I set l= k[y]/(k[n]*k[t]) and d=1/k[t]
getting dY/dT = N-Y
next I look at L1, and divide with k[n]/k[t] on both sides and inserting k[t] = 1/delta ( which I found above)
getting
dN/dT = ((a-b)/(d) - (k*(k[y]*Y))/(d))*N
I now put D = (a-b)/d and k = d/k[y]
getting
dN/dT= (D-Y)*N (which is what I wanted)
Is this the correct way of approaching it?
Thanks Louise
Homework Statement
I need to non-dimensional the following differential equations.
L1:= dn/dt = (a-b-k*y)*n
L2:= dy/dt = l*n-d*y
to
dN/dT= (D-Y)*N
dY/dT = N-Y
knowing that D = (a-b)/d.
The Attempt at a Solution
I start with inserting
n = k[n]*N,
y = k[y]*Y,
t = k[t]*T
L1:= d(k[n]*N))/d(k[t]*T) = (a -b-k*(k[y]*Y)*k[n]*N
L2:= d(k[y]*Y)/d(k[t]*T)= l*(k[n]*N) - d*k[y]*Y
I look at L2 first, and divide with k[y]/k[t] on both sides.
Then I set l= k[y]/(k[n]*k[t]) and d=1/k[t]
getting dY/dT = N-Y
next I look at L1, and divide with k[n]/k[t] on both sides and inserting k[t] = 1/delta ( which I found above)
getting
dN/dT = ((a-b)/(d) - (k*(k[y]*Y))/(d))*N
I now put D = (a-b)/d and k = d/k[y]
getting
dN/dT= (D-Y)*N (which is what I wanted)
Is this the correct way of approaching it?
Thanks Louise
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