Applying Chain Rule to Dimensionless Transformation











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Hello I am trying to show that the equation



$frac{dN}{dt} = rN(1 - frac{N(t - tau)}{K})$



can be rewritten in a dimensionless form as



$frac{dy}{dx} = lambda y(1 - y(x - 1))$
using the transformations $y = frac{N}{K}$ and $x = frac{t}{tau}$.



So far my attempt at the problem is to assume $frac{dy}{dx} = frac{dy}{dN}*frac{dN}{dt}*frac{dt}{dx}$. Using the given information of $y = frac{N}{K}$ and $x = frac{t}{tau}$ it seems



$frac{dy}{dN} = frac{1}{K}$ and $frac{dx}{dt} = frac{1}{tau} Rightarrow frac{dt}{dx} = tau$



I then proceeded to substitute this information into $frac{dy}{dx} = frac{dy}{dN}*frac{dN}{dt}*frac{dt}{dx}$,



$frac{dy}{dx} = frac{1}{K}*(rN(1 - frac{N(t - tau)}{K}))*tau = rtau(frac{N}{K}(1 - frac{N}{K}(t - tau)) =$



$rtau y(1 - y(t - tau))$



At this point I am unsure if I should try to think of a factor to multiply the last step of my simplification so that I can assume $lambda$ is equal to some factor and then match it with what $frac{dy}{dx}$ is supposed to be or if I made a mistake in how I approached the problem so far and it involves a more complicated understanding of how to apply the chain rule. Needless to say in my simplification it should be noted that the inside function of $y = frac{N}{K}$ currently is in the same form of $(t - tau)$ as it is in $frac{dN}{dt}$ but by multiplying by $frac{1}{tau}$ the resultant inside will be of the form $(frac{t}{tau} - frac{tau}{tau}) = (x - 1)$. So my question is what needs to be done to further proceed along this problem? Any help would be much appreciated, thanks.










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  • 1




    $y = N/K$ is a little bit imprecise. More accurate is $y(x) = N(tau x)/K$. Does that help?
    – eyeballfrog
    yesterday










  • @eyeballfrog So should I just distribute $tau$ in the inside function while letting $lambda = r$ to get the desired form or do I need to reassess what $frac{dy}{dN}$ should be using the more precise form of $y(x) = N(tau x)/K$?
    – Jmath99
    yesterday















up vote
0
down vote

favorite












Hello I am trying to show that the equation



$frac{dN}{dt} = rN(1 - frac{N(t - tau)}{K})$



can be rewritten in a dimensionless form as



$frac{dy}{dx} = lambda y(1 - y(x - 1))$
using the transformations $y = frac{N}{K}$ and $x = frac{t}{tau}$.



So far my attempt at the problem is to assume $frac{dy}{dx} = frac{dy}{dN}*frac{dN}{dt}*frac{dt}{dx}$. Using the given information of $y = frac{N}{K}$ and $x = frac{t}{tau}$ it seems



$frac{dy}{dN} = frac{1}{K}$ and $frac{dx}{dt} = frac{1}{tau} Rightarrow frac{dt}{dx} = tau$



I then proceeded to substitute this information into $frac{dy}{dx} = frac{dy}{dN}*frac{dN}{dt}*frac{dt}{dx}$,



$frac{dy}{dx} = frac{1}{K}*(rN(1 - frac{N(t - tau)}{K}))*tau = rtau(frac{N}{K}(1 - frac{N}{K}(t - tau)) =$



$rtau y(1 - y(t - tau))$



At this point I am unsure if I should try to think of a factor to multiply the last step of my simplification so that I can assume $lambda$ is equal to some factor and then match it with what $frac{dy}{dx}$ is supposed to be or if I made a mistake in how I approached the problem so far and it involves a more complicated understanding of how to apply the chain rule. Needless to say in my simplification it should be noted that the inside function of $y = frac{N}{K}$ currently is in the same form of $(t - tau)$ as it is in $frac{dN}{dt}$ but by multiplying by $frac{1}{tau}$ the resultant inside will be of the form $(frac{t}{tau} - frac{tau}{tau}) = (x - 1)$. So my question is what needs to be done to further proceed along this problem? Any help would be much appreciated, thanks.










share|cite|improve this question


















  • 1




    $y = N/K$ is a little bit imprecise. More accurate is $y(x) = N(tau x)/K$. Does that help?
    – eyeballfrog
    yesterday










  • @eyeballfrog So should I just distribute $tau$ in the inside function while letting $lambda = r$ to get the desired form or do I need to reassess what $frac{dy}{dN}$ should be using the more precise form of $y(x) = N(tau x)/K$?
    – Jmath99
    yesterday













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Hello I am trying to show that the equation



$frac{dN}{dt} = rN(1 - frac{N(t - tau)}{K})$



can be rewritten in a dimensionless form as



$frac{dy}{dx} = lambda y(1 - y(x - 1))$
using the transformations $y = frac{N}{K}$ and $x = frac{t}{tau}$.



So far my attempt at the problem is to assume $frac{dy}{dx} = frac{dy}{dN}*frac{dN}{dt}*frac{dt}{dx}$. Using the given information of $y = frac{N}{K}$ and $x = frac{t}{tau}$ it seems



$frac{dy}{dN} = frac{1}{K}$ and $frac{dx}{dt} = frac{1}{tau} Rightarrow frac{dt}{dx} = tau$



I then proceeded to substitute this information into $frac{dy}{dx} = frac{dy}{dN}*frac{dN}{dt}*frac{dt}{dx}$,



$frac{dy}{dx} = frac{1}{K}*(rN(1 - frac{N(t - tau)}{K}))*tau = rtau(frac{N}{K}(1 - frac{N}{K}(t - tau)) =$



$rtau y(1 - y(t - tau))$



At this point I am unsure if I should try to think of a factor to multiply the last step of my simplification so that I can assume $lambda$ is equal to some factor and then match it with what $frac{dy}{dx}$ is supposed to be or if I made a mistake in how I approached the problem so far and it involves a more complicated understanding of how to apply the chain rule. Needless to say in my simplification it should be noted that the inside function of $y = frac{N}{K}$ currently is in the same form of $(t - tau)$ as it is in $frac{dN}{dt}$ but by multiplying by $frac{1}{tau}$ the resultant inside will be of the form $(frac{t}{tau} - frac{tau}{tau}) = (x - 1)$. So my question is what needs to be done to further proceed along this problem? Any help would be much appreciated, thanks.










share|cite|improve this question













Hello I am trying to show that the equation



$frac{dN}{dt} = rN(1 - frac{N(t - tau)}{K})$



can be rewritten in a dimensionless form as



$frac{dy}{dx} = lambda y(1 - y(x - 1))$
using the transformations $y = frac{N}{K}$ and $x = frac{t}{tau}$.



So far my attempt at the problem is to assume $frac{dy}{dx} = frac{dy}{dN}*frac{dN}{dt}*frac{dt}{dx}$. Using the given information of $y = frac{N}{K}$ and $x = frac{t}{tau}$ it seems



$frac{dy}{dN} = frac{1}{K}$ and $frac{dx}{dt} = frac{1}{tau} Rightarrow frac{dt}{dx} = tau$



I then proceeded to substitute this information into $frac{dy}{dx} = frac{dy}{dN}*frac{dN}{dt}*frac{dt}{dx}$,



$frac{dy}{dx} = frac{1}{K}*(rN(1 - frac{N(t - tau)}{K}))*tau = rtau(frac{N}{K}(1 - frac{N}{K}(t - tau)) =$



$rtau y(1 - y(t - tau))$



At this point I am unsure if I should try to think of a factor to multiply the last step of my simplification so that I can assume $lambda$ is equal to some factor and then match it with what $frac{dy}{dx}$ is supposed to be or if I made a mistake in how I approached the problem so far and it involves a more complicated understanding of how to apply the chain rule. Needless to say in my simplification it should be noted that the inside function of $y = frac{N}{K}$ currently is in the same form of $(t - tau)$ as it is in $frac{dN}{dt}$ but by multiplying by $frac{1}{tau}$ the resultant inside will be of the form $(frac{t}{tau} - frac{tau}{tau}) = (x - 1)$. So my question is what needs to be done to further proceed along this problem? Any help would be much appreciated, thanks.







derivatives transformation chain-rule






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Jmath99

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  • 1




    $y = N/K$ is a little bit imprecise. More accurate is $y(x) = N(tau x)/K$. Does that help?
    – eyeballfrog
    yesterday










  • @eyeballfrog So should I just distribute $tau$ in the inside function while letting $lambda = r$ to get the desired form or do I need to reassess what $frac{dy}{dN}$ should be using the more precise form of $y(x) = N(tau x)/K$?
    – Jmath99
    yesterday














  • 1




    $y = N/K$ is a little bit imprecise. More accurate is $y(x) = N(tau x)/K$. Does that help?
    – eyeballfrog
    yesterday










  • @eyeballfrog So should I just distribute $tau$ in the inside function while letting $lambda = r$ to get the desired form or do I need to reassess what $frac{dy}{dN}$ should be using the more precise form of $y(x) = N(tau x)/K$?
    – Jmath99
    yesterday








1




1




$y = N/K$ is a little bit imprecise. More accurate is $y(x) = N(tau x)/K$. Does that help?
– eyeballfrog
yesterday




$y = N/K$ is a little bit imprecise. More accurate is $y(x) = N(tau x)/K$. Does that help?
– eyeballfrog
yesterday












@eyeballfrog So should I just distribute $tau$ in the inside function while letting $lambda = r$ to get the desired form or do I need to reassess what $frac{dy}{dN}$ should be using the more precise form of $y(x) = N(tau x)/K$?
– Jmath99
yesterday




@eyeballfrog So should I just distribute $tau$ in the inside function while letting $lambda = r$ to get the desired form or do I need to reassess what $frac{dy}{dN}$ should be using the more precise form of $y(x) = N(tau x)/K$?
– Jmath99
yesterday















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