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Time between two rates in a ODE system
Clash Royale CLAN TAG#URR8PPP
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I am reading the book on virus dynamics by Novak and May, and it describes a system of ODEs to describe the viral load when antivirals are used. The systems is
$dot y = -ay\dot v = ky-uv$
The solutions of this system are
$y(t)= y^*e^-at\v(t) = v^*(ue^-at-ae^-ut)over (u-a)$.
it is assumed that $u>>a$.
Then it is analysed that the virus load ($v(t)$) start to decline at rate $e^-at$ only after a certain shoulder phase of duration given as $Delta t approx 1/u$ (more precisely, $Delta t= (-1/a) ln (1-a/u) $ ).
Can someone please explain how this $Delta t$ is obtained?
differential-equations dynamical-systems mathematical-modeling
asked Jul 27 at 11:33
clarkson
80411428
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up vote
0
down vote
favorite
I am reading the book on virus dynamics by Novak and May, and it describes a system of ODEs to describe the viral load when antivirals are used. The systems is
$dot y = -ay\dot v = ky-uv$
The solutions of this system are
$y(t)= y^*e^-at\v(t) = v^*(ue^-at-ae^-ut)over (u-a)$.
it is assumed that $u>>a$.
Then it is analysed that the virus load ($v(t)$) start to decline at rate $e^-at$ only after a certain shoulder phase of duration given as $Delta t approx 1/u$ (more precisely, $Delta t= (-1/a) ln (1-a/u) $ ).
Can someone please explain how this $Delta t$ is obtained?
differential-equations dynamical-systems mathematical-modeling
asked Jul 27 at 11:33
clarkson
80411428
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am reading the book on virus dynamics by Novak and May, and it describes a system of ODEs to describe the viral load when antivirals are used. The systems is
$dot y = -ay\dot v = ky-uv$
The solutions of this system are
$y(t)= y^*e^-at\v(t) = v^*(ue^-at-ae^-ut)over (u-a)$.
it is assumed that $u>>a$.
Then it is analysed that the virus load ($v(t)$) start to decline at rate $e^-at$ only after a certain shoulder phase of duration given as $Delta t approx 1/u$ (more precisely, $Delta t= (-1/a) ln (1-a/u) $ ).
Can someone please explain how this $Delta t$ is obtained?
differential-equations dynamical-systems mathematical-modeling
asked Jul 27 at 11:33
clarkson
80411428
I am reading the book on virus dynamics by Novak and May, and it describes a system of ODEs to describe the viral load when antivirals are used. The systems is
$dot y = -ay\dot v = ky-uv$
The solutions of this system are
$y(t)= y^*e^-at\v(t) = v^*(ue^-at-ae^-ut)over (u-a)$.
it is assumed that $u>>a$.
Then it is analysed that the virus load ($v(t)$) start to decline at rate $e^-at$ only after a certain shoulder phase of duration given as $Delta t approx 1/u$ (more precisely, $Delta t= (-1/a) ln (1-a/u) $ ).
Can someone please explain how this $Delta t$ is obtained?
differential-equations dynamical-systems mathematical-modeling
asked Jul 27 at 11:33
clarkson
80411428
asked Jul 27 at 11:33
clarkson
80411428
asked Jul 27 at 11:33
clarkson
80411428
asked Jul 27 at 11:33
clarkson
80411428
80411428
add a comment |Â
add a comment |Â
1 Answer
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To see it, first note that since $ugg a>0$ (assuming they are positive constants), the term $ae^-ut$ will go to $0$ more quickly than $ue^-at$. That is why $v(t)$ can be approximated by $ue^-at/(u-a)v_0$ after a certain time. To compute that time, see that on such a logarithmic scale as in your figure, $ue^-at/(u-a)v_0$ and $e^-atv_0$ correspond to parallel lines. Now since we start at $v_0$, i.e. at $e^-atv_0|_t=0$ and converge to $ue^-at/(u-a)v_0$ then the question can be rephrased to finding the translation from line to line, i.e. solving
beginalign
e^-a(t-Delta t)v_0 &= fracuu-ae^-atv_0\
aDelta t &=mathrmln(u)-mathrmln(u-a)\
Delta t &= -frac1abig(mathrmln(u-a)-mathrmln(u)big)\
&= -frac1amathrmln((u-a)/u)\
&= -frac1amathrmln(1-a/u).
endalign
answered Jul 27 at 12:14
WalterJ
793611
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
To see it, first note that since $ugg a>0$ (assuming they are positive constants), the term $ae^-ut$ will go to $0$ more quickly than $ue^-at$. That is why $v(t)$ can be approximated by $ue^-at/(u-a)v_0$ after a certain time. To compute that time, see that on such a logarithmic scale as in your figure, $ue^-at/(u-a)v_0$ and $e^-atv_0$ correspond to parallel lines. Now since we start at $v_0$, i.e. at $e^-atv_0|_t=0$ and converge to $ue^-at/(u-a)v_0$ then the question can be rephrased to finding the translation from line to line, i.e. solving
beginalign
e^-a(t-Delta t)v_0 &= fracuu-ae^-atv_0\
aDelta t &=mathrmln(u)-mathrmln(u-a)\
Delta t &= -frac1abig(mathrmln(u-a)-mathrmln(u)big)\
&= -frac1amathrmln((u-a)/u)\
&= -frac1amathrmln(1-a/u).
endalign
answered Jul 27 at 12:14
WalterJ
793611
add a comment |Â
up vote
1
down vote
accepted
To see it, first note that since $ugg a>0$ (assuming they are positive constants), the term $ae^-ut$ will go to $0$ more quickly than $ue^-at$. That is why $v(t)$ can be approximated by $ue^-at/(u-a)v_0$ after a certain time. To compute that time, see that on such a logarithmic scale as in your figure, $ue^-at/(u-a)v_0$ and $e^-atv_0$ correspond to parallel lines. Now since we start at $v_0$, i.e. at $e^-atv_0|_t=0$ and converge to $ue^-at/(u-a)v_0$ then the question can be rephrased to finding the translation from line to line, i.e. solving
beginalign
e^-a(t-Delta t)v_0 &= fracuu-ae^-atv_0\
aDelta t &=mathrmln(u)-mathrmln(u-a)\
Delta t &= -frac1abig(mathrmln(u-a)-mathrmln(u)big)\
&= -frac1amathrmln((u-a)/u)\
&= -frac1amathrmln(1-a/u).
endalign
answered Jul 27 at 12:14
WalterJ
793611
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To see it, first note that since $ugg a>0$ (assuming they are positive constants), the term $ae^-ut$ will go to $0$ more quickly than $ue^-at$. That is why $v(t)$ can be approximated by $ue^-at/(u-a)v_0$ after a certain time. To compute that time, see that on such a logarithmic scale as in your figure, $ue^-at/(u-a)v_0$ and $e^-atv_0$ correspond to parallel lines. Now since we start at $v_0$, i.e. at $e^-atv_0|_t=0$ and converge to $ue^-at/(u-a)v_0$ then the question can be rephrased to finding the translation from line to line, i.e. solving
beginalign
e^-a(t-Delta t)v_0 &= fracuu-ae^-atv_0\
aDelta t &=mathrmln(u)-mathrmln(u-a)\
Delta t &= -frac1abig(mathrmln(u-a)-mathrmln(u)big)\
&= -frac1amathrmln((u-a)/u)\
&= -frac1amathrmln(1-a/u).
endalign
answered Jul 27 at 12:14
WalterJ
793611
To see it, first note that since $ugg a>0$ (assuming they are positive constants), the term $ae^-ut$ will go to $0$ more quickly than $ue^-at$. That is why $v(t)$ can be approximated by $ue^-at/(u-a)v_0$ after a certain time. To compute that time, see that on such a logarithmic scale as in your figure, $ue^-at/(u-a)v_0$ and $e^-atv_0$ correspond to parallel lines. Now since we start at $v_0$, i.e. at $e^-atv_0|_t=0$ and converge to $ue^-at/(u-a)v_0$ then the question can be rephrased to finding the translation from line to line, i.e. solving
beginalign
e^-a(t-Delta t)v_0 &= fracuu-ae^-atv_0\
aDelta t &=mathrmln(u)-mathrmln(u-a)\
Delta t &= -frac1abig(mathrmln(u-a)-mathrmln(u)big)\
&= -frac1amathrmln((u-a)/u)\
&= -frac1amathrmln(1-a/u).
endalign
answered Jul 27 at 12:14
WalterJ
793611
edited Jul 27 at 14:05
answered Jul 27 at 12:14
WalterJ
793611
answered Jul 27 at 12:14
WalterJ
793611
answered Jul 27 at 12:14
WalterJ
793611
793611
add a comment |Â
add a comment |Â
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