Justification for integral bound
$begingroup$
I'm currently studying SIR models in mathematical epidemiology, and I'm trying to understand a proof that a certain differential equation on the model tends to $0$ as $t$ goes to infinity. Here's the standard model:
$$left{
begin{array}{lr}
frac{dS}{dt}=-beta SI \
frac{dI}{dt}=beta SI - alpha I \
frac{dR}{dt}=alpha I
end{array}
right.
$$
What I'm trying to justify is why $lim_{ttoinfty}I(t)=0$. Martcheva's book does this as follows by integrating $frac{dS}{dt}$:
$$int_0^inftyfrac{dS}{dt},dt=int_0^infty-beta S(t)I(t),dt$$
$$S_infty-S_0=-betaint_0^infty S(t)I(t),dt$$
$$S_0-S_infty=betaint_0^infty S(t)I(t),dt$$
then since $S$ is strictly decreasing (because $frac{dS}{dt}leq0$ by definition of the model), we have that
$$S_0-S_inftygeqbeta S_infty int_0^infty I(t),dt$$
and so $I(t)$ is integrable in $[0,infty)$, therefore $lim_{ttoinfty} I(t)=0$. Now this last step seems "reasonable" but it isn't proved. My guess is that this is some Lebesgue integrability theorem that I don't know about (my fault for skipping analysis classes...), can anyone point me towards a reference or even come up with a short proof?
real-analysis ordinary-differential-equations
asked Dec 3 '18 at 10:22
AstlyDichrarAstlyDichrar
40528
$endgroup$
add a comment |
$begingroup$
I'm currently studying SIR models in mathematical epidemiology, and I'm trying to understand a proof that a certain differential equation on the model tends to $0$ as $t$ goes to infinity. Here's the standard model:
$$left{
begin{array}{lr}
frac{dS}{dt}=-beta SI \
frac{dI}{dt}=beta SI - alpha I \
frac{dR}{dt}=alpha I
end{array}
right.
$$
What I'm trying to justify is why $lim_{ttoinfty}I(t)=0$. Martcheva's book does this as follows by integrating $frac{dS}{dt}$:
$$int_0^inftyfrac{dS}{dt},dt=int_0^infty-beta S(t)I(t),dt$$
$$S_infty-S_0=-betaint_0^infty S(t)I(t),dt$$
$$S_0-S_infty=betaint_0^infty S(t)I(t),dt$$
then since $S$ is strictly decreasing (because $frac{dS}{dt}leq0$ by definition of the model), we have that
$$S_0-S_inftygeqbeta S_infty int_0^infty I(t),dt$$
and so $I(t)$ is integrable in $[0,infty)$, therefore $lim_{ttoinfty} I(t)=0$. Now this last step seems "reasonable" but it isn't proved. My guess is that this is some Lebesgue integrability theorem that I don't know about (my fault for skipping analysis classes...), can anyone point me towards a reference or even come up with a short proof?
real-analysis ordinary-differential-equations
asked Dec 3 '18 at 10:22
AstlyDichrarAstlyDichrar
40528
$endgroup$
$begingroup$
After some manipulation we obtain that $I$ is monotone, so it has a limit. As $I$ is positive, the limit is either $0$, or a positive real, or else $infty$ (again impossible). If the limit is a positive real then the integral is $infty$, which is impossible.
$endgroup$
– user539887
Dec 3 '18 at 13:07
add a comment |
$begingroup$
I'm currently studying SIR models in mathematical epidemiology, and I'm trying to understand a proof that a certain differential equation on the model tends to $0$ as $t$ goes to infinity. Here's the standard model:
$$left{
begin{array}{lr}
frac{dS}{dt}=-beta SI \
frac{dI}{dt}=beta SI - alpha I \
frac{dR}{dt}=alpha I
end{array}
right.
$$
What I'm trying to justify is why $lim_{ttoinfty}I(t)=0$. Martcheva's book does this as follows by integrating $frac{dS}{dt}$:
$$int_0^inftyfrac{dS}{dt},dt=int_0^infty-beta S(t)I(t),dt$$
$$S_infty-S_0=-betaint_0^infty S(t)I(t),dt$$
$$S_0-S_infty=betaint_0^infty S(t)I(t),dt$$
then since $S$ is strictly decreasing (because $frac{dS}{dt}leq0$ by definition of the model), we have that
$$S_0-S_inftygeqbeta S_infty int_0^infty I(t),dt$$
and so $I(t)$ is integrable in $[0,infty)$, therefore $lim_{ttoinfty} I(t)=0$. Now this last step seems "reasonable" but it isn't proved. My guess is that this is some Lebesgue integrability theorem that I don't know about (my fault for skipping analysis classes...), can anyone point me towards a reference or even come up with a short proof?
real-analysis ordinary-differential-equations
asked Dec 3 '18 at 10:22
AstlyDichrarAstlyDichrar
40528
$endgroup$
I'm currently studying SIR models in mathematical epidemiology, and I'm trying to understand a proof that a certain differential equation on the model tends to $0$ as $t$ goes to infinity. Here's the standard model:
$$left{
begin{array}{lr}
frac{dS}{dt}=-beta SI \
frac{dI}{dt}=beta SI - alpha I \
frac{dR}{dt}=alpha I
end{array}
right.
$$
What I'm trying to justify is why $lim_{ttoinfty}I(t)=0$. Martcheva's book does this as follows by integrating $frac{dS}{dt}$:
$$int_0^inftyfrac{dS}{dt},dt=int_0^infty-beta S(t)I(t),dt$$
$$S_infty-S_0=-betaint_0^infty S(t)I(t),dt$$
$$S_0-S_infty=betaint_0^infty S(t)I(t),dt$$
then since $S$ is strictly decreasing (because $frac{dS}{dt}leq0$ by definition of the model), we have that
$$S_0-S_inftygeqbeta S_infty int_0^infty I(t),dt$$
and so $I(t)$ is integrable in $[0,infty)$, therefore $lim_{ttoinfty} I(t)=0$. Now this last step seems "reasonable" but it isn't proved. My guess is that this is some Lebesgue integrability theorem that I don't know about (my fault for skipping analysis classes...), can anyone point me towards a reference or even come up with a short proof?
real-analysis ordinary-differential-equations
real-analysis ordinary-differential-equations
asked Dec 3 '18 at 10:22
AstlyDichrarAstlyDichrar
40528
asked Dec 3 '18 at 10:22
AstlyDichrarAstlyDichrar
40528
asked Dec 3 '18 at 10:22
AstlyDichrarAstlyDichrar
40528
asked Dec 3 '18 at 10:22
AstlyDichrarAstlyDichrar
40528
asked Dec 3 '18 at 10:22
AstlyDichrarAstlyDichrar
40528
40528
$begingroup$
After some manipulation we obtain that $I$ is monotone, so it has a limit. As $I$ is positive, the limit is either $0$, or a positive real, or else $infty$ (again impossible). If the limit is a positive real then the integral is $infty$, which is impossible.
$endgroup$
– user539887
Dec 3 '18 at 13:07
add a comment |
$begingroup$
After some manipulation we obtain that $I$ is monotone, so it has a limit. As $I$ is positive, the limit is either $0$, or a positive real, or else $infty$ (again impossible). If the limit is a positive real then the integral is $infty$, which is impossible.
$endgroup$
– user539887
Dec 3 '18 at 13:07
$begingroup$
After some manipulation we obtain that $I$ is monotone, so it has a limit. As $I$ is positive, the limit is either $0$, or a positive real, or else $infty$ (again impossible). If the limit is a positive real then the integral is $infty$, which is impossible.
$endgroup$
– user539887
Dec 3 '18 at 13:07
$begingroup$
After some manipulation we obtain that $I$ is monotone, so it has a limit. As $I$ is positive, the limit is either $0$, or a positive real, or else $infty$ (again impossible). If the limit is a positive real then the integral is $infty$, which is impossible.
$endgroup$
– user539887
Dec 3 '18 at 13:07
add a comment |
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$begingroup$
After some manipulation we obtain that $I$ is monotone, so it has a limit. As $I$ is positive, the limit is either $0$, or a positive real, or else $infty$ (again impossible). If the limit is a positive real then the integral is $infty$, which is impossible.
$endgroup$
– user539887
Dec 3 '18 at 13:07