PDE's: Analysis without solving explicitly the equation
$begingroup$
I have the following PDE'S
$$frac{partial}{partial t}u(t,r)=frac{partial}{partial r}big((r-1/2)u(t,r)big)$$
with $u(0, r)=u_0(r)$ compact supported in the interval $[0,1]$ and such that $int_0^1u_0(r)dr=1$.
I know the solution explicitly, Through the characteristic method we know that
that $$u(t,r)=u_0big((r-1/2)e^t+1/2big)e^t$$.
I am wondering about the $lim_{tto +infty}int_0^1u(t,r)G(r) dr$, with $Gin C([0,1], mathbb R)$ a test function.
Since I know the explicit solution it is easy to understand that
$$lim_{tto +infty}int_0^1u(t,r)G(r) dr=G(1/2)$$.
My question is, could I deduce something about the asymptotic behaviour without knowning the explicit solution but just looking at the expression of the PDE?
real-analysis analysis pde asymptotics
asked Jan 10 at 20:30
user268193user268193
718
$endgroup$
add a comment |
$begingroup$
I have the following PDE'S
$$frac{partial}{partial t}u(t,r)=frac{partial}{partial r}big((r-1/2)u(t,r)big)$$
with $u(0, r)=u_0(r)$ compact supported in the interval $[0,1]$ and such that $int_0^1u_0(r)dr=1$.
I know the solution explicitly, Through the characteristic method we know that
that $$u(t,r)=u_0big((r-1/2)e^t+1/2big)e^t$$.
I am wondering about the $lim_{tto +infty}int_0^1u(t,r)G(r) dr$, with $Gin C([0,1], mathbb R)$ a test function.
Since I know the explicit solution it is easy to understand that
$$lim_{tto +infty}int_0^1u(t,r)G(r) dr=G(1/2)$$.
My question is, could I deduce something about the asymptotic behaviour without knowning the explicit solution but just looking at the expression of the PDE?
real-analysis analysis pde asymptotics
asked Jan 10 at 20:30
user268193user268193
718
$endgroup$
$begingroup$
I think you want $G(1/2) int_{-infty}^infty u_0(s); ds$ rather than $G(1/2)$ on the right of your limit.
$endgroup$
– Robert Israel
Jan 10 at 20:55
$begingroup$
You are right, I have the following hypotesis: $int_0^1u_0(r)dr=1$. I added it in the question. Thank you!
$endgroup$
– user268193
Jan 10 at 21:20
2
$begingroup$
One way to understand the result physically is to write it as $dot{rho} + nabla(rho v) = 0$ where the mass density $rho = u$ and the velocity $v=1/2-r$. This is the well-known continuity equation for a fluid describing conservation of mass: $M = int rho,{rm d}r = 1$ is conserved in time. If you look at the velocity field, this is such that matter flows towards $r=1/2$ from both sides. As time goes on we therefore expect all the mass to pile up at this point ($uto Mdelta(r-1/2)$). The integral is the mass-weighted average value of $G$ which will therefore just be $G$ at $1/2$.
$endgroup$
– Winther
Jan 10 at 22:22
add a comment |
$begingroup$
I have the following PDE'S
$$frac{partial}{partial t}u(t,r)=frac{partial}{partial r}big((r-1/2)u(t,r)big)$$
with $u(0, r)=u_0(r)$ compact supported in the interval $[0,1]$ and such that $int_0^1u_0(r)dr=1$.
I know the solution explicitly, Through the characteristic method we know that
that $$u(t,r)=u_0big((r-1/2)e^t+1/2big)e^t$$.
I am wondering about the $lim_{tto +infty}int_0^1u(t,r)G(r) dr$, with $Gin C([0,1], mathbb R)$ a test function.
Since I know the explicit solution it is easy to understand that
$$lim_{tto +infty}int_0^1u(t,r)G(r) dr=G(1/2)$$.
My question is, could I deduce something about the asymptotic behaviour without knowning the explicit solution but just looking at the expression of the PDE?
real-analysis analysis pde asymptotics
asked Jan 10 at 20:30
user268193user268193
718
$endgroup$
I have the following PDE'S
$$frac{partial}{partial t}u(t,r)=frac{partial}{partial r}big((r-1/2)u(t,r)big)$$
with $u(0, r)=u_0(r)$ compact supported in the interval $[0,1]$ and such that $int_0^1u_0(r)dr=1$.
I know the solution explicitly, Through the characteristic method we know that
that $$u(t,r)=u_0big((r-1/2)e^t+1/2big)e^t$$.
I am wondering about the $lim_{tto +infty}int_0^1u(t,r)G(r) dr$, with $Gin C([0,1], mathbb R)$ a test function.
Since I know the explicit solution it is easy to understand that
$$lim_{tto +infty}int_0^1u(t,r)G(r) dr=G(1/2)$$.
My question is, could I deduce something about the asymptotic behaviour without knowning the explicit solution but just looking at the expression of the PDE?
real-analysis analysis pde asymptotics
real-analysis analysis pde asymptotics
asked Jan 10 at 20:30
user268193user268193
718
asked Jan 10 at 20:30
user268193user268193
718
edited Jan 10 at 22:25
user268193
asked Jan 10 at 20:30
user268193user268193
718
asked Jan 10 at 20:30
user268193user268193
718
asked Jan 10 at 20:30
user268193user268193
718
718
$begingroup$
I think you want $G(1/2) int_{-infty}^infty u_0(s); ds$ rather than $G(1/2)$ on the right of your limit.
$endgroup$
– Robert Israel
Jan 10 at 20:55
$begingroup$
You are right, I have the following hypotesis: $int_0^1u_0(r)dr=1$. I added it in the question. Thank you!
$endgroup$
– user268193
Jan 10 at 21:20
2
$begingroup$
One way to understand the result physically is to write it as $dot{rho} + nabla(rho v) = 0$ where the mass density $rho = u$ and the velocity $v=1/2-r$. This is the well-known continuity equation for a fluid describing conservation of mass: $M = int rho,{rm d}r = 1$ is conserved in time. If you look at the velocity field, this is such that matter flows towards $r=1/2$ from both sides. As time goes on we therefore expect all the mass to pile up at this point ($uto Mdelta(r-1/2)$). The integral is the mass-weighted average value of $G$ which will therefore just be $G$ at $1/2$.
$endgroup$
– Winther
Jan 10 at 22:22
add a comment |
$begingroup$
I think you want $G(1/2) int_{-infty}^infty u_0(s); ds$ rather than $G(1/2)$ on the right of your limit.
$endgroup$
– Robert Israel
Jan 10 at 20:55
$begingroup$
You are right, I have the following hypotesis: $int_0^1u_0(r)dr=1$. I added it in the question. Thank you!
$endgroup$
– user268193
Jan 10 at 21:20
2
$begingroup$
One way to understand the result physically is to write it as $dot{rho} + nabla(rho v) = 0$ where the mass density $rho = u$ and the velocity $v=1/2-r$. This is the well-known continuity equation for a fluid describing conservation of mass: $M = int rho,{rm d}r = 1$ is conserved in time. If you look at the velocity field, this is such that matter flows towards $r=1/2$ from both sides. As time goes on we therefore expect all the mass to pile up at this point ($uto Mdelta(r-1/2)$). The integral is the mass-weighted average value of $G$ which will therefore just be $G$ at $1/2$.
$endgroup$
– Winther
Jan 10 at 22:22
$begingroup$
I think you want $G(1/2) int_{-infty}^infty u_0(s); ds$ rather than $G(1/2)$ on the right of your limit.
$endgroup$
– Robert Israel
Jan 10 at 20:55
$begingroup$
I think you want $G(1/2) int_{-infty}^infty u_0(s); ds$ rather than $G(1/2)$ on the right of your limit.
$endgroup$
– Robert Israel
Jan 10 at 20:55
$begingroup$
You are right, I have the following hypotesis: $int_0^1u_0(r)dr=1$. I added it in the question. Thank you!
$endgroup$
– user268193
Jan 10 at 21:20
$begingroup$
You are right, I have the following hypotesis: $int_0^1u_0(r)dr=1$. I added it in the question. Thank you!
$endgroup$
– user268193
Jan 10 at 21:20
2
2
$begingroup$
One way to understand the result physically is to write it as $dot{rho} + nabla(rho v) = 0$ where the mass density $rho = u$ and the velocity $v=1/2-r$. This is the well-known continuity equation for a fluid describing conservation of mass: $M = int rho,{rm d}r = 1$ is conserved in time. If you look at the velocity field, this is such that matter flows towards $r=1/2$ from both sides. As time goes on we therefore expect all the mass to pile up at this point ($uto Mdelta(r-1/2)$). The integral is the mass-weighted average value of $G$ which will therefore just be $G$ at $1/2$.
$endgroup$
– Winther
Jan 10 at 22:22
$begingroup$
One way to understand the result physically is to write it as $dot{rho} + nabla(rho v) = 0$ where the mass density $rho = u$ and the velocity $v=1/2-r$. This is the well-known continuity equation for a fluid describing conservation of mass: $M = int rho,{rm d}r = 1$ is conserved in time. If you look at the velocity field, this is such that matter flows towards $r=1/2$ from both sides. As time goes on we therefore expect all the mass to pile up at this point ($uto Mdelta(r-1/2)$). The integral is the mass-weighted average value of $G$ which will therefore just be $G$ at $1/2$.
$endgroup$
– Winther
Jan 10 at 22:22
add a comment |
1 Answer
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$begingroup$
This is really closely related to the method of characteristics, but doesn't explicitly use the solution.
Consider $$J = J(a(t), b(t), t) = int_{a(t)}^{b(t)} u(t,r); dr $$
where $a(t)$ and $b(t)$ are continuously differentiable functions. By the Fundamental Theorem of Calculus we have
$$ eqalign{dfrac{dJ}{dt} &= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + int_{a(t)}^{b(t)} dfrac{partial u}{partial t}(t,r) ; dr cr
&= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + int_{a(t)}^{b(t)} dfrac{partial}{partial t} left((r-1/2) u(t,r)right); dr cr
&= (b'(t) + b(t) - 1/2) u(t, b(t)) - (a'(t) + a(t) - 1/2) u(t, a(t))cr} $$
In particular, this is $0$ if $b(t)$ and $a(t)$ are solutions of the differential equation $x' + x - 1/2 = 0$. Those solutions all converge to $1/2$ as $t to infty$. Thus if $[a(0), b(0)]$ contains the support of $u_0$, for large $t$
the support of $u(t,cdot)$ is contained in the small interval $[a(t), b(t)]$ near
$1/2$, and $int_{a(t)}^{b(t)} u(t,r); dr = 1$. Take $epsilon > 0$. If the interval is small enough that $|G(r) - G(1/2)|< epsilon$,
we have
$$left|int_0^1 u(t,r) G(r); dr - G(1/2)right| = left|int_{a(t)}^{b(t)} u(t,r) (G(r) - G(1/2)); dr right| < epsilon int_{a(0)}^{b(0)} |u(t,r)|; dr $$
Thus we get $$ int_0^1 u(t,r) G(r); dr to G(1/2) text{as} t to infty$$
answered Jan 10 at 22:32
Robert IsraelRobert Israel
331k23220475
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add a comment |
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$begingroup$
This is really closely related to the method of characteristics, but doesn't explicitly use the solution.
Consider $$J = J(a(t), b(t), t) = int_{a(t)}^{b(t)} u(t,r); dr $$
where $a(t)$ and $b(t)$ are continuously differentiable functions. By the Fundamental Theorem of Calculus we have
$$ eqalign{dfrac{dJ}{dt} &= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + int_{a(t)}^{b(t)} dfrac{partial u}{partial t}(t,r) ; dr cr
&= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + int_{a(t)}^{b(t)} dfrac{partial}{partial t} left((r-1/2) u(t,r)right); dr cr
&= (b'(t) + b(t) - 1/2) u(t, b(t)) - (a'(t) + a(t) - 1/2) u(t, a(t))cr} $$
In particular, this is $0$ if $b(t)$ and $a(t)$ are solutions of the differential equation $x' + x - 1/2 = 0$. Those solutions all converge to $1/2$ as $t to infty$. Thus if $[a(0), b(0)]$ contains the support of $u_0$, for large $t$
the support of $u(t,cdot)$ is contained in the small interval $[a(t), b(t)]$ near
$1/2$, and $int_{a(t)}^{b(t)} u(t,r); dr = 1$. Take $epsilon > 0$. If the interval is small enough that $|G(r) - G(1/2)|< epsilon$,
we have
$$left|int_0^1 u(t,r) G(r); dr - G(1/2)right| = left|int_{a(t)}^{b(t)} u(t,r) (G(r) - G(1/2)); dr right| < epsilon int_{a(0)}^{b(0)} |u(t,r)|; dr $$
Thus we get $$ int_0^1 u(t,r) G(r); dr to G(1/2) text{as} t to infty$$
answered Jan 10 at 22:32
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
This is really closely related to the method of characteristics, but doesn't explicitly use the solution.
Consider $$J = J(a(t), b(t), t) = int_{a(t)}^{b(t)} u(t,r); dr $$
where $a(t)$ and $b(t)$ are continuously differentiable functions. By the Fundamental Theorem of Calculus we have
$$ eqalign{dfrac{dJ}{dt} &= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + int_{a(t)}^{b(t)} dfrac{partial u}{partial t}(t,r) ; dr cr
&= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + int_{a(t)}^{b(t)} dfrac{partial}{partial t} left((r-1/2) u(t,r)right); dr cr
&= (b'(t) + b(t) - 1/2) u(t, b(t)) - (a'(t) + a(t) - 1/2) u(t, a(t))cr} $$
In particular, this is $0$ if $b(t)$ and $a(t)$ are solutions of the differential equation $x' + x - 1/2 = 0$. Those solutions all converge to $1/2$ as $t to infty$. Thus if $[a(0), b(0)]$ contains the support of $u_0$, for large $t$
the support of $u(t,cdot)$ is contained in the small interval $[a(t), b(t)]$ near
$1/2$, and $int_{a(t)}^{b(t)} u(t,r); dr = 1$. Take $epsilon > 0$. If the interval is small enough that $|G(r) - G(1/2)|< epsilon$,
we have
$$left|int_0^1 u(t,r) G(r); dr - G(1/2)right| = left|int_{a(t)}^{b(t)} u(t,r) (G(r) - G(1/2)); dr right| < epsilon int_{a(0)}^{b(0)} |u(t,r)|; dr $$
Thus we get $$ int_0^1 u(t,r) G(r); dr to G(1/2) text{as} t to infty$$
answered Jan 10 at 22:32
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
This is really closely related to the method of characteristics, but doesn't explicitly use the solution.
Consider $$J = J(a(t), b(t), t) = int_{a(t)}^{b(t)} u(t,r); dr $$
where $a(t)$ and $b(t)$ are continuously differentiable functions. By the Fundamental Theorem of Calculus we have
$$ eqalign{dfrac{dJ}{dt} &= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + int_{a(t)}^{b(t)} dfrac{partial u}{partial t}(t,r) ; dr cr
&= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + int_{a(t)}^{b(t)} dfrac{partial}{partial t} left((r-1/2) u(t,r)right); dr cr
&= (b'(t) + b(t) - 1/2) u(t, b(t)) - (a'(t) + a(t) - 1/2) u(t, a(t))cr} $$
In particular, this is $0$ if $b(t)$ and $a(t)$ are solutions of the differential equation $x' + x - 1/2 = 0$. Those solutions all converge to $1/2$ as $t to infty$. Thus if $[a(0), b(0)]$ contains the support of $u_0$, for large $t$
the support of $u(t,cdot)$ is contained in the small interval $[a(t), b(t)]$ near
$1/2$, and $int_{a(t)}^{b(t)} u(t,r); dr = 1$. Take $epsilon > 0$. If the interval is small enough that $|G(r) - G(1/2)|< epsilon$,
we have
$$left|int_0^1 u(t,r) G(r); dr - G(1/2)right| = left|int_{a(t)}^{b(t)} u(t,r) (G(r) - G(1/2)); dr right| < epsilon int_{a(0)}^{b(0)} |u(t,r)|; dr $$
Thus we get $$ int_0^1 u(t,r) G(r); dr to G(1/2) text{as} t to infty$$
answered Jan 10 at 22:32
Robert IsraelRobert Israel
331k23220475
$endgroup$
This is really closely related to the method of characteristics, but doesn't explicitly use the solution.
Consider $$J = J(a(t), b(t), t) = int_{a(t)}^{b(t)} u(t,r); dr $$
where $a(t)$ and $b(t)$ are continuously differentiable functions. By the Fundamental Theorem of Calculus we have
$$ eqalign{dfrac{dJ}{dt} &= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + int_{a(t)}^{b(t)} dfrac{partial u}{partial t}(t,r) ; dr cr
&= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + int_{a(t)}^{b(t)} dfrac{partial}{partial t} left((r-1/2) u(t,r)right); dr cr
&= (b'(t) + b(t) - 1/2) u(t, b(t)) - (a'(t) + a(t) - 1/2) u(t, a(t))cr} $$
In particular, this is $0$ if $b(t)$ and $a(t)$ are solutions of the differential equation $x' + x - 1/2 = 0$. Those solutions all converge to $1/2$ as $t to infty$. Thus if $[a(0), b(0)]$ contains the support of $u_0$, for large $t$
the support of $u(t,cdot)$ is contained in the small interval $[a(t), b(t)]$ near
$1/2$, and $int_{a(t)}^{b(t)} u(t,r); dr = 1$. Take $epsilon > 0$. If the interval is small enough that $|G(r) - G(1/2)|< epsilon$,
we have
$$left|int_0^1 u(t,r) G(r); dr - G(1/2)right| = left|int_{a(t)}^{b(t)} u(t,r) (G(r) - G(1/2)); dr right| < epsilon int_{a(0)}^{b(0)} |u(t,r)|; dr $$
Thus we get $$ int_0^1 u(t,r) G(r); dr to G(1/2) text{as} t to infty$$
answered Jan 10 at 22:32
Robert IsraelRobert Israel
331k23220475
answered Jan 10 at 22:32
Robert IsraelRobert Israel
331k23220475
answered Jan 10 at 22:32
Robert IsraelRobert Israel
331k23220475
answered Jan 10 at 22:32
Robert IsraelRobert Israel
331k23220475
331k23220475
add a comment |
add a comment |
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$begingroup$
I think you want $G(1/2) int_{-infty}^infty u_0(s); ds$ rather than $G(1/2)$ on the right of your limit.
$endgroup$
– Robert Israel
Jan 10 at 20:55
$begingroup$
You are right, I have the following hypotesis: $int_0^1u_0(r)dr=1$. I added it in the question. Thank you!
$endgroup$
– user268193
Jan 10 at 21:20
2
$begingroup$
One way to understand the result physically is to write it as $dot{rho} + nabla(rho v) = 0$ where the mass density $rho = u$ and the velocity $v=1/2-r$. This is the well-known continuity equation for a fluid describing conservation of mass: $M = int rho,{rm d}r = 1$ is conserved in time. If you look at the velocity field, this is such that matter flows towards $r=1/2$ from both sides. As time goes on we therefore expect all the mass to pile up at this point ($uto Mdelta(r-1/2)$). The integral is the mass-weighted average value of $G$ which will therefore just be $G$ at $1/2$.
$endgroup$
– Winther
Jan 10 at 22:22