PDE's: Analysis without solving explicitly the equation












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












2








2


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 22:25







user268193

















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




















  • $begingroup$
    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












1 Answer
1






active

oldest

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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 $$ int_0^1 u(t,r) G(r); dr to G(1/2) text{as} t to infty$$






share|cite|improve this 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 $$ int_0^1 u(t,r) G(r); dr to G(1/2) text{as} t to infty$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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 $$ int_0^1 u(t,r) G(r); dr to G(1/2) text{as} t to infty$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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 $$ int_0^1 u(t,r) G(r); dr to G(1/2) text{as} t to infty$$






        share|cite|improve this answer









        $endgroup$



        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 $$ int_0^1 u(t,r) G(r); dr to G(1/2) text{as} t to infty$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 22:32









        Robert IsraelRobert Israel

        331k23220475




        331k23220475






























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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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            Gyllenstierna

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