weak solution of linear transport (advection) equation












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$begingroup$


How to show that $g(x-at)$ is the weak solution of the initial value problem
$$u_t+au_x=0$$
$$u(x,0)=g(x)$$ where $ g(x)in L^{infty}(mathbb{R})$



Definition: $u$ is said to be the weak solution of the above initial value problem if
$$intlimits_0 ^ {infty} intlimits_{mathbb{R}} ({u{phi}_t+au{phi}_x})dxdt =intlimits_{mathbb{R}}g(x)phi(x,0)dx$$ $forall phi in C_c ^1(mathbb{R} times[0,infty))$






pde hyperbolic-equations transport-equation







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  • 1




    $begingroup$
    You simply need to show that $u(x,t) = g(x-at)$ satisfies the integral equation for all $phiin C_c^1(mathbb{R}times [0,infty))$. Divergence theorem and integration by parts would be useful for you.
    $endgroup$
    – Chee Han
    Jul 27 '18 at 22:54










  • $begingroup$
    Can we apply integration by parts and divergence theorem as g is only $L^{infty}$? Could u please give the complete proof.. I am unable to follow
    $endgroup$
    – Rosy
    Aug 3 '18 at 8:23


















2












$begingroup$


How to show that $g(x-at)$ is the weak solution of the initial value problem
$$u_t+au_x=0$$
$$u(x,0)=g(x)$$ where $ g(x)in L^{infty}(mathbb{R})$



Definition: $u$ is said to be the weak solution of the above initial value problem if
$$intlimits_0 ^ {infty} intlimits_{mathbb{R}} ({u{phi}_t+au{phi}_x})dxdt =intlimits_{mathbb{R}}g(x)phi(x,0)dx$$ $forall phi in C_c ^1(mathbb{R} times[0,infty))$






pde hyperbolic-equations transport-equation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You simply need to show that $u(x,t) = g(x-at)$ satisfies the integral equation for all $phiin C_c^1(mathbb{R}times [0,infty))$. Divergence theorem and integration by parts would be useful for you.
    $endgroup$
    – Chee Han
    Jul 27 '18 at 22:54










  • $begingroup$
    Can we apply integration by parts and divergence theorem as g is only $L^{infty}$? Could u please give the complete proof.. I am unable to follow
    $endgroup$
    – Rosy
    Aug 3 '18 at 8:23
















2












2








2





$begingroup$


How to show that $g(x-at)$ is the weak solution of the initial value problem
$$u_t+au_x=0$$
$$u(x,0)=g(x)$$ where $ g(x)in L^{infty}(mathbb{R})$



Definition: $u$ is said to be the weak solution of the above initial value problem if
$$intlimits_0 ^ {infty} intlimits_{mathbb{R}} ({u{phi}_t+au{phi}_x})dxdt =intlimits_{mathbb{R}}g(x)phi(x,0)dx$$ $forall phi in C_c ^1(mathbb{R} times[0,infty))$






pde hyperbolic-equations transport-equation







share|cite|improve this question











$endgroup$




How to show that $g(x-at)$ is the weak solution of the initial value problem
$$u_t+au_x=0$$
$$u(x,0)=g(x)$$ where $ g(x)in L^{infty}(mathbb{R})$



Definition: $u$ is said to be the weak solution of the above initial value problem if
$$intlimits_0 ^ {infty} intlimits_{mathbb{R}} ({u{phi}_t+au{phi}_x})dxdt =intlimits_{mathbb{R}}g(x)phi(x,0)dx$$ $forall phi in C_c ^1(mathbb{R} times[0,infty))$






pde hyperbolic-equations transport-equation




pde hyperbolic-equations transport-equation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 13:54









Harry49

6,21331132




6,21331132










asked Jul 27 '18 at 19:30









RosyRosy

1076




1076








  • 1




    $begingroup$
    You simply need to show that $u(x,t) = g(x-at)$ satisfies the integral equation for all $phiin C_c^1(mathbb{R}times [0,infty))$. Divergence theorem and integration by parts would be useful for you.
    $endgroup$
    – Chee Han
    Jul 27 '18 at 22:54










  • $begingroup$
    Can we apply integration by parts and divergence theorem as g is only $L^{infty}$? Could u please give the complete proof.. I am unable to follow
    $endgroup$
    – Rosy
    Aug 3 '18 at 8:23
















  • 1




    $begingroup$
    You simply need to show that $u(x,t) = g(x-at)$ satisfies the integral equation for all $phiin C_c^1(mathbb{R}times [0,infty))$. Divergence theorem and integration by parts would be useful for you.
    $endgroup$
    – Chee Han
    Jul 27 '18 at 22:54










  • $begingroup$
    Can we apply integration by parts and divergence theorem as g is only $L^{infty}$? Could u please give the complete proof.. I am unable to follow
    $endgroup$
    – Rosy
    Aug 3 '18 at 8:23










1




1




$begingroup$
You simply need to show that $u(x,t) = g(x-at)$ satisfies the integral equation for all $phiin C_c^1(mathbb{R}times [0,infty))$. Divergence theorem and integration by parts would be useful for you.
$endgroup$
– Chee Han
Jul 27 '18 at 22:54




$begingroup$
You simply need to show that $u(x,t) = g(x-at)$ satisfies the integral equation for all $phiin C_c^1(mathbb{R}times [0,infty))$. Divergence theorem and integration by parts would be useful for you.
$endgroup$
– Chee Han
Jul 27 '18 at 22:54












$begingroup$
Can we apply integration by parts and divergence theorem as g is only $L^{infty}$? Could u please give the complete proof.. I am unable to follow
$endgroup$
– Rosy
Aug 3 '18 at 8:23






$begingroup$
Can we apply integration by parts and divergence theorem as g is only $L^{infty}$? Could u please give the complete proof.. I am unable to follow
$endgroup$
– Rosy
Aug 3 '18 at 8:23












1 Answer
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$begingroup$

Let us make the change of variable $(xi, tau) = (x-at,t)$, so that $partial_t = partial_tau - apartial_xi$ and $partial_x = partial_xi$. Thus, using Fubini's theorem and integration by parts,
begin{aligned}
iint_{Bbb RtimesBbb R_+} u, (phi_t + a phi_x), text d x,text d t
&= iint_{Bbb RtimesBbb R_+} u phi_tau, text d xi,text d tau \
&= iint_{Bbb R_+timesBbb R} u phi_tau, text d tau,text d xi \
&= int_{Bbb R}left[uphiright]_{tauinBbb R_+}text dxi - iint_{Bbb R_+timesBbb R} underbrace{u_tau}_{=0} phi, text d tau,text d xi \
&= -int_{Bbb R}left.(uphi)right|_{t = 0}text dx , .
end{aligned}
We have shown that the definition holds for all smooth $phi$ with compact support. Hence, $u(x,t) = g(x-at)$ is a weak solution to the Cauchy problem of the advection equation. Note that there may be a sign mistake.






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    $begingroup$

    Let us make the change of variable $(xi, tau) = (x-at,t)$, so that $partial_t = partial_tau - apartial_xi$ and $partial_x = partial_xi$. Thus, using Fubini's theorem and integration by parts,
    begin{aligned}
    iint_{Bbb RtimesBbb R_+} u, (phi_t + a phi_x), text d x,text d t
    &= iint_{Bbb RtimesBbb R_+} u phi_tau, text d xi,text d tau \
    &= iint_{Bbb R_+timesBbb R} u phi_tau, text d tau,text d xi \
    &= int_{Bbb R}left[uphiright]_{tauinBbb R_+}text dxi - iint_{Bbb R_+timesBbb R} underbrace{u_tau}_{=0} phi, text d tau,text d xi \
    &= -int_{Bbb R}left.(uphi)right|_{t = 0}text dx , .
    end{aligned}
    We have shown that the definition holds for all smooth $phi$ with compact support. Hence, $u(x,t) = g(x-at)$ is a weak solution to the Cauchy problem of the advection equation. Note that there may be a sign mistake.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let us make the change of variable $(xi, tau) = (x-at,t)$, so that $partial_t = partial_tau - apartial_xi$ and $partial_x = partial_xi$. Thus, using Fubini's theorem and integration by parts,
      begin{aligned}
      iint_{Bbb RtimesBbb R_+} u, (phi_t + a phi_x), text d x,text d t
      &= iint_{Bbb RtimesBbb R_+} u phi_tau, text d xi,text d tau \
      &= iint_{Bbb R_+timesBbb R} u phi_tau, text d tau,text d xi \
      &= int_{Bbb R}left[uphiright]_{tauinBbb R_+}text dxi - iint_{Bbb R_+timesBbb R} underbrace{u_tau}_{=0} phi, text d tau,text d xi \
      &= -int_{Bbb R}left.(uphi)right|_{t = 0}text dx , .
      end{aligned}
      We have shown that the definition holds for all smooth $phi$ with compact support. Hence, $u(x,t) = g(x-at)$ is a weak solution to the Cauchy problem of the advection equation. Note that there may be a sign mistake.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let us make the change of variable $(xi, tau) = (x-at,t)$, so that $partial_t = partial_tau - apartial_xi$ and $partial_x = partial_xi$. Thus, using Fubini's theorem and integration by parts,
        begin{aligned}
        iint_{Bbb RtimesBbb R_+} u, (phi_t + a phi_x), text d x,text d t
        &= iint_{Bbb RtimesBbb R_+} u phi_tau, text d xi,text d tau \
        &= iint_{Bbb R_+timesBbb R} u phi_tau, text d tau,text d xi \
        &= int_{Bbb R}left[uphiright]_{tauinBbb R_+}text dxi - iint_{Bbb R_+timesBbb R} underbrace{u_tau}_{=0} phi, text d tau,text d xi \
        &= -int_{Bbb R}left.(uphi)right|_{t = 0}text dx , .
        end{aligned}
        We have shown that the definition holds for all smooth $phi$ with compact support. Hence, $u(x,t) = g(x-at)$ is a weak solution to the Cauchy problem of the advection equation. Note that there may be a sign mistake.






        share|cite|improve this answer









        $endgroup$



        Let us make the change of variable $(xi, tau) = (x-at,t)$, so that $partial_t = partial_tau - apartial_xi$ and $partial_x = partial_xi$. Thus, using Fubini's theorem and integration by parts,
        begin{aligned}
        iint_{Bbb RtimesBbb R_+} u, (phi_t + a phi_x), text d x,text d t
        &= iint_{Bbb RtimesBbb R_+} u phi_tau, text d xi,text d tau \
        &= iint_{Bbb R_+timesBbb R} u phi_tau, text d tau,text d xi \
        &= int_{Bbb R}left[uphiright]_{tauinBbb R_+}text dxi - iint_{Bbb R_+timesBbb R} underbrace{u_tau}_{=0} phi, text d tau,text d xi \
        &= -int_{Bbb R}left.(uphi)right|_{t = 0}text dx , .
        end{aligned}
        We have shown that the definition holds for all smooth $phi$ with compact support. Hence, $u(x,t) = g(x-at)$ is a weak solution to the Cauchy problem of the advection equation. Note that there may be a sign mistake.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 8 '18 at 9:58









        Harry49Harry49

        6,21331132




        6,21331132






























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