Shock formation condition in IVP of $u_t + uu_x + alpha u = 0$












2












$begingroup$



Consider $u_t + uu_x + alpha u = 0$ for $t > 0$, all $x$ where $alpha > 0$ is a
constant. Find the characteristic equations for the equation with
initial data $u(x, 0) = f(x)$ given. Show that a shock cannot form if $alpha
geq max_{r in H}|f'(r)|$
where $H = {r : f'(r) < 0}$ or if $H$ is empty.




So far, I've found the characteristics by parametrizing
$$begin {cases} x_s=u, x(0,r)=r \ t_s=1,t(0,r)=0 \ u_s = -alpha u, u(0,r)=f(r)end {cases}$$
Then
$frac{du}{ds}=-alpha u Rightarrow u = C_1 e^{-alpha s}$. Considering the initial condition, $u = f(r)e^{-alpha s}.$



$frac{dt}{ds} = 1 Rightarrow t = s$ (since $t(0,r)=0$)



$frac{dx}{ds}=u=f(r)e^{-alpha s} Rightarrow x = -frac{1}{alpha}f(r)e^{-alpha s}+frac{1}{alpha}f(r)+r$ (since $x(0,r)=r$),
i.e. $x = -frac{1}{alpha}f(r)e^{-alpha t}+frac{1}{alpha}f(r)+r$



So how do we show that a shock cannot form?










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$endgroup$












  • $begingroup$
    What do you know about shocks? When do they form?
    $endgroup$
    – Mattos
    Dec 6 '18 at 3:42
















2












$begingroup$



Consider $u_t + uu_x + alpha u = 0$ for $t > 0$, all $x$ where $alpha > 0$ is a
constant. Find the characteristic equations for the equation with
initial data $u(x, 0) = f(x)$ given. Show that a shock cannot form if $alpha
geq max_{r in H}|f'(r)|$
where $H = {r : f'(r) < 0}$ or if $H$ is empty.




So far, I've found the characteristics by parametrizing
$$begin {cases} x_s=u, x(0,r)=r \ t_s=1,t(0,r)=0 \ u_s = -alpha u, u(0,r)=f(r)end {cases}$$
Then
$frac{du}{ds}=-alpha u Rightarrow u = C_1 e^{-alpha s}$. Considering the initial condition, $u = f(r)e^{-alpha s}.$



$frac{dt}{ds} = 1 Rightarrow t = s$ (since $t(0,r)=0$)



$frac{dx}{ds}=u=f(r)e^{-alpha s} Rightarrow x = -frac{1}{alpha}f(r)e^{-alpha s}+frac{1}{alpha}f(r)+r$ (since $x(0,r)=r$),
i.e. $x = -frac{1}{alpha}f(r)e^{-alpha t}+frac{1}{alpha}f(r)+r$



So how do we show that a shock cannot form?










share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you know about shocks? When do they form?
    $endgroup$
    – Mattos
    Dec 6 '18 at 3:42














2












2








2


3



$begingroup$



Consider $u_t + uu_x + alpha u = 0$ for $t > 0$, all $x$ where $alpha > 0$ is a
constant. Find the characteristic equations for the equation with
initial data $u(x, 0) = f(x)$ given. Show that a shock cannot form if $alpha
geq max_{r in H}|f'(r)|$
where $H = {r : f'(r) < 0}$ or if $H$ is empty.




So far, I've found the characteristics by parametrizing
$$begin {cases} x_s=u, x(0,r)=r \ t_s=1,t(0,r)=0 \ u_s = -alpha u, u(0,r)=f(r)end {cases}$$
Then
$frac{du}{ds}=-alpha u Rightarrow u = C_1 e^{-alpha s}$. Considering the initial condition, $u = f(r)e^{-alpha s}.$



$frac{dt}{ds} = 1 Rightarrow t = s$ (since $t(0,r)=0$)



$frac{dx}{ds}=u=f(r)e^{-alpha s} Rightarrow x = -frac{1}{alpha}f(r)e^{-alpha s}+frac{1}{alpha}f(r)+r$ (since $x(0,r)=r$),
i.e. $x = -frac{1}{alpha}f(r)e^{-alpha t}+frac{1}{alpha}f(r)+r$



So how do we show that a shock cannot form?










share|cite|improve this question











$endgroup$





Consider $u_t + uu_x + alpha u = 0$ for $t > 0$, all $x$ where $alpha > 0$ is a
constant. Find the characteristic equations for the equation with
initial data $u(x, 0) = f(x)$ given. Show that a shock cannot form if $alpha
geq max_{r in H}|f'(r)|$
where $H = {r : f'(r) < 0}$ or if $H$ is empty.




So far, I've found the characteristics by parametrizing
$$begin {cases} x_s=u, x(0,r)=r \ t_s=1,t(0,r)=0 \ u_s = -alpha u, u(0,r)=f(r)end {cases}$$
Then
$frac{du}{ds}=-alpha u Rightarrow u = C_1 e^{-alpha s}$. Considering the initial condition, $u = f(r)e^{-alpha s}.$



$frac{dt}{ds} = 1 Rightarrow t = s$ (since $t(0,r)=0$)



$frac{dx}{ds}=u=f(r)e^{-alpha s} Rightarrow x = -frac{1}{alpha}f(r)e^{-alpha s}+frac{1}{alpha}f(r)+r$ (since $x(0,r)=r$),
i.e. $x = -frac{1}{alpha}f(r)e^{-alpha t}+frac{1}{alpha}f(r)+r$



So how do we show that a shock cannot form?







pde characteristics hyperbolic-equations






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edited Dec 6 '18 at 11:34









Harry49

6,17331132




6,17331132










asked Dec 6 '18 at 0:01









dxdydzdxdydz

1769




1769












  • $begingroup$
    What do you know about shocks? When do they form?
    $endgroup$
    – Mattos
    Dec 6 '18 at 3:42


















  • $begingroup$
    What do you know about shocks? When do they form?
    $endgroup$
    – Mattos
    Dec 6 '18 at 3:42
















$begingroup$
What do you know about shocks? When do they form?
$endgroup$
– Mattos
Dec 6 '18 at 3:42




$begingroup$
What do you know about shocks? When do they form?
$endgroup$
– Mattos
Dec 6 '18 at 3:42










2 Answers
2






active

oldest

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1












$begingroup$

This (dissipative) Burgers' equation with relaxation is a typical example of conditional shock formation. The answer follows the steps in this post. The characteristics are the curves
$$
x = -f(x_0)frac{e^{-alpha t} - 1}{alpha} + x_0
$$

along which the solution satisfies
$$
u = fleft(x - frac{e^{alpha t}-1}{alpha} uright)e^{-alpha t} .
$$

Computing $frac{text d x}{text d x_0}$, we find that this derivative vanishes at a given positive time $t$ -i.e. a shock wave forms- if
$$
-frac{ln(1+alpha/f'(x_0))}{alpha} = t >0 .
$$

For this to be possible, the logarithm should be negative, and thus, $f'(x_0)<0$ (in other words, $x_0in H$). However, if $-alpha < f'(x_0) < 0$ for $x_0$ in $H$, the logarithm becomes complex and no shock occurs. Hence the conclusion: a shock cannot occur if $max_{x_0 in H}|f'(x_0)| leq alpha$ or if $H = emptyset$. Alternatively, this condition may be written $inf_{x_0 inBbb R} f'(x_0) geqslant -alpha$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Harry49 already answered to the question.



    I am not standing up here to give another answer. My intervention is a comment, but too long to be edited in the comments section.



    The analytical solving of the PDE with the specified boundary condition is especially interesting as shown below.
    $$u_t+uu_x=-alpha u tag 1$$
    The Charpit-Lagrange equations are :
    $$frac{dt}{1}=frac{dx}{u}=frac{du}{(-alpha u)}$$
    A first family of characteristic curves comes from $frac{dx}{u}=frac{du}{(-alpha u)}$ :
    $$u+alpha x =c_1$$
    A second family of characteristic curves comes from $frac{dt}{1}=frac{du}{(-alpha u)}$ :
    $$ue^{-alpha t}=c_2$$
    The general solution of the PDE Eq.$(1)$ expressed on the form of implicit equation is :
    $$u+alpha x=Phileft(ue^{-alpha t}right) tag 2$$
    where $Phi$ is an arbitrary function (to be determined according to boundary condition).



    Condition : $u(x,0)=f(x)$ with $f(x)$ a known (given) function.



    $$f(x)+alpha x=Phileft(f(x)e^{0}right)=Phileft(f(x)right)$$
    Let $X=f(x)$ and $x=f^{-1}(X)$



    $f^{-1}$ denotes the inverse function of $f$.
    $$Phi(X)=X+alpha f^{-1}(X)$$
    So, the function $Phi$ is determined. We put it into Eq.$(2)$.
    $$u+alpha x=ue^{-alpha t}+alpha f^{-1}left(ue^{-alpha t}right)$$
    $$f^{-1}left(ue^{-alpha t}right)=x+frac{u}{alpha}left(1-e^{-alpha t}right)$$
    $$ue^{-alpha t}=fleft(x+frac{u}{alpha}left(1-e^{-alpha t}right)right) tag 3$$
    Eq.$(3)$ is the implicite form of the analytic solution of $u_t+uu_x+alpha u=0$ with condition $u(x,0)=f(x)$.



    The explicite form of $u(x,t)$ requires to solve Eq.$(3)$ for $u$. The possibility to do it analytically depends on the kind of function $f$.



    Of course, this is only for information without answering to the OP question as mentioned at first place.






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      1












      $begingroup$

      This (dissipative) Burgers' equation with relaxation is a typical example of conditional shock formation. The answer follows the steps in this post. The characteristics are the curves
      $$
      x = -f(x_0)frac{e^{-alpha t} - 1}{alpha} + x_0
      $$

      along which the solution satisfies
      $$
      u = fleft(x - frac{e^{alpha t}-1}{alpha} uright)e^{-alpha t} .
      $$

      Computing $frac{text d x}{text d x_0}$, we find that this derivative vanishes at a given positive time $t$ -i.e. a shock wave forms- if
      $$
      -frac{ln(1+alpha/f'(x_0))}{alpha} = t >0 .
      $$

      For this to be possible, the logarithm should be negative, and thus, $f'(x_0)<0$ (in other words, $x_0in H$). However, if $-alpha < f'(x_0) < 0$ for $x_0$ in $H$, the logarithm becomes complex and no shock occurs. Hence the conclusion: a shock cannot occur if $max_{x_0 in H}|f'(x_0)| leq alpha$ or if $H = emptyset$. Alternatively, this condition may be written $inf_{x_0 inBbb R} f'(x_0) geqslant -alpha$.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        This (dissipative) Burgers' equation with relaxation is a typical example of conditional shock formation. The answer follows the steps in this post. The characteristics are the curves
        $$
        x = -f(x_0)frac{e^{-alpha t} - 1}{alpha} + x_0
        $$

        along which the solution satisfies
        $$
        u = fleft(x - frac{e^{alpha t}-1}{alpha} uright)e^{-alpha t} .
        $$

        Computing $frac{text d x}{text d x_0}$, we find that this derivative vanishes at a given positive time $t$ -i.e. a shock wave forms- if
        $$
        -frac{ln(1+alpha/f'(x_0))}{alpha} = t >0 .
        $$

        For this to be possible, the logarithm should be negative, and thus, $f'(x_0)<0$ (in other words, $x_0in H$). However, if $-alpha < f'(x_0) < 0$ for $x_0$ in $H$, the logarithm becomes complex and no shock occurs. Hence the conclusion: a shock cannot occur if $max_{x_0 in H}|f'(x_0)| leq alpha$ or if $H = emptyset$. Alternatively, this condition may be written $inf_{x_0 inBbb R} f'(x_0) geqslant -alpha$.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          This (dissipative) Burgers' equation with relaxation is a typical example of conditional shock formation. The answer follows the steps in this post. The characteristics are the curves
          $$
          x = -f(x_0)frac{e^{-alpha t} - 1}{alpha} + x_0
          $$

          along which the solution satisfies
          $$
          u = fleft(x - frac{e^{alpha t}-1}{alpha} uright)e^{-alpha t} .
          $$

          Computing $frac{text d x}{text d x_0}$, we find that this derivative vanishes at a given positive time $t$ -i.e. a shock wave forms- if
          $$
          -frac{ln(1+alpha/f'(x_0))}{alpha} = t >0 .
          $$

          For this to be possible, the logarithm should be negative, and thus, $f'(x_0)<0$ (in other words, $x_0in H$). However, if $-alpha < f'(x_0) < 0$ for $x_0$ in $H$, the logarithm becomes complex and no shock occurs. Hence the conclusion: a shock cannot occur if $max_{x_0 in H}|f'(x_0)| leq alpha$ or if $H = emptyset$. Alternatively, this condition may be written $inf_{x_0 inBbb R} f'(x_0) geqslant -alpha$.






          share|cite|improve this answer











          $endgroup$



          This (dissipative) Burgers' equation with relaxation is a typical example of conditional shock formation. The answer follows the steps in this post. The characteristics are the curves
          $$
          x = -f(x_0)frac{e^{-alpha t} - 1}{alpha} + x_0
          $$

          along which the solution satisfies
          $$
          u = fleft(x - frac{e^{alpha t}-1}{alpha} uright)e^{-alpha t} .
          $$

          Computing $frac{text d x}{text d x_0}$, we find that this derivative vanishes at a given positive time $t$ -i.e. a shock wave forms- if
          $$
          -frac{ln(1+alpha/f'(x_0))}{alpha} = t >0 .
          $$

          For this to be possible, the logarithm should be negative, and thus, $f'(x_0)<0$ (in other words, $x_0in H$). However, if $-alpha < f'(x_0) < 0$ for $x_0$ in $H$, the logarithm becomes complex and no shock occurs. Hence the conclusion: a shock cannot occur if $max_{x_0 in H}|f'(x_0)| leq alpha$ or if $H = emptyset$. Alternatively, this condition may be written $inf_{x_0 inBbb R} f'(x_0) geqslant -alpha$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 '18 at 12:48

























          answered Dec 6 '18 at 11:33









          Harry49Harry49

          6,17331132




          6,17331132























              0












              $begingroup$

              Harry49 already answered to the question.



              I am not standing up here to give another answer. My intervention is a comment, but too long to be edited in the comments section.



              The analytical solving of the PDE with the specified boundary condition is especially interesting as shown below.
              $$u_t+uu_x=-alpha u tag 1$$
              The Charpit-Lagrange equations are :
              $$frac{dt}{1}=frac{dx}{u}=frac{du}{(-alpha u)}$$
              A first family of characteristic curves comes from $frac{dx}{u}=frac{du}{(-alpha u)}$ :
              $$u+alpha x =c_1$$
              A second family of characteristic curves comes from $frac{dt}{1}=frac{du}{(-alpha u)}$ :
              $$ue^{-alpha t}=c_2$$
              The general solution of the PDE Eq.$(1)$ expressed on the form of implicit equation is :
              $$u+alpha x=Phileft(ue^{-alpha t}right) tag 2$$
              where $Phi$ is an arbitrary function (to be determined according to boundary condition).



              Condition : $u(x,0)=f(x)$ with $f(x)$ a known (given) function.



              $$f(x)+alpha x=Phileft(f(x)e^{0}right)=Phileft(f(x)right)$$
              Let $X=f(x)$ and $x=f^{-1}(X)$



              $f^{-1}$ denotes the inverse function of $f$.
              $$Phi(X)=X+alpha f^{-1}(X)$$
              So, the function $Phi$ is determined. We put it into Eq.$(2)$.
              $$u+alpha x=ue^{-alpha t}+alpha f^{-1}left(ue^{-alpha t}right)$$
              $$f^{-1}left(ue^{-alpha t}right)=x+frac{u}{alpha}left(1-e^{-alpha t}right)$$
              $$ue^{-alpha t}=fleft(x+frac{u}{alpha}left(1-e^{-alpha t}right)right) tag 3$$
              Eq.$(3)$ is the implicite form of the analytic solution of $u_t+uu_x+alpha u=0$ with condition $u(x,0)=f(x)$.



              The explicite form of $u(x,t)$ requires to solve Eq.$(3)$ for $u$. The possibility to do it analytically depends on the kind of function $f$.



              Of course, this is only for information without answering to the OP question as mentioned at first place.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Harry49 already answered to the question.



                I am not standing up here to give another answer. My intervention is a comment, but too long to be edited in the comments section.



                The analytical solving of the PDE with the specified boundary condition is especially interesting as shown below.
                $$u_t+uu_x=-alpha u tag 1$$
                The Charpit-Lagrange equations are :
                $$frac{dt}{1}=frac{dx}{u}=frac{du}{(-alpha u)}$$
                A first family of characteristic curves comes from $frac{dx}{u}=frac{du}{(-alpha u)}$ :
                $$u+alpha x =c_1$$
                A second family of characteristic curves comes from $frac{dt}{1}=frac{du}{(-alpha u)}$ :
                $$ue^{-alpha t}=c_2$$
                The general solution of the PDE Eq.$(1)$ expressed on the form of implicit equation is :
                $$u+alpha x=Phileft(ue^{-alpha t}right) tag 2$$
                where $Phi$ is an arbitrary function (to be determined according to boundary condition).



                Condition : $u(x,0)=f(x)$ with $f(x)$ a known (given) function.



                $$f(x)+alpha x=Phileft(f(x)e^{0}right)=Phileft(f(x)right)$$
                Let $X=f(x)$ and $x=f^{-1}(X)$



                $f^{-1}$ denotes the inverse function of $f$.
                $$Phi(X)=X+alpha f^{-1}(X)$$
                So, the function $Phi$ is determined. We put it into Eq.$(2)$.
                $$u+alpha x=ue^{-alpha t}+alpha f^{-1}left(ue^{-alpha t}right)$$
                $$f^{-1}left(ue^{-alpha t}right)=x+frac{u}{alpha}left(1-e^{-alpha t}right)$$
                $$ue^{-alpha t}=fleft(x+frac{u}{alpha}left(1-e^{-alpha t}right)right) tag 3$$
                Eq.$(3)$ is the implicite form of the analytic solution of $u_t+uu_x+alpha u=0$ with condition $u(x,0)=f(x)$.



                The explicite form of $u(x,t)$ requires to solve Eq.$(3)$ for $u$. The possibility to do it analytically depends on the kind of function $f$.



                Of course, this is only for information without answering to the OP question as mentioned at first place.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Harry49 already answered to the question.



                  I am not standing up here to give another answer. My intervention is a comment, but too long to be edited in the comments section.



                  The analytical solving of the PDE with the specified boundary condition is especially interesting as shown below.
                  $$u_t+uu_x=-alpha u tag 1$$
                  The Charpit-Lagrange equations are :
                  $$frac{dt}{1}=frac{dx}{u}=frac{du}{(-alpha u)}$$
                  A first family of characteristic curves comes from $frac{dx}{u}=frac{du}{(-alpha u)}$ :
                  $$u+alpha x =c_1$$
                  A second family of characteristic curves comes from $frac{dt}{1}=frac{du}{(-alpha u)}$ :
                  $$ue^{-alpha t}=c_2$$
                  The general solution of the PDE Eq.$(1)$ expressed on the form of implicit equation is :
                  $$u+alpha x=Phileft(ue^{-alpha t}right) tag 2$$
                  where $Phi$ is an arbitrary function (to be determined according to boundary condition).



                  Condition : $u(x,0)=f(x)$ with $f(x)$ a known (given) function.



                  $$f(x)+alpha x=Phileft(f(x)e^{0}right)=Phileft(f(x)right)$$
                  Let $X=f(x)$ and $x=f^{-1}(X)$



                  $f^{-1}$ denotes the inverse function of $f$.
                  $$Phi(X)=X+alpha f^{-1}(X)$$
                  So, the function $Phi$ is determined. We put it into Eq.$(2)$.
                  $$u+alpha x=ue^{-alpha t}+alpha f^{-1}left(ue^{-alpha t}right)$$
                  $$f^{-1}left(ue^{-alpha t}right)=x+frac{u}{alpha}left(1-e^{-alpha t}right)$$
                  $$ue^{-alpha t}=fleft(x+frac{u}{alpha}left(1-e^{-alpha t}right)right) tag 3$$
                  Eq.$(3)$ is the implicite form of the analytic solution of $u_t+uu_x+alpha u=0$ with condition $u(x,0)=f(x)$.



                  The explicite form of $u(x,t)$ requires to solve Eq.$(3)$ for $u$. The possibility to do it analytically depends on the kind of function $f$.



                  Of course, this is only for information without answering to the OP question as mentioned at first place.






                  share|cite|improve this answer











                  $endgroup$



                  Harry49 already answered to the question.



                  I am not standing up here to give another answer. My intervention is a comment, but too long to be edited in the comments section.



                  The analytical solving of the PDE with the specified boundary condition is especially interesting as shown below.
                  $$u_t+uu_x=-alpha u tag 1$$
                  The Charpit-Lagrange equations are :
                  $$frac{dt}{1}=frac{dx}{u}=frac{du}{(-alpha u)}$$
                  A first family of characteristic curves comes from $frac{dx}{u}=frac{du}{(-alpha u)}$ :
                  $$u+alpha x =c_1$$
                  A second family of characteristic curves comes from $frac{dt}{1}=frac{du}{(-alpha u)}$ :
                  $$ue^{-alpha t}=c_2$$
                  The general solution of the PDE Eq.$(1)$ expressed on the form of implicit equation is :
                  $$u+alpha x=Phileft(ue^{-alpha t}right) tag 2$$
                  where $Phi$ is an arbitrary function (to be determined according to boundary condition).



                  Condition : $u(x,0)=f(x)$ with $f(x)$ a known (given) function.



                  $$f(x)+alpha x=Phileft(f(x)e^{0}right)=Phileft(f(x)right)$$
                  Let $X=f(x)$ and $x=f^{-1}(X)$



                  $f^{-1}$ denotes the inverse function of $f$.
                  $$Phi(X)=X+alpha f^{-1}(X)$$
                  So, the function $Phi$ is determined. We put it into Eq.$(2)$.
                  $$u+alpha x=ue^{-alpha t}+alpha f^{-1}left(ue^{-alpha t}right)$$
                  $$f^{-1}left(ue^{-alpha t}right)=x+frac{u}{alpha}left(1-e^{-alpha t}right)$$
                  $$ue^{-alpha t}=fleft(x+frac{u}{alpha}left(1-e^{-alpha t}right)right) tag 3$$
                  Eq.$(3)$ is the implicite form of the analytic solution of $u_t+uu_x+alpha u=0$ with condition $u(x,0)=f(x)$.



                  The explicite form of $u(x,t)$ requires to solve Eq.$(3)$ for $u$. The possibility to do it analytically depends on the kind of function $f$.



                  Of course, this is only for information without answering to the OP question as mentioned at first place.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 9 '18 at 8:46

























                  answered Dec 9 '18 at 8:37









                  JJacquelinJJacquelin

                  43.2k21852




                  43.2k21852






























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