Finding the time when the speed of discontinuity becomes time-dependent in traffic flow












2












$begingroup$



I am trying to use the following conservation law:



$$u_t+f(u)_x=0 text{where} f(u)=u(1-u).$$
IC: $u(x,0)=frac{1}{4}$ for BC: $u(0,t)=1$ for $t>0$.




I found the solution as $u(x,t)=begin{cases} frac{1}{4} & x<t\
1 & x>t end{cases}$
.



Where my confusion lied was how you used s to get the bounds above.



Then we are tasked with the following questions:




Suppose that at $t_0=1$ the red light turns green - where t=0 the traffic is stopped - with a solution of,
$u(x,t)=begin{cases} frac{1}{4} & x<t\
1 & 0>x>t\
0 & x>0end{cases}$

where the boundaries come from the first part.




At the first discontinuity there is a constant speed. I need to find the time at which the speed is non-constant.



Find the curve $y(t)$ describing the movement.



Now for the first part I have done the following. Let $u(x,t)=v(x/t)=v(z)$. Then I substituted this into the first equation to get,



$$frac{xv'}{t^2}+frac{v'}{t}-frac{2vv'}{t}=0,$$ noting that we have used the chain rule to elimnate $v_t,v_x$. Then we can solve this ODE to get $$v=frac{x+t}{2t}.$$ Then, from the Riemann Problem we have,
$$u(x,t)=begin{cases}u_l & f'(u_l)t\
v(z) & f'(u_l)t<x<f'(u_r)t\
u_r & x>f'(u_r)tend{cases} = begin{cases}1 & x<-t\
frac{x+t}{2t} & -t<x<t\
0 & x>t end{cases}$$



From here I am confused on how to find $t_1$ and then how to progress onto finding the curve. Any tips would be appreciated.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Could you give more details how "the light is green/red" is reflected in your equation? It seems that you are computing shock/dispersion waves, does the Rankine-Hugoniot equation appear in your calculus?
    $endgroup$
    – LutzL
    Jan 8 at 12:19










  • $begingroup$
    @LutzL this is part of where my confusion lies - I am unsure how the redgreen light part works in relation to the equations. Yes I originally thought of using the Rankine-Hugoniot Equation to get $s=y'=frac{u_r+u_l}{2}$ but I was unsure on where to apply it.
    $endgroup$
    – KieranSQ
    Jan 8 at 12:45










  • $begingroup$
    You are working from the slides macs.hw.ac.uk/~lb138/slides_ch6_extra.pdf or something similar?
    $endgroup$
    – LutzL
    Jan 8 at 12:53










  • $begingroup$
    I’m working from something similar - notes directly from my lecture.
    $endgroup$
    – KieranSQ
    Jan 8 at 12:55










  • $begingroup$
    The model is explained in macs.hw.ac.uk/~lb138/slides_ch6.pdf.
    $endgroup$
    – LutzL
    Jan 8 at 13:03


















2












$begingroup$



I am trying to use the following conservation law:



$$u_t+f(u)_x=0 text{where} f(u)=u(1-u).$$
IC: $u(x,0)=frac{1}{4}$ for BC: $u(0,t)=1$ for $t>0$.




I found the solution as $u(x,t)=begin{cases} frac{1}{4} & x<t\
1 & x>t end{cases}$
.



Where my confusion lied was how you used s to get the bounds above.



Then we are tasked with the following questions:




Suppose that at $t_0=1$ the red light turns green - where t=0 the traffic is stopped - with a solution of,
$u(x,t)=begin{cases} frac{1}{4} & x<t\
1 & 0>x>t\
0 & x>0end{cases}$

where the boundaries come from the first part.




At the first discontinuity there is a constant speed. I need to find the time at which the speed is non-constant.



Find the curve $y(t)$ describing the movement.



Now for the first part I have done the following. Let $u(x,t)=v(x/t)=v(z)$. Then I substituted this into the first equation to get,



$$frac{xv'}{t^2}+frac{v'}{t}-frac{2vv'}{t}=0,$$ noting that we have used the chain rule to elimnate $v_t,v_x$. Then we can solve this ODE to get $$v=frac{x+t}{2t}.$$ Then, from the Riemann Problem we have,
$$u(x,t)=begin{cases}u_l & f'(u_l)t\
v(z) & f'(u_l)t<x<f'(u_r)t\
u_r & x>f'(u_r)tend{cases} = begin{cases}1 & x<-t\
frac{x+t}{2t} & -t<x<t\
0 & x>t end{cases}$$



From here I am confused on how to find $t_1$ and then how to progress onto finding the curve. Any tips would be appreciated.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Could you give more details how "the light is green/red" is reflected in your equation? It seems that you are computing shock/dispersion waves, does the Rankine-Hugoniot equation appear in your calculus?
    $endgroup$
    – LutzL
    Jan 8 at 12:19










  • $begingroup$
    @LutzL this is part of where my confusion lies - I am unsure how the redgreen light part works in relation to the equations. Yes I originally thought of using the Rankine-Hugoniot Equation to get $s=y'=frac{u_r+u_l}{2}$ but I was unsure on where to apply it.
    $endgroup$
    – KieranSQ
    Jan 8 at 12:45










  • $begingroup$
    You are working from the slides macs.hw.ac.uk/~lb138/slides_ch6_extra.pdf or something similar?
    $endgroup$
    – LutzL
    Jan 8 at 12:53










  • $begingroup$
    I’m working from something similar - notes directly from my lecture.
    $endgroup$
    – KieranSQ
    Jan 8 at 12:55










  • $begingroup$
    The model is explained in macs.hw.ac.uk/~lb138/slides_ch6.pdf.
    $endgroup$
    – LutzL
    Jan 8 at 13:03
















2












2








2


1



$begingroup$



I am trying to use the following conservation law:



$$u_t+f(u)_x=0 text{where} f(u)=u(1-u).$$
IC: $u(x,0)=frac{1}{4}$ for BC: $u(0,t)=1$ for $t>0$.




I found the solution as $u(x,t)=begin{cases} frac{1}{4} & x<t\
1 & x>t end{cases}$
.



Where my confusion lied was how you used s to get the bounds above.



Then we are tasked with the following questions:




Suppose that at $t_0=1$ the red light turns green - where t=0 the traffic is stopped - with a solution of,
$u(x,t)=begin{cases} frac{1}{4} & x<t\
1 & 0>x>t\
0 & x>0end{cases}$

where the boundaries come from the first part.




At the first discontinuity there is a constant speed. I need to find the time at which the speed is non-constant.



Find the curve $y(t)$ describing the movement.



Now for the first part I have done the following. Let $u(x,t)=v(x/t)=v(z)$. Then I substituted this into the first equation to get,



$$frac{xv'}{t^2}+frac{v'}{t}-frac{2vv'}{t}=0,$$ noting that we have used the chain rule to elimnate $v_t,v_x$. Then we can solve this ODE to get $$v=frac{x+t}{2t}.$$ Then, from the Riemann Problem we have,
$$u(x,t)=begin{cases}u_l & f'(u_l)t\
v(z) & f'(u_l)t<x<f'(u_r)t\
u_r & x>f'(u_r)tend{cases} = begin{cases}1 & x<-t\
frac{x+t}{2t} & -t<x<t\
0 & x>t end{cases}$$



From here I am confused on how to find $t_1$ and then how to progress onto finding the curve. Any tips would be appreciated.










share|cite|improve this question











$endgroup$





I am trying to use the following conservation law:



$$u_t+f(u)_x=0 text{where} f(u)=u(1-u).$$
IC: $u(x,0)=frac{1}{4}$ for BC: $u(0,t)=1$ for $t>0$.




I found the solution as $u(x,t)=begin{cases} frac{1}{4} & x<t\
1 & x>t end{cases}$
.



Where my confusion lied was how you used s to get the bounds above.



Then we are tasked with the following questions:




Suppose that at $t_0=1$ the red light turns green - where t=0 the traffic is stopped - with a solution of,
$u(x,t)=begin{cases} frac{1}{4} & x<t\
1 & 0>x>t\
0 & x>0end{cases}$

where the boundaries come from the first part.




At the first discontinuity there is a constant speed. I need to find the time at which the speed is non-constant.



Find the curve $y(t)$ describing the movement.



Now for the first part I have done the following. Let $u(x,t)=v(x/t)=v(z)$. Then I substituted this into the first equation to get,



$$frac{xv'}{t^2}+frac{v'}{t}-frac{2vv'}{t}=0,$$ noting that we have used the chain rule to elimnate $v_t,v_x$. Then we can solve this ODE to get $$v=frac{x+t}{2t}.$$ Then, from the Riemann Problem we have,
$$u(x,t)=begin{cases}u_l & f'(u_l)t\
v(z) & f'(u_l)t<x<f'(u_r)t\
u_r & x>f'(u_r)tend{cases} = begin{cases}1 & x<-t\
frac{x+t}{2t} & -t<x<t\
0 & x>t end{cases}$$



From here I am confused on how to find $t_1$ and then how to progress onto finding the curve. Any tips would be appreciated.







pde characteristics 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 Jan 17 at 12:47







KieranSQ

















asked Jan 8 at 11:41









KieranSQKieranSQ

969




969












  • $begingroup$
    Could you give more details how "the light is green/red" is reflected in your equation? It seems that you are computing shock/dispersion waves, does the Rankine-Hugoniot equation appear in your calculus?
    $endgroup$
    – LutzL
    Jan 8 at 12:19










  • $begingroup$
    @LutzL this is part of where my confusion lies - I am unsure how the redgreen light part works in relation to the equations. Yes I originally thought of using the Rankine-Hugoniot Equation to get $s=y'=frac{u_r+u_l}{2}$ but I was unsure on where to apply it.
    $endgroup$
    – KieranSQ
    Jan 8 at 12:45










  • $begingroup$
    You are working from the slides macs.hw.ac.uk/~lb138/slides_ch6_extra.pdf or something similar?
    $endgroup$
    – LutzL
    Jan 8 at 12:53










  • $begingroup$
    I’m working from something similar - notes directly from my lecture.
    $endgroup$
    – KieranSQ
    Jan 8 at 12:55










  • $begingroup$
    The model is explained in macs.hw.ac.uk/~lb138/slides_ch6.pdf.
    $endgroup$
    – LutzL
    Jan 8 at 13:03




















  • $begingroup$
    Could you give more details how "the light is green/red" is reflected in your equation? It seems that you are computing shock/dispersion waves, does the Rankine-Hugoniot equation appear in your calculus?
    $endgroup$
    – LutzL
    Jan 8 at 12:19










  • $begingroup$
    @LutzL this is part of where my confusion lies - I am unsure how the redgreen light part works in relation to the equations. Yes I originally thought of using the Rankine-Hugoniot Equation to get $s=y'=frac{u_r+u_l}{2}$ but I was unsure on where to apply it.
    $endgroup$
    – KieranSQ
    Jan 8 at 12:45










  • $begingroup$
    You are working from the slides macs.hw.ac.uk/~lb138/slides_ch6_extra.pdf or something similar?
    $endgroup$
    – LutzL
    Jan 8 at 12:53










  • $begingroup$
    I’m working from something similar - notes directly from my lecture.
    $endgroup$
    – KieranSQ
    Jan 8 at 12:55










  • $begingroup$
    The model is explained in macs.hw.ac.uk/~lb138/slides_ch6.pdf.
    $endgroup$
    – LutzL
    Jan 8 at 13:03


















$begingroup$
Could you give more details how "the light is green/red" is reflected in your equation? It seems that you are computing shock/dispersion waves, does the Rankine-Hugoniot equation appear in your calculus?
$endgroup$
– LutzL
Jan 8 at 12:19




$begingroup$
Could you give more details how "the light is green/red" is reflected in your equation? It seems that you are computing shock/dispersion waves, does the Rankine-Hugoniot equation appear in your calculus?
$endgroup$
– LutzL
Jan 8 at 12:19












$begingroup$
@LutzL this is part of where my confusion lies - I am unsure how the redgreen light part works in relation to the equations. Yes I originally thought of using the Rankine-Hugoniot Equation to get $s=y'=frac{u_r+u_l}{2}$ but I was unsure on where to apply it.
$endgroup$
– KieranSQ
Jan 8 at 12:45




$begingroup$
@LutzL this is part of where my confusion lies - I am unsure how the redgreen light part works in relation to the equations. Yes I originally thought of using the Rankine-Hugoniot Equation to get $s=y'=frac{u_r+u_l}{2}$ but I was unsure on where to apply it.
$endgroup$
– KieranSQ
Jan 8 at 12:45












$begingroup$
You are working from the slides macs.hw.ac.uk/~lb138/slides_ch6_extra.pdf or something similar?
$endgroup$
– LutzL
Jan 8 at 12:53




$begingroup$
You are working from the slides macs.hw.ac.uk/~lb138/slides_ch6_extra.pdf or something similar?
$endgroup$
– LutzL
Jan 8 at 12:53












$begingroup$
I’m working from something similar - notes directly from my lecture.
$endgroup$
– KieranSQ
Jan 8 at 12:55




$begingroup$
I’m working from something similar - notes directly from my lecture.
$endgroup$
– KieranSQ
Jan 8 at 12:55












$begingroup$
The model is explained in macs.hw.ac.uk/~lb138/slides_ch6.pdf.
$endgroup$
– LutzL
Jan 8 at 13:03






$begingroup$
The model is explained in macs.hw.ac.uk/~lb138/slides_ch6.pdf.
$endgroup$
– LutzL
Jan 8 at 13:03












1 Answer
1






active

oldest

votes


















2












$begingroup$

The Lagrange equations are $frac{dt}1=frac{dx}{1-2z}=frac{dz}0$ with $z=u(x(t),t)$ constant along the characteristic curves. So that $$z=phi(x-(1-2z)t)~~text{ or }~~ x+(2z-1)t=psi(z).$$





For $x<0$, $t>0$ there are two values for $z$ giving possibly characteristic curves going through each point $(x,t)$,




  • $z=1$ from the vertical boundary at $x_{init}=0$, leading to $x+t=t_{init}>0$, $x=t_{init}-t$ for $t>t_{init}$, and


  • $z=frac14$ for $t_{init}=0$ giving $x-t/2=x_{init}<0$, $x=x_{init}+t/2$ for $t<-2x_{init}$.



These two phases collide, starting at $(x,t)=(0,0)$. The phase boundary $x=v(t)$ has, by the Rankine-Hugoniot condition for a shock wave, as speed the mean of the speeds of the phases, $$frac{dv}{dt}=frac{-1+1/2}2=-frac14.$$ With that the solution of the PDE is
$$
u(x,t)=begin{cases}frac14,&x<-frac t4,\ 1, & -frac t4 < x < 0.end{cases}
$$





Then at $t=1$ the block at $x=0$, that is the boundary condition fixing the value there, is removed. The existing solution for $x<0$ is extended to the whole real axis via $u(x,1)=0$ for $x>0$. This gives a shock from the existing phase boundary starting at $a_0=-frac14$ and a rarefaction wave at $x=0$. The available values of $z$ are




  • $z=frac14$ for $t_{init}=1$, $x_{init}<a_0=-frac14$ giving $x-t/2=x_{init}-t_{init}/2$, $x=x_{init}+(t-1)/2$,


  • $z=1$ for $a_0<x_{init}<0$, leading to $x+t=x_{init}+t_{init}>0$, $x=x_{init}+1-t$, and


  • $zin [0,1]$ along the curves $x=(t-1)(1-2z)$,


  • $z=0$ for $t-1>x>0$.



The remainder of the phase $z=1$ is used up where its left boundary $x=1-t$ for $x_{init}=0$ meets the shock wave at the right boundary $x=-frac t4$. This happens at $t=frac43$. After that the phase $z=frac14$ collides with the rarefaction wave, the mean of the speeds at $(x,t)=(v(t),t)$ is
$$
v'(t)=frac{frac12+frac{v(t)}{t-1}}2iff left(frac{v}{sqrt{t-1}}right)'=frac1{4sqrt{t-1}}
\
implies frac{v(t)}{sqrt{t-1}}=C+frac{sqrt{t-1}}2
$$

and thus with $v(frac43)=-frac13$
$$
C=-frac{sqrt3}{2},~~ boxed{v(t)=-frac{sqrt{3(t-1})}2+frac{t-1}2}.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how did you get that $x-frac{t}{2}=x_i-frac{t_i}{2}$? I followed in the first part but now I am stuck. Thanks in advance!
    $endgroup$
    – KieranSQ
    Jan 9 at 18:16








  • 1




    $begingroup$
    We have $z=frac14$ and $x+(2z-1)t=const.$ so that $x-frac12t$ is constant and thus also the same where the characteristic curve crosses the boundary.
    $endgroup$
    – LutzL
    Jan 9 at 19:40










  • $begingroup$
    My last question, how do you get $u_r=frac{u}{1-t}$? I cannot see why - apologies for all of the questions.
    $endgroup$
    – KieranSQ
    Jan 9 at 21:57






  • 1




    $begingroup$
    The line $x+(2z-1)t=c$ that goes through the rarefaction center $(x,t)=(0,1)$ requires $c=(2z-1)$, so that $x=-(2z-1)(t-1)$, the slope of that line is thus $dot x=1-2z=frac{x}{t-1}$. I see that re-using $u$ is a bad idea, I'll change that to $v$.
    $endgroup$
    – LutzL
    Jan 9 at 22:06











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

The Lagrange equations are $frac{dt}1=frac{dx}{1-2z}=frac{dz}0$ with $z=u(x(t),t)$ constant along the characteristic curves. So that $$z=phi(x-(1-2z)t)~~text{ or }~~ x+(2z-1)t=psi(z).$$





For $x<0$, $t>0$ there are two values for $z$ giving possibly characteristic curves going through each point $(x,t)$,




  • $z=1$ from the vertical boundary at $x_{init}=0$, leading to $x+t=t_{init}>0$, $x=t_{init}-t$ for $t>t_{init}$, and


  • $z=frac14$ for $t_{init}=0$ giving $x-t/2=x_{init}<0$, $x=x_{init}+t/2$ for $t<-2x_{init}$.



These two phases collide, starting at $(x,t)=(0,0)$. The phase boundary $x=v(t)$ has, by the Rankine-Hugoniot condition for a shock wave, as speed the mean of the speeds of the phases, $$frac{dv}{dt}=frac{-1+1/2}2=-frac14.$$ With that the solution of the PDE is
$$
u(x,t)=begin{cases}frac14,&x<-frac t4,\ 1, & -frac t4 < x < 0.end{cases}
$$





Then at $t=1$ the block at $x=0$, that is the boundary condition fixing the value there, is removed. The existing solution for $x<0$ is extended to the whole real axis via $u(x,1)=0$ for $x>0$. This gives a shock from the existing phase boundary starting at $a_0=-frac14$ and a rarefaction wave at $x=0$. The available values of $z$ are




  • $z=frac14$ for $t_{init}=1$, $x_{init}<a_0=-frac14$ giving $x-t/2=x_{init}-t_{init}/2$, $x=x_{init}+(t-1)/2$,


  • $z=1$ for $a_0<x_{init}<0$, leading to $x+t=x_{init}+t_{init}>0$, $x=x_{init}+1-t$, and


  • $zin [0,1]$ along the curves $x=(t-1)(1-2z)$,


  • $z=0$ for $t-1>x>0$.



The remainder of the phase $z=1$ is used up where its left boundary $x=1-t$ for $x_{init}=0$ meets the shock wave at the right boundary $x=-frac t4$. This happens at $t=frac43$. After that the phase $z=frac14$ collides with the rarefaction wave, the mean of the speeds at $(x,t)=(v(t),t)$ is
$$
v'(t)=frac{frac12+frac{v(t)}{t-1}}2iff left(frac{v}{sqrt{t-1}}right)'=frac1{4sqrt{t-1}}
\
implies frac{v(t)}{sqrt{t-1}}=C+frac{sqrt{t-1}}2
$$

and thus with $v(frac43)=-frac13$
$$
C=-frac{sqrt3}{2},~~ boxed{v(t)=-frac{sqrt{3(t-1})}2+frac{t-1}2}.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how did you get that $x-frac{t}{2}=x_i-frac{t_i}{2}$? I followed in the first part but now I am stuck. Thanks in advance!
    $endgroup$
    – KieranSQ
    Jan 9 at 18:16








  • 1




    $begingroup$
    We have $z=frac14$ and $x+(2z-1)t=const.$ so that $x-frac12t$ is constant and thus also the same where the characteristic curve crosses the boundary.
    $endgroup$
    – LutzL
    Jan 9 at 19:40










  • $begingroup$
    My last question, how do you get $u_r=frac{u}{1-t}$? I cannot see why - apologies for all of the questions.
    $endgroup$
    – KieranSQ
    Jan 9 at 21:57






  • 1




    $begingroup$
    The line $x+(2z-1)t=c$ that goes through the rarefaction center $(x,t)=(0,1)$ requires $c=(2z-1)$, so that $x=-(2z-1)(t-1)$, the slope of that line is thus $dot x=1-2z=frac{x}{t-1}$. I see that re-using $u$ is a bad idea, I'll change that to $v$.
    $endgroup$
    – LutzL
    Jan 9 at 22:06
















2












$begingroup$

The Lagrange equations are $frac{dt}1=frac{dx}{1-2z}=frac{dz}0$ with $z=u(x(t),t)$ constant along the characteristic curves. So that $$z=phi(x-(1-2z)t)~~text{ or }~~ x+(2z-1)t=psi(z).$$





For $x<0$, $t>0$ there are two values for $z$ giving possibly characteristic curves going through each point $(x,t)$,




  • $z=1$ from the vertical boundary at $x_{init}=0$, leading to $x+t=t_{init}>0$, $x=t_{init}-t$ for $t>t_{init}$, and


  • $z=frac14$ for $t_{init}=0$ giving $x-t/2=x_{init}<0$, $x=x_{init}+t/2$ for $t<-2x_{init}$.



These two phases collide, starting at $(x,t)=(0,0)$. The phase boundary $x=v(t)$ has, by the Rankine-Hugoniot condition for a shock wave, as speed the mean of the speeds of the phases, $$frac{dv}{dt}=frac{-1+1/2}2=-frac14.$$ With that the solution of the PDE is
$$
u(x,t)=begin{cases}frac14,&x<-frac t4,\ 1, & -frac t4 < x < 0.end{cases}
$$





Then at $t=1$ the block at $x=0$, that is the boundary condition fixing the value there, is removed. The existing solution for $x<0$ is extended to the whole real axis via $u(x,1)=0$ for $x>0$. This gives a shock from the existing phase boundary starting at $a_0=-frac14$ and a rarefaction wave at $x=0$. The available values of $z$ are




  • $z=frac14$ for $t_{init}=1$, $x_{init}<a_0=-frac14$ giving $x-t/2=x_{init}-t_{init}/2$, $x=x_{init}+(t-1)/2$,


  • $z=1$ for $a_0<x_{init}<0$, leading to $x+t=x_{init}+t_{init}>0$, $x=x_{init}+1-t$, and


  • $zin [0,1]$ along the curves $x=(t-1)(1-2z)$,


  • $z=0$ for $t-1>x>0$.



The remainder of the phase $z=1$ is used up where its left boundary $x=1-t$ for $x_{init}=0$ meets the shock wave at the right boundary $x=-frac t4$. This happens at $t=frac43$. After that the phase $z=frac14$ collides with the rarefaction wave, the mean of the speeds at $(x,t)=(v(t),t)$ is
$$
v'(t)=frac{frac12+frac{v(t)}{t-1}}2iff left(frac{v}{sqrt{t-1}}right)'=frac1{4sqrt{t-1}}
\
implies frac{v(t)}{sqrt{t-1}}=C+frac{sqrt{t-1}}2
$$

and thus with $v(frac43)=-frac13$
$$
C=-frac{sqrt3}{2},~~ boxed{v(t)=-frac{sqrt{3(t-1})}2+frac{t-1}2}.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how did you get that $x-frac{t}{2}=x_i-frac{t_i}{2}$? I followed in the first part but now I am stuck. Thanks in advance!
    $endgroup$
    – KieranSQ
    Jan 9 at 18:16








  • 1




    $begingroup$
    We have $z=frac14$ and $x+(2z-1)t=const.$ so that $x-frac12t$ is constant and thus also the same where the characteristic curve crosses the boundary.
    $endgroup$
    – LutzL
    Jan 9 at 19:40










  • $begingroup$
    My last question, how do you get $u_r=frac{u}{1-t}$? I cannot see why - apologies for all of the questions.
    $endgroup$
    – KieranSQ
    Jan 9 at 21:57






  • 1




    $begingroup$
    The line $x+(2z-1)t=c$ that goes through the rarefaction center $(x,t)=(0,1)$ requires $c=(2z-1)$, so that $x=-(2z-1)(t-1)$, the slope of that line is thus $dot x=1-2z=frac{x}{t-1}$. I see that re-using $u$ is a bad idea, I'll change that to $v$.
    $endgroup$
    – LutzL
    Jan 9 at 22:06














2












2








2





$begingroup$

The Lagrange equations are $frac{dt}1=frac{dx}{1-2z}=frac{dz}0$ with $z=u(x(t),t)$ constant along the characteristic curves. So that $$z=phi(x-(1-2z)t)~~text{ or }~~ x+(2z-1)t=psi(z).$$





For $x<0$, $t>0$ there are two values for $z$ giving possibly characteristic curves going through each point $(x,t)$,




  • $z=1$ from the vertical boundary at $x_{init}=0$, leading to $x+t=t_{init}>0$, $x=t_{init}-t$ for $t>t_{init}$, and


  • $z=frac14$ for $t_{init}=0$ giving $x-t/2=x_{init}<0$, $x=x_{init}+t/2$ for $t<-2x_{init}$.



These two phases collide, starting at $(x,t)=(0,0)$. The phase boundary $x=v(t)$ has, by the Rankine-Hugoniot condition for a shock wave, as speed the mean of the speeds of the phases, $$frac{dv}{dt}=frac{-1+1/2}2=-frac14.$$ With that the solution of the PDE is
$$
u(x,t)=begin{cases}frac14,&x<-frac t4,\ 1, & -frac t4 < x < 0.end{cases}
$$





Then at $t=1$ the block at $x=0$, that is the boundary condition fixing the value there, is removed. The existing solution for $x<0$ is extended to the whole real axis via $u(x,1)=0$ for $x>0$. This gives a shock from the existing phase boundary starting at $a_0=-frac14$ and a rarefaction wave at $x=0$. The available values of $z$ are




  • $z=frac14$ for $t_{init}=1$, $x_{init}<a_0=-frac14$ giving $x-t/2=x_{init}-t_{init}/2$, $x=x_{init}+(t-1)/2$,


  • $z=1$ for $a_0<x_{init}<0$, leading to $x+t=x_{init}+t_{init}>0$, $x=x_{init}+1-t$, and


  • $zin [0,1]$ along the curves $x=(t-1)(1-2z)$,


  • $z=0$ for $t-1>x>0$.



The remainder of the phase $z=1$ is used up where its left boundary $x=1-t$ for $x_{init}=0$ meets the shock wave at the right boundary $x=-frac t4$. This happens at $t=frac43$. After that the phase $z=frac14$ collides with the rarefaction wave, the mean of the speeds at $(x,t)=(v(t),t)$ is
$$
v'(t)=frac{frac12+frac{v(t)}{t-1}}2iff left(frac{v}{sqrt{t-1}}right)'=frac1{4sqrt{t-1}}
\
implies frac{v(t)}{sqrt{t-1}}=C+frac{sqrt{t-1}}2
$$

and thus with $v(frac43)=-frac13$
$$
C=-frac{sqrt3}{2},~~ boxed{v(t)=-frac{sqrt{3(t-1})}2+frac{t-1}2}.
$$






share|cite|improve this answer











$endgroup$



The Lagrange equations are $frac{dt}1=frac{dx}{1-2z}=frac{dz}0$ with $z=u(x(t),t)$ constant along the characteristic curves. So that $$z=phi(x-(1-2z)t)~~text{ or }~~ x+(2z-1)t=psi(z).$$





For $x<0$, $t>0$ there are two values for $z$ giving possibly characteristic curves going through each point $(x,t)$,




  • $z=1$ from the vertical boundary at $x_{init}=0$, leading to $x+t=t_{init}>0$, $x=t_{init}-t$ for $t>t_{init}$, and


  • $z=frac14$ for $t_{init}=0$ giving $x-t/2=x_{init}<0$, $x=x_{init}+t/2$ for $t<-2x_{init}$.



These two phases collide, starting at $(x,t)=(0,0)$. The phase boundary $x=v(t)$ has, by the Rankine-Hugoniot condition for a shock wave, as speed the mean of the speeds of the phases, $$frac{dv}{dt}=frac{-1+1/2}2=-frac14.$$ With that the solution of the PDE is
$$
u(x,t)=begin{cases}frac14,&x<-frac t4,\ 1, & -frac t4 < x < 0.end{cases}
$$





Then at $t=1$ the block at $x=0$, that is the boundary condition fixing the value there, is removed. The existing solution for $x<0$ is extended to the whole real axis via $u(x,1)=0$ for $x>0$. This gives a shock from the existing phase boundary starting at $a_0=-frac14$ and a rarefaction wave at $x=0$. The available values of $z$ are




  • $z=frac14$ for $t_{init}=1$, $x_{init}<a_0=-frac14$ giving $x-t/2=x_{init}-t_{init}/2$, $x=x_{init}+(t-1)/2$,


  • $z=1$ for $a_0<x_{init}<0$, leading to $x+t=x_{init}+t_{init}>0$, $x=x_{init}+1-t$, and


  • $zin [0,1]$ along the curves $x=(t-1)(1-2z)$,


  • $z=0$ for $t-1>x>0$.



The remainder of the phase $z=1$ is used up where its left boundary $x=1-t$ for $x_{init}=0$ meets the shock wave at the right boundary $x=-frac t4$. This happens at $t=frac43$. After that the phase $z=frac14$ collides with the rarefaction wave, the mean of the speeds at $(x,t)=(v(t),t)$ is
$$
v'(t)=frac{frac12+frac{v(t)}{t-1}}2iff left(frac{v}{sqrt{t-1}}right)'=frac1{4sqrt{t-1}}
\
implies frac{v(t)}{sqrt{t-1}}=C+frac{sqrt{t-1}}2
$$

and thus with $v(frac43)=-frac13$
$$
C=-frac{sqrt3}{2},~~ boxed{v(t)=-frac{sqrt{3(t-1})}2+frac{t-1}2}.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 9 at 22:08

























answered Jan 8 at 13:57









LutzLLutzL

57.4k42054




57.4k42054












  • $begingroup$
    how did you get that $x-frac{t}{2}=x_i-frac{t_i}{2}$? I followed in the first part but now I am stuck. Thanks in advance!
    $endgroup$
    – KieranSQ
    Jan 9 at 18:16








  • 1




    $begingroup$
    We have $z=frac14$ and $x+(2z-1)t=const.$ so that $x-frac12t$ is constant and thus also the same where the characteristic curve crosses the boundary.
    $endgroup$
    – LutzL
    Jan 9 at 19:40










  • $begingroup$
    My last question, how do you get $u_r=frac{u}{1-t}$? I cannot see why - apologies for all of the questions.
    $endgroup$
    – KieranSQ
    Jan 9 at 21:57






  • 1




    $begingroup$
    The line $x+(2z-1)t=c$ that goes through the rarefaction center $(x,t)=(0,1)$ requires $c=(2z-1)$, so that $x=-(2z-1)(t-1)$, the slope of that line is thus $dot x=1-2z=frac{x}{t-1}$. I see that re-using $u$ is a bad idea, I'll change that to $v$.
    $endgroup$
    – LutzL
    Jan 9 at 22:06


















  • $begingroup$
    how did you get that $x-frac{t}{2}=x_i-frac{t_i}{2}$? I followed in the first part but now I am stuck. Thanks in advance!
    $endgroup$
    – KieranSQ
    Jan 9 at 18:16








  • 1




    $begingroup$
    We have $z=frac14$ and $x+(2z-1)t=const.$ so that $x-frac12t$ is constant and thus also the same where the characteristic curve crosses the boundary.
    $endgroup$
    – LutzL
    Jan 9 at 19:40










  • $begingroup$
    My last question, how do you get $u_r=frac{u}{1-t}$? I cannot see why - apologies for all of the questions.
    $endgroup$
    – KieranSQ
    Jan 9 at 21:57






  • 1




    $begingroup$
    The line $x+(2z-1)t=c$ that goes through the rarefaction center $(x,t)=(0,1)$ requires $c=(2z-1)$, so that $x=-(2z-1)(t-1)$, the slope of that line is thus $dot x=1-2z=frac{x}{t-1}$. I see that re-using $u$ is a bad idea, I'll change that to $v$.
    $endgroup$
    – LutzL
    Jan 9 at 22:06
















$begingroup$
how did you get that $x-frac{t}{2}=x_i-frac{t_i}{2}$? I followed in the first part but now I am stuck. Thanks in advance!
$endgroup$
– KieranSQ
Jan 9 at 18:16






$begingroup$
how did you get that $x-frac{t}{2}=x_i-frac{t_i}{2}$? I followed in the first part but now I am stuck. Thanks in advance!
$endgroup$
– KieranSQ
Jan 9 at 18:16






1




1




$begingroup$
We have $z=frac14$ and $x+(2z-1)t=const.$ so that $x-frac12t$ is constant and thus also the same where the characteristic curve crosses the boundary.
$endgroup$
– LutzL
Jan 9 at 19:40




$begingroup$
We have $z=frac14$ and $x+(2z-1)t=const.$ so that $x-frac12t$ is constant and thus also the same where the characteristic curve crosses the boundary.
$endgroup$
– LutzL
Jan 9 at 19:40












$begingroup$
My last question, how do you get $u_r=frac{u}{1-t}$? I cannot see why - apologies for all of the questions.
$endgroup$
– KieranSQ
Jan 9 at 21:57




$begingroup$
My last question, how do you get $u_r=frac{u}{1-t}$? I cannot see why - apologies for all of the questions.
$endgroup$
– KieranSQ
Jan 9 at 21:57




1




1




$begingroup$
The line $x+(2z-1)t=c$ that goes through the rarefaction center $(x,t)=(0,1)$ requires $c=(2z-1)$, so that $x=-(2z-1)(t-1)$, the slope of that line is thus $dot x=1-2z=frac{x}{t-1}$. I see that re-using $u$ is a bad idea, I'll change that to $v$.
$endgroup$
– LutzL
Jan 9 at 22:06




$begingroup$
The line $x+(2z-1)t=c$ that goes through the rarefaction center $(x,t)=(0,1)$ requires $c=(2z-1)$, so that $x=-(2z-1)(t-1)$, the slope of that line is thus $dot x=1-2z=frac{x}{t-1}$. I see that re-using $u$ is a bad idea, I'll change that to $v$.
$endgroup$
– LutzL
Jan 9 at 22:06


















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