Inviscid Burgers' equation with sinusoidal initial data
$begingroup$
I have a question about the following Burgers' equation.
$u_t + (frac12u^2)_x = 0 $
with $u(x,0) = sin(x)$ on $[0,2pi]$ and periodic boundary conditions.
When I studied this equation numerically, I notice that once the shock forms, the shock stays still in the same place, yet the magnitude of the function decreases as time passes. That is, the initial function $sin(x)$ ranges from $-1$ to $1$, but as time passes, the magnitude of the maximum and minimum of the function decays.
Is this what is supposed to happen analytically?
pde hyperbolic-equations
edited Dec 10 '18 at 9:10
Harry49
6,85631237
asked Nov 15 '14 at 5:18
monononomononono
883615
$endgroup$
add a comment |
$begingroup$
I have a question about the following Burgers' equation.
$u_t + (frac12u^2)_x = 0 $
with $u(x,0) = sin(x)$ on $[0,2pi]$ and periodic boundary conditions.
When I studied this equation numerically, I notice that once the shock forms, the shock stays still in the same place, yet the magnitude of the function decreases as time passes. That is, the initial function $sin(x)$ ranges from $-1$ to $1$, but as time passes, the magnitude of the maximum and minimum of the function decays.
Is this what is supposed to happen analytically?
pde hyperbolic-equations
edited Dec 10 '18 at 9:10
Harry49
6,85631237
asked Nov 15 '14 at 5:18
monononomononono
883615
$endgroup$
add a comment |
$begingroup$
I have a question about the following Burgers' equation.
$u_t + (frac12u^2)_x = 0 $
with $u(x,0) = sin(x)$ on $[0,2pi]$ and periodic boundary conditions.
When I studied this equation numerically, I notice that once the shock forms, the shock stays still in the same place, yet the magnitude of the function decreases as time passes. That is, the initial function $sin(x)$ ranges from $-1$ to $1$, but as time passes, the magnitude of the maximum and minimum of the function decays.
Is this what is supposed to happen analytically?
pde hyperbolic-equations
edited Dec 10 '18 at 9:10
Harry49
6,85631237
asked Nov 15 '14 at 5:18
monononomononono
883615
$endgroup$
I have a question about the following Burgers' equation.
$u_t + (frac12u^2)_x = 0 $
with $u(x,0) = sin(x)$ on $[0,2pi]$ and periodic boundary conditions.
When I studied this equation numerically, I notice that once the shock forms, the shock stays still in the same place, yet the magnitude of the function decreases as time passes. That is, the initial function $sin(x)$ ranges from $-1$ to $1$, but as time passes, the magnitude of the maximum and minimum of the function decays.
Is this what is supposed to happen analytically?
pde hyperbolic-equations
pde hyperbolic-equations
edited Dec 10 '18 at 9:10
Harry49
6,85631237
asked Nov 15 '14 at 5:18
monononomononono
883615
edited Dec 10 '18 at 9:10
Harry49
6,85631237
asked Nov 15 '14 at 5:18
monononomononono
883615
edited Dec 10 '18 at 9:10
Harry49
6,85631237
edited Dec 10 '18 at 9:10
Harry49
6,85631237
edited Dec 10 '18 at 9:10
Harry49
6,85631237
6,85631237
asked Nov 15 '14 at 5:18
monononomononono
883615
asked Nov 15 '14 at 5:18
monononomononono
883615
asked Nov 15 '14 at 5:18
monononomononono
883615
883615
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The method of characteristics gives $u=sin(x-ut)$. This implicit equation can be used to approach the solution, until a shock wave is met by the characteristics. Here is a sketch of the characteristics in the $x$-$t$ plane:
The shock is formed when characteristics intersect for the first time, i.e. at the breaking time
begin{aligned}
t_B &= frac{-1}{min partial_x u(x,0)} =1 , ,
end{aligned}
as shown in the figure.
The value $u_L$ of $u$ on the left of the shock is deduced from the characteristic curves
$$
u_L(x_s(t),t) = sinleft[x_s(t)-u_L(x_s(t),t), tright]
$$
passing through the shock located at $x_s(t)$.
The value $u_R$ on the right is obtained similarly. The position $x_s(t)$ of the shock wave is given by the Rankine-Hugoniot condition $x'_s(t) = frac{1}{2} (u_L+u_R)$, with initial position $x_s(1) = pi$. Therefore, using symmetry, $u_R=-u_L$ and the shock wave stays at the same place. Its magnitude $u_L$ increases and then vanishes in time, as shown in the figure below, where $u_L$ --obtained numerically-- is plotted with respect to $t$:
answered Dec 5 '17 at 12:48
Harry49Harry49
6,85631237
$endgroup$
add a comment |
$begingroup$
No, analytically that is not what happens. The equation in your question is discussed in Miller's Applied Asymptotic Analysis, and I quote from the middle of p. 78:
Once shocks form, there is no longer a global unique solution of the nonlinear initial-value problem.
It turns out that it's possible to modify the equation in such a way that the shocks are stable (though still move as time progresses, if I recall correctly) by adding a diffusion term. I definitely recommend reading section 3.6 of Miller's book.
answered Nov 15 '14 at 6:57
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
1
$begingroup$
@mononono Note that this chapter of Miller is about the method of characteristics (with implicit solution $u=sin(x−ut)$), which is known to work only until a shock wave is encountered (see my answer for details).
$endgroup$
– Harry49
Dec 5 '17 at 13:14
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The method of characteristics gives $u=sin(x-ut)$. This implicit equation can be used to approach the solution, until a shock wave is met by the characteristics. Here is a sketch of the characteristics in the $x$-$t$ plane:
The shock is formed when characteristics intersect for the first time, i.e. at the breaking time
begin{aligned}
t_B &= frac{-1}{min partial_x u(x,0)} =1 , ,
end{aligned}
as shown in the figure.
The value $u_L$ of $u$ on the left of the shock is deduced from the characteristic curves
$$
u_L(x_s(t),t) = sinleft[x_s(t)-u_L(x_s(t),t), tright]
$$
passing through the shock located at $x_s(t)$.
The value $u_R$ on the right is obtained similarly. The position $x_s(t)$ of the shock wave is given by the Rankine-Hugoniot condition $x'_s(t) = frac{1}{2} (u_L+u_R)$, with initial position $x_s(1) = pi$. Therefore, using symmetry, $u_R=-u_L$ and the shock wave stays at the same place. Its magnitude $u_L$ increases and then vanishes in time, as shown in the figure below, where $u_L$ --obtained numerically-- is plotted with respect to $t$:
answered Dec 5 '17 at 12:48
Harry49Harry49
6,85631237
$endgroup$
add a comment |
$begingroup$
The method of characteristics gives $u=sin(x-ut)$. This implicit equation can be used to approach the solution, until a shock wave is met by the characteristics. Here is a sketch of the characteristics in the $x$-$t$ plane:
The shock is formed when characteristics intersect for the first time, i.e. at the breaking time
begin{aligned}
t_B &= frac{-1}{min partial_x u(x,0)} =1 , ,
end{aligned}
as shown in the figure.
The value $u_L$ of $u$ on the left of the shock is deduced from the characteristic curves
$$
u_L(x_s(t),t) = sinleft[x_s(t)-u_L(x_s(t),t), tright]
$$
passing through the shock located at $x_s(t)$.
The value $u_R$ on the right is obtained similarly. The position $x_s(t)$ of the shock wave is given by the Rankine-Hugoniot condition $x'_s(t) = frac{1}{2} (u_L+u_R)$, with initial position $x_s(1) = pi$. Therefore, using symmetry, $u_R=-u_L$ and the shock wave stays at the same place. Its magnitude $u_L$ increases and then vanishes in time, as shown in the figure below, where $u_L$ --obtained numerically-- is plotted with respect to $t$:
answered Dec 5 '17 at 12:48
Harry49Harry49
6,85631237
$endgroup$
add a comment |
$begingroup$
The method of characteristics gives $u=sin(x-ut)$. This implicit equation can be used to approach the solution, until a shock wave is met by the characteristics. Here is a sketch of the characteristics in the $x$-$t$ plane:
The shock is formed when characteristics intersect for the first time, i.e. at the breaking time
begin{aligned}
t_B &= frac{-1}{min partial_x u(x,0)} =1 , ,
end{aligned}
as shown in the figure.
The value $u_L$ of $u$ on the left of the shock is deduced from the characteristic curves
$$
u_L(x_s(t),t) = sinleft[x_s(t)-u_L(x_s(t),t), tright]
$$
passing through the shock located at $x_s(t)$.
The value $u_R$ on the right is obtained similarly. The position $x_s(t)$ of the shock wave is given by the Rankine-Hugoniot condition $x'_s(t) = frac{1}{2} (u_L+u_R)$, with initial position $x_s(1) = pi$. Therefore, using symmetry, $u_R=-u_L$ and the shock wave stays at the same place. Its magnitude $u_L$ increases and then vanishes in time, as shown in the figure below, where $u_L$ --obtained numerically-- is plotted with respect to $t$:
answered Dec 5 '17 at 12:48
Harry49Harry49
6,85631237
$endgroup$
The method of characteristics gives $u=sin(x-ut)$. This implicit equation can be used to approach the solution, until a shock wave is met by the characteristics. Here is a sketch of the characteristics in the $x$-$t$ plane:
The shock is formed when characteristics intersect for the first time, i.e. at the breaking time
begin{aligned}
t_B &= frac{-1}{min partial_x u(x,0)} =1 , ,
end{aligned}
as shown in the figure.
The value $u_L$ of $u$ on the left of the shock is deduced from the characteristic curves
$$
u_L(x_s(t),t) = sinleft[x_s(t)-u_L(x_s(t),t), tright]
$$
passing through the shock located at $x_s(t)$.
The value $u_R$ on the right is obtained similarly. The position $x_s(t)$ of the shock wave is given by the Rankine-Hugoniot condition $x'_s(t) = frac{1}{2} (u_L+u_R)$, with initial position $x_s(1) = pi$. Therefore, using symmetry, $u_R=-u_L$ and the shock wave stays at the same place. Its magnitude $u_L$ increases and then vanishes in time, as shown in the figure below, where $u_L$ --obtained numerically-- is plotted with respect to $t$:
answered Dec 5 '17 at 12:48
Harry49Harry49
6,85631237
edited Dec 8 '17 at 12:55
answered Dec 5 '17 at 12:48
Harry49Harry49
6,85631237
answered Dec 5 '17 at 12:48
Harry49Harry49
6,85631237
answered Dec 5 '17 at 12:48
Harry49Harry49
6,85631237
6,85631237
add a comment |
add a comment |
$begingroup$
No, analytically that is not what happens. The equation in your question is discussed in Miller's Applied Asymptotic Analysis, and I quote from the middle of p. 78:
Once shocks form, there is no longer a global unique solution of the nonlinear initial-value problem.
It turns out that it's possible to modify the equation in such a way that the shocks are stable (though still move as time progresses, if I recall correctly) by adding a diffusion term. I definitely recommend reading section 3.6 of Miller's book.
answered Nov 15 '14 at 6:57
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
1
$begingroup$
@mononono Note that this chapter of Miller is about the method of characteristics (with implicit solution $u=sin(x−ut)$), which is known to work only until a shock wave is encountered (see my answer for details).
$endgroup$
– Harry49
Dec 5 '17 at 13:14
add a comment |
$begingroup$
No, analytically that is not what happens. The equation in your question is discussed in Miller's Applied Asymptotic Analysis, and I quote from the middle of p. 78:
Once shocks form, there is no longer a global unique solution of the nonlinear initial-value problem.
It turns out that it's possible to modify the equation in such a way that the shocks are stable (though still move as time progresses, if I recall correctly) by adding a diffusion term. I definitely recommend reading section 3.6 of Miller's book.
answered Nov 15 '14 at 6:57
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
1
$begingroup$
@mononono Note that this chapter of Miller is about the method of characteristics (with implicit solution $u=sin(x−ut)$), which is known to work only until a shock wave is encountered (see my answer for details).
$endgroup$
– Harry49
Dec 5 '17 at 13:14
add a comment |
$begingroup$
No, analytically that is not what happens. The equation in your question is discussed in Miller's Applied Asymptotic Analysis, and I quote from the middle of p. 78:
Once shocks form, there is no longer a global unique solution of the nonlinear initial-value problem.
It turns out that it's possible to modify the equation in such a way that the shocks are stable (though still move as time progresses, if I recall correctly) by adding a diffusion term. I definitely recommend reading section 3.6 of Miller's book.
answered Nov 15 '14 at 6:57
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
No, analytically that is not what happens. The equation in your question is discussed in Miller's Applied Asymptotic Analysis, and I quote from the middle of p. 78:
Once shocks form, there is no longer a global unique solution of the nonlinear initial-value problem.
It turns out that it's possible to modify the equation in such a way that the shocks are stable (though still move as time progresses, if I recall correctly) by adding a diffusion term. I definitely recommend reading section 3.6 of Miller's book.
answered Nov 15 '14 at 6:57
Antonio VargasAntonio Vargas
20.7k245111
answered Nov 15 '14 at 6:57
Antonio VargasAntonio Vargas
20.7k245111
answered Nov 15 '14 at 6:57
Antonio VargasAntonio Vargas
20.7k245111
answered Nov 15 '14 at 6:57
Antonio VargasAntonio Vargas
20.7k245111
20.7k245111
1
$begingroup$
@mononono Note that this chapter of Miller is about the method of characteristics (with implicit solution $u=sin(x−ut)$), which is known to work only until a shock wave is encountered (see my answer for details).
$endgroup$
– Harry49
Dec 5 '17 at 13:14
add a comment |
1
$begingroup$
@mononono Note that this chapter of Miller is about the method of characteristics (with implicit solution $u=sin(x−ut)$), which is known to work only until a shock wave is encountered (see my answer for details).
$endgroup$
– Harry49
Dec 5 '17 at 13:14
1
1
$begingroup$
@mononono Note that this chapter of Miller is about the method of characteristics (with implicit solution $u=sin(x−ut)$), which is known to work only until a shock wave is encountered (see my answer for details).
$endgroup$
– Harry49
Dec 5 '17 at 13:14
$begingroup$
@mononono Note that this chapter of Miller is about the method of characteristics (with implicit solution $u=sin(x−ut)$), which is known to work only until a shock wave is encountered (see my answer for details).
$endgroup$
– Harry49
Dec 5 '17 at 13:14
add a comment |
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