Mixing
Numerical Solution to Advection Equation
$begingroup$
I was wondering if classic Schemes like Lax-Friedrichs, Lax-Wendroff or Upwind Schemes work for the following PDE
$$dfrac{partial u}{partial t}+e^{-x}(cos(t)+2)dfrac{partial u}{partial x}=1+u^2$$ with initial condition
$$u(x,0) = e^{-x^2}$$
I've tried solving them with the schemes mentioned above with different time step $$Delta t leq dfrac{Delta x}{max|a(x,t)|}$$ where $a(x,t)=e^{-x}(cos(t)+2)$ but I always end up with unstable solutions. If anyone could point me in a direction or a source that would be great!
numerical-methods computational-mathematics hyperbolic-equations
edited Aug 8 '18 at 19:50
Harry49
6,44631133
asked Jun 2 '18 at 19:06
Ken CaluyaKen Caluya
412
$endgroup$
add a comment |
$begingroup$
I was wondering if classic Schemes like Lax-Friedrichs, Lax-Wendroff or Upwind Schemes work for the following PDE
$$dfrac{partial u}{partial t}+e^{-x}(cos(t)+2)dfrac{partial u}{partial x}=1+u^2$$ with initial condition
$$u(x,0) = e^{-x^2}$$
I've tried solving them with the schemes mentioned above with different time step $$Delta t leq dfrac{Delta x}{max|a(x,t)|}$$ where $a(x,t)=e^{-x}(cos(t)+2)$ but I always end up with unstable solutions. If anyone could point me in a direction or a source that would be great!
numerical-methods computational-mathematics hyperbolic-equations
edited Aug 8 '18 at 19:50
Harry49
6,44631133
asked Jun 2 '18 at 19:06
Ken CaluyaKen Caluya
412
$endgroup$
$begingroup$
What is the spatial domain? In any case, along the characteristics you will indeed see finite time blowup because of the $1+u^2$ on the right hand side.
$endgroup$
– Ian
Jun 2 '18 at 19:10
$begingroup$
The spatial domain is from [0,5] and the solution using the method of characteristics is u(x,t) = tan(t+arctan(e^{-[ln(e^x-(sin(t)+2t))]^2})
$endgroup$
– Ken Caluya
Jun 2 '18 at 19:17
$begingroup$
The solution is of the form $tan(t+b(x,t))$ where $b$ is a bounded function. This $tan$ is where the blowup is coming from.
$endgroup$
– Ian
Jun 2 '18 at 19:20
add a comment |
$begingroup$
I was wondering if classic Schemes like Lax-Friedrichs, Lax-Wendroff or Upwind Schemes work for the following PDE
$$dfrac{partial u}{partial t}+e^{-x}(cos(t)+2)dfrac{partial u}{partial x}=1+u^2$$ with initial condition
$$u(x,0) = e^{-x^2}$$
I've tried solving them with the schemes mentioned above with different time step $$Delta t leq dfrac{Delta x}{max|a(x,t)|}$$ where $a(x,t)=e^{-x}(cos(t)+2)$ but I always end up with unstable solutions. If anyone could point me in a direction or a source that would be great!
numerical-methods computational-mathematics hyperbolic-equations
edited Aug 8 '18 at 19:50
Harry49
6,44631133
asked Jun 2 '18 at 19:06
Ken CaluyaKen Caluya
412
$endgroup$
I was wondering if classic Schemes like Lax-Friedrichs, Lax-Wendroff or Upwind Schemes work for the following PDE
$$dfrac{partial u}{partial t}+e^{-x}(cos(t)+2)dfrac{partial u}{partial x}=1+u^2$$ with initial condition
$$u(x,0) = e^{-x^2}$$
I've tried solving them with the schemes mentioned above with different time step $$Delta t leq dfrac{Delta x}{max|a(x,t)|}$$ where $a(x,t)=e^{-x}(cos(t)+2)$ but I always end up with unstable solutions. If anyone could point me in a direction or a source that would be great!
numerical-methods computational-mathematics hyperbolic-equations
numerical-methods computational-mathematics hyperbolic-equations
edited Aug 8 '18 at 19:50
Harry49
6,44631133
asked Jun 2 '18 at 19:06
Ken CaluyaKen Caluya
412
edited Aug 8 '18 at 19:50
Harry49
6,44631133
asked Jun 2 '18 at 19:06
Ken CaluyaKen Caluya
412
edited Aug 8 '18 at 19:50
Harry49
6,44631133
edited Aug 8 '18 at 19:50
Harry49
6,44631133
edited Aug 8 '18 at 19:50
Harry49
6,44631133
6,44631133
asked Jun 2 '18 at 19:06
Ken CaluyaKen Caluya
412
asked Jun 2 '18 at 19:06
Ken CaluyaKen Caluya
412
asked Jun 2 '18 at 19:06
Ken CaluyaKen Caluya
412
412
$begingroup$
What is the spatial domain? In any case, along the characteristics you will indeed see finite time blowup because of the $1+u^2$ on the right hand side.
$endgroup$
– Ian
Jun 2 '18 at 19:10
$begingroup$
The spatial domain is from [0,5] and the solution using the method of characteristics is u(x,t) = tan(t+arctan(e^{-[ln(e^x-(sin(t)+2t))]^2})
$endgroup$
– Ken Caluya
Jun 2 '18 at 19:17
$begingroup$
The solution is of the form $tan(t+b(x,t))$ where $b$ is a bounded function. This $tan$ is where the blowup is coming from.
$endgroup$
– Ian
Jun 2 '18 at 19:20
add a comment |
$begingroup$
What is the spatial domain? In any case, along the characteristics you will indeed see finite time blowup because of the $1+u^2$ on the right hand side.
$endgroup$
– Ian
Jun 2 '18 at 19:10
$begingroup$
The spatial domain is from [0,5] and the solution using the method of characteristics is u(x,t) = tan(t+arctan(e^{-[ln(e^x-(sin(t)+2t))]^2})
$endgroup$
– Ken Caluya
Jun 2 '18 at 19:17
$begingroup$
The solution is of the form $tan(t+b(x,t))$ where $b$ is a bounded function. This $tan$ is where the blowup is coming from.
$endgroup$
– Ian
Jun 2 '18 at 19:20
$begingroup$
What is the spatial domain? In any case, along the characteristics you will indeed see finite time blowup because of the $1+u^2$ on the right hand side.
$endgroup$
– Ian
Jun 2 '18 at 19:10
$begingroup$
What is the spatial domain? In any case, along the characteristics you will indeed see finite time blowup because of the $1+u^2$ on the right hand side.
$endgroup$
– Ian
Jun 2 '18 at 19:10
$begingroup$
The spatial domain is from [0,5] and the solution using the method of characteristics is u(x,t) = tan(t+arctan(e^{-[ln(e^x-(sin(t)+2t))]^2})
$endgroup$
– Ken Caluya
Jun 2 '18 at 19:17
$begingroup$
The spatial domain is from [0,5] and the solution using the method of characteristics is u(x,t) = tan(t+arctan(e^{-[ln(e^x-(sin(t)+2t))]^2})
$endgroup$
– Ken Caluya
Jun 2 '18 at 19:17
$begingroup$
The solution is of the form $tan(t+b(x,t))$ where $b$ is a bounded function. This $tan$ is where the blowup is coming from.
$endgroup$
– Ian
Jun 2 '18 at 19:20
$begingroup$
The solution is of the form $tan(t+b(x,t))$ where $b$ is a bounded function. This $tan$ is where the blowup is coming from.
$endgroup$
– Ian
Jun 2 '18 at 19:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We consider here a balance law of the form
$$
partial_t u + a partial_x u = r (u)
$$
where $a = e^{-x}(cos t + 2)$ and $r (u) = 1+u^2$. As indicated in the comments, the solution from the method of characteristics blows up in finite time. Nevertheless, let us consider numerical integration up to this blow-up time. We write an explicit finite-volume discretization using the Lax-Friedrichs method:
$$
u_i^{n+1} = frac{1}{2}(u_{i-1}^n+u_{i+1}^n) - frac{Delta t}{2Delta x}a_i^n left(u_{i+1}^n- u_{i-1}^nright) + Delta t, r(u_i^n), ,
$$
where $u_i^nsimeq u(iDelta x, nDelta t)$ and $a_i^n = a(iDelta x, nDelta t)$.
This method is stable under the condition
$$
Delta t leq min_i min leftlbrace frac{Delta x}{|a_i^n|}, frac{2}{|r'(u_i^n)|}rightrbrace .
$$
The first stability restriction corresponds to the Courant-Friedrichs-Lewy (CFL) condition, while the second stability restriction is due to the first-order explicit time integration (forward Euler integration). Depending on the value of $r(u_i^n)$, this may be penalizing.
An alternative strategy relies on operator splitting, and unconditionally stable integration of the relaxation term. A Godunov splitting scheme based on the Lax-Friedrichs method and the backward Euler method reads
begin{aligned}
u_i^{*} &= frac{1}{2}(u_{i-1}^n+u_{i+1}^n) - frac{Delta t}{2Delta x}a_i^n left(u_{i+1}^n- u_{i-1}^nright) \
u_i^{n+1} &= u_i^* + Delta t, r(u_i^{n+1})
end{aligned}
which is stable under the CFL condition.
answered Aug 8 '18 at 20:29
Harry49Harry49
6,44631133
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
We consider here a balance law of the form
$$
partial_t u + a partial_x u = r (u)
$$
where $a = e^{-x}(cos t + 2)$ and $r (u) = 1+u^2$. As indicated in the comments, the solution from the method of characteristics blows up in finite time. Nevertheless, let us consider numerical integration up to this blow-up time. We write an explicit finite-volume discretization using the Lax-Friedrichs method:
$$
u_i^{n+1} = frac{1}{2}(u_{i-1}^n+u_{i+1}^n) - frac{Delta t}{2Delta x}a_i^n left(u_{i+1}^n- u_{i-1}^nright) + Delta t, r(u_i^n), ,
$$
where $u_i^nsimeq u(iDelta x, nDelta t)$ and $a_i^n = a(iDelta x, nDelta t)$.
This method is stable under the condition
$$
Delta t leq min_i min leftlbrace frac{Delta x}{|a_i^n|}, frac{2}{|r'(u_i^n)|}rightrbrace .
$$
The first stability restriction corresponds to the Courant-Friedrichs-Lewy (CFL) condition, while the second stability restriction is due to the first-order explicit time integration (forward Euler integration). Depending on the value of $r(u_i^n)$, this may be penalizing.
An alternative strategy relies on operator splitting, and unconditionally stable integration of the relaxation term. A Godunov splitting scheme based on the Lax-Friedrichs method and the backward Euler method reads
begin{aligned}
u_i^{*} &= frac{1}{2}(u_{i-1}^n+u_{i+1}^n) - frac{Delta t}{2Delta x}a_i^n left(u_{i+1}^n- u_{i-1}^nright) \
u_i^{n+1} &= u_i^* + Delta t, r(u_i^{n+1})
end{aligned}
which is stable under the CFL condition.
answered Aug 8 '18 at 20:29
Harry49Harry49
6,44631133
$endgroup$
add a comment |
$begingroup$
We consider here a balance law of the form
$$
partial_t u + a partial_x u = r (u)
$$
where $a = e^{-x}(cos t + 2)$ and $r (u) = 1+u^2$. As indicated in the comments, the solution from the method of characteristics blows up in finite time. Nevertheless, let us consider numerical integration up to this blow-up time. We write an explicit finite-volume discretization using the Lax-Friedrichs method:
$$
u_i^{n+1} = frac{1}{2}(u_{i-1}^n+u_{i+1}^n) - frac{Delta t}{2Delta x}a_i^n left(u_{i+1}^n- u_{i-1}^nright) + Delta t, r(u_i^n), ,
$$
where $u_i^nsimeq u(iDelta x, nDelta t)$ and $a_i^n = a(iDelta x, nDelta t)$.
This method is stable under the condition
$$
Delta t leq min_i min leftlbrace frac{Delta x}{|a_i^n|}, frac{2}{|r'(u_i^n)|}rightrbrace .
$$
The first stability restriction corresponds to the Courant-Friedrichs-Lewy (CFL) condition, while the second stability restriction is due to the first-order explicit time integration (forward Euler integration). Depending on the value of $r(u_i^n)$, this may be penalizing.
An alternative strategy relies on operator splitting, and unconditionally stable integration of the relaxation term. A Godunov splitting scheme based on the Lax-Friedrichs method and the backward Euler method reads
begin{aligned}
u_i^{*} &= frac{1}{2}(u_{i-1}^n+u_{i+1}^n) - frac{Delta t}{2Delta x}a_i^n left(u_{i+1}^n- u_{i-1}^nright) \
u_i^{n+1} &= u_i^* + Delta t, r(u_i^{n+1})
end{aligned}
which is stable under the CFL condition.
answered Aug 8 '18 at 20:29
Harry49Harry49
6,44631133
$endgroup$
add a comment |
$begingroup$
We consider here a balance law of the form
$$
partial_t u + a partial_x u = r (u)
$$
where $a = e^{-x}(cos t + 2)$ and $r (u) = 1+u^2$. As indicated in the comments, the solution from the method of characteristics blows up in finite time. Nevertheless, let us consider numerical integration up to this blow-up time. We write an explicit finite-volume discretization using the Lax-Friedrichs method:
$$
u_i^{n+1} = frac{1}{2}(u_{i-1}^n+u_{i+1}^n) - frac{Delta t}{2Delta x}a_i^n left(u_{i+1}^n- u_{i-1}^nright) + Delta t, r(u_i^n), ,
$$
where $u_i^nsimeq u(iDelta x, nDelta t)$ and $a_i^n = a(iDelta x, nDelta t)$.
This method is stable under the condition
$$
Delta t leq min_i min leftlbrace frac{Delta x}{|a_i^n|}, frac{2}{|r'(u_i^n)|}rightrbrace .
$$
The first stability restriction corresponds to the Courant-Friedrichs-Lewy (CFL) condition, while the second stability restriction is due to the first-order explicit time integration (forward Euler integration). Depending on the value of $r(u_i^n)$, this may be penalizing.
An alternative strategy relies on operator splitting, and unconditionally stable integration of the relaxation term. A Godunov splitting scheme based on the Lax-Friedrichs method and the backward Euler method reads
begin{aligned}
u_i^{*} &= frac{1}{2}(u_{i-1}^n+u_{i+1}^n) - frac{Delta t}{2Delta x}a_i^n left(u_{i+1}^n- u_{i-1}^nright) \
u_i^{n+1} &= u_i^* + Delta t, r(u_i^{n+1})
end{aligned}
which is stable under the CFL condition.
answered Aug 8 '18 at 20:29
Harry49Harry49
6,44631133
$endgroup$
We consider here a balance law of the form
$$
partial_t u + a partial_x u = r (u)
$$
where $a = e^{-x}(cos t + 2)$ and $r (u) = 1+u^2$. As indicated in the comments, the solution from the method of characteristics blows up in finite time. Nevertheless, let us consider numerical integration up to this blow-up time. We write an explicit finite-volume discretization using the Lax-Friedrichs method:
$$
u_i^{n+1} = frac{1}{2}(u_{i-1}^n+u_{i+1}^n) - frac{Delta t}{2Delta x}a_i^n left(u_{i+1}^n- u_{i-1}^nright) + Delta t, r(u_i^n), ,
$$
where $u_i^nsimeq u(iDelta x, nDelta t)$ and $a_i^n = a(iDelta x, nDelta t)$.
This method is stable under the condition
$$
Delta t leq min_i min leftlbrace frac{Delta x}{|a_i^n|}, frac{2}{|r'(u_i^n)|}rightrbrace .
$$
The first stability restriction corresponds to the Courant-Friedrichs-Lewy (CFL) condition, while the second stability restriction is due to the first-order explicit time integration (forward Euler integration). Depending on the value of $r(u_i^n)$, this may be penalizing.
An alternative strategy relies on operator splitting, and unconditionally stable integration of the relaxation term. A Godunov splitting scheme based on the Lax-Friedrichs method and the backward Euler method reads
begin{aligned}
u_i^{*} &= frac{1}{2}(u_{i-1}^n+u_{i+1}^n) - frac{Delta t}{2Delta x}a_i^n left(u_{i+1}^n- u_{i-1}^nright) \
u_i^{n+1} &= u_i^* + Delta t, r(u_i^{n+1})
end{aligned}
which is stable under the CFL condition.
answered Aug 8 '18 at 20:29
Harry49Harry49
6,44631133
edited Jan 13 at 16:59
answered Aug 8 '18 at 20:29
Harry49Harry49
6,44631133
answered Aug 8 '18 at 20:29
Harry49Harry49
6,44631133
answered Aug 8 '18 at 20:29
Harry49Harry49
6,44631133
6,44631133
add a comment |
add a comment |
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$begingroup$
What is the spatial domain? In any case, along the characteristics you will indeed see finite time blowup because of the $1+u^2$ on the right hand side.
$endgroup$
– Ian
Jun 2 '18 at 19:10
$begingroup$
The spatial domain is from [0,5] and the solution using the method of characteristics is u(x,t) = tan(t+arctan(e^{-[ln(e^x-(sin(t)+2t))]^2})
$endgroup$
– Ken Caluya
Jun 2 '18 at 19:17
$begingroup$
The solution is of the form $tan(t+b(x,t))$ where $b$ is a bounded function. This $tan$ is where the blowup is coming from.
$endgroup$
– Ian
Jun 2 '18 at 19:20