Message transfered on Computational Science - Solving advection equation - periodic conditions

Message transfered on Computational Science - Solving advection equation - periodic conditions - using roll...

Message transfered on Computational Science :

cf https://scicomp.stackexchange.com/questions/30635/solving-advection-equation-periodic-conditions-using-roll-python-function

I have to solve numerically the advection equation with periodic boundaries conditions : u(t,0) = u(t,L) with L the length of system to solve.

I start also with u(0,x) = uexacte(0,x) = sin(2*pi*x/L)

Here the main part of the code with loop time (we use here FTCS scheme) :

V=1

L=1

# analytical solution --------------------------

def uexacte(t,x):

    return sin(2*pi*(x-V*t)/L)



# 1. Centre FTCS (Forward Time Centered Space)

cfl = 0.25

nx = 10

tend = 1



#

dx = L/(nx-1.)

dt = cfl*dx/V

nt = int(tend/dt)+1

print "CFL=%5.2f tend=%4.1f --> %i iterations en temps"%(cfl,tend,nt)



# Arrays

x = linspace(0,L,nx)

# Bounadry condition

u0 = uexacte(0,x)



# Starting solution

t=0.0 ; u=copy(u0)



# Time loop

for i in range(1,nt):

    # FTCS

    #u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])

    # Using roll

    u = u + - cfl/2*(roll(u,-1)- roll(u,1))

    # Update time

    t = t+dt

I don't understand the solution given by teacher who uses the python function roll in this way :

# Using roll

u = u - cfl/2*(roll(u,-1)- roll(u,1))

One says that with the using of roll, we are sure to respect the periodic boundary conditions but I don't understand why ?

Indeed, my first approach was to do :

u[0] = u[nx-1]

u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])

but this doesn't work and I don't know how to implement this periodic conditions in this way (without using roll function).

If someone could explain this matter and the trick with roll function, this would be nice to tell it.

UPDATE 1 :

I tried with classical approach (simple recurrence formula ) like this :

# Time loop

for i in range(1,nt):

    # FTCS

    u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])



    # Try to impose periodic boundary conditions but without success

    u[0] = u[0] - cfl/2*(u[0] - u[nx-1])



    # Update time

    t = t+dt

Indeed, the result is bad (values for each side are not the same). I could impose the theorical values at each step but in practise, we don't know always the analytical solution.

What is the trick to impose this periodic boundary condition on the numerical solution at each step of time ?

edited Dec 3 '18 at 10:45

asked Nov 27 '18 at 16:30

youpilat13

8221441

add a comment |

Message transfered on Computational Science :

cf https://scicomp.stackexchange.com/questions/30635/solving-advection-equation-periodic-conditions-using-roll-python-function

I have to solve numerically the advection equation with periodic boundaries conditions : u(t,0) = u(t,L) with L the length of system to solve.

I start also with u(0,x) = uexacte(0,x) = sin(2*pi*x/L)

Here the main part of the code with loop time (we use here FTCS scheme) :

V=1

L=1

# analytical solution --------------------------

def uexacte(t,x):

    return sin(2*pi*(x-V*t)/L)



# 1. Centre FTCS (Forward Time Centered Space)

cfl = 0.25

nx = 10

tend = 1



#

dx = L/(nx-1.)

dt = cfl*dx/V

nt = int(tend/dt)+1

print "CFL=%5.2f tend=%4.1f --> %i iterations en temps"%(cfl,tend,nt)



# Arrays

x = linspace(0,L,nx)

# Bounadry condition

u0 = uexacte(0,x)



# Starting solution

t=0.0 ; u=copy(u0)



# Time loop

for i in range(1,nt):

    # FTCS

    #u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])

    # Using roll

    u = u + - cfl/2*(roll(u,-1)- roll(u,1))

    # Update time

    t = t+dt

I don't understand the solution given by teacher who uses the python function roll in this way :

# Using roll

u = u - cfl/2*(roll(u,-1)- roll(u,1))

One says that with the using of roll, we are sure to respect the periodic boundary conditions but I don't understand why ?

Indeed, my first approach was to do :

u[0] = u[nx-1]

u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])

but this doesn't work and I don't know how to implement this periodic conditions in this way (without using roll function).

If someone could explain this matter and the trick with roll function, this would be nice to tell it.

UPDATE 1 :

I tried with classical approach (simple recurrence formula ) like this :

# Time loop

for i in range(1,nt):

    # FTCS

    u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])



    # Try to impose periodic boundary conditions but without success

    u[0] = u[0] - cfl/2*(u[0] - u[nx-1])



    # Update time

    t = t+dt

Indeed, the result is bad (values for each side are not the same). I could impose the theorical values at each step but in practise, we don't know always the analytical solution.

What is the trick to impose this periodic boundary condition on the numerical solution at each step of time ?

edited Dec 3 '18 at 10:45

asked Nov 27 '18 at 16:30

youpilat13

8221441

add a comment |

Message transfered on Computational Science :

cf https://scicomp.stackexchange.com/questions/30635/solving-advection-equation-periodic-conditions-using-roll-python-function

I have to solve numerically the advection equation with periodic boundaries conditions : u(t,0) = u(t,L) with L the length of system to solve.

I start also with u(0,x) = uexacte(0,x) = sin(2*pi*x/L)

Here the main part of the code with loop time (we use here FTCS scheme) :

V=1

L=1

# analytical solution --------------------------

def uexacte(t,x):

    return sin(2*pi*(x-V*t)/L)



# 1. Centre FTCS (Forward Time Centered Space)

cfl = 0.25

nx = 10

tend = 1



#

dx = L/(nx-1.)

dt = cfl*dx/V

nt = int(tend/dt)+1

print "CFL=%5.2f tend=%4.1f --> %i iterations en temps"%(cfl,tend,nt)



# Arrays

x = linspace(0,L,nx)

# Bounadry condition

u0 = uexacte(0,x)



# Starting solution

t=0.0 ; u=copy(u0)



# Time loop

for i in range(1,nt):

    # FTCS

    #u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])

    # Using roll

    u = u + - cfl/2*(roll(u,-1)- roll(u,1))

    # Update time

    t = t+dt

I don't understand the solution given by teacher who uses the python function roll in this way :

# Using roll

u = u - cfl/2*(roll(u,-1)- roll(u,1))

One says that with the using of roll, we are sure to respect the periodic boundary conditions but I don't understand why ?

Indeed, my first approach was to do :

u[0] = u[nx-1]

u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])

but this doesn't work and I don't know how to implement this periodic conditions in this way (without using roll function).

If someone could explain this matter and the trick with roll function, this would be nice to tell it.

UPDATE 1 :

I tried with classical approach (simple recurrence formula ) like this :

# Time loop

for i in range(1,nt):

    # FTCS

    u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])



    # Try to impose periodic boundary conditions but without success

    u[0] = u[0] - cfl/2*(u[0] - u[nx-1])



    # Update time

    t = t+dt

Indeed, the result is bad (values for each side are not the same). I could impose the theorical values at each step but in practise, we don't know always the analytical solution.

What is the trick to impose this periodic boundary condition on the numerical solution at each step of time ?

edited Dec 3 '18 at 10:45

asked Nov 27 '18 at 16:30

youpilat13

8221441

Message transfered on Computational Science :

cf https://scicomp.stackexchange.com/questions/30635/solving-advection-equation-periodic-conditions-using-roll-python-function

I have to solve numerically the advection equation with periodic boundaries conditions : u(t,0) = u(t,L) with L the length of system to solve.

I start also with u(0,x) = uexacte(0,x) = sin(2*pi*x/L)

Here the main part of the code with loop time (we use here FTCS scheme) :

V=1

L=1

# analytical solution --------------------------

def uexacte(t,x):

    return sin(2*pi*(x-V*t)/L)



# 1. Centre FTCS (Forward Time Centered Space)

cfl = 0.25

nx = 10

tend = 1



#

dx = L/(nx-1.)

dt = cfl*dx/V

nt = int(tend/dt)+1

print "CFL=%5.2f tend=%4.1f --> %i iterations en temps"%(cfl,tend,nt)



# Arrays

x = linspace(0,L,nx)

# Bounadry condition

u0 = uexacte(0,x)



# Starting solution

t=0.0 ; u=copy(u0)



# Time loop

for i in range(1,nt):

    # FTCS

    #u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])

    # Using roll

    u = u + - cfl/2*(roll(u,-1)- roll(u,1))

    # Update time

    t = t+dt

I don't understand the solution given by teacher who uses the python function roll in this way :

# Using roll

u = u - cfl/2*(roll(u,-1)- roll(u,1))

One says that with the using of roll, we are sure to respect the periodic boundary conditions but I don't understand why ?

Indeed, my first approach was to do :

u[0] = u[nx-1]

u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])

but this doesn't work and I don't know how to implement this periodic conditions in this way (without using roll function).

If someone could explain this matter and the trick with roll function, this would be nice to tell it.

UPDATE 1 :

I tried with classical approach (simple recurrence formula ) like this :

# Time loop

for i in range(1,nt):

    # FTCS

    u[1:nx-1] = u[1:nx-1] - cfl/2*(u[2:nx] - u[0:nx-2])



    # Try to impose periodic boundary conditions but without success

    u[0] = u[0] - cfl/2*(u[0] - u[nx-1])



    # Update time

    t = t+dt

Indeed, the result is bad (values for each side are not the same). I could impose the theorical values at each step but in practise, we don't know always the analytical solution.

What is the trick to impose this periodic boundary condition on the numerical solution at each step of time ?

python boundary periodicity

edited Dec 3 '18 at 10:45

asked Nov 27 '18 at 16:30

youpilat13

8221441

edited Dec 3 '18 at 10:45

asked Nov 27 '18 at 16:30

youpilat13

8221441

edited Dec 3 '18 at 10:45

asked Nov 27 '18 at 16:30

youpilat13

8221441

asked Nov 27 '18 at 16:30

youpilat13

8221441

asked Nov 27 '18 at 16:30

youpilat13

8221441

add a comment |

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