Applying neumann boundary conditions to diffusion equation solution in python
For the diffusion equation
$$frac{partial u(x,t)}{partial t}=Dfrac{partial^2 u(x,t)}{partial x^2}+Cu(x,t)$$
with the boundary conditions $u(−frac{L}{2},t)=u(frac{L}{2},t)=0$ I've programmed the numerical solution into python correctly (I think).
import numpy as np
import matplotlib.pyplot as plt
L=np.pi # value chosen for the critical length
s=101 # number of steps in x
t=10002 # number of timesteps
ds=L/(s-1) # step in x
dt=0.0001 # time step
D=1 # diffusion constant, set equal to 1
C=1 # creation rate of neutrons, set equal to 1
Alpha=(D*dt)/(ds*ds) # constant for diffusion term
Beta=C*dt # constant for u term
x = np.linspace(-L/2, 0, num=51)
x = np.concatenate([x, np.linspace(x[-1] - x[-2], L/2, num=50)]) # setting x in the specified interval
u=np.zeros(shape=(s,t)) #setting the function u
u[50,0]=1/ds # delta function
for k in range(0,t-1):
u[0,k]=0 # boundary conditions
u[s-1,k]=0
for i in range(1,s-1):
u[i,k+1]=(1+Beta-2*Alpha)*u[i,k]+Alpha*u[i+1,k]+Alpha*u[i-1,k] # numerical solution
if k == 50 or k == 100 or k == 250 or k == 500 or k == 1000 or k == 10000: # plotting at times
plt.plot(x,u[:,k])
plt.title('Numerical Solution of the Diffusion equation over time')
plt.xlabel('x')
plt.ylabel('u(x,t)')
plt.show()
However now I have to change the right boundary condition into $u_x(frac{L}{2})=0$ and I'm not really sure how to change my code to reflect this. If I do this the critical length should decrease and the function should start increasing exponentially, but everything I've tried usually does nothing to my plot - is there something wrong with my original code possibly? Any help is really appreciated, I've been trying for ages but can't seem to get it! Thanks!
differential-equations numerical-methods partial-derivative boundary-value-problem python
asked Nov 27 at 16:58
mcaiojethewo
356
add a comment |
For the diffusion equation
$$frac{partial u(x,t)}{partial t}=Dfrac{partial^2 u(x,t)}{partial x^2}+Cu(x,t)$$
with the boundary conditions $u(−frac{L}{2},t)=u(frac{L}{2},t)=0$ I've programmed the numerical solution into python correctly (I think).
import numpy as np
import matplotlib.pyplot as plt
L=np.pi # value chosen for the critical length
s=101 # number of steps in x
t=10002 # number of timesteps
ds=L/(s-1) # step in x
dt=0.0001 # time step
D=1 # diffusion constant, set equal to 1
C=1 # creation rate of neutrons, set equal to 1
Alpha=(D*dt)/(ds*ds) # constant for diffusion term
Beta=C*dt # constant for u term
x = np.linspace(-L/2, 0, num=51)
x = np.concatenate([x, np.linspace(x[-1] - x[-2], L/2, num=50)]) # setting x in the specified interval
u=np.zeros(shape=(s,t)) #setting the function u
u[50,0]=1/ds # delta function
for k in range(0,t-1):
u[0,k]=0 # boundary conditions
u[s-1,k]=0
for i in range(1,s-1):
u[i,k+1]=(1+Beta-2*Alpha)*u[i,k]+Alpha*u[i+1,k]+Alpha*u[i-1,k] # numerical solution
if k == 50 or k == 100 or k == 250 or k == 500 or k == 1000 or k == 10000: # plotting at times
plt.plot(x,u[:,k])
plt.title('Numerical Solution of the Diffusion equation over time')
plt.xlabel('x')
plt.ylabel('u(x,t)')
plt.show()
However now I have to change the right boundary condition into $u_x(frac{L}{2})=0$ and I'm not really sure how to change my code to reflect this. If I do this the critical length should decrease and the function should start increasing exponentially, but everything I've tried usually does nothing to my plot - is there something wrong with my original code possibly? Any help is really appreciated, I've been trying for ages but can't seem to get it! Thanks!
differential-equations numerical-methods partial-derivative boundary-value-problem python
asked Nov 27 at 16:58
mcaiojethewo
356
add a comment |
For the diffusion equation
$$frac{partial u(x,t)}{partial t}=Dfrac{partial^2 u(x,t)}{partial x^2}+Cu(x,t)$$
with the boundary conditions $u(−frac{L}{2},t)=u(frac{L}{2},t)=0$ I've programmed the numerical solution into python correctly (I think).
import numpy as np
import matplotlib.pyplot as plt
L=np.pi # value chosen for the critical length
s=101 # number of steps in x
t=10002 # number of timesteps
ds=L/(s-1) # step in x
dt=0.0001 # time step
D=1 # diffusion constant, set equal to 1
C=1 # creation rate of neutrons, set equal to 1
Alpha=(D*dt)/(ds*ds) # constant for diffusion term
Beta=C*dt # constant for u term
x = np.linspace(-L/2, 0, num=51)
x = np.concatenate([x, np.linspace(x[-1] - x[-2], L/2, num=50)]) # setting x in the specified interval
u=np.zeros(shape=(s,t)) #setting the function u
u[50,0]=1/ds # delta function
for k in range(0,t-1):
u[0,k]=0 # boundary conditions
u[s-1,k]=0
for i in range(1,s-1):
u[i,k+1]=(1+Beta-2*Alpha)*u[i,k]+Alpha*u[i+1,k]+Alpha*u[i-1,k] # numerical solution
if k == 50 or k == 100 or k == 250 or k == 500 or k == 1000 or k == 10000: # plotting at times
plt.plot(x,u[:,k])
plt.title('Numerical Solution of the Diffusion equation over time')
plt.xlabel('x')
plt.ylabel('u(x,t)')
plt.show()
However now I have to change the right boundary condition into $u_x(frac{L}{2})=0$ and I'm not really sure how to change my code to reflect this. If I do this the critical length should decrease and the function should start increasing exponentially, but everything I've tried usually does nothing to my plot - is there something wrong with my original code possibly? Any help is really appreciated, I've been trying for ages but can't seem to get it! Thanks!
differential-equations numerical-methods partial-derivative boundary-value-problem python
asked Nov 27 at 16:58
mcaiojethewo
356
For the diffusion equation
$$frac{partial u(x,t)}{partial t}=Dfrac{partial^2 u(x,t)}{partial x^2}+Cu(x,t)$$
with the boundary conditions $u(−frac{L}{2},t)=u(frac{L}{2},t)=0$ I've programmed the numerical solution into python correctly (I think).
import numpy as np
import matplotlib.pyplot as plt
L=np.pi # value chosen for the critical length
s=101 # number of steps in x
t=10002 # number of timesteps
ds=L/(s-1) # step in x
dt=0.0001 # time step
D=1 # diffusion constant, set equal to 1
C=1 # creation rate of neutrons, set equal to 1
Alpha=(D*dt)/(ds*ds) # constant for diffusion term
Beta=C*dt # constant for u term
x = np.linspace(-L/2, 0, num=51)
x = np.concatenate([x, np.linspace(x[-1] - x[-2], L/2, num=50)]) # setting x in the specified interval
u=np.zeros(shape=(s,t)) #setting the function u
u[50,0]=1/ds # delta function
for k in range(0,t-1):
u[0,k]=0 # boundary conditions
u[s-1,k]=0
for i in range(1,s-1):
u[i,k+1]=(1+Beta-2*Alpha)*u[i,k]+Alpha*u[i+1,k]+Alpha*u[i-1,k] # numerical solution
if k == 50 or k == 100 or k == 250 or k == 500 or k == 1000 or k == 10000: # plotting at times
plt.plot(x,u[:,k])
plt.title('Numerical Solution of the Diffusion equation over time')
plt.xlabel('x')
plt.ylabel('u(x,t)')
plt.show()
However now I have to change the right boundary condition into $u_x(frac{L}{2})=0$ and I'm not really sure how to change my code to reflect this. If I do this the critical length should decrease and the function should start increasing exponentially, but everything I've tried usually does nothing to my plot - is there something wrong with my original code possibly? Any help is really appreciated, I've been trying for ages but can't seem to get it! Thanks!
differential-equations numerical-methods partial-derivative boundary-value-problem python
differential-equations numerical-methods partial-derivative boundary-value-problem python
asked Nov 27 at 16:58
mcaiojethewo
356
asked Nov 27 at 16:58
mcaiojethewo
356
asked Nov 27 at 16:58
mcaiojethewo
356
asked Nov 27 at 16:58
mcaiojethewo
356
asked Nov 27 at 16:58
mcaiojethewo
356
356
add a comment |
add a comment |
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