Combining Crank-Nicolson and Runge-Kutta to Solve a Reaction-Diffusion System

We have already derived the Crank-Nicolson method to integrate the following reaction-diffusion system numerically:

$$\frac{\partial u}{\partial t} = D \frac{\partial^2 u}{\partial x^2} + f(u),$$

$$\frac{\partial u}{\partial x}\Bigg|_{x = 0, L} = 0.$$

Please refer to the earlier blog post for details.

In our previous derivation, we constructed the following stencil that we would go on to rearrange into a system of linear equations that we needed to solve every time step:

$$\frac{U_j^{n+1} - U_j^n}{\Delta t} = \frac{D}{2 \Delta x^2} \left( U_{j+1}^n - 2 U_j^n + U_{j-1}^n + U_{j+1}^{n+1} - 2 U_j^{n+1} + U_{j-1}^{n+1}\right) + f(U_j^n),$$

where $j$ and $n$ are space and time grid points respectively.

Rearranging the above set of equations, we effectively integrate the reaction part with the explicit Euler method like so:

$$U_j^{n+1} = U_j^n + \Delta t f(U_j^n).$$

For functions $f$ that change rapidly for small changes in their input (stiff equations), using the explicit Euler method may pose stability problems unless we choose a sufficiently small $\Delta t$.

Therefore, I have been wondering if it would be possible to use a more sophisticated and stable numerical scheme to integrate the reaction part in the context of our Crank-Nicolson scheme.

For instance, to integrate the reaction part with the classical Runge-Kutta method, we would write out the following set of equations instead of the aforementioned one:

$$\frac{U_j^{n+1} - U_j^n}{\Delta t} = \frac{D}{2 \Delta x^2} \left( U_{j+1}^n - 2 U_j^n + U_{j-1}^n + U_{j+1}^{n+1} - 2 U_j^{n+1} + U_{j-1}^{n+1}\right) + \frac{1}{6} \left(k_1 + 2 k_2 + 2 k_3 + k_4 \right),$$

where

$$k_1 = f(U_j^n),$$

$$k_2 = f\left( U_j^n + \frac{\Delta t}{2} k_1 \right),$$

$$k_3 = f\left( U_j^n + \frac{\Delta t}{2} k_2 \right),$$

$$k_4 = f\left( U_j^n + \Delta t k_3 \right).$$

Whether or not doing this makes sense theoretically I am not certain. But going ahead and implementing this to the numerical example we discussed earlier seems to suggest that this does work.

In the following Python code that is mostly a copy of our previous code we compare the time behaviour and accuracy (measured by mass conservation as our reaction diffusion system preserves mass) of the explicit Euler and Runge-Kutta 4 reaction integration.

We realize that the differences between the obtained numerical results are negligible and we shall compare both approaches with a stiffer reaction term another time.

We shall also take a look at more sophisticated measures of numerical stability another time.

In [ ]:
%matplotlib inline

import numpy
from matplotlib import pyplot
In [ ]:
numpy.set_printoptions(precision=3)
In [ ]:
L = 1.
J = 200
dx = float(L)/float(J-1)
x_grid = numpy.array([j*dx for j in range(J)])
In [ ]:
T = 500
N = 1000
dt = float(T)/float(N-1)
t_grid = numpy.array([n*dt for n in range(N)])
In [ ]:
D_v = float(10.)/float(100.)
D_u = 0.01 * D_v

k0 = 0.067
f = lambda u, v: dt*(v*(k0 + float(u*u)/float(1. + u*u)) - u)
g = lambda u, v: -f(u,v)
 
sigma_u = float(D_u*dt)/float((2.*dx*dx))
sigma_v = float(D_v*dt)/float((2.*dx*dx))

total_protein = 2.26
In [ ]:
no_high = 10
U =  numpy.array([0.1 for i in range(no_high,J)] + [2. for i in range(0,no_high)])
V = numpy.array([float(total_protein-dx*sum(U))/float(J*dx) for i in range(0,J)])

Let us take a look at the inhomogeneous initial condition:

In [ ]:
pyplot.ylim((0., 2.1))
pyplot.xlabel('x')
pyplot.ylabel('concentration')
pyplot.plot(x_grid, U)
pyplot.plot(x_grid, V)
pyplot.show()

These are the matrices of our system of linear equations whose derivation we described earlier.

In [ ]:
A_u = numpy.diagflat([-sigma_u for i in range(J-1)], -1) +\
      numpy.diagflat([1.+sigma_u]+[1.+2.*sigma_u for i in range(J-2)]+[1.+sigma_u]) +\
      numpy.diagflat([-sigma_u for i in range(J-1)], 1)
        
B_u = numpy.diagflat([sigma_u for i in range(J-1)], -1) +\
      numpy.diagflat([1.-sigma_u]+[1.-2.*sigma_u for i in range(J-2)]+[1.-sigma_u]) +\
      numpy.diagflat([sigma_u for i in range(J-1)], 1)
        
A_v = numpy.diagflat([-sigma_v for i in range(J-1)], -1) +\
      numpy.diagflat([1.+sigma_v]+[1.+2.*sigma_v for i in range(J-2)]+[1.+sigma_v]) +\
      numpy.diagflat([-sigma_v for i in range(J-1)], 1)
        
B_v = numpy.diagflat([sigma_v for i in range(J-1)], -1) +\
      numpy.diagflat([1.-sigma_v]+[1.-2.*sigma_v for i in range(J-2)]+[1.-sigma_v]) +\
      numpy.diagflat([sigma_v for i in range(J-1)], 1)

Function f_vec_ee returns the explicit Euler time step vector while f_vec_rk returns the vector obtained applying the Runge-Kutta 4 method.

In [ ]:
def f_vec_ee(U,V):
    return numpy.multiply(dt, numpy.subtract(numpy.multiply(V, 
           numpy.add(k0, numpy.divide(numpy.multiply(U,U), numpy.add(1., numpy.multiply(U,U))))), U))
In [ ]:
def f_vec_rk(U, V):
    f_vec = lambda U, V: numpy.subtract(numpy.multiply(V, 
                         numpy.add(k0, numpy.divide(numpy.multiply(U,U), numpy.add(1., numpy.multiply(U,U))))), U)
    k1 = f_vec(U, V)
    k2 = f_vec(U + numpy.multiply(dt/2., k1), V - numpy.multiply(dt/2., k1))
    k3 = f_vec(U + numpy.multiply(dt/2., k2), V - numpy.multiply(dt/2., k2))
    k4 = f_vec(U + numpy.multiply(dt, k3), V - numpy.multiply(dt, k3))
    
    return numpy.multiply(dt/6., k1 + numpy.multiply(2., k2) + numpy.multiply(2., k3) + k4)
In [ ]:
U_record_ee = numpy.empty(shape=(N,J))
V_record_ee = numpy.empty(shape=(N,J))

U_record_rk = numpy.empty(shape=(N,J))
V_record_rk = numpy.empty(shape=(N,J))

U_record_ee[0][:] = U[:]
V_record_ee[0][:] = V[:]

U_record_rk[0][:] = U[:]
V_record_rk[0][:] = V[:]

for ti in range(1,N):
    U_record_ee[ti][:] = numpy.linalg.solve(A_u, B_u.dot(U_record_ee[ti-1][:]) +
                                            f_vec_ee(U_record_ee[ti-1][:],V_record_ee[ti-1][:]))
    V_record_ee[ti][:] = numpy.linalg.solve(A_v, B_v.dot(V_record_ee[ti-1][:]) - 
                                            f_vec_ee(U_record_ee[ti-1][:],V_record_ee[ti-1][:]))
    
    U_record_rk[ti][:] = numpy.linalg.solve(A_u, B_u.dot(U_record_rk[ti-1][:]) +
                                            f_vec_rk(U_record_rk[ti-1][:],V_record_rk[ti-1][:]))
    V_record_rk[ti][:] = numpy.linalg.solve(A_v, B_v.dot(V_record_rk[ti-1][:]) - 
                                            f_vec_rk(U_record_rk[ti-1][:],V_record_rk[ti-1][:]))

The initial protein mass in our system:

In [ ]:
print 'Explicit Euler', numpy.sum(numpy.multiply(dx, U_record_ee[0]) + numpy.multiply(dx, V_record_ee[0]))
print 'Runge-Kutta 4 ', numpy.sum(numpy.multiply(dx, U_record_ee[0]) + numpy.multiply(dx, V_record_ee[0]))

Since our reaction-diffusion system preserves mass, we should retain the same protein mass at steady-state for both numerical approaches:

In [ ]:
print 'Explicit Euler %.14f' % numpy.sum(numpy.multiply(dx, U_record_ee[-1]) + numpy.multiply(dx, V_record_ee[-1]))
print 'Runge-Kutta 4  %.14f' % numpy.sum(numpy.multiply(dx, U_record_rk[-1]) + numpy.multiply(dx, V_record_rk[-1]))

We realize that the difference between the two numerical methods is neglibible and we shall compare both approaches for a stiffer system another time.

A plot of the steady-state concentration profiles confirms that we cannot observe a significant differences between the results generated by both methods (varying J and N paints the same pictures).

In [ ]:
pyplot.ylim((0., 2.1))
pyplot.xlabel('x')
pyplot.ylabel('concentration')
pyplot.plot(x_grid, U_record_ee[-1])
pyplot.plot(x_grid, V_record_ee[-1])
pyplot.plot(x_grid, U_record_rk[-1])
pyplot.plot(x_grid, V_record_rk[-1])
pyplot.show()

Kymograph of U integrated with the explicit Euler method.

In [ ]:
fig, ax = pyplot.subplots()
pyplot.xlabel('x')
pyplot.ylabel('t')
pyplot.ylim((0., T))
heatmap = ax.pcolormesh(x_grid, t_grid, U_record_ee, vmin=0., vmax=1.2)
colorbar = pyplot.colorbar(heatmap)
colorbar.set_label('concentration U')

Kymograph of U integrated with the Runge-Kutta 4 method.

In [ ]:
fig, ax = pyplot.subplots()
pyplot.xlabel('x')
pyplot.ylabel('t')
pyplot.ylim((0., T))
heatmap = ax.pcolormesh(x_grid, t_grid, U_record_rk, vmin=0., vmax=1.2)
colorbar = pyplot.colorbar(heatmap)
colorbar.set_label('concentration U')