[Math] Stability and convergence for Heun’s method

numerical methodsordinary differential equationsstability-in-odes

Let
\begin{align}
y'(t) &= -150y(t)+49-150t, ~~~t\in[0,1]\\
y(0) &= 1/3+0.1
\end{align}
I've know the solution: $y(t) = 0.1·e^{-150t}-t+1/3$.

I'm testing a couple different numerical methods to determine which is the best (i.e., Euler's forward/backward, trapezoidal, Runge-Kutta, Heun). Can anyone help me determine the stability for the Heun method?

Heun's method: $\begin{align}y_{n+1}&=y_n+h\phi(t_n,y_n,h)\\[0.3em]\phi(t,u,h) &= \tfrac{1}{2}[f(t,u)+f(t+h,u+hf(t,u))]\end{align}$

Best Answer

Stability of explicit Runge-Kutta methods around a fixed point or a slow moving equilibrium is obtained when $Lh\sim 1$ with $L$ the $y$-Lipschitz constant and $h$ the step size. The exact constant on the right side depends on the order of the method, usually it is between 1 and 5. For first and second order methods that is only a guideline as the stability region does not contain a semi-circle around the origin in the negative complex half-plane.

To get a practical feeling for this, just solve your problem for various step sizes over the integration interval from $0$ to $1/3$, as $y(1/3)=0.1e^{-50}$ is a really small number, relative to the initial value and floating point precision.

def Heun(f,t0,y0,tf,h):
    def HeunStep(t,y,h): 
        k1 = h*f(t,y); 
        k2 = h*f(t+h,y+k1); 
        return y+0.5*(k1+k2);
    t, y = t0, y0
    while t+1.1*h<tf: t, y = t+h, HeunStep(t,y,h)
    return HeunStep(t,y,tf-t)

def f(t,y): return 49-150*(t+y)
t0,y0 = 0.0,1./3+0.1
for h in np.linspace(3.5,0.5,16)/150:
    print "h=%.4e 150*h=%.4e, y(1/3)=%.12g"%(h,150*h, Heun(f,t0,y0,1./3,h))

with the results

h=2.3333e-02 150*h=3.5000e+00, y(1/3)=3382845.45109
h=2.2000e-02 150*h=3.3000e+00, y(1/3)=1820496.14682
h=2.0667e-02 150*h=3.1000e+00, y(1/3)=558708.429462
h=1.9333e-02 150*h=2.9000e+00, y(1/3)=79763.1519923
h=1.8000e-02 150*h=2.7000e+00, y(1/3)=9204.09082254
h=1.6667e-02 150*h=2.5000e+00, y(1/3)=1648.41784102
h=1.5333e-02 150*h=2.3000e+00, y(1/3)=37.6118108918
h=1.4000e-02 150*h=2.1000e+00, y(1/3)=0.740437945724
h=1.2667e-02 150*h=1.9000e+00, y(1/3)=0.00432803787737
h=1.1333e-02 150*h=1.7000e+00, y(1/3)=1.0688786734e-05
h=1.0000e-02 150*h=1.5000e+00, y(1/3)=1.14794370277e-08
h=8.6667e-03 150*h=1.3000e+00, y(1/3)=5.57823318786e-12
h=7.3333e-03 150*h=1.1000e+00, y(1/3)=2.8015784137e-15
h=6.0000e-03 150*h=9.0000e-01, y(1/3)=2.34187669257e-17
h=4.6667e-03 150*h=7.0000e-01, y(1/3)=-1.38777878078e-17
h=3.3333e-03 150*h=5.0000e-01, y(1/3)=2.60208521397e-17

As the stability region suggests, $0<h<\frac2{150}$ are valid choices in the table above, giving results close to zero. Sensible results that are practially zero as floating point numbers (and relative to the initial value) are only obtained for about $h<\frac{1.5}{150}$.

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