# Runge–Kutta methods¶

We come now to one of the major and most-used types of methods for initial-value problems: Runge–Kutta (RK) methods.1 They are one-step methods in the sense of (170), though they are not often written in that form. RK methods boost the accuracy past first order by evaluating the ODE function \(f(t,u)\) more than once per time step.

## A second-order method¶

Consider a series expansion of the exact solution to \(u'=f(t,u)\),

If we replace \(\hat{u}'\) by \(f\) and keep only the first two terms on the right-hand side, we would obtain the Euler method. To get more accuracy we will need to compute or estimate the third term too. Note that

where we have applied the multidimensional chain rule to the derivative, because both of the arguments to \(f\) depend on \(t\). Using this expression in (178), we obtain

We have no desire to calculate and then code those partial derivatives of \(f\) directly; an approximate approximation is called for. Observe that

Matching this expression to the term in brackets in (179), it seems natural to select \(\alpha = h/2\) and \(\beta = \frac{1}{2}hf\bigl(t_i,\hat{u}(t_i)\bigr)\). Doing so, we find

Truncating results in the one-step formula

We call this the **improved Euler** method. Thanks to the definitions above of \(\alpha\) and \(\beta\), the omitted terms are of size

Therefore \(h\tau_{i+1}=O(h^3)\), and the order of accuracy of improved Euler is two. We refer to it as **IE2**.

## Implementation¶

Runge–Kutta methods are called **multistage** methods. We can see why if we interpret (181) from the inside out. In the first stage, the method takes an Euler half-step to time \(t_i+h/2\):

The second stage employs an Euler-style strategy over the whole time step, but using the value from the first stage to get the slope, rather than using \(f(t_i,u_i)\):

Our implementation of IE2 is shown in ie2.

**Improved Euler method for an IVP.**

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 | ```
"""
ie2(ivp,n)
Apply the Improved Euler method to solve the given IVP using `n`
time steps. Returns a vector of times and a vector of solution
values.
"""
function ie2(ivp,n)
# Time discretization.
a,b = ivp.tspan
h = (b-a)/n
t = [ a + i*h for i in 0:n ]
# Initialize output.
u = fill(float(ivp.u0),n+1)
# Time stepping.
for i in 1:n
uhalf = u[i] + h/2*ivp.f(u[i],ivp.p,t[i]);
u[i+1] = u[i] + h*ivp.f(uhalf,ivp.p,t[i]+h/2);
end
return t,u
end
``` |

## More Runge–Kutta methods¶

The idea of matching Taylor expansions can be generalized to higher orders of accuracy. To do so, however, we must introduce additional stages, each having free parameters so that more terms in the series may be matched. The amount of algebra grows rapidly in size and complexity, though there is a sophisticated theory for keeping track of it. We do not give the derivation details.

There are many known Runge–Kutta methods. We present a generic \(s\)-stage method in the form

This recipe is completely determined by the number of stages \(s\) and the constants \(a_{ij}\), \(b_j\), and \(c_i\). Often an RK method is presented as just a table of these numbers, as in

For example, IE2 is given by

Here are two more 2-stage, second-order methods, **modified Euler** and **Heun’s**, respectively:

The most commonly used RK method, and perhaps the most popular IVP method of all, is the fourth-order one given by

This formula is often called “the” fourth-order RK method—even though there are others—and we shall refer to it as RK4. Written out, the recipe is

Our implementation is given in rk4.

**Fourth-order Runge-Kutta for an IVP.**

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 | ```
"""
rk4(ivp,n)
Apply "the" Runge-Kutta 4th order method to solve the given IVP
using `n` time steps. Returns a vector of times and a vector of
solution values.
"""
function rk4(ivp,n)
# Time discretization.
a,b = ivp.tspan
h = (b-a)/n
t = [ a + i*h for i in 0:n ]
# Initialize output.
u = fill(float(ivp.u0),n+1)
# Time stepping.
for i in 1:n
k1 = h*ivp.f( u[i], ivp.p, t[i] )
k2 = h*ivp.f( u[i]+k1/2, ivp.p, t[i]+h/2 )
k3 = h*ivp.f( u[i]+k2/2, ivp.p, t[i]+h/2 )
k4 = h*ivp.f( u[i]+k3, ivp.p, t[i]+h )
u[i+1] = u[i] + (k1 + 2*(k2 + k3) + k4)/6
end
return t,u
end
``` |

## Efficiency¶

As with rootfinding and integration, the usual point of view is that evaluations of \(f\) are the only significant computations and are therefore to be minimized in number. One of the most important characteristics of a multistage method is that each stage requires an evaluation of \(f\); that is, a single time step of an \(s\)-stage method requires \(s\) evaluations of \(f\).

The error decreases *geometrically* as the number of stages grows algebraically, so trading a stage for an increase in order is a good deal. But \(s=5\), \(6\), or \(7\) gives a maximal order of accuracy of \(s-1\); this decreases to \(s-2\) for \(s=8\) and \(s=9\), etc. Fourth order is considered adequate and the “sweet spot” for many applications.

## Exercises¶

✍ For each IVP, write out (possibly using a calculator) the first time step of the improved Euler method with \(h=0.2\).

**(a)**\(u' = -2t u, \ 0 \le t \le 2, \ u(0) = 2;\ \hat{u}(t) = 2e^{-t^2}\)**(b)**\(u' = u + t, \ 0 \le t \le 1, \ u(0) = 2;\ \hat{u}(t) = -1-t+3e^t\)**(c)**\((1+x^3)uu' = x^2,\ 0 \le x \le 3, \ u(0) = 1;\ \hat{u}(x) = [1+(2/3)\ln (1+x^3)]^{1/2}\)✍ Use the modified Euler method to solve the problems in the preceding exercise.

✍ Use Heun’s method to solve the problems in exercise 1 above.

✍ Use RK4 to solve the problems in exercise 1 above.

✍ Using (179) and (180), show that Heun’s method has order of accuracy at least two.

✍ Using (179) and (180), show that the modified Euler method has order of accuracy at least two.

⌨ For each IVP, compute the solution using rk4. (i) Plot the solution for \(n=300\). (ii) For \(n=100,200,300,\ldots,1000\), compute the error at the final time and make a log–log convergence plot, including a reference line for 4th-order convergence.

**(a)**\(u''+ 9u = 9t, \: 0< t< 2\pi, \: u(0) = 1,\: u'(0) = 1; \: \hat{u}(t) = t+\cos (3t)\)**(b)**\(u''+ 9u = \sin(2t), \: 0< t< 2\pi, \: u(0) = 2,\: u'(0) = 1\);\(\quad \hat{u}(t) = (1/5) \sin(3t) + 2 \cos (3t)+ (1/5) \sin (2t)\)

**(c)**\(u''- 9u = 9t, \: 0< t< 1, \: u(0) = 2,\: u'(0) = -1; \: \hat{u}(t) = e^{3t} + e^{-3t}-t\)**(d)**\(u''+ 4u'+ 4u = t, \: 0< t< 4, \: u(0) = 1,\: u'(0) = 3/4; \: \hat{u}(t) = (3t+5/4)e^{-2t} + (t-1)/4\)**(e)**\(x^2 y'' +5xy' + 4y = 0,\: 1<x<e^2, \: y(1) = 1, \: y'(1) = -1, \: \hat{y}(x) = x^{-2}( 1 + \ln x)\)**(f)**\(2 x^2 y'' +3xy' - y = 0,\: 1<x<16, \: y(1) = 4, \: y'(1) = -1, \: \hat{y}(x) = 2(x^{1/2} + x^{-1})\)**(g)**\(x^2 y'' -xy' + 2y = 0,\: 1<x<e^{\pi}, \: y(1) = 3, \: y'(1) = 4\);\(\quad \hat{y}(x) = x \left[ 3 \cos \left( \ln x \right)+\sin \left( \ln x \right) \right]\)

**(h)**\(x^2 y'' + 3xy' + 4y = 0,\: e^{\pi/12} < x < e^{\pi}, \: y(e^{\pi/12}) = 0, \: y'(e^{\pi/12}) = -6\);\(\quad \hat{y}(x) = x^{-1} \left[ 3 \cos \left( 3 \ln x \right)+\sin \left( 3 \ln x \right) \right]\)

⌨ Do the SIR model exercise, but using rk4 instead of

`solve`

.⌨ Do the FitzHugh--Nagumo model exercise, but using rk4 instead of

`solve`

.✍ Consider the problem \(u'=ku\), \(u(0) = 1\) for constant \(k\) and \(t>0\).

**(a)**Find an explicit formula in terms of \(h\) and \(k\) for \(u_{i+1}/u_i\) in the modified Euler solution.**(b)**Find values of \(k\) and \(h\) such that \(|u_i|\to\infty\) as \(i\to\infty\) while the exact solution \(\hat{u}(t)\) is bounded as \(t\to\infty\).⌨ Modify rk4 to implement Heun’s method. Test your function on \(2 x^2 y'' +3x y' - y = 0\) over \([1,16]\), with \(y(1) = 4\), \(y'(1) = -1\). Use the exact solution \(\hat{y}(x) = 2(x^{1/2} + x^{-1})\) to show that the error at \(x=16\) converges at second order.

- 1
Americans tend to pronounce these German names as “run-ghuh kut-tah.”