↜ Back to index Introduction to Numerical Analysis 1

# Lecture 4

In today’s lecture we will look at Runge–Kutta methods:

## Runge-Kutta methods

We consider the ODE

\[ \left\{ \begin{aligned} x'(t) &= f(x(t), t), && t > 0,\\ x(0) &= x_0. \end{aligned} \right. \]

The Euler method that we covered in the previous lecture has a very slow convergence: the global truncation error is proportional to \(h\), the time-step. We can improve this significantly by using more function evaluations of the right-hand side \(f(x,t)\) to approximate the value of the solution \(x\) at time \(t + h\) from the value at \(t\).

This is the idea of Runge-Kutta methods. The extra work that these function evaluations require is usually worth the improved convergence rate and smaller error achieved with fewer time steps (larger \(h\)).

### Midpoint method

Let us set \(t_i = ih\). In the Euler method we use the value of the right-hand side \(f(x,t)\) at \((x_{i-1}, t_{i-1})\) to find the approximate solution \(x_i\). However, the function \(f(x(t),t)\) changes with time and therefore this is not very optimal. Instead, we should try to use the value of \(f(x,t)\) at the middle of the time step \(f(x(t_{i-1} + \frac h2), t_{i-1} + \frac h2)\). Unfortunately, we do not know the value \(x(t_{i-1} + \frac h2)\). But we can approximate it by using the Euler method for time step of length \(\frac h2\). This gives us \(x(t_{i-1} + \frac h2) \sim x_{i-1} + \frac h2 f(x_{i-1}, t_{i-1})\) and we obtain the formula for the **midpoint method**

\[ x_i = x_{i-1} + h f(x_{i-1} + \frac h2 f(x_{i-1}, t_{i-1}), t_{i-1} + \frac h2) \]

See wikipedia for more details. This is an explicit method of **order 2**.

*Exercise*: Apply the midpoint method to the ODE

\[ \left\{ \begin{aligned} x' &= x(1+t), && t > 0,\\ x(0) &= 1. \end{aligned} \right. \]

The exact solution is \(x(t) = \exp(t (t + 2) / 2)\).

The code is implemented in the file `midpoint.f90`

.

*Exercise:* Run the code in `midpoint.f90`

and use gnuplot to plot the resulting solution. Compare it to the exact solution \(\exp(x)\).

*Exercise:* Use the code in `midpoint-error.f90`

to find the global truncation error at \(t = 1\) for various \(h\). Use gnuplot to estimate the order of the midpoint method. (Recall the logarithmic scale from last lecture.)

### Classical fourth order Runge-Kutta method

If we perform 4 functional evaluations for each time step, we can increase the order of the method to 4. This leads to the classical **fourth order Runge-Kutta method**:

\[ \begin{aligned} k_1 &= f(x_{i-1}, t),\\ k_2 &= f(x_{i-1} + \frac h2 k_1, t + \frac h2),\\ k_3 &= f(x_{i-1} + \frac h2 k_2, t + \frac h2),\\ k_4 &= f(x_{i-1} + h k_3, t + h),\\ x_i &= x_{i-1} + \frac h6 (k_1 + 2 k_2 + 2 k_3 + k_4). \end{aligned} \]

See more details about this method on wikipedia.

*Exercise:* Implement the fourth order Runge-Kutta method to solve the same ODE as in the case of the midpoint method. Estimate its order using gnuplot as before.

## Higher accuracy floating point numbers

We have been so far using only *single-precision* floating point numbers. Therefore we have not been able to obtain errors smaller than around 10^-7. To increase the accuracy of floating point computation, we can use the *double-precision* floating point numbers by defining the variables as:

`to be figured out...`