Stability

4 Stability

There is a stability constraint that is common to many methods for obtaining numerical solutions of the wave equation. Issues relating to stability of numerical methods can be extremely complicated, but the following Key Point is enough for our purposes.

Key Point 25

The numerical method seen in this Section requires that

$μ \leq 1 that is, \frac{c δ t}{δ x} \leq 1$

for solutions not to grow unrealistically with $n$ .

This is called the CFL condition (named after an acronym of three mathematicians Courant, Friedrichs and Lewy).

Exercises

Suppose that $u = u (x, t)$ satisfies the wave equation $u_{t t} = c^{2} u_{x x}$ in $t > 0$ and $0 < x < 0.6$ . It is given that $u$ satisfies boundary conditions $u (0, t) = u (0.6, t) = 0$ $(t > 0)$ and initial conditions that need not be stated for the purposes of this question. The application is such that the wave speed $c = 1.4$ .

The numerical method
$u_{j}^{n + 1} = 2 u_{j}^{n} - u_{j}^{n + 1} + μ^{2} (u_{j + 1}^{n} - 2 u_{j}^{n} + u_{j - 1}^{n})$

where $μ = c δ t ∕ δ x$ , is implemented using $δ x = 0.15$ and $δ t = 0.1$ .

Suppose that, after 7 time-steps, the following data forms part of the numerical solution:

$\begin{matrix} u_{0}^{6} & = & 0.0000 & u_{0}^{7} & = & 0.0000 \\ u_{1}^{6} & = & 0.1024 & u_{1}^{7} & = & 0.0997 \\ u_{2}^{6} & = & 0.1986 & u_{2}^{7} & = & 0.1730 \\ u_{3}^{6} & = & 0.2361 & u_{3}^{7} & = & 0.1169 \\ u_{4}^{6} & = & 0.0000 & u_{4}^{7} & = & 0.0000 \end{matrix}$

Carry out the next time-step so as to find an approximation to $u$ at $t = 8 δ t$ .
Suppose that $u = u (x, t)$ satisfies the wave equation $u_{t t} = c^{2} u_{x x}$ in $t > 0$ and $0 < x < 1$ . It is given that $u$ satisfies boundary conditions $u (0, t) = u (1, t) = 0$ $(t > 0)$ . The initial elevation may be summarised as
$\begin{matrix} f_{0} & = & 0.0000 & f_{1} & = & 0.7812 & f_{2} & = & 0.2465 \\ f_{3} & = & - 0.1209 & f_{4} & = & 0.0000 \end{matrix}$

and the string is initially at rest (that is, $g (x) = 0$ ). The application is such that the wave speed $c = 1$ .

Carry out the first two time-steps of the numerical method

$u_{j}^{n + 1} = 2 u_{j}^{n} - u_{j}^{n - 1} + μ^{2} (u_{j + 1}^{n} - 2 u_{j}^{n} + u_{j - 1}^{n})$

where $μ = c δ t ∕ δ x$ in which $δ x = 0.25$ and $δ t = 0.2$ .

In this case $μ = 1.4 \times 0.1 ∕ 0.15 = 0.93333$ and the required time-step is carried out as follows: $\begin{array}{rcl} u_{0}^{8} & = & 0 from the boundary condition \\ u_{1}^{8} = 2 u_{1}^{7} - u_{1}^{6} + μ^{2} (u_{2}^{7} - 2 u_{1}^{7} + u_{0}^{7}) & = & 0.0740 \\ u_{2}^{8} = 2 u_{2}^{7} - u_{2}^{6} + μ^{2} (u_{3}^{7} - 2 u_{2}^{7} + u_{1}^{7}) & = & 0.0347 \\ u_{3}^{8} = 2 u_{3}^{7} - u_{3}^{6} + μ^{2} (u_{4}^{7} - 2 u_{3}^{7} + u_{2}^{7}) & = & - 0.0552 \\ u_{4}^{8} & = & 0 from the boundary condition \end{array}$
to 4 decimal places and these are the approximations to $u (0, 8 δ t)$ , $u (0.15, 8 δ t)$ , $u (0.3, 8 δ t)$ , $u (0.45, 8 δ t)$ and $u (0.6, 8 δ t)$ , respectively.
In this case $μ = 1 \times 0.2 ∕ 0.25 = 0.8$ and the first time-step is carried out as follows: $\begin{array}{rcl} u_{0}^{1} & = & 0 from the boundary condition \\ u_{1}^{1} = \frac{1}{2} μ^{2} (f_{0} + f_{2}) + (1 - μ^{2}) f_{1} + δ t g_{1} & = & 0.3601 \\ u_{2}^{1} = \frac{1}{2} μ^{2} (f_{1} + f_{3}) + (1 - μ^{2}) f_{2} + δ t g_{2} & = & 0.3000 \\ u_{3}^{1} = \frac{1}{2} μ^{2} (f_{2} + f_{4}) + (1 - μ^{2}) f_{3} + δ t g_{3} & = & 0.0354 \\ u_{4}^{1} & = & 0 from the boundary condition \end{array}$
to 4 decimal places.

The second time-step is as follows:
$\begin{array}{rcl} u_{0}^{2} & = & 0 from the boundary condition \\ u_{1}^{2} = 2 u_{1}^{1} - u_{1}^{0} + μ^{2} (u_{2}^{1} - 2 u_{1}^{1} + u_{0}^{1}) & = & - 0.3299 \\ u_{2}^{2} = 2 u_{2}^{1} - u_{2}^{0} + μ^{2} (u_{3}^{1} - 2 u_{2}^{1} + u_{1}^{1}) & = & 0.2226 \\ u_{3}^{2} = 2 u_{3}^{1} - u_{3}^{0} + μ^{2} (u_{4}^{1} - 2 u_{3}^{1} + u_{2}^{1}) & = & 0.3384 \\ u_{4}^{2} & = & 0 from the boundary condition \end{array}$
to 4 decimal places.

4 Stability

Key Point 25

Exercises

Answer