Solution Notes -- Example Problem Set 9

(A) (i): When numerically computing the solution to an ODE that involves higher-than first-order derivatives, the higher-order terms must be reduced to 1st-order terms. This can be accomplished by introducing new substituted variables such as

$$z(x) = \frac{dy(x)}{dx}$$

which adds another equation into the coupled system to be solved, but now we have terms like ${\displaystyle \frac{dz(x)}{dx}}$ instead of ${\displaystyle \frac{dy^{2}(x)}{dx^{2}}}$.

(ii) Thus the example ODE,

$$\frac{d^{2}y}{dx^{2}} + q(x)\frac{dy}{dx} = r(x)$$

would be replaced by the following 1st-order system:

$$\frac{dy(x)}{dx} = z(x)$$

$$\frac{dz(x)}{dx} = r(x) - q(x)z(x)$$

(B) (i) The incrementing rule for the Euler method of numerical integration is:

$$f(x_{n+1}) = f(x_{n}) + h f^{\prime}(x_{n})$$

(ii) If the stepsize h is too large, this piecewise-linear approximation to the solution will be inaccurate where the true solution has significant second derivatives (which this method ignores). The solution may ``blow up."

(iii) If the stepsize h is made too small, the computation may take too long and the round-off error (truncation of floating-point numbers depending on the word length of the machine) will accumulate too much, making the solution inaccurate or even unstable.

(iv) The primary advantage of the fourth-order Runge-Kutta method over the Euler method for numerical integration of ODEs is that its accumulated error depends on the stepsize h as its 4th power, h⁴, whereas in the Euler method the accumulated error is simply linear in h. Thus, for example, reducing the stepsize h by half would only reduce the accumulated error by half in the Euler method, but it would reduce it 16-fold in the Runge-Kutta method.