Math428: Problem Set #6 solution guide

1.

If y(x) is a three times differentiable function, prove that

$$y_k' - \frac{y_{k+1}-y_{k-1}}{2h} = O(h^2),$$

where y_k = y(kh). Hint: Use Taylor series, of course.

Here, we just make the substitutions

$$\begin{eqnarray*}y_{k+1} & = & y_k + h y_k' + \frac{1}{2} h^2 y_k'' + \frac{1}{... ... y_k' + \frac{1}{2} h^2 y_k'' - \frac{1}{6} h^3 y_k'''+ O(h^4). \end{eqnarray*}$$

After making the substitutions and cancelling everything out, you end up with a remainder

$$\frac{1}{6} h^2 y_k'' + O(h^3) = O(h^2).$$

2.

If y(x) is a four times differentiable function, prove that

$$y_k'' - \frac{y_{k+1}-2 y_k+y_{k-1}}{h^2} = O(h^2).$$

This is very similar to the previous problem, except that one expands one step further.

$$\begin{eqnarray*}y_{k+1} & = & y_k + h y_k' + \frac{1}{2} h^2 y_k'' + \frac{1}{... ...' - \frac{1}{6} h^3 y_k''' + \frac{1}{24} h^4 y_k'''' + O(h^5). \end{eqnarray*}$$

After making all the cancellations, one ends up with a remainder like

$$\frac{1}{12} h^2 y_k'''' + O(h^3) = O(h^2)$$

3.

Derive a second order finite difference for y_k' with variable step size using y_k+1, y_k, and y_k-1. To do this, find coefficients A₁, A₂ and A₃ such that

$$y_k' - \frac{A_1 y_{k+1}+A_2 y_k + A_3 y_{k-1}}{h_1 + h_2} = O(h^2),$$

where h₁ = x_k+1-x_k and h₂ = x_k-x_k-1. Use Taylor series again. Expect the A's will depend upon the h's. You can check your answer by seeing if you recover the usual centered difference formula when h₁=h₂.

First, we need to expand y_k+1 and y_k-1.

$$\begin{eqnarray*}y_{k+1} & = & y_k + h_1 y_k' + \frac{1}{2} h_1^2 y_k'' + O(h^3)... ...y_{k-1} & = & y_k - h_2 y_k' + \frac{1}{2} h_2^2 y_k'' + O(h^3). \end{eqnarray*}$$

Next, we need to set up the linear system that eliminate terms with y_k, y_k' and y_k''.

$$\begin{eqnarray*}A_1 + A_2 + A_3 & = & 0, \\ \frac{h_1 A_1 - h_2 A_3}{h_1+h_2} & = & 1, \\ h_1^2 A_1 + h_2^2 A_3 & = & 0. \end{eqnarray*}$$

If we solve this system, we find that

$$\begin{eqnarray*}A_1 & = & \frac{h_1}{h_2}, \\ A_2 & = & \frac{h_2^2 - h_1^2}{h_1 h_2}, \\ A_3 & = & - \frac{h_2}{h_1}. \end{eqnarray*}$$

4.

[Based on KC 8.9: 1] Solve the two point boundary value problem

$$y'' = e^x + y^2 \cos x - (x-1) y', \ \ \ y(0) = 1,\ \ \ y(1) = 3,$$

using the shooting method with the secant rule.

See matlab m-file and diary for this one. I implemented a system for y and $\frac{\partial y}{\partial z}$ because I knew I would need it for the next problem. One can see that the secant method finds a correct value of y'(0) (0.17978) in a few steps, so that y(1) is 3 to 5 digits of precision. Of course, one could continue the process to gain whatever precision one needs.

5.

[Based on KC 8.9: 1] Solve the two point boundary value problem

$$y'' = e^x + y^2 \cos x - (x-1) y', \ \ \ y(0) = 1,\ \ \ y(1) = 3,$$

using the shooting method with Newton's Method.

This is the same as the previous problem with similar results except we use Newton's method.

6.

[Based on KC 8.9: 1] Solve the two point boundary value problem

$$y'' = e^x + y \cos x - (x-1) y', \ \ \ y(0) = 1,\ \ \ y(1) = 3,$$

using finite differences. Note that the differential equation in this problem is different from the previous two.

For this problem, we set up a linear system by writing out each derivative and second derivative as a second order finite difference. Then, we solve the linear system using Gaussian elimination. The matlab diary and m files are posted on the web.