Linear Algebra and Its Applications, Exercise 1.7.2

Exercise 1.7.2. Given the differential equation

-\frac{d^2u}{dx^2} + u = x, \quad u(0) = u(1) = 0

what is the finite difference matrix for h = \frac{1}{4} (corresponding to a 3 by 3 matrix)?

Answer: We have

-\frac{d^2u}{dx^2} + u \approx - [u(x-h) - 2u(x) + u(x+h)]/h^2 + u(x)

so that the equation above can be expressed in finite-difference terms as

-[u_{j-1} - 2u_j + u_{j+1} + u_j]/h^2 + u_j = jh

\rightarrow -u_{j-1} + 2u_j - u_{j+1} + h^2 u_j = h^2 jh

\rightarrow -u_{j-1} + (2 + h^2)u_j - u_{j+1} = h^3 j

replacing continuous values of x by a set of discrete values x = jh, so tht u_j is the approximation of u(x) at the point x = jh.

In this case we have h = \frac{1}{4} and are thus considering meshpoints x = jh, 0 \le j \le 4. The above equation can then be rewritten as

-u_{j-1} + (2 + \frac{1}{4}^2) u_j - u_{j+1} = (\frac{1}{4})^3 j

This equation can be simplified as follows:

-u_{j-1} + (2 + \frac{1}{16})u_j - u_{j+1} = \frac{1}{64}j

\rightarrow -u_{j-1} + \frac{33}{16}u_j - u_{j+1} = \frac{1}{64} j

We already know that u(0) = u(1) = 0 so that we have u_0 = u_4 = 0. For j = 1 the above equation becomes

-u_0 + \frac{33}{16} u_1 - u_2 = 1 \cdot \frac{1}{64} \rightarrow \frac{33}{16} u_1 - u_2 = \frac{1}{64}

\rightarrow 33 u_2 - 16 u_3 = \frac{1}{4}

For j = 2 the above equation becomes

-u_1 +\frac{33}{16} u_2 - u_3 = 2 \cdot \frac{1}{64} \rightarrow -u_1 + \frac{33}{16} u_2 - u_3 = \frac{2}{64}

\rightarrow -16u_1 + 33u_2 - 16u_3 = \frac{1}{2}

and for j = 3 the above equation becomes

-u_2 + \frac{33}{16} u_3 - u_4 = 3 \cdot \frac{1}{64} \rightarrow -u_2 + \frac{33}{64} u_1 = \frac{3}{64}

\rightarrow -16u_2 + 33u_3 = \frac{3}{4}

The corresponding 3 by 3 finite-difference matrix is then

\begin{bmatrix} 33&-16& \\ -16&33&-16 \\ &-16&33 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} \\ \frac{1}{2} \\ \frac{3}{4} \end{bmatrix}

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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