## Linear Algebra and Its Applications, Exercise 1.7.5

Exercise 1.7.5. What would the difference matrix in equation (6) look like if the boundary conditions were $u(0) = 1$ and $u(1) = 0$ (instead of $u(0) = 0$ and $u(1) = 0$)?

Answer: The finite difference equation would still be as given in equation (5): $-u_{j+1} + 2u_j - u_{j-1} = h^2f(jh)$

For $j = 1$ this equation would become $-u_2 + 2u_1 - u_0 = h^2 f(jh) \rightarrow -u_2 + 2u_1 - 1 = h^2 f(jh)$ $\rightarrow -u_2 + 2u_1 = h^2 f(jh) + 1 = h^2 [f(jh) + \frac{1}{h^2}]$

Since $h = \frac{1}{6}$ the corresponding 5 by 5 finite-difference matrix is then $\begin{bmatrix} 2&-1&&& \\ -1&2&-1&& \\ &-1&2&-1& \\ &&-1&2&-1 \\ &&&-1&2 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5 \end{bmatrix} = \frac{1}{36} \begin{bmatrix} f(\frac{1}{6}) + 36 \\ f(\frac{2}{6}) \\ f(\frac{3}{6}) \\ f(\frac{4}{6}) \\ f(\frac{5}{6}) \end{bmatrix}$

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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