# How to Find a Finite-Difference Matrix

I have read several websites trying to explain finite-differential equations, but I haven't been able to find one that explains how it's put into the matrix form.

$f(x) = -\frac{d^2u}{dx^2}$ where $u(0) = 0$ and $u(1) = 0$ becomes

$\begin{bmatrix} 2 & -1 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & -1 & 2 & -1 \\ 0 & 0 & 0 & -1 & 2 \end{bmatrix}$

multiplied by one column of values between $u_1$ to $u_5$ is equal to

$h^2$$\begin{bmatrix} f(h) \\ f(2h) \\ f(3h) \\ f(4h) \\ f(5h) \end{bmatrix} Looking at the equation that causes this matrix, it confuses me. The difference equation is$$-u_{j+1}+2u_j-u_{j-1}=h^2f(jh)$$The initial term, when j=1, should make it so that the equation is:$$u_2+2u_1-u_0$$Would this not cause the first row to become (-1, 2, 0, 0, 0) because u_0 has been defined as 0? I know that my thinking is wrong, since the book tells me so, but I don't understand how the first and last row is determined. On a related note, how does the matrix equation if the boundaries are changed? For an example, if u_0 = 1 and u_1 = 2 on the original equation. Will the answer become \begin{bmatrix} 2 & 0 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & -1 & 2 & -1 \\ 0 & 0 & 0 & -1 & 2 \end{bmatrix} • You might want to learn more about the finite difference methods. I am sure there are enough textbooks on the same that explain the process in detail. – user14082 Sep 22 '12 at 18:08 ## 2 Answers The first and last row are determined by boundary conditions. For instance, the matrix you've described arises if u_0 and u_6 are taken to be zero (though other BC might give rise to the same matrix). As for your second question, we don't get to set the values of u_1,\ldots,u_5. Your matrix is multiplying the vector {\bf u} to give the set of finite difference equations. However, u_0=0 gives a constraint$$h^2 f(h)=-u_0+2u_1-u_2=2u_1-u_2$$which is correctly reproduced by the first matrix in your post. • do u have a reference? – Charlie Parker Feb 1 '18 at 20:00 Just adding to the answer, summarizing Jonathan. \begin{eqnarray} \begin{pmatrix} 2 & 0 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & -1 & 2 & -1 \\ 0 & 0 & 0 & -1 & 2 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5 \end{pmatrix} = \begin{pmatrix} f(h) \\ f(2h) \\ f(3h) \\ f(4h) \\ f(5h) \\ \end{pmatrix} \end{eqnarray} Only$ u_1 $to$u_5$goes in the matrix due the Boundary Condition$u_0=u_6=0\$, that explains why your logic about the first and last rows were wrong. Just remember the multiplication logic for matrix/array.

• do u have a reference? – Charlie Parker Feb 1 '18 at 20:01