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To solve ${d^2y \over dx^2} =f(x)$, $0<x<1$ with $y(0)=\alpha, y(1) = \beta$. We can get a finite difference approximation by taking $$\frac{y_{j+1}-2y_j+y_{j-1}}{h^2} =f_j \\\Rightarrow \frac{1}{2}y_{j+1}-y_j+\frac{1}{2}y_{j-1} =\frac{h^2}{2}f_j$$

Then we get a system of linear equations which can be written as $\left(\begin{array}{ccccc} -1 & 1/2 & 0 &0 &0 \\ 1/2 & -1 & 1/2 &0 &0\\ 0 & 1/2 & -1 &1/2&0\\ 0 & 0 & 1/2&-1 &1/2\\ 0 & 0 & 0 &1/2 &-1\end{array} \right) \cdot \left(\begin{array}{c} y_1 \\ y_2 \\ y_3 \\ y_4 \\ y_5 \end{array} \right) = \left(\begin{array}{c} \frac{h^2}{2}f_1 -\frac{\alpha}{2}\\ \frac{h^2}{2}f_2 \\ \frac{h^2}{2}f_3 \\ \frac{h^2}{2}f_4 \\ \frac{h^2}{2}f_5 -\frac{\beta}{2} \end{array} \right)$

which we can solve by gaussian elimination.

I got the idea from

I understand the above, but how do I calculate/find/write out $f_2$, $f_3$, $f_4$ etc.? What if my DE is$$-{d^2y\over dx^2} + {dy\over dx} =x \Leftrightarrow {d^2y\over dx^2}={dy\over dx}-x $$ with $u(0)=\alpha, u(1) = \beta$?

Edit: I rearranged the equation and got


$j=1 \Rightarrow f_{2}(1-\frac{h}{2})+y_{0}(1+\frac{h}{2})+y_1(-2)=h^2f_1$.

So does $h^2f_1 = \frac{1}{36}\cdot(-\frac{1}{6})?$ And does $h^2f_2 = \frac{1}{36}\cdot(-\frac{2}{6})$?

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The answer to your first question is hidden in the PDF file you link to. The $f_j$ are defined by $f_j = f(x_j)$ with $x_j = jh = j/(N+1)$ where $N$ in the number of grid points. In your example with the 5-by-5 matrix, $h=\frac16$ so $f_j = f(j/h)$.

For your second question, you need to take care of the extra term $\frac{dy}{dx}$. One possibility (the most obvious one for me) is to use the approximation $$ \frac{dy}{dx} \approx \frac{y_{j+1} - y_{j-1}}{2h}, $$ compare with equation (2) in the PDF file. This will lead to a similar matrix equation as the one you wrote down.

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Thanks @JitseNiesen The main part I was confused on is how to calculate the $f_j$. Could you see if what I've done is correct? I rearranged the equation and got $f_{j+1}(1-\frac{h}{2})+y_{j-1}(1+\frac{h}{2})+y_j(-2)=h^2f_j$. For example, $j=1 \Rightarrow f_{2}(1-\frac{h}{2})+y_{0}(1+\frac{h}{2})+y_1(-2)=h^2f_1$. So does $h^2f_1 = \frac{1}{36}\cdot(-\frac{1}{6})?$ And does $h^2f_2 = \frac{1}{36}\cdot(-\frac{2}{6})$? – Richard May 10 '12 at 8:14
@Richard Yes, that's correct, except for a typo (you wrote $f_{j+1}$ where it should be $y_{j+1}$). – Jitse Niesen May 10 '12 at 13:45

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