1
$\begingroup$

The domain we want to obtain the solution on is $x \in [0,1]$.

Let's write the second difference equation corresponding to this differential equation problem $\frac{y_{i-1}-2y_i+y_{i+1}}{h^2} = 1$ (is that step valid by the way?). The first condition is $ y_0 = 0 $. For the second condition we use the first forward difference: $\frac{y_1-y_0}{h} = 0$, so $y_1 = y_0 = 0$. We choose $h = 0.1111$ for the grid. Then we solve $Ay = b$,

where the matrix $A = \begin{bmatrix} 1 & 0 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & 0 & \dots & 0 \\ 1 & -2 & 1 & 0 & \dots & 0\\ 0 & 1 & -2 & 1 & \dots & 0 \\ \vdots & & \ddots & \ddots & \ddots & \vdots \\ 0 & \dots & 0 & 1 & -2 & 1 \end{bmatrix},$ $b = [0, 0, h^2, h^2, \dots,h^2]^T.$

This is a script (you could play with it), solving this problem:

k = 10;
x = linspace(0, 1, k);
h = (x(end) - x(1)) / (k - 1);
D = sparse(1 : k, 1 : k, 1 * ones(1, k), k, k);
Elow = sparse(2 : k, 1 : k-1, -2 *ones(1, k-1), k, k);
Elow2 = sparse(3 : k, 1 : k - 2, ones(1, k - 2), k, k);
S = Elow + D + Elow2;
S(1,1) = 1;
S(1,2) = 0;
S(2,1) = 0; S(2,2) = 1; S(2,3) = 0;
Sf = full(S);
b = h^2 * ones(1,k)'; b(1) = 0; b(2) = 0;
u = S \ b;

Increasing the number of $k$ we can see that finite difference solution looks like the analytical solution $y = 0.5 x^2$

On the other hand we can solve $y_{i-1}-2y_i+y_{i+1} = 1, y_0 = 0, y_1 =0$ directly as 2nd order nonhomogeneous recurrence equation, the solution would be $y_i = 0.5(i^2 - i)$.

ps I know that the RK4 method is OK for such problem, but I'm trying to study different numerical methods, so I took finite difference method for this equation to be sure that it's working. enter image description here

Summing up, I have several questions: is this the case when my finite difference scheme should converge to actual solution? why the recurrence equation solution gives so bad result? Is it correct actually?

$\endgroup$
  • $\begingroup$ I wanted to make a link to my quesion on quora, where the idea of comparing with the recurrence solution was appeared, but it was prohibited by rules so I give it here(link) $\endgroup$ – lR55 Jan 25 '17 at 20:05
1
$\begingroup$

The problem with a direct recurrence is that you write the recurrence wrong. It is $$y_{i-1}-2y_i+y_{i+1} = h^2$$

therefore the solution is $y_i = 0.5(i^2 - i)h^2$. Now recall that $x = ih$, or $i = \frac{x}{h}$, so $y(x) = 0.5(x^2 - hx)$, and as $h$ tends to $0$, the approximation tends to the solution as expected.

$\endgroup$
  • $\begingroup$ Thank's a lot! That silly mistake has really tied me into knots about the methods I used. $\endgroup$ – lR55 Jan 25 '17 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.