| 71 | [[latex($\triangledown ^2 U = 6x$)]] |
| 72 | |
| 73 | gives the exact solution. This is the same as saying |
| 74 | |
| 75 | [[latex($A\vec{U} = 6\vec{x}$)]], |
| 76 | |
| 77 | that is the LTE is zero (i.e. discretization of the equation leads to the exact solution). |
| 78 | |
| 79 | To check this I wrote out the form for the LTE, which is the same as given above but with f(xi) replaced with 6x(i). Using the exact solution in this formula gives, |
| 80 | |
| 81 | [[latex($\tau_i = U''_i + \frac{1}{12}h^2 U''''_i + O(h^4) - 6x_i$)]] |
| 82 | |
| 83 | The ODE is easy to solve by hand, and gives a general solution of, |
| 84 | |
| 85 | [[latex($U(x) = x^3 + cx + d$)]] |
| 86 | |
| 87 | As you can see, the truncation error requires 2nd derivatives and ''higher'', and so the LTE is exactly zero as I wanted to verify. |
| 88 | |
| 89 | |
| 90 | |
| 91 | |
| 92 | |
| 93 | |