| | 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 | |
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| | 93 | |