By E. van Groesen (auth.), Adriaan van der Burgh, Juriaan Simonis (eds.)
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0) 0 o ...... (40) 0 Next we add the contribution of e2 which contributes to giving: 0 0 2/h -l/h 0 0 -l/h l/h 0 0 0 0 811,812,821 and 822 (41) 0 Continuing in this way we finally obtain 2 -1 -1 S=! 0 0 2 -1 0 0 0 1 h = -A h (42) -1 2 -1 1 0 -1 in which A is just the matrix obtained by FDM and FVM methods. This is certainly encouraging. 4 Construction of f by element vectors In the same way we may consider the components of the right hand side II:: (43) (44) Again we have broken up the vector component I" into contributions of the various elements ej.
In an appropriate way, it is possible to arrive at the equation (40) in the scaled variables. A KALEIDOSCOPIC EXCURSION INTO DIFFERENTIAL EQUATIONS 25 (37) of a Hamiltonian system. This may lead immediately to the idea how consistent discretisations for (39) can be found using Fourier truncation as before. Indeed, based on this Hamiltonian structure, we could exploit the observation made for the linear equation. That is, substitution of the truncation for u in fl(u) will produce a function of the Fourier coefficients, the truncated energy as in (38).
It will be clear, that with these definitions u cannot satisfy equation (1) in any classical sense. The first derivative of u is constant on an element and jumps across element boundaries, the second derivative is 0 on the interior of the elements and does not exist on element boundaries. We shall try to reformulate problem 1 in such a way that it makes sense for functions of type U. 1 J. VANKAN Weak formulation of the problem = Let us take an arbitrary smooth function 1/J, with 1/J(0) 0, multiply equation (1) left and right with 1/J and integrate from 0 to 1.