Question

From physics, the total energy for this system is called the Hamiltonian and given by H(,vF(s) ds 0 In general, a numerical scheme will not conserve the energy of the system, i.e. H will not be constant. Suppose however that a trapezoidal method can be modified to produce an energy-conserving scheme. In particular assume the numerical scheme has the form where W is a yet-to-be-determined function, h is the time-step, and (yjl, {vj) denote the discretized position and velocity. Using the Hamiltonian given above, show that Uj+1 F(s) ds Conclude the numerical scheme (1) conserves energy if Uj+1 F(s) ds

Answer 1

Given: Hamiltonian H(y, v)--mv2-8F(s)ds Numerical scheme has the form: 77n The Hamiltonian for jth component can be written by substituting the values. Then for (j+1)th component, 1+1 Substituting the values from numerical scheme, Simplifying this, we get 1-yj)-F(s)ds j+1 Fs)ds Where we have split the integral into two parts.

yj+1 The first two terms on RHS are same as equation (1). Using that value, yj+1 This gives: yj+1 F(s)ds Solving RHS yj+1 F (s)ds +1 F (s)ds Using numerical scheme, the 15 term in LHS can be simplified as: H1 (Vj-1-yj)

For the numerical scheme to conserve energy, This can happen when the RHS of simplified equation becomes zero. i.e. ryj+1 F(s)ds 0 F(s)ds +1-y)y