Negative semi definite

The expansion in eigenfunctions leads to expressions of the various quantities pertaining to the stochastic process - as in equations (7.13) through (7.16). It also simplifies some of the derivations, in particular the proof of the approach to equilibrium. In fact, according to (7.13) it is sufficient to prove that all X other than X = 0 are positive, i.e., that W is negative semi-definite. In the same notation as in V.5 one has for any vector pn = x pl in the Hilbert space... 

Exercise. Consider the right-hand side of (3.5) as a linear operator W acting on the space of functions P(X) defined for 0 < X < oo and obeying (3.6). Verify that it has the symmetry property (V.7.5) and is negative semi-definite, the only eigenfunction with zero eigenvalue being (3.7). 

Hence, V0 is negative semi-definite this guarantees the boundedness of and 90. By invoking Barbalat s lemma [3], it can be recognized that V0 -> 0, which implies the global uniform convergence to 0 of as t -> oo, while 9a is only guaranteed to be uniformly bounded. ... 

The function V is guaranteed to be negative semi-definite if the arbitrary positive constant q is chosen so as to satisfy the inequality... 

The rhs of this equation is negative semi-definite (Appendix) and we have the result... 

The matrices B and C are both positive definite and so is their product, the matrix A, When (X — X ) is not zero, the rhs of (A5) is negative definite and when it is zero, that is at the stationary state, the rhs of (A5) is zero. Therefore the rhs of (A5) is negative semi-definite. Thus we have shown that... 

Exercise. Define jump moments for the case that Y has more components. Show that the matrix a (ye) must be negative definite, or at least semi-definite. 

The left-hand side of the above expression is a quadratic form in the variables N, A 2 A n 22, 13 and 23- It is therefore associated with a unique symmetric matrix, and for this quadratic form to be positive semi-definite it is necessary and sufficient that all the leading minors (determinants of the principal submatrices) of the associated matrix be non-negative [2, pp.479-480]. It is clear from the form of (4.89) that the coefficients of and A must be non-negative, while the first two lines on the left-hand side are associated with the two symmetric matrices... 

What initially seems to be an advantage of the semi-implicit methods is really a negative. In the semi-implicit methods, the Jacobian is directly within the definition of the same method-, in the implicit methods, it is used (if we use a Newton method for the nonlinear system solution) only indirectly, solely for the solution of the nonlinear system. Thus, the semi-implicit methods have the following disadvantages. 

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