singular matrix- integral
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From: Magdalena Moczydlowska on 13 May 2010 14:10 Hi! I have one question about the specific integral. Assume that A is singular so it is not invertible ( we also kno that A is stiff and semi-definite matrix so ||e^{hA}||<1). I have in this moment no idea how to calculate Int_{0}^{h} ( A* exp(h-s)A ) ds Int-integral Thank you for help ! Magdalena
From: Raymond Manzoni on 13 May 2010 15:56 Magdalena Moczydlowska a �crit : > Hi! > > I have one question about the specific integral. > Assume that A is singular so it is not invertible ( we also kno that A > is stiff and semi-definite matrix so ||e^{hA}||<1). > > I have in this moment no idea how to calculate > > Int_{0}^{h} ( A* exp(h-s)A ) ds > > Int-integral > > Thank you for help ! > > Magdalena Well d/dh int_0^h exp((h-s)A) ds should be : exp((h-h)A) + int_0^h A*exp((h-s)A) ds so... Hoping this helped, Raymond
From: Raymond Manzoni on 13 May 2010 16:19 Raymond Manzoni a �crit : > Magdalena Moczydlowska a �crit : >> Hi! >> >> I have one question about the specific integral. >> Assume that A is singular so it is not invertible ( we also kno that A >> is stiff and semi-definite matrix so ||e^{hA}||<1). >> >> I have in this moment no idea how to calculate >> >> Int_{0}^{h} ( A* exp(h-s)A ) ds >> >> Int-integral >> >> Thank you for help ! >> >> Magdalena > > > Well d/dh int_0^h exp((h-s)A) ds should be : > > exp((h-h)A) + int_0^h A*exp((h-s)A) ds so... > further : d/dh int_0^h exp((h-s)A) ds = d/dh int_0^h exp(t A) dt = exp(h A) Hoping it's not too wrong... Raymond
From: Ray Vickson on 13 May 2010 17:18 On May 13, 11:10 am, Magdalena Moczydlowska <magdamoczydlow...(a)gmail.com> wrote: > Hi! > > I have one question about the specific integral. > Assume that A is singular so it is not invertible ( we also kno that A > is stiff and semi-definite matrix so ||e^{hA}||<1). > > I have in this moment no idea how to calculate > > Int_{0}^{h} ( A* exp(h-s)A ) ds > > Int-integral > > Thank you for help ! > > Magdalena As Raymond Manzoni has indicated, all you need to do is compute exp(t*A), then integrate it (element-by-element) to get int_{t=0}^h exp(t*A) dt. As for getting F(t) = exp(t*A), see, eg., http://www.inf.ethz.ch/personal/cellier/Lect/NMC/nmc_ln8.pdf or http://www.cs.cornell.edu/cv/ResearchPDF/19ways+.pdf (update of the previous article). If the eigenvalues of A are r1,..,r1, r2,...,r2 ..., rk ,...,rk with multiplicity(r1) = m1,..., multiplicity(rk) = mk, then for any analytic function f(x) we have f(A) = sum_{j=1}^k sum_{i=1}^(mj) E_{ij} *f^(i-1)(rj), where f^(p)(x) = d^p f(x)/dx^p and the E_{ij} are some constant matrices. We can determine the E_{ij} by looking at simple functions of the form f(x) = x^0 = i (f(A) = I), f(x) = x (f(A) = A), f(x) = x^2 (f(A) = A^2), ... . Note that the usual way is to find the Jordan canonical form and to then determine f(J) for each Jordan block. However, we don't need to actually do this, at least not always. As an example, consider a matrix with eigenvalue 1 of multiplicity 3. We have Case 1: A = [[1 1 0], [0,1 0], [0,0 1]] = diag(J1, J2), where J1 = [[1 1],[0 1]] and J2 = [1] are two Jordan blocks. We have f(A) = E11*f(1) + E12*f'(1) + E21*f(1), where the first two terms correspond to Jordan block J1 and the last to J2. However, writing E1 = E11 + E21 we have, simply, f(A) = E1 * f(1) + E12 * f'(1). Case 2: the identity, in which case f(A) = I*f(1). Case 3: a single Jordan block A = [[1 1 0],[0 1 1], [0 0 1]], with f(A) = E11*f(1) + E12 * f'(1) + E13 * f''(1). Without knowing the Jordan form, how can we tell which form is correct? Well, it does not matter: in all cases we can write f(A) = E1 * f(1) + E2 * f'(1) + E3 * f''(1). In Case 1, E1 = E11 + E21, E2 = E12 and E3 = 0. In Case 2, E1 = I, E2 = E3 = 0. In Case 3, E1 = E11, E2 = E12 and E3 = E13. We can find the "coefficients" E1, E2 and E3 from computations of A^0 = E1*1, and A^1 = E1*1 + E2 * 1 , A^2 = E1 *1 + 2*E2 * 1 + 2*E3 * 1; we don't need to know the E_{ij}. This will work whenever using the A^k leads to a unique solution. In other cases (if they occur) we might, indeed, need to know the actual Jordan canonical form. R.G. Vickson
From: Raymond Manzoni on 13 May 2010 18:51
Raymond Manzoni a �crit : > Raymond Manzoni a �crit : >> Magdalena Moczydlowska a �crit : >>> Hi! >>> >>> I have one question about the specific integral. >>> Assume that A is singular so it is not invertible ( we also kno that A >>> is stiff and semi-definite matrix so ||e^{hA}||<1). >>> >>> I have in this moment no idea how to calculate >>> >>> Int_{0}^{h} ( A* exp(h-s)A ) ds >>> >>> Int-integral >>> >>> Thank you for help ! >>> >>> Magdalena >> >> >> Well d/dh int_0^h exp((h-s)A) ds should be : >> >> exp((h-h)A) + int_0^h A*exp((h-s)A) ds so... >> > further : > d/dh int_0^h exp((h-s)A) ds = d/dh int_0^h exp(t A) dt = exp(h A) > > Hoping it's not too wrong... > Raymond the most direct derivation is of course : int_0^h A*exp((h-s)A) ds = int_0^h A*exp(t A) dt = int_0^h d(exp(t A))/dt dt = exp(h A) - exp(0 A) but I'll let you judge which method is easiest to justify! Raymond |