stability analysis, routh-hurwitz criterion, jacobi matrix [urgent] (repost)
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From: Arno Narque on 26 Jul 2010 15:38 Hello, Ok, sorry guys this is a repost. There were display problems in my last message, so i hope it's better now. I have got a quite urgent questions regarding stability analysis of a three dimensional of differential equations. I don't quite understand how this works in that case or more which method the author uses. I have got the Jacobi matrix (of the first order conditions, so actually it's a Hessian) of the dynamic system which is linearized around a fixed point (steady state): r+wh_k & wh_q-c_q & 0 \\ -qr_k - qr_hh_k & -qr_hh_q & 0 \\ .. & . & p\\ The author of the paper says that the first two variables in the third line are positive and real (.) but can be omitted since they are not involved in the calculation of the eigenvalues. Ok now I calculate the eigenvalues: det(M-\lambdaI)= [r + wh_k - \lambda_1][-qr_hh_q - \lambda_2][p - \lambda_3] - [wh_q - c_q][-qr_k - qr_hh_k][p - \lambda_3] = 0 One of the eigenvalues is obvious: p = \lambda_3 where by definition p>0. The author now says that for the system to be stable, he says that is a sufficient condition, it must be shown that the eigen values \lambda_1 and \lambda_2 have the opposite sign, therefore they are negative. To show this, the aouthor says, it is enough to show that the first determinant of the second order of the jacobi matrix is positive. Therefore that det(�r+wh_k & wh_q-c_q \\ -qr_k - qr_hh_k & -qr_hh_q \\)>0 I don't get on which method this stability analysis is based. Is it some sort of the Routh-Hurwitz criterion? The paper where I found this problem is: Optimal Taxation of Capital Income in General Equilibrium with Infinite Lives: http://ideas.repec.org/a/ecm/emetrp/v54y1986i3p607-22.html . This question refers to the appendix. It would be so great if somebody of you could help me out!!! I'd be so grateful! It would be also cool if somebody could suggest me something like "stability analysis for dummies ;-)" Thank you in advance, Yours, Arno Is it better now? It should ....
From: Robert Israel on 26 Jul 2010 16:10 > Hello, > > Ok, sorry guys this is a repost. There were display problems in my last > message, so i hope it's better now. I have got a quite urgent questions > regarding stability analysis of a three dimensional of differential > equations. I don't quite understand how this works in that case or more > which method the author uses. > > I have got the Jacobi matrix (of the first order conditions, so > actually it's a Hessian) of the dynamic system which is linearized > around a fixed point (steady state): > > r+wh_k & wh_q-c_q & 0 \\ > -qr_k - qr_hh_k & -qr_hh_q & 0 \\ > . & . & p\\ > The author of the paper says that the first two variables in the third > line are positive and real (.) but can be omitted since they are not > involved in the calculation of the eigenvalues. Ok now I calculate the > eigenvalues: > > det(M-\lambdaI)= [r + wh_k - \lambda_1][-qr_hh_q - \lambda_2][p - > \lambda_3] - [wh_q - c_q][-qr_k - qr_hh_k][p - \lambda_3] = 0 No need for subscripts on those lambdas. > One of the eigenvalues is obvious: p = \lambda_3 where by definition > p>0. The author now says that for the system to be stable, he says that > is a sufficient condition, it must be shown that the eigen values > \lambda_1 and \lambda_2 have the opposite sign, therefore they are > negative. I guess I don't understand something about the context here: if this is the Hessian of the dynamic system X' = F(X) about a fixed point, stability would require all eigenvalues to have non-positive real part, including p which you say is positive. > To show this, the aouthor says, it is enough to show that the > first determinant of the second order of the jacobi matrix is positive. > Therefore that > > det(�r+wh_k & wh_q-c_q \\ -qr_k - qr_hh_k & -qr_hh_q \\)>0 The other two eigenvalues are the roots of the quadratic (r + wh_k - lambda)(-qr_hh_q - lambda) - (wh_q - c_q)(-qr_k - qr_hh_k) = lambda^2 + A lambda + B where A = qr_hh_q - r - wh_k and B = (r + wh_k)(-qr_hh_q) - (wh_q - c_q)(-qr_k - qr_hh_k) (which is that "first determinant of the second order"). Those roots are (-A +/- sqrt(A^2 - 4 B))/2. If B > 0, we either have two real roots of the same sign or two complex conjugate roots. The sign of the real part, in any case, is the opposite of the sign of A. It's certainly not enough to just show B > 0 without knowing about A. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Arno Narque on 27 Jul 2010 14:08 On 2010-07-26 22:10:54 +0200, Robert Israel said: > on't understand something about the context here: if this is > the Hessian of the dynamic system X' = F(X) about a fixed point, stability > would require all eigenvalues to have non-positive real part, including > p which you say is positive. Hi Robert! Thank you very much for your answer. I was wrong and I completely misunderstood the purpose of all that. It's true it is a Hessian, the Jacobi of the first order conditions, and the purpose was to show that the equilibrium exhibits saddle point stability. The procedure is then very clear. I have the eigenvalue \lambda_3 p, which is p, positive and real. For the system to be a saddle point in the n=3 Hessian, there have to be two positive and one negative eigenvalue. So I can show that the principal minor, so the n=2 matrix in the upper left corner, has a det<0 and since the det of a matrix always equals the product of its eigenvalues it is therefore shown that the other two eigenvalues are positive and negative respectively. The system therefore is on a saddle point. Would you say this argumentation is correct? Thanks for your help and best regards, Arno
From: Arno Narque on 27 Jul 2010 14:10
On 2010-07-26 22:10:54 +0200, Robert Israel said: >> >> Hello, >> >> Ok, sorry guys this is a repost. There were display problems in my last >> message, so i hope it's better now. I have got a quite urgent questions >> regarding stability analysis of a three dimensional of differential >> equations. I don't quite understand how this works in that case or more >> which method the author uses. >> >> I have got the Jacobi matrix (of the first order conditions, so >> actually it's a Hessian) of the dynamic system which is linearized >> around a fixed point (steady state): >> >> r+wh_k & wh_q-c_q & 0 \\ >> -qr_k - qr_hh_k & -qr_hh_q & 0 \\ >> . & . & p\\ > >> The author of the paper says that the first two variables in the third >> line are positive and real (.) but can be omitted since they are not >> involved in the calculation of the eigenvalues. Ok now I calculate the >> eigenvalues: >> >> det(M-\lambdaI)= [r + wh_k - \lambda_1][-qr_hh_q - \lambda_2][p - >> \lambda_3] - [wh_q - c_q][-qr_k - qr_hh_k][p - \lambda_3] = 0 > > No need for subscripts on those lambdas. > >> One of the eigenvalues is obvious: p = \lambda_3 where by definition >> p>0. The author now says that for the system to be stable, he says that >> is a sufficient condition, it must be shown that the eigen values >> \lambda_1 and \lambda_2 have the opposite sign, therefore they are >> negative. > > I guess I don't understand something about the context here: if this is > the Hessian of the dynamic system X' = F(X) about a fixed point, stability > would require all eigenvalues to have non-positive real part, including > p which you say is positive. > >> To show this, the aouthor says, it is enough to show that the >> first determinant of the second order of the jacobi matrix is positive. >> Therefore that >> >> det(�r+wh_k & wh_q-c_q \\ -qr_k - qr_hh_k & -qr_hh_q \\)>0 > > The other two eigenvalues are the roots of the quadratic > > (r + wh_k - lambda)(-qr_hh_q - lambda) - (wh_q - c_q)(-qr_k - qr_hh_k) > = lambda^2 + A lambda + B > > where A = qr_hh_q - r - wh_k and > B = (r + wh_k)(-qr_hh_q) - (wh_q - c_q)(-qr_k - qr_hh_k) > (which is that "first determinant of the second order"). > > Those roots are (-A +/- sqrt(A^2 - 4 B))/2. If B > 0, we either have > two real roots of the same sign or two complex conjugate roots. The > sign of the real part, in any case, is the opposite of the sign of A. > It's certainly not enough to just show B > 0 without knowing about A. Hi Robert! Thank you very much for your answer. I was wrong and I completely misunderstood the purpose of all that. It's true it is a Hessian, the Jacobi of the first order conditions, and the purpose was to show that the equilibrium exhibits saddle point stability. The procedure is then very clear. I have the eigenvalue \lambda_3 p, which is p, positive and real. For the system to be a saddle point in the n=3 Hessian, there have to be two positive and one negative eigenvalue. So I can show that the principal minor, so the n=2 matrix in the upper left corner, has a det<0 and since the det of a matrix always equals the product of its eigenvalues it is therefore shown that the other two eigenvalues are positive and negative respectively. The system therefore is on a saddle point. Would you say this argumentation is correct? Thanks for your help and best regards, Arno |