D. Keffer, ChE 55,Universit of Tennessee, Ma, 999 Analtial Stud of Stabilit of Sstems of ODEs David Keffer Department of Chemial Engineering Universit of Tennessee, Knoxville date begun: September 999 updated: Otober, 5 Table of Contents I. linear ODEs with onstant oeffiients II. N linear ODEs with onstant oeffiients 4 III. Linear ODEs with variable oeffiients 7 IV. Non-Linear ODEs i
D. Keffer, ChE 55,Universit of Tennessee, September, 999 I. linear ODEs with onstant oeffiients The material in Setion I of these leture notes orresponds to the ontents of Chapter Setions,. and.4, pages 6-75, Kreszig, 8 th Edition. This material applies to sstems of linear ODEs with onstant oeffiients. This is a ver limited subset of problems but it is worth seeing what rigorous riteria for stabilit we an ahieve. We will proeed b examining a sstem of linear ODEs. The work will appl for a sstem of n linear ODEs but, it is muh easier to visualize in two dimensions. We have a sstem of ODEs of the form d A + b (I.) and we have an initial ondition of the form: (x x ) (I.) o o This sstem of equations has a ritial point at ( x ) d d whih ields, where satisfies the ondition: d a + a + b (I.) d a + a + b A b b a a ab a a det ab a b det ( A) b ( ) A (I.4) or, equivalentl A b b a a ab a a det ab a b det ( A) b ( ) A (I.5) Clearl the ritial points are the values of when the derivatives of are zero. However, when the derivatives are zero, the funtion is onstant, so these are the stead-state, or equilibrium, or long-time (depending on the problem) solutions to the ODE. All trajetories pass through the ritial point, inluding the eigenvetors.
D. Keffer, ChE 55,Universit of Tennessee, September, 999 Five Tpes of Critial Points There are five tpes of ritial points. Improper nodes Proper nodes Saddle points Centers Spiral points Kreszig gives the following wa to determine whih tpe of ritial point a sstem has. For a x matrix, write the harateristi equation: ( λ)( a λ) a a a (I.6) with eigenvalues given b ( λ)( a λ) a a a (I.6) ( + a ) λ + det(a) λ a (I.7) ( a + a ) ± ( a + a ) 4 det(a) λ (I.8) Kreszig hooses to write this as p ± Δ λ (I.9) where p a + a, q det(a), Δ p 4q Case. If q > and Δ We have either a proper or improper node. We have eigenvalues that are purel real. Case. If q < (foring Δ > ) We have a saddle point. We have eigenvalues that are purel real. Case. If p and q > (foring Δ < ) We have a enter. We have eigenvalues that are purel imaginar. Case 4. If p and Δ < We have a spiral point. We have eigenvalues that are omplex (both nonzero real and imaginar parts).
D. Keffer, ChE 55,Universit of Tennessee, September, 999 Kreszig has examples of eah of the five tpes of ritial points on pages 64-66. Stabilit In addition to the tpe of ritial point, we might like to know the stabilit of the ritial point. Case. If p < and q > The ritial point is stable and attrative. We have eigenvalues with negative real parts. The imaginar part an be zero (giving a purel real eigenvalue) or positive and negative (giving omplex onjugates for the eigenvalues). The ritial point is either a node or a spiral point. Case. If p and q > The ritial point is stable. We have eigenvalues with zero real parts. The eigenvalues are purel imaginar omplex onjugates. The ritial point is a enter. Case. If p > or q < The ritial point is unstable. We have at least one eigenvalue with positive real part. The ritial point is a spiral point, node, or saddle point. See Figure 88, on page 7 of Kreszig. What s the point? The point of studing ritial point tpe and stabilit is to determine a priori what sort of qualitative behavior ou an expet from the solution to a sstem of ODEs. Starting at an initial ondition, ou an expet the solution to move through x toward a stable attrator but awa from an unstable point. For example, ou an expet the solution to move in a spiral toward the ritial point if the eigenvalues have negative real parts and nonzero imaginar parts. This onept will be illustrated in the homework assignments.
D. Keffer, ChE 55,Universit of Tennessee, September, 999 II. N linear ODEs with onstant oeffiients When we unambiguousl outlined stabilit analsis for a sstem of linear ODEs with onstant oeffiients. What if we have n linear ODEs? We have some immediate diffiulties. What is the definition of a ritial point? For a sstem of ODEs, it is d d d (II.) d B analog, the definition of a ritial point for a sstem of n linear ODEs is d i for all i n (II.) Let s take the x example. A b (II.) So that we see that a sstem of linear equations, regardless of the size, has onl one ritial point. In order to determine the tpe and stabilit of the ritial point, we an no longer use Kreszig s, p, q, and Δ riteria beause the harateristi equation is no longer quadrati. However, we an still use the eigenvalue riteria. If the eigenvalues all have negative real parts, the ritial point is a stable attrator. If an one of the eigenvalues has a positive real part, the ritial point is unstable. If all of the eigenvalues have zero real parts, the ritial point is a enter. If an of the eigenvalues have nonzero imaginar parts, there will be a spiraling nature to the ritial point. Example: Consider the sstem of linear ODEs: d A + b (I.) where 4
D. Keffer, ChE 55,Universit of Tennessee, September, 999 A and b The eigenvalues and eigenvetors of A are: + i Λ i and W i i All of the real parts of the eigenvalues are negative. Therefore, we should expet an attrator. Some of the eigenvalues have nonzero imaginar parts, therefore we should expet a spiral. We see that the ritial point is given as: Some trajetories in various planes of -D spae are given below. variable ().5.5.5 -.5 - variable ().5.5.5 -.5 -.5 - -.5.5.5.5.5 variable () - - -.5.5.5.5 variable () 5
D. Keffer, ChE 55,Universit of Tennessee, September, 999.5.5 variable ().5 -.5 - - -.5.5.5.5 variable () One an see that we have spiraling, attrative behavior in three-dimensions, as we expeted from the qualitative features of the eigenvalues. 6
D. Keffer, ChE 55,Universit of Tennessee, September, 999 III. Linear ODEs with variable oeffiients We have a sstem of ODEs of the form d ( x) + b( x) A (III.) and we have an initial ondition of the form: (x x ) (III.) o o Again, we will have a ritial point at ( x ) d d whih ields, where satisfies the ondition: d a( x) + a( x) + b( x) (III.) d a( x) + a( x) + b( x) ( x) A b b a ( x) ( x) a a a ( x) a ( ) ( ) ( ) ( ) x b x a x b x ( x) det( A( x) ) ( ) ( ) ( ) ( ) ( ( )) x b x a x b x det A x (III.4) So it beomes lear that for ever value of the dependent variable x, there is a ritial point that is a funtion of x. Likewise, the eigenvalues and eigenfuntions are now funtions of x. At an instant in x, the riteria for tpes and stabilit of ritial points holds. If the tpe and stabilit hold for a range of x, then we an expet the assoiated behavior of the solution to hold over that range. Example: A sin(x) b The eigenvalues of A are Λ 4 and the eigenvetors W 7
D. Keffer, ChE 55,Universit of Tennessee, September, 999 The eigenvalues are purel real, and negative. Therefore, we expet a stable attrator, with node-like behavior. Using equation (III.4) we find that the ritial points as a funtion of x look like:.4.5..5 ritial points.8.6.4. ritial point ().5.5 -. -.5 -.4 5 5 x - - -.5.5.5.5.5 ritial point () In the figure on the left, we have plotted the () omponent of the ritial point as a funtion of x in red and the () omponent of the ritial point as a funtion of x in blue. In the figure on the right, we have plotted () vs () as parametri funtions of x. Let s all this urve of ritial points the ritial path. Below we show several solutions to the ODE, starting from different initial onditions. The starting points of eah line are indiated b green squares. The ending points are indiated b red irles..5.5 ritial point ().5.5 -.5 - - -.5.5.5.5.5 ritial point () We see that the solution is indeed an attrator. All points lead to the ritial path. Moreover, the lead to the ritial path in a node-like wa, without spiraling. The urious behavior is at the 8
D. Keffer, ChE 55,Universit of Tennessee, September, 999 enter where we do find a lial stead state. This lial stead state is due to the sine funtion in b(x). It is interesting that the trajetories never atuall fall on the blak line indiated b the ritial path, but rather form a le about it. This must be due to the fat that the ODEs at time x are heading toward a solution defined b b(x). However, at some inremental time later, x, the solution has now moved and is defined b b(x ). Thus, the path of the ODE must be altered. The solution an be said to lag behind the ritial path. All solutions find the same lag, as indiated b the fat that regardless of the initial ondition, the final position (in this ase plotted at x5) is the same. Example: x A b The eigenvalues and eigenvetors are now funtions of x, as shown below: eigenvalues - - - -4-5 -6-7 4 6 8 4 6 8 x eigenvetor omponent.9.8.7.6.5.4... -.8 -.6 -.4 -...4.6.8 eigenvetor omponent We an see that the eigenvalues start small and negative; one linearl dereases and the other appears to exponentiall derease. The eigenvetors appear to be approahing asmptoti values of [,] and [,] respetivel. The point is that the eigenvalues are alwas negative and alwas purel real. Plots of the ritial path are given below, both as a funtion of x and parametriall. 9
D. Keffer, ChE 55,Universit of Tennessee, September, 999.4.4.5.. ritial point ().5..5 ritial points.8.6..4.5...4.6.8..4 ritial point () 4 6 8 4 6 8 x Lastl, we show trajetories for various initial onditions..5.5 ritial point ().5.5 -.5 - - 4 5 ritial point () This figure shows that we have a stable node for a ritial point. We would expet this beause our eigenvalues are real and negative. The onl differene between this ase and the ase where the matrix A is onstant is that now our ritial point is mobile. The trajetories follow along behind it.
D. Keffer, ChE 55,Universit of Tennessee, September, 999 IV. Non-Linear ODEs Most of the world s problems are non-linear. What good does our previous analsis of linear ODEs do for us? Well, it tells us the five tpes of ritial points. It tells us that the ritial points are the stead-state solutions. It tells us that we an understand the behavior of a sstem of ODEs b looking at a phase portrait. With the generalized definition of the ritial points, d i for all i n (II.) we an still determine ritial points in the nonlinear ase. Example: d (x) (x) 4 d (x) (x) + A ritial point of this sstem of equations is.444.7 In order to determine the eigenvalues, we need a matrix. In this ase, sine the ODEs are nonlinear, we annot diretl write the ODEs in matrix form. Instead, we will take a Talor Series expansion of the ODEs about the ritial point. d d ( ( x), ( x) ) ( x) ( x) 4 f ( ( x), ( x) ) ( x) + ( x) f We expand the two funtions in a Talor series about the ritial point and trunate after the linear terms, f ( ( x), ( x) ) f (, ) + ( ( x) ) + ( ( x ) ) d d f ( ( x), ( x) ) f (, ) + ( ( x) ) + ( ( x ) ) d d We have linearized the funtions and the linearized ODEs are now,
D. Keffer, ChE 55,Universit of Tennessee, September, 999 ( ) ( ) ( ) x d x d f d ) ( ) (, + + ( ) ( ) ( ) x d x d f d ) ( ) (, + + We an write this in matrix notation as b J d + where J is the Jaobian, whih has the definition, d d d d J and b is a vetor of onstants, ( ) ( ) d d f d d f b,, The Jaobian used here is the exat same Jaobian that is used in the Newton-Raphson method. One we have linearized the ODE, we an use the straightforward proedure to determine the eigenvalues and eigenvetors of the Jaobian. ( ) det λ I J We now proeed as before in the eigenanalsis. For this partiular, example problem, we have -.8488 J The eigenvalues of the Jaobian matrix are
D. Keffer, ChE 55,Universit of Tennessee, September, 999 -.66 Λ.48 The eigenvetors of the Jaobian matrix are W -.845 -.599 -.599.845 The eigenvalues are real, therefore we either have a node or a saddle point. The eigenvalues are not all of the same sign. Therefore, we have a saddlepoint. All saddlepoints are unstable. We should expet that this problem has a ritial point that behaves like a saddle point. In truth, this behavior should be observed ver lose to the ritial point. It ma extend beond that. Plots of trajetories starting from initial onditions near the ritial point ield: 4 ritial point () - -4-6 -8 - - -5 5 5 5 5 ritial point () Clearl, this is an unstable ritial point. The starting points (green squares) all lead awa from the ritial point (blak star) to their respetive ending points at x. (red irles). The eigenvetors leading awa from the ritial point are not straight lines sine the problem is nonlinear. Example: d (x) + (x) + d (x) (x)
A ritial point of this sstem of equations is D. Keffer, ChE 55,Universit of Tennessee, September, 999.68.68 In order to determine the eigenvalues, we again linearize the sstem of ODEs with a Talor series expansion. The Jaobian of the linearized problem is J d d d d -.6 The eigenvalues of the Jaobian matrix are -.8 +.99i Λ -.8 -.99i The eigenvetors of the Jaobian matrix are W.77.84 +.7i.77.84 -.7i The eigenvalues are omplex, therefore we have a spiral point. The real omponent of the eigenvalues are less than zero. Therefore, the point is stable. We should observe the behavior of a stable spiral point, at least near the ritial point. Plots of trajetories starting from initial onditions near the ritial point ield: -.4 -.45 -.5 ritial point () -.55 -.6 -.65 -.7 -.75 -.8 -.85.4.45.5.55.6.65.7.75.8.85.9 ritial point () 4
Clearl, this is a stable attrator with a spiraling nature. Non-linear equations an have more than one ritial point. Example: D. Keffer, ChE 55,Universit of Tennessee, September, 999 x (); % extent of reation T (); % Temperature K Cin.; % inlet onentration mol/l C Cin*(-x); % onentration Q 6e-; % volumetri flowrate l/s R 8.4; % gas onstant J/mol/K Ea 68; % ativation energ J/mol ko 4.48e+6; % reation rate prefator /s k ko*exp(-ea/(r*t)); % reation rate onstant /s V 8; % reator volume l Cp 4.9e; % heat apait J/kg/K Tin 98; % inlet feed temperature K Tref 98; % thermodnami referene temperature K DHr -.9e5; % heat of rxn J/mol rho.; % densit kg/l ddt() /V*(Q*Cin - Q*C - k*c*v); % mass balane mol/s ddt() /(Cp*rho*V)*(Q*Cp*rho*Tin - Q*Cp*rho*T - DHr*k*C*V); % NRG balane J/s ddt() -/Cin*ddt(); % onvert from onentration to extent These are the design equations for a ontinuousl stirred-tank reator with a single first-order exothermi reation, operating under adiabati onditions. The unknowns are the extent of reation and the the temperature. This problem has three stead states. The ritial points of this sstem of equations are x.59.5.988, T,,.4,, and 47.9, 445. In order to determine the eigenvalues, we again linearize the sstem of ODEs with a Talor series expansion. In this ase, sine we have more than one ritial point, we must evaluate the Jaobian at eah of the stead states. The Jaobian of the linearized problem is J d d d d dt dt dc dc dc dc dt dt 5
D. Keffer, ChE 55,Universit of Tennessee, September, 999 J Q k V ΔH rcink C pρ Q V ke a Cin RT ΔH rcke C ρrt p a We evaluate this Jaobian at eah ritial point and get the eigenvalues. For the first ritial point, we have J -.4 -.8. -.7 -.4 Λ -.7 W.866.996. -. The eigenvalues of the first ritial point are real and negative. Therefore, the first ritial point will behave like a stable improper node. For the seond ritial point, we have J -.5 -.495..7 -.4 Λ.6 W -.45 -.999 -. -. The eigenvalues of the seond ritial point are real. One is positive and one is negative. Therefore, the seond ritial point will behave like a saddle point, whih is alwas unstable. For the third and final ritial point, we have J -.944-8.5878.4.54 -.895 +.47i Λ -.895.47i 6
D. Keffer, ChE 55,Universit of Tennessee, September, 999 W.7 -.85i.7 +.85i.. The eigenvalues of the third ritial point are omplex. The real omponents of the eigenvalues are negative. Therefore, the third ritial point will behave like a stable spiral point. Some trajetories, based on different initial onditions (different initial onentrations and reator temperatures) are shown below. The trajetories start at green squares and end at red irles. The time that transpired along eah trajetor is minute..9.8.7 extent of reation.6.5.4... 5 4 45 5 55 6 65 temperature (K) (a) initial temperatures < T < 5 initial extent of reations < x < duration of trajetor se (larger version of plot available on last page of this setion) From the trajetor plots given above we an determine the nature of the ritial points (stead state solutions in this example). The low-onversion/low-temperature and the highonversion/high-temperature solutions are stable attrators. The intermediate solution is an unstable node. The eigenvalues assoiated with the attrators are less than zero. At least one eigenvalue assoiated with the unstable node is negative. We an also see some qualitative information about the sstem. We an define roughl the basins of attration for the two attrators. For the oarse grid we used, an initial temperature of 8 K or higher onverged to the high ritial point. An initial temperature of 4 K or lower onverged to the low ritial point. For initial temperatures of 6K, those with high initial extents of reation proeeded to the low root; those with low initial extents of reation onverged to the high root. The trajetories that led to the low root, approahed with a final tangents that appeared to be nearl parallel to the differene vetor between the low and middle root. The trajetories that led to the high root, approahed with two different final tangents. The first seemed to be nearl 7
D. Keffer, ChE 55,Universit of Tennessee, September, 999 parallel to the differene vetor between the high and middle root. The seond, whih most of the trajetories followed, appeared to be roughl perpendiular to the the first. Some initial onditions with low initial extent of reation and low temperature, proeeded through temperatures higher than the high root on their wa to the high root. This is beause the reator is full of unreated produt. It reats initiall, whih, sine the reation is exothermi, heats up the reator. It then takes some time for new feed to enter and ool the reator to its stead state temperature. 8
D. Keffer, ChE 55,Universit of Tennessee, September, 999.9.8.7 extent of reation.6.5.4... 5 4 45 5 55 6 65 temperature (K)