MEC702_Wk_9_Lect_1 - MEC702 Wk 9 Lect 1 Criteria for...

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Unformatted text preview: MEC702 Wk 9 Lect 1 Criteria for Critical Points. Stability. Critical Points of a System of ODE • For the system of ODEs y10 = a11 y1 + a12 y2 y20 = a21 y1 + a22 y2 , we have the y1 y2 -plane (phase plane) and can obtain the derivative y 0 dt dy2 = 20 dy1 y1 dt = = y20 y10 a21 y1 + a22 y2 . a11 y1 + a12 y2 • This associates every point P : (y1 , y2 ) with a unique tangent direction dy1 dy2 of the trajectory passing through P , except some point P0 . • The point P0 , at which point of the system. dy2 dy1 becomes undetermined, Five Types of Critical Points 1. Improper Nodes 2. Proper Nodes 3. Saddle Points 4. Centres 5. Spiral Points 1 0 0 is called a critical Improper Node An improper node is a critical point P0 at which all the trajectories, except for two of them, have the same limiting direction of the tangent. The other two trajectories also have a limiting direction of the tangent at P0 which, however, is dierent. Example. The system  −3 ~y = A~y = 1 0  y10 = −3y1 + y2 1 ~y =⇒ −3 y20 = y1 − 3y2 has an improper node at (y1 , y2 ) = (0, 0). Solving we have λ1 = −2 =⇒ λ2 = −4 =⇒   1 1   1 ~x2 = −1 ~x1 = and general solution ~y = c1     1 −2t 1 e + c2 e−4t 1 −1 =⇒ y1 = c1 e−2t + c2 e−4t y2 = c1 e−2t − c2 e−4t . The common limiting direction at 0 is that of the eigenvector ~x1 because e−4t tends to 0 faster than e−2t as t increases. 2 Proper Node A proper node is a critical point P0 at which every trajectory has a denite limiting direction and for any given direction d~ at P0 , there is a trajectory havingd~ as its limiting direction. Example. The system  1 ~y = A~y = 0 0  y10 = y1 0 ~y =⇒ 0 1 y2 = y2 has a proper node at (y1 , y2 ) = (0, 0). Solving we have λ1 = 1 =⇒ ~x1 =   0 0 meaning that there are no relationship between x and y in ~x. This implies that any ~x which is not the zero vector ~0 is an eigenvector. So let us take the standard basis of R2     1 0 ~x1 = and ~x2 = . 0 1 Hence a general solution is ~y = c1     1 t 0 t e + c2 e 0 1 y 1 = c1 e t =⇒ y 2 = c2 e t =⇒ This is a family of straight lines that go through the origin. 3 y2 = c1 y1 . c2 Saddle Point A saddle point is a critical point P0 at which there are two incoming trajectories, two outgoing trajectories, and all other trajectories in a neighbourhood of P0 bypass P0 . Example. The system  1 ~y = A~y = 0 0  0 ~y −1 =⇒ y10 = y1 y20 = −y2 . has a saddle point at the point P0 = (0, 0). Solving we have λ1 = 1 =⇒ λ2 = −1 =⇒   1 0   0 ~x2 = 1 ~x1 = and the general solution ~y = c1     1 t 0 −t e + c2 e 0 1 =⇒ y1 = c1 et y2 = c2 e−t =⇒ where c varies throughout R, creating a family of curves. 4 y1 y2 = c Centre A centre is a critical point that is enclosed by innitely many trajectories. Example. The system  0 ~y = −4 0  1 ~y 0 =⇒ y10 = y2 y20 = −4y1 , has a centre at the origin (0, 0). Solving we have λ1 = 2i =⇒ λ2 = −2i =⇒   1 2i   1 ~x2 = −2i ~x1 = and a general solution is     1 2it 1 ~y = c1 e + c2 e−2it 2i −2i =⇒ y1 = c1 e2it + c2 e−2it y2 = 2ic1 e2it − 2ic2 e−2it . We can transform this into a real solution using Euler's formula (eit = cos t + i sin t). However, without going that far, notice that from the system, 1 4y1 y10 = −y2 y20 =⇒ 2y12 + y22 = c 2 for constant c. This is a family of ellipses enclosing the centre at the origin (0, 0). 5 Spiral Point A spiral point is a critical point P0 about which the trajectories spiral, approaching P0 as t → ∞ (or tracing these spirals in the opposite sense, away from P0 ). Example. The system ~y 0 =  −1 −1  1 ~y −1 =⇒ y10 = −y1 + y2 y20 = −y1 − y2 . has a spiral point at the origin. Solving λ1 = −1 + i =⇒ λ2 = −1 − i =⇒   1 i   1 ~x2 = −i ~x1 = and the general solution is ~y = c1     1 (−1+i)t 1 (−1−i)t e + c2 e . i −i Instead of expressing our solution as a general real solution, we take a shortcut using the equations of the system. Multiply the rst equation of the system by y1 , the second by y2 , then add to get y1 y10 + y2 y20 = −(y12 + y22 ). We switch to polar co-ordinates (r, θ), where r2 = y12 + y22 . Dierentiating this last equation with respect to θ gives 2rr0 = 2y1 y10 + 2y2 y20 =⇒ rr0 = −r2 For each c ∈ R, this is a spiral. 6 =⇒ r0 = −r =⇒ r = ce−θ . Criteria for Critical Points The values assisting in classifying Critical Points are p = λ1 + λ2 q = λ1 λ2 ∆ = (λ1 − λ2 )2 for the eigenvalues λ1 and λ2 of the system where the table points out the criteria for each type of critical point. Name (a) Node (b) Saddle point (c) Center (d) Spiral Point p =0 6= 0 q >0 <0 >0 ∆ ≥0 <0 7 Comments on λ1 , λ2 Real, same signs Real, opposite signs Pure imaginary Complex, not pure imaginary Stability Critical points may also be classied in terms of their stability. Stability concepts are basic in engineering and other application. In a nutshell: The 'stability' of a system means that a small change (i.e., a small disturbance) at some instant changes the behaviour of the system only slightly at all future times t. Denition. A critical point P0 of the system of ODEs is • stable if, roughly, all trajectories of the system that at some instant are close to P0 remain close to P0 at all future times t; precisely, if for every disk D of radius  > 0 with center P0 , there is a disk Dδ of radius δ > 0 with center P0 such that every trajectory of the system that has a point P1 (corresponding to t = t1 ) in Dδ has all its points corresponding to t ≥ t1 in D . • unstable if P0 is not stable. • stable and attractive (or asymptotically stable ) if P0 is stable and every trajectory that has a point in Dδ approaches P0 as t → ∞. Stability Criteria for Critical Points Type of Stability (a) Stable and Attractive (b) Stable (c) Unstable p = λ1 q = λ1 λ2 p < 0 and q > 0 p ≤ 0 and q > 0 p > 0 and q < 0 Example. The system ~y 0 =  −3 1  1 y −3 has eigenvalues λ1 = −2 and λ2 = −4. Hence p = λ1 + λ2 = −6 q = λ1 λ2 = 8 ∆ = (λ1 − λ2 )2 = 4 therefore the (improper) node at ~0 = (0, 0) is stable and unattractive. 8 ...
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