Main topics and some repetition exercises for the course MMG511/MVE161 ODE and mathematical modeling in year Main topics in the course:

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1 Main topics and some repetition exercises for the course MMG5/MVE6 ODE and mathematical modeling in year 04. Main topics in the course:. Banach fixed point principle. Picard- Lindelöf theorem. Lipschitz functions.. Dependence of solutions on initial data and right hand side. Gronwall inequality. 3. Extensibility of solution. 4. General linear systems. Superposition principle. Uniquness of solutions. Dimension of the solutions space. Infinite existence time. Fundamental matrix solution. 5. Linear systems with constant matrix. Generalized eigenvectors. Jordan matrix. General solution. Phase portraits in plane. 6. Periodic linear systems. Floquet theory. 7. Stability by linearization (Lyapunovs first method). Grobman - Hartmann theorem. 8. Stability by Lyapunovs functions (Lyapunovs second method and Krasovsii- La Salle theorem). 9. Limit sets, invariant sets. General properties of limit sets. 0. Properties of limit sets in plane. Poincare Bendixson theory. Exercises Banach fixed point principle. Consider the following operator Kx = K(x)(t) = 0 B(t, s) [x(s)] ds + g(t), Show that it acts from V V, where V is the Banach space C([0, ]) of continuous functions with norm x = sup x(t). Here B(t, s) and g(t) are continuous functions and B(t, s) < 0.5 for all t, s [0, ]. Estimate the norm K(x) of K(x)(t). Find requirements on the function g(t) such that Banach s contraction principle implies that K(x)(t) has a fixed point. Dependence of solutions on initial data and right hand side. Gronwall inequality. Compare solutions to two similar equations x = f (t, x ),and x = f (t, x ) with initial data and right hand sides satisfying: x (a) x (a) δ, f (t, x) f (t, x) ε, f (t, x) f (t, x) L x x.

2 Extensibility of solution. Apriori estimate implies infinite existence time for solutions. Linear growth of the right hand side implies an estimate by Gronwall inequality and therefore solvability on infinite interval of time. Consider { the system of ODE: x = x sin(x ) + x cos(x ) x = x x /( + x ) + t Prove that all solutions to initial value problems with x(0) = x 0, y(0) = y 0 can be extended to an arbitrary time interval [0, T ]. Linear systems with constant matrix. Find general solutions to[ following ] ODEs: r = Ar with A = Exercise. Show that if a matrix A has complex eigenvalues λ, = α ± iω then the transformation B AB by matrix B = [Re v, Im v] with complex eigenvector vector ] v corresponding to one of eigenvalues gives matrix B AB = [ α ω. ω α Lyapunov method. (Hsu: Exercise 5.8) Show instability of the origin. Give complete argument for your proof. Check that variational matrix in the origin is degenerate. x = x 3 + yx y = y + x 3 (Hsu: Example 5..9.) Consider Van der Pol equation x + ε(x )x + x = 0 and its equivalent Lienard form. x = y ε( x3 3 x) y = x Find region where there is no periodic solution around the origin (domain of repulsion) From an old exam. Consider the following system of ODE and investigate stability of the fixed point in the origin. { x = y 3 x 5 y = x y 3 + y 5

3 Poincare Bendixson theory Show that there is a µ 0 such that the following system x = µx + y + xy xy y = x + µy x y 3 has at least one limit set that is a periodic orbit. Answers and hints. Banach fixed point principle. We like to have estimate K(x) K(y) θ x y for x, y in some closed subset B of C([0, ]). K(x) K(y) sup ([x(s)] 0 B(t, s) [y(s)] ) ds = sup 0 B(t, s) (x(s) y(s)) (x(s) + y(s)) ds sup B(t, s)ds 0 x y x + y x y x + y x y ( x + y ) t,s [0,] We can choose a ball B C([0, ]) such that for any x, y C it follows x + y θ <, for example B can be taken as a set of functions with x /4. On this set K will be a contraction because K(x) K(y) x y (0.5). To apply Banachs principle we need also that K maps B into itselḟ, namely that K(x) /4 for x < /4. It gives requirement on function g(t). Estimate the operator K : K(x) sup ([x(s)] 0 B(t, s) ) ds + sup g(t) x + g Taking for example a ball x < /4 in V we like to show that the map of this ball by operator K will be inside the the same ball, namely that K(x) /4. Using the estimate above we observe that it will be valid if K(x) /6 + g /4. This estimate is satisfied for g 3/6. Therefore for g = sup g(s) 3/6 the operator K satisfies conditions of the Banach fixed point theory and has a unique fixed point in the ball x < /4. Extensibility of solution. Observe that the right hand side of the system x = F (t, x) above has a linear growth, namely that for any T > 0 there are constants M(T ) and L(T ) such that F (t, x) M(T ) + L(T ) x, for (t, x) [ T, T ] R Linear systems with constant matrix. 3

4 We use the general formula (3.9) from the book for solutions ti linear systems x = Ax with constant coefficients with initial data x 0 = s j= x0,j where x 0,j M(λ j, A) - are projections of x 0 to all generalized eigenspaces M(λ j, A) of the matrix A corresponding to each eigenvector. Each of x 0,j can be a linear combination of eigenvectors (can be several) and associated generalized eigenvectors corresponding to the eigenvalue λ j. Point out that the expression (A λ j I) k acting on a generalized eigenvector will be another generalized eigenvector with lower index, or corresponding eigenvector (zero terms are skipped in the sum). x(t) = e At x 0 = s j= ([ nj k=0 ] ) (A λ j I) k t k x 0,j e λjt k! where s is the number of distinct eigenvalues to A and n j is the algebraic multiplicity of the eigenvalue λ j. General solution for real eigenvalues λ j can be expressed by finding basis of R n in terms of eigenvectors v j and generalized eigenvectors v (k) j k =,...n j corresponding to all eigenvalues λ j, j =,...s, to A in the form x 0 =...C p v j + C p+ v () j + C p+ v () j... Substituting this expression for x 0 in to the general formula above and carrying out all matrix-matrix and matrix-vector, multiplications one gets a general solution. Keep in mind that (A λ j I) v j = 0 and (A λ j I) v 0, j = 0 e.t.c., so some [ terms in the expression nj ] k=0 (A λ ji) k t k k! x 0,j for x 0,j = C p v j + C p+ v () j + C p+ v () j +... are zero Answer: x (t) = (C + C t)e t ; x (t) = (C + C + C t)e t. (not unique [ expression) ] 3 A =, characteristic polynomial: λ +λ+, λ = - multiplicity [ ]. Eigenvector v =.Generalized eigenvector satisfies the equation (A [ ] [ ] [ ] x λi)v () = v or = y [ ] 0 x + y =. We choose v () = The solution with x 0 = C v + C v () will be x(t) = e t C [ v + e] t C v () [ + te] t (A + I) v [ () = ] e t C v + e t C v () + 0 te t C v = e t C + e t C + e t tc Exercise. Hint: consider a complex eigenvalue /λ and eigenvector v and the complex solution x (t) = ve λt.use Re x (t) and Im x (t) ac basis for the solution space. Lyapunov method. 4

5 (Hsu: Exercise 5.8) ( Hint: Use the test function V (x, y) = x y ). (Hsu: Example 5..9.) Hint: Consider backward movement in time ( (leading to change ε < 0 in the equation) and the test function V (x, y) = x + y ) V = εx ( x3 3 ) V 0 for x < 3. Region {V 3/} is bounded and V is a Lyapunovs function in G. Therefore periodic solution must be outside this circle. From an old exam. Origin is asymptotically stable. Hint. Choose a test function L(x, y) = x + y 4. 5