Modeling and Simulation with ODE for MSE

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1 Zentrum Mathematik Technische Universität München Prof. Dr. Massimo Fornasier WS 6/7 Dr. Markus Hansen Sheet 7 Modeling and Simulation with ODE for MSE The exercises can be handed in until Wed, 4..6,. in the post box located in the MI basement! Exercise (Sheet 6 Continued) Solve the inhomogeneous system of ODEs where using the results from sheet 6. Solution: ẏ = Ay + b, y() = y A = 4 4, b =, y = 4, (i) general solution of the homogenous system of ODEs Using the eigenvalues and eigenvectors which were already calculated in Sheet 6, we immediately obtain ( ) y h (t) = c e λt v + c e λt + c e λ t h () + tv = c + c e t + c e t + t, with constant c, c, c C. (ii) stationary solutions y s (t) of the inhomogenous system of ODEs When searching for a particular solution of the inhomogeneous system we particularly can try to find stationary (constant) solutions (note that in every component the inhomogeneity b is a polynomial of degree, this is in analogy to the Method of Undetermined Coefficients). Any constant solution y s of the inhomogeneous problem has to fulfill = 4 4 y s (t) +. Solving this linear system of algebraic equations we find that all solutions are of the form y s = + µ, µ C. Note that necessarily there are infinitely many constant solutions, since already the homogeneous ODE systemhadconstant solutions. We specifically point out the particular solution y p (t) =.

2 (iii) general solution of the inhomogenous system of ODEs As usual the general solution of the inhomogeneous problem is given as the sum of a particular solution of the inhomogeneous problem and the general solution of the homogeneous problem, y(t) = y h (t)+y p (t) = c +c e t +c e t + t +, c j C. (iv) solution of the IVP Evaluating the general solution at t = we obtain a linear system of algebraic equations to determine the parameters c, c and c, 4 = y = c +c +c + with solution c c =. c This results in the final solution for the initial value problem, y(t) = + e t + 6t. + t c c c = 4, Exercise ) (inverse problem) Let the function y(t) = c e t + c e t + c te t with real parameters c, c, c be given. To which system of ODEs is y(t) the general solution? Solution: Step : This time the fist step is to compare the given function (i.e. the solution of the ODE system we are tasked to reconstruct) with the general form for solutions of ODEs of first order with constant coefficients. It becomes clear that we can immediately identify eigenvalues and (generalized) eigenvectors of the coefficient matrix. The eigenvalues can be read off from the exponents of the exponential functions: We have a single eigenvector λ =, and a double eigenvalue λ = (that this has to be a double eigenvalue follows from the presence of a term te t ). Next, the general theory tells us that for single eigenvalues λ with eigenvector v we find the solution e λt v. Vice versa this means that v = is the eigenvector belonging to λ.

3 Moreover, the presence of the term te t also tells us that the eigenspace belonging to λ = is deficient, i.e. it is one-dimensional. In that case, i.e. for a double eigenvalue λ with onedimensional eigenspace spanned by v, the general theory yields that e λt v and (h + tv)e λt are linearly independent solutions, where h is the generalized eigenvector. Conversely, this means that v = (the vector belonging to the term te t ) is the eigenvector belonging to λ =. Finally, we are left with the term proportional to e t, from which we would have to determine the generalized eigenvector belonging to λ =. BUT: According to the general theory, the term needs to be of the form (c v + c h)e t, which contradicts our current situation. In other words: The function y(t) cannot be the solution of a homogeneous system of linear first order ODEs. Step : correction of the initially given solution Let s find a correction to y(t) which CAN be identified as the solution of a homogeneous system. We already found v and v, which particularly means that c v e t should be a solution of the ODE system. In other words, what is missing is a term c he t with a vector h which can be chosen freely at this point, as long as {v, v, h} are linearly independent. Since it s already contained in the initial function y, let s choose h =. Putting T = (v, v, h) =, we find det T = = 9, i.e. T is a regular matrix and {v, v, h} are linearly independent (i.e. h is a viable choice as generalized eigenvector). Since we further already know the Jordan normal form J of the coefficient matrix A of the ODE system, we now find from J = T AT A = T JT = = 9 Summarizing, we have found that ỹ(t) = c e t + is a solution of the ODE system = c + c e t + c te t y = 9 y 4 9 y 5 9 y y = 9 9 y + 9 y y y = 9 y 6 9 y 6 9 y

4 Remark: To avoid fractional coefficients one often compensates factors like our 9 (which originally stemmed from inverting T ) by modifying J, as long as the Jordan normal form is a diagonal matrix (which in turn leads to larger eigenvalues exactly that s the reason for the eigenvalues 9 and ±8 in problem of sheet 6). That this simplifying of the coefficient matrix doesn t work in case of a non-diagonal Jordan normal form, in turn was overlooked in the preparation of the third exercise of this sheet (I apologize for that). Exercise ) (System of ODEs) Solve the initial value problem ẍ = 4ẋ 7 x 8 7 x, ẋ = x + ẋ x, x () =, ẋ () =, x () =. Solution: As a first step, we convert the first equation into two first order equations. Renaming y = x and y = x, we obtain the system ẏ = y, ẏ = 7 y + 4y 8 7 y, ẏ = y + y y. We hence have to consider the coefficient matrix Next step thus is determining eigenvalues and eigenvectors. The characteristic polynomial for A is given by p(λ) = det(a λi) = λ(4 λ)( λ) ( λ) 7 7 λ = λ + λ 4. The first root for this polynomial is λ =, so that we obtain λ λ + 4 = λ (λ + ) 4λ + 4 = (λ 4λ)(λ + ) + 4λ + 4 = (λ 4λ + 4)(λ + ). It follows that the characteristic polynomial has a double root in λ =. As for eigenvectors, the linear system (A + I)v = has the solution v =, whereas (A I)v has the solution v = 4

5 (the eigenspace is only one-dimensional). This means, we additionally need a generalized eigenvector, which is given as the solution of (A I)h = v, which results in h =. Putting everything together we arrive at the solution y (t) y(t) = y (t) = c e t + c e t 4 + c e t + t 4 y (t) for the ODE system of first order, and hence ( ) ( ) ( ) ( ) x (t) = c x (t) e t + c e t + c te t is the desired solution of the original system of ODEs. Information and material related to the lecture can be found at the lecture webpage

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