MODULE 13. Topics: Linear systems

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1 Topics: Linear systems MODULE 13 We shall consider linear operators and the associated linear differential equations. Specifically we shall have operators of the form i) Lu u A(t)u where A(t) is an n n matrix with continuous functions of t in its entries and u = (u 1,..., u n ). ii) Lu n j=0 a j(t)u (j) (t). In most applications this L will operate on a single function u(t). We assume that the coefficient of the jth derivative is continuous in t, and that a n (t) 0 for any t in the interval of interest. Hence without loss of generality we may assume that a n (t) 1. Both operators have in common that they are linear, i.e., L(u + αv) = Lu + αlv. In either case we shall consider the differential equation Lu = 0 In other words, we are going to examine the null space of L in the domain of L consisting of all sufficiently smooth functions. if we set As we noted before, the nth order equation can be transformed into a first order system v 1 = u v 2 = u v n = u (n 1) then the first order system equivalent to the single nth order scalar equation is Lv v A(t)v = 0 where A(t) = a 0 a 1 a 2 a n 1 83

2 The equivalence between the nth order scalar equation and the first order system will be exploited time and again. We also note that the equation Lu = 0 for a first order system will usually be written as u = A(t)u so that we can apply the above existence and uniqueness theorem. Theorem: dim N (L) = n. Proof. Consider the n initial value problems u i = A(t)u i u i (0) = ê i, i = 1,..., n. It follows from our existence and uniqueness theorem that the functions {u i (t)} exist, at least in a neighborhood of t = 0. Without proof we shall accept that if the coefficients are bounded for all t then these solutions in fact exist for all t. These solutions are linearly independent, for if n α j u j (t) = U(t) α = 0 j=1 where U is the matrix whose ith column is u i (t), then also U(0) α = 0. But U(0) = I, hence α = (α 1,..., α n ) = 0. Now suppose that w(t) is any vector N (L). Then by linearity the function v(t) = U(t)w(0) satisfies v (t) = A(t)v(t) v(0) = w(0). Uniqueness guarantees that v(t) w(t). Hence any function in the null space of L is representable as a linear combination of the solutions {u i }. Hence these n functions form a basis of N (L). 84

3 Example: If then By inspection, u 1 (t) = u = 0 1 u 1 0 cosh t sinh t, u sinh t 2 (t) =. cosh t ( e t w(t) = e t ) N (L) and 2w(t) = u 1 (t) u 2 (t). What does this theorem tell us about n Lu a j (t)u (j) (t) = 0 j=0 The corresponding vector system v = A(t)v has n linearly independent solutions v i (t), i = 1,..., n. The first component of each vector v i (t) is a solution of Lu = 0. The remaining components are the derivatives of u. If we set u i (t) = first component of v i (t) then Lu i (t) = 0 and u (k) i (0) = δ i,k+1. In other words, u 1 (0) = 1 and all derivatives up to order (n 1) are zero at t = 0, u 2 (0) = 0, u 2(0) = 1, and all other derivatives of u 2 are zero at t = 0. If we look at the Wronskian of these n functions we see from W (t) = det u 1(t) u 2 (t) u n (t) = det(v 1 v 2... v n ) u (n 1) 1 (t) u (n 1) 2 (t) u (n 1) n (t) 85

4 that W (0) = 1 so that the functions {u i (t)} are linearly independent. The uniqueness of solutions for an initial value problem can then be used to demonstrate that these functions are a basis of N (L). We summarize: i) The equation u = A(t)u has n linearly independent solutions {u i } and any solution of this system can be written as a linear combination of the {u i }. ii) The equation n Lu = a j (t)u (j) (t) = 0 j=0 has n linearly independent solutions {u i } and any solution of Lu = 0 can be written as a linear combination of the {u i }. Example: Consider the boundary value problem Lu u + u = 0 u(0) + u(1) = 1 u (2) = 2. We verify that two linearly independent solutions of Lu = 0 are u 1 (t) = cos t and u 2 (t) = sin t. A solution u(t) of our boundary value problem, if it exists, must belong to the span{u 1 (t), u 2 (t)}, i.e., We satisfy the boundary conditions if u(t) = α 1 u 1 (t) + α 2 u 2 (t). 1 + cos 1 sin 1 α1 = sin 2 cos 2 α The system is not singular. α 1 and α 2 can be found, hence a solution u(t) exists. Moreover, this solution is unique because if there were two solutions v(t) and w(t) then the function u(t) = v(t) w(t) 86

5 would have to satisfy Lu = 0 u(0) + u(1) = 0 u (2) = 0. But the only function in the span of u 1 (t) and u 2 (t) which satisfies these boundary conditions is the zero function. The dominant practical difficulty in solving linear equations is the calculation of the basis functions. Only for constant coefficient systems is there a consistent way of generating the basis. This topic will be studied at length a little later. However, given one element in the null space of an nth order linear operator one can often find a second linearly independent element from a related simpler problem. We shall discuss only the case of a second order equation of the form Lu a 2 (t)u + a 1 (t)u + a 0 (t)u = 0 where the coefficients are continuous in t on some interval. Here the computation proceeds as follows: Let u 1 (t) be an element in N (L), i.e., Lu 1 (t) = 0 then one can find a second element N (L) of the form u 2 (t) = φ(t)u 1 (t) where φ is found from (essentially) a first order separable equation. Since we want Lu 2 = 0 we find that φ must be chosen so that a 2 (t)[φ u 1 + 2φ u 1 + φu 1] + a 1 (t)[φ u 1 + φu 1] + a 0 (t)φu 1 = 0. Since Lu 1 = 0 this equation simplifies to a 2 (t)u 1 φ + [2a 2 (t)u 1 + a 1 (t)u 1 ]φ = 0. 87

6 We now have a first order separable equation in ψ φ which we can write as ψ ψ = [2a 2(t)u 1(t) + a 1 (t)u 1 (t)] a 2 (t)u 1 (t) and which has a non-zero exponential solution. The function φ is then obtained by integrating ψ. Example: Consider Lu t 2 u + 3tu + u = 0. It is known that one can find a solution of the form u 1 (t) = t α. If we substitute this function into Lu = 0 we find that α must be chosen such that [α(α 1) + 3α + 1]t α = 0, i.e., α 2 + 2α + 1 = 0, which yields the solution α = 1 and u 1 (t) = 1 t in any interval not including the origin. Since the above quadratic in α has only one repeated root we do not obtain two different elements in N (L). Let us then find a second element of the form u 2 (t) = φ(t) 1 t If we substitute u 2 into the differential equation we obtain [ t 2 φ 1 t 1 2φ 2 + φ 2 ] [ + 3t φ 1 3 t φ 1 ] + φ 1 2 t = 0 so that tφ + φ = 0. It follows that φ (t) = K t 88

7 and hence that φ(t) = K ln t + c. Hence a second element in N (L) is u 2 (t) = 1 t ln t. By inspection u 2 is not a scalar multiple of u 1 so we have a basis for N (L). Finally we observe that the element φ generated in this way is always linearly independent from u 1. We simply look at the Wronskian: u1 (t) φ(t)u det 1 (t) u 1(t) φ (t)u 1 (t) + φ(t)u = u 2 1(t)φ (t) 1(t) which is non-zero at some point because φ is an exponential function. 89

8 Homework Module 13 1) Consider Lu u + 3tu 4 sin u = f(t). i) Prove or disprove: L is linear on C 2 [0, 1]. ii) Determine f(t) such that u(t) = 5 t is a solution of Lu = f(t). 2) Consider Lu = u 2u + u = te t. i) Show that {e t, te t } is a basis of N (L). ii) Determine a particular integral of the form u p (t) = (a 0 + a 1 t + a 2 t 2 + a 3 t 3 )e t iii) Find a solution of Lu = te t, u(1) = 1, u (2) = 2. iv) Write a first order system equivalent to Lu = te t, u(1) = 1, u (2) = 2 and give its solution. 3) Find a basis of N (L) when Lu = u + 4u. Solve Lu = 0 subject to π 0 or explain why no such solution exists. 4) Find a solution of the form u(t) = t α for u(t)dt = 1 Lu t 2 u + 5tu + 4u = 0 then find a basis for N (L). 90