Linear ODE Theory (Sections 3.2 and 3.3)

Linear ODE Theory (Sections 3.2 and 3.3) In the following table, take L to be the linear operator: L(y) = y (n) + p1 (t)y (n−1) + . . . + pn−1 (t)y 0 + pn (t)y where y (k) denotes the k th derivative of y with respect to t. In Linear Algebra: In Differential Equations: If T : IRn → IRm is a linear operator, then T (x) = b can be written as If L : C n [0, 1] → C[0, 1] is a linear operator, the second order linear diff. equation can be written as: Ax = b L(y) = f (t) The Nullspace of A is important, because all solutions are written in the form: The Nullspace of L is important, because all solutions are written in the form: x = xh + xp y(t) = yh + yp where xh is the homogeneous part of the solution, Axh = 0. xp is called the particular part of the solution. where yh solves: L(y) = 0, and is called the homogeneous part of the solution. yp is called the particular solution. The dimension of the nullspace tells us how many ”pieces” there are to xh , in the sense that: The dimension of the nullspace tells us how many ”pieces” there are to yh , in the sense that: xh = c1 v1 + . . . + cp v p yh = c1 y1 + . . . + cp yp where the vi are linearly independent. (This is the Principle of Superposition) where the yi are “linearly independent” functions. (This is the Principle of Superposition) The dimension of the nullspace of L is n, the order of the differential equation. (Proved later). p Definition: {v i }i=1 are linearly independent iff the only solution to: Definition: {y1 (t), . . . , yp (t)} are linearly independent functions iff the only solution to: c1 v 1 + . . . + cp v p = 0 c1 y1 (t) + . . . + cp yp (t) = 0 is c1 = c2 = . . . = cn = 0. is c1 = c2 = . . . = cn = 0. The same constants must work for all time. Checking for linear independence of n solutions: Check that the Wronskian (at some t0 ) is non-zero (this is an n × n matrix): Checking for linear independence of n vectors in IRn : Check that det[v 1 , . . . , v n ] 6= 0 y1 y10 .. . W (y1 , . . . , yn ) = (n−1) y1 1 y2 y20 .. . (n−1) y2 ... ... yn yn0 .. . (n−1) . . . yn 6= 0 We now get more specific. We are interested in solving: ay 00 + by 0 + cy = f (t) or L(y) = f (t) (1) 1. All solutions to Equation (1) are of the form: y(t) = yh (t) + yp (t) where yh is called the homogeneous part of the solution, and yp is called the particular part of the solution. 2. The dimension of the nullspace is two (proved in class). Therefore, yh = c1 y1 + c2 y2 where y1 and y2 are linearly independent solutions to L(y) = 0. Definition: A set of linearly independent solutions is said to be a fundamental set of solutions if every solution to the initial value problem, L(y) = 0, y(0) = y0 is a linear combination of the fundamental set (therefore, a fundamental set is another name for a basis). 3. Are y1 and y2 linearly independent? From the definition, we would need to know that c1 y1 + c2 y2 = 0 In the case of two functions, this is easy to check. Two functions are linearly independent iff they are not constant multiples of each other. For the more general case of k functions, we need to compute what’s called the Wronskian. See your text for more details. 4. Construction of the Fundamental Set, y1 and y2 : How can we construct a fundamental set of solutions? We solve the following initial value problems: ay 00 + by 0 + cy = 0, ay 00 + by 0 + cy = 0, y(0) = 1, y 0 (0) = 0 y(0) = 0, y 0 (0) = 1 (2) (3) Let the solution to IVP (2) be y1 (t), and the solution to IVP (3) be y2 (t). Then the initial conditions guarantee that y1 and y2 are linearly independent, and since we know that there are only two solutions in the fundamental set, we have constructed the entire fundamental set. Using this fundamental set, the solution to an arbitrary initial value problem: ay 00 + by 0 + cy = 0, y(0) = y0 , y 0 (0) = v0 is: y(t) = y0 y1 (t) + v0 y2 (t) 5. Putting it all together: Given ay 00 + by 0 + cy = 0, we solve by assuming that y(t) = ert . This leads us to solving the characteristic equation: ar2 + br + c = 0 for r. We solve this by using the quadratic formula. We have seen that we obtain a fundamental set of solutions if we have two real solutions to the characteristic equation. The rest of chapter three discusses how we obtain a fundamental set of solutions if we only have one real root, and how to interpret the solutions if we have two complex conjugate roots. 2