newell_lec11-15_100713

I vectors writing any function which is piecewise

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: refore the supposition that W (y1 , y2 ) = 0 at some point on I is false and W (y1 , y2 ) ￿= 0 for points on I . We now know what we mean by essentially different. Let y1 (t), y2 (t) be two l.i. solutions of (6.15). Then the linear combination y (t) = Ay1 (t) + By2 (t) satisfies ￿ (6.15) and can satisfy the initial conditions y (t0 ) = y0 , dy (t0 ) = y0 for all t0 ∈ I dt ￿ and all choices of real numbers y0 , y0 . Let us summarize: For this course, our main interest is in identifying linearly independent solutions y1 (t), y2 (t) of (6.15) with coefficients p(t), q (t) continuous on the interval I of interest. For this purpose, the linear independence of y1 (t), y2 (t) on I is equivalent to the non-vanishing of the Wronskian on I . The equivalence is broken if our interval includes points at which p(t) and/or q (t) are discontinuous. Example: Take f (t) = t, g (t) = t2 on the interval I = (−1, 1). Then t and t2 are certainly l.i. on (−1, 1). [Since k1 t + k2 t2 = 0 for all t ∈ (−1, 1) implies that k1 = k2 = 0. Put for example t = 1/2 or t = 1/4.] They are also solutions to 2 2 y ￿￿ y ￿ + 2 y = 0. t t But the Wronskian, W (f, g ) = t2 , vanishes at t = 0. Note that p, q are discontinuous ￿t 2 at t = 0 and exp t0 t￿ dt￿ is not defined if t and t0 have opposite signs. Example: Take f (t) = t3 , g (t) = t2 |t| on the interval I = (−1, 1). Then W ≡ 0 for all t ∈ I yet t3 and t2 |t| are l.i. on I . [Take k1 t3 + k2 t2 |t| = 0 for 20 all t ∈ (−1, 1). Put t = 1/2, −1/2 ⇒ k1 = k2 = 0.] Also t3 and t2 |t| are solutions to 3 3 y ￿￿ − y ￿ + 2 y = 0 t t ￿t 3 But again p, q are discontinuous at t = 0 and exp t0 t￿ dt￿ is not defined if t and t0 have opposite signs. In many other contexts (e.g. writing any vector in R3 as a linear combination of three l.i. vectors; writing any function which is piecewise continuous on (−L, L) as π π a linear combination of the l.i. basis ‘vectors’, {cos nLx , sin nLx }∞ , we will need n=0 the notion of linear independence. The ‘Wronskian’ will not always be the most useful way to test. 21 Homework Lectures 11-15 Question 1: For each of the following equations, give the (a) order and say whether each is (b) linear or nonlinear (c) autonomous or nonautonomous (i) dy dt + 3ty = 5 dy dx (ii) + y 2 = 3y + 10 (iii) d3 y dt3 + y dy + cos t = 0 dt (iv) d2 y dt2 + 0.2 dy + y = cos ω t dt (v) d2 y dx2 (vi) dy dx (vii) + xy = 0 = −x y d4 y dt4 =0 Question 2: Solve the following initial value problems: (i) dy dt (ii) + y = e−t , dy dt = y (1 − y ), (iii) d2 y dt2 (iv) dy dx (v) dy dx y (0) = 1 (a) y (0) = 2; (b) y (0) = 1 . 2 + 2 dy + 5y = cos 2t, dt = −x, y √ = + y, y (0) = dy dt (0) = 0. y (1) = 1. y (0) = 0. Is there more than one solution? 23 Question 3: A tank initially contains 100 gallons of brine with a concentration of 1/2 lb. per gallon. At time t = 0, an inlet valve is turned on which carries a new brine solution into the tank at a rate of 2 gallons per mi...
View Full Document

This note was uploaded on 01/22/2014 for the course MATH 254 taught by Professor Indik during the Fall '08 term at Arizona.

Ask a homework question - tutors are online