Existence and Uniqueness Theorem for Linear Equations

Linear ODEs We discussed first-order linear ODEs in Section 2.3, which had the following form. dy + p(x) y = q(x) dx Now, we are going to generalize this definition to higher-order ODEs. Definition Let p0 (x), p1 (x), . . . , pn−1 (x), and q(x) be single variable functions of a real variable x. An nth-order linear ODE has the following form. y (n) +pn−1 (x) y (n−1) +pn−2 (x) y (n−2) +· · ·+p1 (x) y 0 +p0 (x) y = q(x) An nth-order linear IVP must also have n points of the form: y (x0 ) = y0 , y 0 (x0 ) = y1 , y 00 (x0 ) = y2 , .. . y (n−1) (x0 ) = yn−1 . Definition (ALTERNATE) Let a0 (x), a1 (x), . . . , an (x), and f (x) be single variable functions of a real variable x. An nth-order linear ODE has the following form. an (x) y (n) + an−1 (x) y (n−1) + an−2 (x) y (n−2) + · · · + a1 (x) y 0 + a0 (x) y = f (x) An nth-order linear IVP must also have n points of the form: y (x0 ) = y0 , y 0 (x0 ) = y1 , y 00 (x0 ) = y2 , .. . y (n−1) (x0 ) = yn−1 . Characterizing Linear Equations For an nth-order ODE to be linear, the following two conditions must be satisfied. 1. No derivatives of y are inside of any functions (such as trig functions, logarithms, exponentials, power functions etc.) Each derivative of y is being raised to the first power only (including the zero-th derivative of y ). 2. No derivatives of y can be multiplied by any other derivatives of y (including the zero-th derivative of y ). Note that y (0) is the zero-th derivative of y , meaning no derivatives of y have been taken. Thus, we have y (0) = y . First-order Linear Equation If n = 1 then p0 (x) and q(x) are the only functions we have and we only need one initial value. y 0 + p0 (x) y = q(x), y (x0 ) = y0 Second-order Linear Equation When n = 2, we now have p1 (x), p0 (x) and q(x) and we need two initial values. y 00 + p1 (x) y 0 + p0 (x) y = q(x), y (x0 ) = y0 , y 0 (x0 ) = y1 Third-order Linear Equation For n = 3, we have p2 (x), p1 (x), p0 (x) and q(x) and we need three initial values. y 000 + p2 (x) y 00 + p1 (x) y 0 + p0 (x) y = q(x), y (x0 ) = y0 , y 0 (x0 ) = y1 , y 00 (x0 ) = y2 Existence and Uniqueness Theorem for Linear Equations (EUTLE) Consider the nth order linear IVP below. y (n) + pn−1 (x) y (n−1) + pn−2 (x) y (n−2) + · · · + p1 (x) y 0 + p0 (x) y = q(x) y (x0 ) = y0 , y 0 (x0 ) = y1 , y 00 (x0 ) = y2 , · · · , y (n−1) (x0 ) = yn−1 . Let I be an open interval containing x0 on which q(x) and pi (x) (for 0 ≤ i ≤ n − 1) are ALL continuous. Then, the IVP has a unique solution on I . Finding the Largest Interval Suppose that we are tasked to find the largest open interval I such that EUTLE guarantees a unique solution to an nth-order IVP. I Find the intervals where p0 , p1 , . . . , pn−1 are continuous. I Find the interval where q(x) is continuous. I Draw a number line and mark x0 on the number line and indicate the intervals where the various functions are continuous. I The largest interval CONTAINING x0 on which all of n + 1 functions are continuous IS the largest interval on which EUTLE applies. Example Find the largest interval on which EULTLE guarantees us a unique solution to the IVP. p y 000 + sin(x) y 00 + x 2 − 4 y 0 + log(4 − x) y = 5, y (3) = 1, y 0 (3) = 1015 , y 00 (3) = 0 Example Find the largest interval on which EULTLE guarantees us a unique solution to the IVP. p y 000 + sin(x) y 00 + x 2 − 4 y 0 + log(4 − x) y = 5, y (−3) = 1, y 0 (−3) = 1015 , y 00 (−3) = 0 Example Find the largest interval on which EULTLE guarantees us a unique solution to the IVP. y 000 + 1 1 1 1 y 00 + y0 + 2 y= , x −1 x +1 x −9 x −5 y (2) = 0, y 0 (2) = 0, y 00 (2) = 0 Example Find the largest interval on which EULTLE guarantees us a unique solution to the IVP. y 000 + 1 1 1 1 y 00 + y0 + 2 y= , x −1 x +1 x −9 x −5 y (3) = 0, y 0 (3) = 0, y 00 (3) = 0 Example Find the largest interval on which EULTLE guarantees us a unique solution to the IVP. y (4) + y 000 + 2x y 00 + (x 2 + 3) y 0 + sin(2x − 5) y = cos(x), y (2) = 0, y 0 (2) = 0, y 00 (2) = 0, y 000 (2) = 0