The Euler Equidimensional Equation ( 3.2)

Transcription

1 The Euler Equidimensional Equation ( 3.) The Euler Equidimensional Equation ( 3.)

2 The Euler Equidimensional Equation Definition The Euler equidimensional equation for the unknown function y with singular point at x 0 R is given by the equation below, where p 0, q 0 are real constants, Remarks: (x x 0 ) y + p 0 (x x 0 ) y + q 0 y = 0. The Euler equation has variable coefficients. Functions y(x) = e rx are not solutions of the Euler equation. The point x 0 R is a singular point of the equation, since y + p 0 (x x 0 ) y + The particular case x 0 = 0 is is given by q 0 (x x 0 ) y = 0. x y + p 0 x y + q 0 y = 0. The Euler Equidimensional Equation ( 3.)

3 Main Idea to Get Solutions Recall: In.3 we got solutions to the equation y + a 1 y + a 0 y = 0. We looked for functions of the form y(x) = e rx. The exponential cancels out from the equation and we obtain an equation only for r without x, ( r + a 1 r + a 0 ) e rx = 0 ( r + a 1 r + a 0 ) = 0. (1) In the case of the Euler equidimensional equation x y + p 0 x y + q 0 y = 0 the exponential function e rx does not have the property given in Eq. (1), since ( x r + p 0 x r + q 0 ) e rx = 0 x r + p 0 x r + q 0 = 0, and the later equation still involves the variable x. Main Idea to Get Solutions Remark: Look for solutions like y(x) = x r. These function have the following property: y (x) = r x r 1 x y (x) = r x r ; y (x) = r(r 1) x r x y (x) = r(r 1) x r. Introduce y = x r into Euler s equation x y + p 0 x y + q 0 y = 0, for x 0 we obtain r(r 1) + p0 r + q 0 x r = 0 r(r 1) + p 0 r + q 0 = 0. The last equation involves only r, not x. This equation is called the indicial equation, and is also called the Euler characteristic equation.

4 The Euler Equidimensional Equation ( 3.) The Main Result Theorem (Euler Equidimensional Equation with x 0 = 0) Given p 0, q 0 R, consider the Euler equation x y + p 0 x y + q 0 y = 0, x 0. () Let r +, r - be solutions of r(r 1) + p 0 r + q 0 = 0. (a) If r + r -, real or complex, then a general solution of Eq. () is y(x) = c 1 x r + + c x r -, x 0, c 1, c R (or C). (b) If r + = r - = r 0, then a real-valued general solution of Eq. () is y(x) = ( c 1 + c ln x ) x r 0, x 0, c 1, c R. Given x 1 0, y 0, y 1 R, there is a unique solution to the IVP x y + p 0 x y + q 0 y = 0, y(x 1 ) = y 0, y (x 1 ) = y 1.

5 The Main Result Theorem (Case x 0 0) Given p 0, q 0, x 0 R, consider the Euler equation (x x 0 ) y + p 0 (x x 0 ) y + q 0 y = 0, x x 0. (3) Let r +, r - be solutions of r(r 1) + p 0 r + q 0 = 0. (a) If r + r -, real or complex, then a general solution of Eq. (3) is y(x) = c 1 x x 0 r + + c x x 0 r -, x x 0, c 1, c R (or C). (b) If r + = r - = r 0, then a real-valued general solution of Eq. (3) is y(x) = ( c 1 + c ln x x 0 ) x x 0 r 0, x x 0, c 1, c R. Given x 1 x 0, y 0, y 1 R, there is a unique solution to the IVP (x x 0 ) y + p 0 (x x 0 ) y + q 0 y = 0, y(x 1 ) = y 0, y (x 1 ) = y 1. The Euler Equidimensional Equation ( 3.)

6 The Roots of the Indicial Polynomial Example (Different Real Roots) Find the general solution of the Euler equation x y + 4x y + y = 0, x > 0. Solution: We look for solutions of the form y(x) = x r, x y (x) = rx r, x y (x) = r(r 1) x r. Introduce y(x) = x r into Euler equation, r(r 1) + 4r + x r = 0 r(r 1) + 4r + = 0. The solutions of r + 3r + = 0 are given by r ± = 1 3 ± 9 8 r + = 1 r - =. The general solution for x > 0 is y(x) = c 1 x 1 + c x. The Roots of the Indicial Polynomial Example (Repeated Roots) Find the general solution of x y 3x y + 4 y = 0, x > 0. Solution: We look for solutions of the form y(x) = x r, x y (x) = rx r, x y (x) = r(r 1) x r. Introduce y(x) = x r into Euler equation, r(r 1) 3r + 4 x r = 0 r(r 1) 3r + 4 = 0. The solutions of r 4r + 4 = 0 are given by r ± = 1 4 ± r + = r - =. Two linearly independent solutions are y 1 (x) = x, y = x ln(x). The general solution for x > 0 tis y(x) = c 1 x + c x ln(x).

7 The Roots of the Indicial Polynomial Example (Complex Roots) Find the general solution of the Euler equation x y 3x y + 13 y = 0, x 0. Solution: We look for solutions of the form y(x) = x r, x y (x) = rx r, x y (x) = r(r 1) x r. Introduce y(x) = x r into Euler equation r(r 1) 3r + 13 x r = 0 r(r 1) 3r + 13 = 0. The solutions of the indicial equation r 4r + 13 = 0 are r ± = 1 4 ± 16 5 r± = 1 4 ± 36 { r+ = + 3i r - = 3i. The general solution is y(x) = c 1 x (+3i) + c x ( 3i), x 0. The Euler Equidimensional Equation ( 3.)

8 Real Solutions for Complex Roots Theorem (Real Fundamental Solutions with x 0 = 0) If p 0, q 0 R satisfy that (p 0 1) 4q 0 < 0, then the indicial polynomial p(r) = r(r 1) + p 0 r + q 0 of the Euler equation x y + p 0 x y + q 0 y = 0, x 0 (4) has complex roots r + = α + iβ and r - = α iβ, where α = (p 0 1), β = 1 4q 0 (p 0 1). A complex-valued fundamental solutions pair to Eq. (4) is ỹ 1 (x) = x (α+iβ), ỹ (x) = x (α iβ). A real-valued fundamental solutions pair to Eq. (4) is y 1 (x) = x α cos ( β ln x ), y (x) = x α sin ( β ln x ). Real Solutions for Complex Roots Proof: Given ỹ 1 = x (α+iβ) and ỹ = x (α iβ), introduce y 1 = 1 ) (ỹ1 + ỹ, y = 1 ) (ỹ1 ỹ. i Use another Euler equation to rewrite ỹ 1 and ỹ, ỹ 1 = x (α+iβ) = x α x iβ = x α e ln( x iβ) = x α e iβ ln( x ). ỹ 1 = x α cos ( β ln x ) + i sin ( β ln x ), ỹ = x α cos ( β ln x ) i sin ( β ln x ). We conclude that y 1 (x) = x α cos ( β ln x ), y (x) = x α sin ( β ln x ).

9 Real Solutions for Complex Roots Example (Real Solutions for Complex Roots) Find a real-valued general solution of the Euler equation x y 3x y + 13 y = 0, x 0. Solution: The indicial equation is r(r 1) 3r + 13 = 0. The solutions of the indicial equations are r 4r + 13 = 0 r + = + 3i, r - = 3i. A complex-valued general solution for x 0 is y(x) = c 1 x (+3i) + c x ( 3i) c 1, c C. A real-valued general solution for x 0 is y(x) = c 1 x cos ( 3 ln x ) + c x sin ( 3 ln x ), c 1, c R.