Mathematics 22: Lecture 12

Transcription

1 Mathematics 22: Lecture 12 Second-order Linear Equations Dan Sloughter Furman University January 28, 2008 Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

2 Definition We call an equation of the form d 2 u dt 2 + p(t)du + q(t)u = g(t) dt a second-order linear differential equation. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

3 Definition We call an equation of the form d 2 u dt 2 + p(t)du + q(t)u = g(t) dt a second-order linear differential equation. If g(t) 0, we say the equation is homogeneous. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

4 Theorem Suppose p, q, and g are continuous on (α, β) and t 0 is a point in (α, β). Then there exists one, and only one, function u(t) satisfying the equation d 2 u dt 2 + p(t)du + q(t)u = g(t) dt with initial conditions u(t 0 ) = a and u (t 0 ) = b. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

5 Theorem Suppose p, q, and g are continuous on (α, β) and t 0 is a point in (α, β). Then there exists one, and only one, function u(t) satisfying the equation d 2 u dt 2 + p(t)du + q(t)u = g(t) dt with initial conditions u(t 0 ) = a and u (t 0 ) = b. Note: in particular, if g(t) 0, the only solution satisfying the equation with the initial conditions u(t 0 ) = 0 and u (t 0 ) = 0 is the trivial solution u(t) 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

6 Operators We may associate with the equation an operator d 2 u dt 2 + p(t)du + q(t)u = g(t) dt L[u(t)](t) = d 2 u dt 2 + p(t)du dt + q(t)u. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

7 Operators We may associate with the equation an operator d 2 u dt 2 + p(t)du + q(t)u = g(t) dt L[u(t)](t) = d 2 u dt 2 + p(t)du dt + q(t)u. A solution of the equation is then a function which satisfies L[u(t)](t) = g(t). Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

8 Linearity If c is a constant, then, for any function u, L[cu(t)](t) = c d 2 u dt 2 + cp(t)du + cq(t)u = cl[u(t)](t). dt Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

9 Linearity If c is a constant, then, for any function u, L[cu(t)](t) = c d 2 u dt 2 + cp(t)du + cq(t)u = cl[u(t)](t). dt If u 1 and u 2 are two functions, then L[u 1 (t) + u 2 (t)](t) = d 2 dt 2 (u 1(t) + u 2 (t)) + p(t) d dt (u 1(t) + u 2 (t)) + q(t)(u 1 (t) + u 2 (t)) ( d 2 = dt 2 u 1(t) + p(t) d ) dt u 1(t) + q(t)u 1 (t) ( d 2 + dt 2 u 2(t) + p(t) d ) dt u 2(t) + q(t)u 2 (t) = L[u 1 (t)](t) + L[u 2 (t)](t). Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

10 Linearity (cont d) We say L is a linear operator. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

11 Linearity (cont d) We say L is a linear operator. Note: it follows that if L[u 1 (t)](t) = 0 and L[u 2 (t)](t) = 0, then, for any constants c 1 and c 2, L[c 1 u 1 (t) + c 2 u 2 (t)](t) = c 1 L[u 1 (t)](t) + c 2 L[u 2 (t)](t) = 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

12 Linearity (cont d) We say L is a linear operator. Note: it follows that if L[u 1 (t)](t) = 0 and L[u 2 (t)](t) = 0, then, for any constants c 1 and c 2, L[c 1 u 1 (t) + c 2 u 2 (t)](t) = c 1 L[u 1 (t)](t) + c 2 L[u 2 (t)](t) = 0. That is: if u 1 and u 2 are solutions of a second-order linear homogeneous differential equation, then u(t) = c 1 u 1 (t) + c 2 u 2 (t) is also a solution for any constants c 1 and c 2. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

13 Toward the general solution Now suppose u 1 and u 2 are solutions of the same linear second-order homogeneous differential equation. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

14 Toward the general solution Now suppose u 1 and u 2 are solutions of the same linear second-order homogeneous differential equation. Let u be any other solution of the equation. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

15 Toward the general solution Now suppose u 1 and u 2 are solutions of the same linear second-order homogeneous differential equation. Let u be any other solution of the equation. Question: can we find c 1 and c 2 such that u(t) = c 1 u 1 (t) + c 2 u 2 (t)? Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

16 Toward the general solution Now suppose u 1 and u 2 are solutions of the same linear second-order homogeneous differential equation. Let u be any other solution of the equation. Question: can we find c 1 and c 2 such that Let a = u(t 0 ) and let b = u (t 0 ). u(t) = c 1 u 1 (t) + c 2 u 2 (t)? Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

17 Toward the general solution Now suppose u 1 and u 2 are solutions of the same linear second-order homogeneous differential equation. Let u be any other solution of the equation. Question: can we find c 1 and c 2 such that Let a = u(t 0 ) and let b = u (t 0 ). u(t) = c 1 u 1 (t) + c 2 u 2 (t)? We would like to find find c 1 and c 2 such that a = c 1 u 1 (t 0 ) + c 2 u 2 (t 0 ) b = c 1 u 1(t 0 ) + c 2 u 2(t 0 ). Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

18 The general solution By Cramer s rule, we must have c 1 = c 2 = au 2 (t 0) bu 2 (t 0 ) u 1 (t 0 )u 2 (t 0) u 2 (t 0 )u 1 (t 0) bu 1 (t 0 ) au 1 (t 0) u 1 (t 0 )u 2 (t 0) u 2 (t 0 )u 1 (t 0) provided u 1 (t 0 )u 2 (t 0) u 2 (t 0 )u 1 (t 0) 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

19 The general solution By Cramer s rule, we must have c 1 = c 2 = au 2 (t 0) bu 2 (t 0 ) u 1 (t 0 )u 2 (t 0) u 2 (t 0 )u 1 (t 0) bu 1 (t 0 ) au 1 (t 0) u 1 (t 0 )u 2 (t 0) u 2 (t 0 )u 1 (t 0) provided u 1 (t 0 )u 2 (t 0) u 2 (t 0 )u 1 (t 0) 0. In that case u(t) and c 1 u 1 (t) + c 2 u 2 (t) are two solutions of the equation satisfying the same initial conditions. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

20 The general solution By Cramer s rule, we must have c 1 = c 2 = au 2 (t 0) bu 2 (t 0 ) u 1 (t 0 )u 2 (t 0) u 2 (t 0 )u 1 (t 0) bu 1 (t 0 ) au 1 (t 0) u 1 (t 0 )u 2 (t 0) u 2 (t 0 )u 1 (t 0) provided u 1 (t 0 )u 2 (t 0) u 2 (t 0 )u 1 (t 0) 0. In that case u(t) and c 1 u 1 (t) + c 2 u 2 (t) are two solutions of the equation satisfying the same initial conditions. It follows, by the uniqueness part of the theorem above, that u(t) = c 1 u 1 (t) + c 2 u 2 (t). Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

21 The general solution (cont d) It follows, under the above condition, that u(t) = c 1 u 1 (t) + c 2 u 2 (t) is the general solution of the equation. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

22 Definition Given differentiable functions u 1 and u 2, we call W (t) = u 1 (t)u 2(t) u 1(t)u 2 (t) = u 1(t) u 2 (t) u 1 (t) u 2 (t) the Wronskian of u 1 and u 2. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

23 Theorem Suppose p and q are continuous on an interval (α, β). If u 1 and u 2 are solutions of d 2 u dt 2 + p(t)du + q(t)u = 0, dt then either W (t) 0 for all t in (α, β) or for all t in (α, β). W (t) = 0 Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

24 Definition We say functions u 1 and u 2 are linearly dependent on an interval (α, β) if there exist constants c 1 and c 2, not both 0, for which for all t in (α, β). c 1 u 1 (t) + c 2 u 2 (t) = 0 Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

25 Definition We say functions u 1 and u 2 are linearly dependent on an interval (α, β) if there exist constants c 1 and c 2, not both 0, for which for all t in (α, β). c 1 u 1 (t) + c 2 u 2 (t) = 0 If u 1 and u 2 are not linearly dependent, we say they are linearly independent. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

26 Theorem Suppose p and q are continuous on an interval (α, β) and u 1 and u 2 are solutions of d 2 u dt 2 + p(t)du + q(t)u = 0. dt If W (t 0 ) = 0 for some t 0 in (α, β), then u 1 and u 2 are linearly dependent. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

27 Proof If W (t 0 ) = 0, then there exist constants c 1 and c 2, not both 0, for which c 1 u 1 (t 0 ) + c 2 u 2 (t 0 ) = 0 c 1 u 1(t 0 ) + c 2 u 2(t 0 ) = 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

28 Proof If W (t 0 ) = 0, then there exist constants c 1 and c 2, not both 0, for which c 1 u 1 (t 0 ) + c 2 u 2 (t 0 ) = 0 c 1 u 1(t 0 ) + c 2 u 2(t 0 ) = 0. If follows that u(t) = c 1 u 1 (t) + c 2 u 2 (t) is a solution of the equation satisfying the initial conditions u(t 0 ) = 0 and u (t 0 ) = 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

29 Proof If W (t 0 ) = 0, then there exist constants c 1 and c 2, not both 0, for which c 1 u 1 (t 0 ) + c 2 u 2 (t 0 ) = 0 c 1 u 1(t 0 ) + c 2 u 2(t 0 ) = 0. If follows that u(t) = c 1 u 1 (t) + c 2 u 2 (t) is a solution of the equation satisfying the initial conditions u(t 0 ) = 0 and u (t 0 ) = 0. Hence, by the uniqueness and existence theorem, u(t) = 0 for all t in (α, β). Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14

30 Proof If W (t 0 ) = 0, then there exist constants c 1 and c 2, not both 0, for which c 1 u 1 (t 0 ) + c 2 u 2 (t 0 ) = 0 c 1 u 1(t 0 ) + c 2 u 2(t 0 ) = 0. If follows that u(t) = c 1 u 1 (t) + c 2 u 2 (t) is a solution of the equation satisfying the initial conditions u(t 0 ) = 0 and u (t 0 ) = 0. Hence, by the uniqueness and existence theorem, u(t) = 0 for all t in (α, β). That is, c 1 u 1 (t) + c 2 u 2 (t) = 0 for all t in (α, β). That is, u 1 and u 2 are linearly dependent. Dan Sloughter (Furman University) Mathematics 22: Lecture 12 January 28, / 14