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Sect10.1 - Section 10.1 Problem 2 Problem Either Solve the...

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Unformatted text preview: Section 10.1, Problem 2 _ Problem. Either Solve the boundary value problem y” + 2y = 0, y’(0) = 1, y’ = 0, or else show that it has no solution. Solution. The general solution to the differential equation y” + 2y = 0 is y(t) =,C'1cos(\/§t) + Czsz'n(\/§t); the only challenge is whether 0'1 and 02 can be chosen to satisfy the boundary conditions. . From the formulas, one knows y’ (t) = —x/§Clsin(x/§t) + x/ngcosh/fit), so that y'(0) = 001 + x/QCz. Since one knows y’ (0) = 1, then 02 = 1. Evaluating 3/ (7r), one finds that —\/§Clsin(\/§7r) +\/§Cgcos(\/§7r) = 0. (Note that sin(\/§7r) 75 0 — this is an easy mistake to make in a problem like this.) As one knows that 02 = 1, then 01 = = cot(\/§7r). Section 10.1, Problem 4 Problem. Either solve the boundary value problem y” + y = 0, y’(0) = 1, y’ (L) = 0, or else show that it has no solution. . 1 Solution. The general solution of the differential equation y" + y = 0 is y(t) = Clcos(t) + Czsin(t); the only challenge is Whether 01 and 02 can be chosen to satisfy the given boundary conditions. L 1‘ ' As 3/ (t) = —Clsz'n(t) + 0200303), then y’(0) = 02 = 1, so that 02 = 1. Also, y’(L) = —C’13in(L) + C2COS(L) = —Cls1ln(L) + 003(L) = 0. Then 01 = ~Zifzgl), whichis fine for most values of L. There is one problem: if the cosine of L is zero, then 01 is undefined. I Therefore this boundary value problem has a solution only if cos(L) aé 0, that is, if there is no integer k for which L = (2k — So there is a solution if and only ifLaé (2k—1)§. “0.1.0.1’ Section 10.1, Problem 7 _ Problem. Either solve the given boundary value problem y” + 4y = cos(a:), y(0) = 0, y(7r) = 0, or show that it has no solution. I Solution. This is a nonhomogenous second order linear diiferential equation, and must be solved with a homogenous solution (let yH(t) = Clsz'n(2t) +Czcos(2t)) and a particular solution (let yp(t) = Acos(t) + Bsin(t)i and find the values A and B must have). In this case, A = %, and B : 0. d Then y(t) = yp(t)+yfi(t) = %CO$(t)+018in(2t)+02008(2t). y(0) = %+02 = 0, sothat 02 = —%. From the other initial condition, y(7r) = —% + 0'2 = 0, and so 02 = +%. This is an inconsistency; there is no value of 02 Which satisfies both these initial conditions and so the boundary value problem has no Solution. 110.1!» 2 Section 10.1, Problem 11 Problem. Find the eigenvalues and the eigenfunctions of the boundary value problem y” + Ay = 0, y(0) = 0, y’ (1r) 2 0. Assume that the eigenvalues are real. Solution. There are two different ways the problem can break down, depend- ing on whether A is positive or negative, so it is easiest to consider positive and values of A separately. I Suppose A is positive. Then A = ,u2 for some positive number ,u. Then the differential equation y” + My = 0 has the general solution = 1cos(ut) + Czsinout), and y’ (t) = —p013in(pt) + MCgcosmt). I _ As y(0) = 01 = 0, then 01 = 0..Then y’(7r) = uCgcosUm) = 0. For the equation to have any solutions, either 02 equals zero (meaning the solution is y(t) = O, a boring case known as the trivial solution), or else cos(mr) = 0. cos(,u7r) can equal zero only if a is some integer plus % (as cos(%7r) = 0, and cos(%7r) = 0, and cos(-g-7r) = 0, et cetera). So u = 213:1, with I: an integer, and A = The eigenfunction for eigenvalue A;c = is yk(t) = sin(-2—k—;—1t), for k = 1, 2, 3, Now Suppose that A is negative. This means that A = —u2 for some positive number p, and the differential equation is y” — [By = 0. The general solution for this I problem is written with the hyperbolic trigOnometr-ic functions sinh and cosh, which may be worth some review if they are unfamiliar. (They’re discussed, briefly, in the freshman calculus books.) The general solution is y(t) = Clcosh(,ut) Cgsinhmt), and y’ = uClsz'nhMt) + M02008h(,l1.t). A So y(0) = Clcosh(0) + Cgsi'nh(0) = 01 - 1 + 02 - 0 = 0, and so 01 = 0. Then y’ (7r) = Czucos(,u7r) = Cgcosh(mr) = 0-. However, the hyperbolic cosine of t is never zero. Therefore 0'2 = 0 and the only solution of this system is the trivial solution, y(t) = 0 for all t. The conclusion is there are no negative values of A which are eigenfunctionsfor this problem; they’ve all been found by assuming A > 0, above. There is one unaddressed case — What if A = 0? In that case, the differential equation is y” = 0, for which the solution must be that y(t) = C’lt + 02. Then 3/ (t) = 01. So y(0) = 02 = 0, and y’(7r) = 01 = 0, and the solution is again the trivial solution, y(t) = 0 for all t. m Section 10.1, Problem 12 Problem. Find the eigenvalues and eigenfunctions of the boundary value problem 3;” + Ay = 0, y’(0) = 0, y(7r) = 0.' Assume that all eigenvalues are real. Solution. There are two different ways the problem can break down, depend- ing on whether A is positive or negative, so it is easiest to consider positive and values of A separately. Suppose A is positive. Then A = p2 for some positive number ,u. Then the differential equation 3;” + ,u2y = 0 has the general solution y(t) = Clcosmt) + Cgsin(,ut), and y’(t) = —,uC'132'n(,ut) +ngcos(ut). y’(0) = #02,: 0, and so 02 = 0. As y(7r) = Clcos(mr) = 0, then either 01 = 0, or else cos(mr) = 0. If 01 = 0 then y(t) = 0 for all t, a boring case known as the trivial solution. To avoid this one must have that Mr = ig—ln' for some integer It. So for a solution to exist, ,u = 31°25;- for some integer k, and the eigenvalue A = “2‘: So the eigenvalues are Ak = (MT—U2, and the corresponding eigenfunctions are yk(t) = 005(2’3—"123 for k = 1,2,3, A Now suppose that A is negative. This means that A = -,u2 for some positive number n, and the differential equation is y” — My = 0. The general solution for this > problem is written with. the hyperbolic trigonometric functions sinh and cosh, which may be worth some review if they are unfamiliar. (They’re discussed, briefly, in the freshman calculus books.) The general solution is y(t) = C'lcoshmt) + Cgsz'nhmt). So y' (t) = ,uCflsz'nhMt) + ,uC'zcoshmt). ' Then y’(0) = ungstht) = 0, and 02 == 0. So y(7r) = Clcoshflm) = 0, but the hyperbolic cosine of t is never zero. This requires that 01 = 0, which means the only solution is the trivial solution, y(t) = 0 for all t. From this one concludes there are no negative eigenvalues or corresponding eigenfunctions. There is one unaddressed case — what if A = 0? In that case, the differential equation is y” = 0, for which the solution must be that y(t) = Clt + 02. Then 3/ (t) = 01. So = 02 = 0, and 3/ (7r) 2 01 = 0, and the solution is again the trivial solution, y(t) = 0 for all t. Section 10.1, Problem 16 Problem. Find the eigenvalues and eigenfunctions of the boundary value problem 3;” — Ay = 0, y(0) = 0, y’ (L) = 0. Assume that all eigenvalues are real. Solution. There are two different ways the problem can break down, depend— ing on whether A is positive or negative, so it is easiest to consider positive and values of A separately. Suppose A is positive. Then A = M2 for some positive number M. Then the differential equation y” — My 2 0 has the general solution y(t) = C’lcosh(,ut) + Czsinh(ut), Where cosh and sinh are the hyperbolic trigonometric functions, which may be worth some review if they are unfamiliar. (They are addreSSed in any freshman Calculus textbook.) 3/ (t) = pClsz'nhQut) + ,uCzcoshQit). y(0) =. 01 = 0, and so 01 = 0. As 3/ (L) = Czcosh(,uL) = 0, then either 02 = 0, or else cosh(;uL) =0. But the hyperbolic cosine. is never zero, so 02 = 0 and the only solution is y(t) = 0 for all t, a boring case known as the trivial solution. The conclusion one should draw is there are no positive eigenvalues. Now suppose that A is negative. This means that A = —p2 for some positive number ,u, andthe _differential,equationvis. y” + My =_ “0. The ,generalwsolution is y(t) = Clcos(,u,t) + 023mm). So y’ (t) = -,uC'1sin(,ut) + “0200.5(pt). Then y(0) = Clcos(0) =0, and 01 = 0. So y’ (L) = Cgcos(/.LL) = 0. To avoid the trivial solution (02 = 0), then oos(,uL) = 0, whichrequires that ML = 2’“T‘lfl' for some integer k. So u. = $511.}, and the eigenvalues are Ak = flag = —(L2'1%)2, for any integer k. The corresponding eigenfunctions are yk = There is one unaddressed case — What if A = 0? In that case, the differential equation is y” = O, for which the solution must be that y(t) = Clt + 02. Then 3/ (t) = 01. So y(0) = 02 = O, and'y’ (7r) = 01 = 0, and the solution is again the trivial solution, y(t) = 0 for all t. ...
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