Question

3. (10 pts) Use the method of separating variables to solve the problem utt = curr u(0,t) = 0 = u(l,t) ur. 0) = 3.7 - 4, u(3,0) = 0 for 0 <r<l, t>0 fort > 0 for 0 <r<1

This is a partial differential equations question. Please help me solve for u(x,t):

Find the eigenvalues/eigenfunction and then use the initial conditions/boundary conditions to find Fourier coefficients for the equation.

Answer 1

domu (x, t) = c2 ** (2,t) for all 0 < x <l and t > 0 (1) The conditions that the left and right hand ends are held at height zero are encoded in the "boundary conditions" (2) u(0,t) = 0 ull, t) = 0) for all t > 0 for all t > 0 (3) As we have been told the position and speed of the string at time 0, there are given functions f(x) and g(x) such that the "initial conditions" u(x,0) = f(x) ut(x,0) = g(x) for all 0 < x <l for all 0 < x <e (༡) are satisfied. The problem is to determine u(x, t) for all r and t. Outline of the Method of Separation of Variables We are going to solve this problem in three steps. Step 1 In the first step, we find all solutions of (1) that are of the special form u(x, t) = X(X)T(t) for some function X(x) that depends on x but not t and some function T(t) that depends on t but not x. This is where the name "separation of variables" comes from. It is of course too much to expect that all solutions of (1) are of this form. But if we find a bunch of solutions X (3)T;(t) of this form, then since (1) is a linear equation, a;X;(2)T;(t) is also a solution for any choice of the constants ai. (Check this yourself!) If we are lucky (and we shall be lucky), we will be able to choose the constants a; so that the other conditions (2-5) are also satisfied. Step 2 We impose the boundary conditions (2) and (3). Step 3 We impose the initial conditions (4) and (5). The First Step – Finding Factorized Solutions The factorized function u(x, t) = X(X)T(t) is a solution to the wave equation (1) if and only if X (a)T"(t) = c+X"(x)T(t) = 6 = 210)

The left hand side is independent of t. So the right hand side, which is equal to the left hand side, must be independent of t too. The right hand side is independent of x. So the left hand side must be independent of x too. So both sides must be independent of both x and t. So both sides must be constant. Let's call the constant o. So we have 2 T0 = 0 T"(t) – coT(t) = 0 (6) X"(x) – oX(x) = 0) We now have two constant coefficient ordinary differential equations, which we solve in the usual way. We try X(x) = e" and T(t) = est for some constants r and s to be determined. These are solutions if and only 1 onze"– oe"x = 0 test - c?oest = 0 (72 - o)e"? = 0) (s2 – c0)est = 0) p2 – 0 = 0 sa - co = 0 r=#vo s = cvo If o #0, we now have two independent solutions, namely eVor and e-Vor, for X (x) and two independent solutions, namely ecvot and e-cvot, for T(t). If o +0, the general solution to (6) is X(x) = djevor + d2e-Vor T(t) = dze Vot + dae-cvot for arbitrary constants di, d2, d3 and 24. Il y = 0, the equations (6) simplify to X"(x) = 0) T"(t) = 0) and the general solution is X(X) = dı + d20 T(t) = d3 + det for arbitrary constants di, d2, d3 and d4. We have now found a huge number of solutions to the wave equation (1). Namely u(x, t) = (djevox + d2e-Vox) (dze Vöt +d4e-cvot) for arbitrary o #0 and arbitrary dı,d2, d3, da u(x, t) = (dı + d2x) (d3 + dat) for arbitrary dị, d2, d3, d4 The Second Step - Imposition of the Boundary Conditions If X ( T;(t), i = 1,2,3,... all solve the wave equation (1), then i a;X;( T;(t) is also a solution for any choice of the constants a;. This solution satisfies the boundary condition (2) if and only if a;X;(0)T;(t) = 0 for all t > 0 This will certainly be the case if X:(0) = 0) for all i. In fact, if the ai's are nonzero and the Ti(t)'s are independent, then (2) is satisfied if and only if all of the X (0)'s are zero. For us, it will be good enough to simply restrict our attention to Xi's for which X:(0) = 0, so I am not even going to define what "independent" means(1). Similarly, u(x, t) = Lia;X;(x)T;(t) satisfies the boundary condition (3) if and only if a;X;(4)T;(t) = 0 for all t > 0

and this will certainly be the case if X;(l) = 0 for all i. We are now going to go through the solutions that we found in Step 1 and discard all of those that fail to satisfy X(0) = X(C) = 0. First, consider 0 = 0 so that X (2) = di +d2x. The condition X (0) = 0 is satisfied if and only if di = 0. The condition X(C) = 0) is satisfied if and only if dı + ld2 = 0. So the conditions X (0) = X(C) = 0) are both satisfied only if di = d2 = 0, in which case X (2) is identically zero. There is nothing to be gained by keeping an identically zero X (x), so we discard o = 0 completely. Next, consider o + 0 so that dieVox + d2e-Vox. The condition X (0) = 0 is satisfied if and only if di + d2 = 0. So we require that d2 = -d1. The condition X (C) = 0) is satisfied if and only if () = djevol + d2e-Vol = dı (evol - e-vol) If dị were zero, then X(x) would again be identically zero and hence useless. So instead, we discard any o that does not obey evol-e-Vol = 0) 6 evol = e-völ evol = 1 In the last step, we multiplied both sides of evol = e-vol by evol. One o that obeys e2vol = 1 is o = 0). But we are now considering only o + 0. Fortunately, there are infinitely many complex numbers(2) that work. In fact evol = 1 if and only if there is an integer k such that 2Vol = 2kri Vork, o=-622 With Vo =kr and d2 = -dı, X(x)T(t) = dı (e281 - 2 2 2) (dze?"kat + dae-2017) = 2rdį sin ( 82 x) [(d3 + da) cos (cka t) + 2(dz – da) sin ( ckt t)] = sin ( 64 x) [ak: cos (ckIt) + Bk sin (6kA )] where ok = 2ndı(d3 + da) and Bk = -2d1(d3 – da). Note that, to this point, dı,dz and dų are allowed to be any complex numbers so that ak and Bk are allowed to be any complex numbers. The Third Step – Imposition of the Initial Conditions We now know that u(x, t) = { sin ( 49 x[ax cos (sku t) + Bx sin (skt 1)] k=1 obeys the wave equation (1) and the boundary conditions (2) and (3), for any choice of the constants ak, Bk. It remains only to see if we can choose the ak's and Bk's to satisfy k=1 f(x) = u(a,0) = { an sin (4) g(x) = u(a,0) = Ž *** sin(+1) k=1 But any reasonably smooth) function, h(c), defined on the interval 0 < x < l, has a unique representation) h(x) = bx sin kama k=1

as a linear combination of sin kna's and we also know the formula Bk = § | h(x) sin km2 dx for the coefficients. We can make (7) match (4') by choosing h(x) = f(x) and bk = 0k. This tells us that ak = į S. f(x) sin kær dx. Similarly, we can make (7) match (5') by choosing h(x) = g(x) and bk = BACKA. This tells us that ckt Bk = So g(x) sin kar dx. So we have a solution: |u(x, t) = Ž sin (423[ax cos (45 t) + Bx sin (44F 4)] with fax = /* $(a) sin kau de Bx = ca*9(a) sin kqz dr