Question

l, t)4u (x, t), 0<x< L, 0 <t Evaluate u(1.1; 0.3) where u(x, t) u(0, 1)= u(L, t)- 0v1> 0 u(x, 0)= f(x), u,(x, 0)- g(x), 0<x< L L=T al f(x) 3sin 2x, g(x)=-2sin 3x b/ For f(x)-xn-x & g(x)-0, approximate numerically u(x, t) by the first term. L-S c/f(x)=-3sin g(x)- 5 2sin d/ f(x)-0, g()= .3 x +1 approximate numerically u(x, t) by the first term c/ f(x)-2(5-xx, g(x) x+1 3 approximate numerically u(x, t) by the first couple 4

Answer 1

First, I am giving the solution of a general question and then solved your problem.

0 x<L, t>0, . Пп — п ,р (Wave equation) и(0, г) %3D 0, and u(L,) %3D 0, (Boundary conditions) и(х, 0) —/ (x), and и (х, 0) 3 g(x). (Initial conditions) We first let u(x, t) = X(x)T(t) and separate the wave equation into two ordinary differential equations. Substituting u = X" T and u XT" into the wave equation, it becomes a? х"Т-XT".

Dividing both sides by a X T X" T" х ат As for the heat conduction equation, it is customary to consider the constant a as a function oft and group it with the rest of t-terms. Insert the constant of separation and break apart the equation: X" Т" х ат X" X 0 X"-X Т" T"%3D -а?1T T" + а?.T%3D0. The boundary conditions also separate: u(0,t) 0 - u(L,t) 0 X(L)T(t) = 0 -> X(L) 0 X(0)T(t)0 - T(t) 0 T(t) 0 or or

The boundary conditions also separate: X(0) 0 u(0,t) 0 X(0)T(f)= 0 u(L,t) 0 X(L)T(t)= 0 -» X() = 0 T(t) = 0 Tt) 0 or or As usual, in order to obtain nontrivial solutions, we need to choose X(0) 0 and XL) 0 as the new boundary conditions. The result after separation of variables, is the following simultaneous system of ordinary differential equations, with a set of boundary conditions: X"X 0 and XL) 0, T" aT 0

The next step is to solve the eigenvalue problem X(0) 0,XL) 0. X" X0 We have already solved this eigenvalue problem, recall. The solutions are n 1,2, 3,. Eigenvalues: L2 n t x Xsin L n 1,2, 3, .. Eigenfunctions: Next, substitute the eigenvalues found above into the second equation to find T). After putting eigenvalues A into it, the equation of T becomes 2 2 n T" a2 T 0 It is a second order homogeneous linear equation with constant coefficients It's characteristic have a pair of purely imaginary complex conjugate roots:

It is a second order homogeneous linear equation with constant coefficients It's characteristic have a pair of purely imaginary complex conjugate roots: an r =t L Thus, the solutions are simple harmonic: апnt + В, sin L апat T() A cos- n 1, 2, 3, . Multiplying each pair of X and T together and sum them up, we find the general solution of the one-dimensional wave equation, with both ends fixed, to be

апnt an tt +B sin L плх sin L u(х,1) —— A cos n n 1 There are two sets of (infinitely many) arbitrary coefficients. We can solve for them using the two initial conditions Set 0 and apply the first initial condition, the initial (vertical) displacement of the string ux,0)f(x), we have n u(x,0) (4 cos(0)B sin(0)) sin" L n-1 nTx A, sin = f(x) n-1 Therefore, we see that the initial displacementf(x) needs to be a Fourier sine series. Since f(x) can be an arbitrary function, this usually means that we need to expand it into its odd periodic extension (of period 2L). The

Therefore we see that the initial displacementf(x) needs to be a Fourier sine series. Since f(x) can be an arbitrary function, this usually means that we need to expand it into its odd periodic extension (of period 2L). The coefficients An are then found by the relation An bn, where bn are the corresponding Fourier sine coefficients of f(x). That is 24 Ab плх f(x)sin dx n L Notice that the entire sequence of the coefficients A, are determined exactly by the initial displacement. They are completely independent of the other sequence of coefficients Bn, which are determined solely by the second initial condition, the initial (vertical) velocity of the string. To find B, differentiate ux, t) with respect to t and apply the initial velocity, иАх, 0) %3 g(x). we

an tt annt COS L an t an t n Tx sin - A L u, (x,t) -sin + B_ L L n-1 Set t 0 and equate it with g(x) nTx an t u,(x,0 B sin = g(x) L n-1 We see that g(x) needs also be a Fourier sine series. Expand it into its odd periodic extension (period 2L), if necessary. Once g(x) is written into a sine series, the previous equation becomes - ant nTx nTx sin L u, (x,0) B L g(x)b, sin L n-1 n-1 Compare the coefficients of the like sine terms, we see 2 ntx an t B =b g(x)sin IL -dx

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fix)= T-X Ja) = o E An Ces 2nt Sinm u(xt) nei where An 2. f(2) = - 3 Sin T L=5 gla) = 2 Sinmx Bn Sin 2nnSn nm 5 A 22nt ua,t) = n Sm2 2nnt/s An Sin (2nnt) 4n(x, t) 2nm 2nm ner -3 Sin TX A, 3 ne(z,0) 2 Sinmx Sinnm Bn 2nT eruall ane Sin deems At 4T B5 x Bs 2XSx T 2 = 4 Sin 2.mt Sinm 3 Ces2nt Sin m ua t) 7oom to 1:1