Suggested solutions, TMA4125 Calculus 4N

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1 Suggested solutions, TMA5 Calculus N Charles Curry May 9th 07. The graph of g(x) is displayed below. We have b n = = = 0 [ nπ = nπ ( x) nπx dx nπx dx cos nπx ] x nπx dx [ nπx x cos nπ ] ( cos nπ + cos nπ + cos nπ cos nπ = cos nπ + nπ (nπ) nπ + nπ ) + cos nπx dx [ nπx (nπ) ] Now cos nπ = ( ) n, and moreover nπ (k+)π is zero if n is even, whilst ( ) k, we may write the Fourier series of g as = g(x) = π ( ) n n= n nπx + π k= ( ) k (k + )πx (k + )

2 . We notice the equation may be written as y = t + y cos t Now L(y ) = sy (s) y(0) = sy. Ug the tables of Laplace transforms for t and cos t and the result L(y cos t) = Y L(cos t), we find We rearrange to find sy = s + + Y s s + s Y ( s + ) = s + s +, from which we deduce Y (s) = s + s. Ug the table of inverse Laplace transforms we find y(t) = + t. a) The idea is to use the identity F(f (x)) = iw ˆf(w). For this purpose, we define f(x) = C + x g(x)dx, where C is an appropriately chosen constant. Then f (x) = g(x), and hence ˆf(w) = ĝ(w), from which we find iw (ĝ(w) ) f(x) = F = ĝ(w) iw π iw eiwx dw In particular we have b a g(x)dx = f(b) f(a) = π b) Taking Fourier transforms with respect to x gives ĝ(w)(e iwb e iwa ) dw iw û tt (w, t) = c w û(w, t), û(w, 0) = 0, û t (x, 0) = ĝ(w), where we have also taken Fourier transforms of the initial conditions. As there are no derivatives with respect to w in the equation, we can solve like an ODE, obtaining û = A(w) cos cwt + B(w) cwt

3 Now û(w, 0) = 0 A = 0. We then differentiate: and hence û t (w, 0) = cwb. initial condition then gives û t = cwb cos cwt Setting cwb(w) = ĝ(w) to satisfy the other û = ĝ(w) cw cwt c) We use the inverse Fourier transform to find u(x, t) = π Ug the identity cwt = eicwt e icwt, we find i u(x, t) = π Applying the identity from part a) then gives u(x, t) = c ĝ(w) cw (cwt) eiwx dw ĝ(w)(e iw(x+ct) e iw(x ct) ) dw icw x+ct. a) We write u = F (x)g(y), and thus Rearranging gives x ct g(x )dx u xx + u yy = F G + F G = 0, F F = G G, and as the left hand side is a function of x alone, and the right hand side is a function of y alone, they must both be equal to some constant, k, i.e F kf = 0, G + kg = 0 The boundary conditions give (for non-trivial solutions) F (0) = F () = 0. We solve the equation for F subject to these conditions. There are three possible cases:. k = p > 0. Then F (x) = A cosh px + B h px, whereupon F (0) = 0 A = 0. It follows that F (x) = Bp cosh px. As cosh is never zero, the condition F () = 0 gives B = 0, and hence we obtain only the trivial solution.

4 . k = 0. Then F (x) = Ax + B and F (0) = 0 gives B = 0. However, F (x) = A, and hence F () = 0 forces A = 0, giving again only the trivial solution. k = p < 0. Now F (x) = A cos px + B px As F (0) = 0 we have A = 0. We then differentiate to find F (x) = Bp cos px. To satisfy F () = 0, we require cos p = 0 p = (n + )π, n =,,,... We now solve which has a general solution G p G = 0 G n = A n cosh (n + )πy + B n h (n + )πy We therefore have the general solution u(x, t) = n= b) Differentiation gives ( (n + )πy A n cosh + B n h ) (n + )πy (n + )πx u y (x, 0) = n= (n + )πb n (n + )πx = 0, hence B n = 0 for all n. We then have u(x, ) = A n cosh n= (n + )π (n + )πx = 50 5πx The middle term is the Fourier series for the rightmost term, and comparing the two we see that A n = 0 for n, whilst A = 50 cosh 5π

5 We then have u(x, t) = 50 5πx cosh 5π cosh 5πy 5. We rearrange the equations to give The Gauss-Seidel iteration is then x = ( x ) x = ( x ) x = ( x x ) x n+ = ( xn ) x n+ = ( xn ) Setting in x 0 = x 0 = x 0 = 0, we have x n+ = ( xn+ x n+ ) x =, x =, x = (. ( )) = Repeating the procedure we have x = ( ) = 5 8, x = ( ) = 5 8, x = ( 5 8 ( 5 )) = 8 6. a) We have 0 tan(x)dx / (tan 0 + tan + tan + tan + tan ) = b) We now use the formula ɛ h 80 max f () (x) = 5N, where N is the number of integration subintervals (which must be even for Simpson s rule). In particular, to ensure ɛ 0 6, we require N

6 taking logarithms (we could also take the fourth root) gives N = exp( log( 5 06 )) = as N must be an even integer, we require N = 8 subintervals to ensure the requested accuracy. 7. The backward Euler formula gives y = y 0 + ht y setting in h =,y 0 = and t =, we find y 8 y = 0, which is a nonlinear equation for y. We use Newton s method to find a numerical solution to this equation, as specified in the problem. Now if we set f(y ) equal to the left hand side above, we find f (y ) = 8 cos y Write y n for the nth Newton iterate; Newton s method is then y n+ = y n yn 8 y n 8 cos y n Setting in y 0 =, we find Repeating the procedure gives y = y = cos = (.066) 8 cos(.066) = The Crank-Nicolson scheme discretizes first in space ug the usual finite difference formulae, then in time ug the trapezium rule. Letting u m (t) u(mh, t), we have t u m = h (u m u m + u m+ ) + 8u m = f m (u) This is a system of ODEs to which we apply the Trapezium rule: u n+ = u n + k ( f(u n ) + f(u n+ ) ), 6

7 where u = (u,..., u N ) and f(u) = (f (u),..., f N (u)) T. We then obtain the approximations u n m u(mh, nk), which follow u n+ m = u n m+ k h (un m u n m+u n m++u n+ m u n+ m +u n+ m )+ k (8un m+8u n+ m ) Setting in k = h =, this becomes u n+ m = u n m + (u n m u n m + u n m+ + u n+ m u n+ m + u n+ m ) + u n m + u n+ m We then bring all the terms in u n+ to the left, and all those in u n to the right: u n+ m + u n+ m u n+ m+ = u n m u n m + u n m+ The boundary conditions are u n 0 =, u n = 0, which can be combined with the above to give the following system 0 u n+ 0 u n u n+ = The initial conditions are u n+ u 0 = (u 0, u 0, u 0 ) = (,, ), inserting this gives the linear system 0 u u = 0 u Solving this system by, e.g. Gaussian elimination gives u = (u, u, u ) = ( 7 8,, 5 8 ) and for the next time step we have the linear system 0 u 9 = 0 Solving this system we obtain u u u = ( 9 6, 5, 6 ) u n u n 7