Mathematical Methods and its Applications (Solution of assignment-12) Solution 1 From the definition of Fourier transforms, we have.
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1 For 2 weeks course only Mathematical Methods and its Applications (Solution of assignment-2 Solution From the definition of Fourier transforms, we have F e at2 e at2 e it dt e at2 +(it/a dt ( setting ( a t + i 2a τ. e 2 /(4a τ 2 dτ e, a 2e 2 /4a a e τ 2 dτ Option a is correct. Solution 2 e 2 /4a a Here, option b is correct. z /2 e z dz (putting τ z e 2 /4a a Γ(/2 F c xe ax π a e 2 /4a. xe ax cos x dx Re xe (i ax dx Re xe (i ax dx xe (i ax Re i a Re e(i ax (i a 2, (i a 2 (since lim xe (i ax lim e (i ax, as a >. x x Re 2 + a 2 2ai a2 2 (a
2 Solution 3 F (2 + 9i 2 ( ( F 5 + i 4 + i Since F e 5t u (t and F e 4t u (t, therefore by Convolution theorem for Fourier transforms, we 5 + i 4 + i get F (2 + 9i 2 ( e 4t e τ u (tu(t τdτ t e 4t e τ dτ, since u (tu(t τ e 5τ u (te 4(t τ u(t τdτ { if τ t, otherwise. Option c is correct. e 4t e 5t, t. Solution 4 F s t Put t z dt dz. Therefore, we have F s t sin t dt t ( sin z z dz So, option d is correct sin z z dz π 2. Solution 5 So, option c is correct. F c f(t t cos t dt + f(t cos t dt 2 2 cos ( cos. 2 (2 t cos t dt 2
3 Solution 6 Let u(x, t F (xg(t. Then 2 u t 2 F (xg (t and 2 u x 2 F (xg(t. Substituting these in the given differential equation, we get F F G 9G Therefore, we obtain the equations Using the boundary conditions, we get Consider the following three cases. Case I k. From ( we get k (say, where k constant. F kf and G 9kG. ( u(, t F (G(t for all t F (. u(2, t F (2G(t for all t F (2. F (x, whose solution is F (x Ax + B. Using conditions F ( F (2, we get A and B. Therefore, Hence, k. F (x, which gives u(x, t. Case II k p 2. From ( we get F p 2 F. The solution is given as F (x C e px + C 2 e px. This implies F ( C + C 2 and F (2 C e 2p + C 2 e 2p. Solving, we get C C 2 which gives u(x, t. 3
4 Hence, k is not a positive quantity. Case III k p 2. From ( we get On solving, we obtain F + p 2 F. F (x C cos px + C 2 sin px. Applying the boundary conditions, we get F ( C and F (2 C 2 sin 2p. To obtain nontrivial solutions, we need C 2. Therefore, sin 2p sin nπ, n I p nπ 2, n I Thus, we obtain Again, from (, we get F n C n sin(nπx/2. G + 9p 2 G. The solution is given as G(t D cos(3pt + E sin(3pt D cos(3nπt/2 + E sin(3nπt/2. Hence, the solution of the given equation is u n (x, t F n G n sin(nπx/2d n cos(3nπt/2 + E n sin(3nπt/2. Using the superposition principle, we get where u(x, t u n (x, t u(x, Integrating, we get sin(nπx/2d n cos(3nπt/2 + E n sin(3nπt/2. D n sin(nπx/2 D n x sin(nπx/2 dx + { x if x, 2 x if x 2, u(x, sin(nπx/2 dx 2 D n 8 n 2 π 2 sin(nπ/2. 4 (2 x sin(nπx/2 dx.
5 Again, u t (x, where E n(3nπ/2 sin(nπx/2, E n(3nπ/ E n. Hence, the solution of the given equation is u t (x, sin(nπx/2 dx u(x, t 8 π 2 sin(nπ/2 cos(3nπt/2 sin(nπx/2. n2 Option d is correct. Solution 7 Applying Fourier sine transforms to the given differential equation, we have u 2 u F s F s t x 2 d dt F s(, t 2 F s (, t + u(, t 2 F s (, t, (since u(, t. The solution of this linear, first order equation is where k( is an arbitrary function of. Taking sine transform of the boundary condition, we get F s (, t k(e 2t, (2 F s u(x, G( Putting t in (2, we get u(x, sin(t dt sin(t dt ( cos(t cos. F s (, k( G(. 5
6 Therefore, ( cos F s (, t G(e 2t e 2t. Taking inverse Fourier sine transforms, we obtain Option c is correct. u(x, t 2 π 2 π ( cos F s (, t sin(x d e 2t sin(x d. Solution 8 Let u(x, t F (xg(t. Then u t F (xg (t and 2 u x 2 F (xg(t. Substituting these in the given differential equation, we get Therefore, F F G G Using the boundary conditions, we get k (say, k constant. F kf and G kg. (3 u(, t F (G(t for all t F (. u(4, t F (4G(t for all t F (4. Now, for k and k p 2, the solution is not possible (as shown in solution 6. Therefore, the only case possible is k p 2. From (3, we get F + p 2 F. On solving, we obtain Applying the boundary conditions, we get F (x C cos px + C 2 sin px. F ( C and F (4 C 2 sin 4p. To obtain nontrivial solutions, we need C 2. Therefore, sin 4p sin nπ, n I p nπ 4, n I Thus, we obtain F n C n sin(nπx/4. 6
7 Again, from (3, we get The solution is given as G + p 2 G. Therefore, we get G(t Ae p2t Ae (n2 π 2 t/6. G n (t A n e (n2 π 2 t/6. Hence, the solution of the given equation is ( u n (x, t F n G n C n sin(nπx/4 A n e (n2 π 2t/6 D n sin(nπx/4e (n2 π 2t/6, where D n C n A n. Further, using the superposition principle, we get Now, where u(x, t u n (x, t D n 2 4 D n sin(nπx/4e (n2 π 2t/6, < x < 4. u(x, 2x 4 D n sin(nπx/4 2x, u(x, sin(nπx/4 dx Integrating, we obtain D n 6 nπ ( n+. Hence, the solution of the given equation is Option d is correct. u(x, t 4 x sin(nπx/4 dx. 6 nπ ( n+ sin(nπx/4e (n2 π2 t/6. Solution 9 Since the domain of the bar is < x <, therefore, we take Fourier transform with respect to x. Let F u(x, y F (, y. Taking Fourier transforms on both sides with respect to x, we obtain 2 u 2 u F + F x 2 y 2 2 F (, y + d2 F (, y dy2 (D 2 2 F (, y, D d dy. 7
8 The solution of this ordinary differential equation is F (, y A cosh(y + B sinh(y. (4 Taking Fourier transforms on the boundary condition, we obtain (since F e a x F u(x, F (, F e 2 x , 2a, < x <, a >. For y from (4, we have a F (, A cosh + B sinh (5 Again, taking Fourier transform on the other boundary condition, we get u F y (x, d F (,. (6 dy Differentiating both sides of (4 with respect to y, we get Using (6 and putting y, we obtain d F (, y A sinh(y + B cosh(y. dy d F (, B dy Therefore, from (5, we get A B. 4 ( cosh. Hence, from (4, we obtain F (, y Taking inverse Fourier transforms, 2π u(x, y 2π 2π 4 cosh(y ( cosh. ( 4 cosh(y ( cosh F (, ye ix d ( 4 cosh(y ( cosh 8 e ix d (cos(x + i sin(x d.
9 Here, i ( 4 cosh(y sin(x d, since the integral is odd in. 2π ( cosh Therefore, the solution is given by Option a is correct. u(x, y 2π 4 cosh(y (4 + 2 cosh cos(xd. 9
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