Consider the following boundary value problem (F) : Fu Fu I>0, t>0..(1) H² u(r,0) = 0, (r,0) = 1, x >0--- (2) %3D u(0, t) = sin t, t >0… (3), Let U(z, s) be the Laplace transform of u(r,t) acting to the variable t i.e: U(r, 8) = L(u(x,t), and suppose that U(r,s) is bounded as s→ +o0. (1) By apply the Laplace transform to the equation(1) of (F) and by using the equation (2) of (F), we obtain the following ODE: a. Uzz(r, s) – s°U(1, 8) = –1 b. Uz(r, s) – s°U(x, s) = -1 c. Uzr(r, 8) – s²Uz(r, s) = -1 d. None of the above (2) The solution of the ODE obtained in part (1) is: a. U(r,s)= A(s)e¬sz + B(s)e*z + ! b. U(7, s) = A(s)e¬sz + B(s)e*z +4 c. U(1, s) = A(s)e-" + B(s)e*r d. None of the above (3) Using the fact that, the Laplace transform of u(x, t) is bounded as s → +∞o and using the equation(3) of (F), we obtain: a. U(r,8) = (T-)e== +! b. U(r,s) = (Tr c. U(z,s) = ( - )e-s" + d. None of the above (4) The general solution of (F) is: (H(t – a) is the unit step function) a. u(x,t) =t – (t – r)H(t – 1) + sin(t – x)H(t – x) b. u(x, t) = sin(t – 1)H(t – x) c. u(r,t) =t – (t – 2)H(t – z) d. None of the above %3D
Consider the following boundary value problem (F) : Fu Fu I>0, t>0..(1) H² u(r,0) = 0, (r,0) = 1, x >0--- (2) %3D u(0, t) = sin t, t >0… (3), Let U(z, s) be the Laplace transform of u(r,t) acting to the variable t i.e: U(r, 8) = L(u(x,t), and suppose that U(r,s) is bounded as s→ +o0. (1) By apply the Laplace transform to the equation(1) of (F) and by using the equation (2) of (F), we obtain the following ODE: a. Uzz(r, s) – s°U(1, 8) = –1 b. Uz(r, s) – s°U(x, s) = -1 c. Uzr(r, 8) – s²Uz(r, s) = -1 d. None of the above (2) The solution of the ODE obtained in part (1) is: a. U(r,s)= A(s)e¬sz + B(s)e*z + ! b. U(7, s) = A(s)e¬sz + B(s)e*z +4 c. U(1, s) = A(s)e-" + B(s)e*r d. None of the above (3) Using the fact that, the Laplace transform of u(x, t) is bounded as s → +∞o and using the equation(3) of (F), we obtain: a. U(r,8) = (T-)e== +! b. U(r,s) = (Tr c. U(z,s) = ( - )e-s" + d. None of the above (4) The general solution of (F) is: (H(t – a) is the unit step function) a. u(x,t) =t – (t – r)H(t – 1) + sin(t – x)H(t – x) b. u(x, t) = sin(t – 1)H(t – x) c. u(r,t) =t – (t – 2)H(t – z) d. None of the above %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
laplace transform pde part 1 2 3
Transcribed Image Text:Consider the following boundary value problem (F) :
Pu
r> 0, t>0. (1)
H2
du
u(r,0) = 0,
a (r, 0) = 1, r>0.. (2)
%3D
u(0, t) = sin t, t>0• (3),
Let U(r, s) be the Laplace transform of u(x,t) acting to the variable t i.e:
U (r, s) = L(u(x, t),
and suppose that U(r, s) is bounded as s→ +o.
(1) By apply the Laplace transform to the equation(1) of (F) and by using the equation (2) of
(F), we obtain the following ODE:
a. Urz(x, s) – s®U (1, s) = –1
b. Uz(r, s) – s°U(x, s) = -1
c. Uzz(x, 8) – s²U#(r, s) = –1
d. None of the above
(2) The solution of the ODE obtained in part (1) is:
a. U(r, s) = A(s)e-s" + B(s)e® +:
b. U(r, s) = A(s)e-s + B(s)e + ;
c. U(r, s) = A(s)e¬s" + B(s)e*¤
d. None of the above
(3) Using the fact that, the Laplace transform of u(x, t) is bounded as s → +0 and using the
equation(3) of (F), we obtain:
a. U(r, 8) = ( - )e-sz + !
b. U(r, 8) = ( - )e
c. U(r, 8) = ( -)e=s= +
d. None of the above
(4) The general solution of (F) is: (H(t – a) is the unit step function)
a. u(r, t) = t – (t – x)H(t – x) + sin(t – x)H(t – x)
b. u(x, t) = sin(t – x)H(t – x)
c. u(x, t) =t – (t – 2)H(t – x)
d. None of the above
1
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