Solve du/dt (x,t) = d^2u/dx^2 (x,t) + x(x-1)e^(-t) for 0<= x<=1 t>=0 u(0,t) = 0

Question

Solve

du/dt (x,t) = d^2u/dx^2 (x,t) + x(x-1)e^(-t)
for 0<= x<=1 t>=0

u(0,t) = 0
u(1,t) = 0

u(x,0) = sin 3x for 0 <=x<=1

Thanks

Explanation / Answer

Try separation of variables. Let u(x,t) = X(x) T(t). So, X T' = X'' T - XT ==> T' / T = X''/X - 1 = k - 1 for some constant k. Moreover, the boundary conditions transform as follows: u_x (0, t) = 0 ==> X'(0) = 0 u(1,t) = 2 cosh 1 ==> X(1) = 2 cosh 1. We obtain 2 ODE's T' - (k-1) T = 0 X'' - kX = 0 (that's why I used k - 1 above). -------------- Solving T' - kT = 0 yields T(t) = Ae^((k-1)t) for some constant A. Solving X'' - kX = 0: If k > 0, then we have characteristic equation r^2 - k = 0 ==> r = ±vk. So, X = B cosh(xvk) + C sinh(xvk). Using the boundary conditions: X'(0) = 0 ==> C = 0. X(1) = 2 cosh 1 ==> B cosh(1vk) = 2 cosh 1 ==> B = 2 and k = 1. So, X = 2 cosh x. (We could look at the other cases of k, but given the rest of the solution below, this seems very unnecessary...) For k = 1, we have T(t) = Ae^0 = A. ---------------- So, u(x,t) = X(x) T(t) = 2A cosh x. Finally, u(x,0) = 2 cosh x ==> A = 1. Thus, the desired solution is u(x,t) = 2 cosh x.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.