Question

Mark which statements below are true, using the following Consider the diffusion problem u(0,t)=0, u(L,t)=50 where FER is a constant, forcing term Any attempt to solve this using separation of variables fails. This is because the PDE is not homogeneous. A more fruitful approach arises from splitting the solution into the sum of two u(z,t) = X(z)T(t) + us(z), where the subscript designates the function as the steady limit and does not represent a derlvative. BEWARE: MARKING A STATEMENT TRUE THAT IS ACTUALLY FALSE RECEIVES A NEGATIVE MARK! A. cannot satisfy u(0, t)=0, u(L,t)=50. For the functions of the set (sin(|n 1,2,3,...) zero and negative values of are not considered because they are not orthogonal The full solution is u(x,t) us(z)+> Cn sin( 1n where 50T F 0 D. The time part of the solution is obtained by solving the following BVP OE. im u(x,t)-0 t-00 F. The functions of the set (sin(TE)|n - 1,2,3,...) are orthogonal on To, . G. Upon separation of variables, the time problem is given by T(t) + aT(t)0 H. The steady state solution is 50 u,(z + Fr(x - L) O I u(x,t) - Us() cannot be represented by a Fourier series because normal separability does not work. J. Substituting the alternative split form of a solution with transient and steady parts gives td(z) + X"(z)T(t) = X(z)T'(t)

Answer 1

let u(xt)(x.t)+u, (x) then where u, (x) is a solution for (m), =-F sang yong u, (0) = 0and us (L) = 50 since ", (0) = 0 ,-0 u, (x)-to since u, (L)-5050- 50 FL L 2 50 L now v(x,t) be the solutionof (Lt)Lt)-u, (L) 50-50 0 굿 (50, FL the suitable solution is v(x,1)-(acos px +bsin px) since v(0,1)-0 (a)e-r-o: a:0 v(x,t)= b sin pxe-,, since v(L,1) = 0 → b sin pL e-rt-0 sin pL = 0 p = _ for n = 1,2,3, then v(x,t)= sin - 보-1 굿 (50FL since v(x,0)=f(x)-F +-|x , 50 FL 2 L 2 MTx hence 50 FL hence the inal solutionis 50 FL 2 L 2 ow A is false, Bistrue, C is false, Dis false, E is true Fis true, Gis true, H is false, I is false, J is false