Question

Question (2.5)

In Problems 2.4-2.9, construct the weak forms and, whenever possible, the associated quadratic functionals (I). A linear differential equation: 2.4 du + u x dx dx du u(0) =1, dx =2 2.5 A nonlinear equation: du +f=0 for 0 <x<1 dx dx dx = 0 u(1)-V2

Answer 1

A non-linear equation -d/dx (u du/dx) + f = 0 for 0 < x < 1 (u du/dx)_x = 0 u(1) = squareroot 2 solution To convert into weak form (1) multiply one by a virtual function v^V (x) satisfying u^v(0) = 0 (-u u_x)_x u^v = -fu^v shift derivation (-u u_e U^u_x congruent (-u u_x) u^x + (-u.u_x) U^u_x by replacing LHS (-u u_x u u^v)_x - (-u. u_x) u^v_x = fu^v integrate internal^1_0 (-u u_x u^v)_x dx - integral^1_0 (-u u_x) u^v_x dx = -integral^1_0 f u^v (x) dx [-u u_x) u^v_x dx = -integral^1_0 f u^v (x) dx now apply B.C -u(1). U_x(1) , u^v (1) - 0 + integral^1_0 (u, u_x)u^v_x dx = -integral^1_0 f u^v (x) dx arrived at weak statement find u(x) where u(0_= 0 s.t integral^1_0 (u u_x)u^v_x dx = squareroot 2 u^v (1) - integral^1_0 f u^v (x) dx forall u^v (x) u^v (0) = 0