w corresponds to how much we can change our integral... Also called the variation of our solution function. If we know exact value of our solution function at any point, i.e. BCs, we cannot vary it. Hence we took w=del u every where, but at x=0, the BCs implied w=0 - which is still del u. Just del u =0 .
while doing second set of BC's you missed wu term in Bilinear functional
why choose w = delta u only in the second set of boundary conditions?
w corresponds to how much we can change our integral... Also called the variation of our solution function. If we know exact value of our solution function at any point, i.e. BCs, we cannot vary it. Hence we took w=del u every where, but at x=0, the BCs implied w=0 - which is still del u. Just del u =0
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How solve: y"-2*y^3=0 ;
y (-1)=1/2 , y (0)=1/3
Jo aap likh rhe hai bo pta hi nhi chl rha tb koi kaise pdgega
thank u so much sir
copy and pasting and that too in poor way.
kya ha ye