Boundary value Problem

[Sarahisme]

hey

 

i am having a bit of trouble with this problem, as the course textbook has nothing on doing these problems using green's functions (all we have is a few slides of dodgey lecture notes)...

 

anyways... the problem..

 

 

 

so to start with, for part (a)

 

now from what i can gather, the BVP has a unique solution if and only if the corresponding HBVP (homogeneous boundary value problem)

y''=0, y(0)=0, y(1) = 0

 

has no solution other than the trivial one, y(x) = 0.

 

 

i guess then the associated HLDE (homogenous linear differential equation) needs to be solved...

 

so

 

y''(0) = 0

 

so we look for answers of the form

 

y(x) = Ax + B

 

then the left boundary condition (y(0) = 0 ) leads to

u(x) = Ax

 

 

is this the right way to go about things?

 

cheers

-Sarah

[Sarahisme]

hmm ok, how does this look..?

 

for the 'appropritate Green's Function' for part (a):

 

The left boundary condition gives:

 

u(x) = x

 

and the right boundary condition gives:

 

v(x) = x - 1

 

so the wronskian = W(x) = uv' - u'v = x - (x-1) = 1

 

so u,v are linearly independent.

 

then the greens function is

 

[math] G(x,e) = x(\epsilon - 1) [/tex] for [tex] 0 \leq x \leq \epsilon [/math]

 

[math] G(x,e) = \epsilon(x - 1) [/tex] for [tex] \epsilon \leq x \leq 1 [/math]

 

 

So the solution to the associated boundary value problem with homogenous conditions is:

 

[math] y(x) = \int_0^1G(x,\epsilon)f(\epsilon)d\epsilon [/math]

 

 

so then is the solution to the orginal BVP something like

 

[math] y(x) = \int_0^1G(x,\epsilon)f(\epsilon)d\epsilon + A +Bx [/math]

 

?????

 

this is where i think i get quite lost (maybe before here if i have already made a giant mistake somewhere! )