Find the lowest order solution for a boundary value problem

[yhasky]

Hello,
I need help in solving the problem:

" find the lowest order uniform approximation to the boundary value problem εy''+y'sinx+ysin(2x)=0. y(0)=(pi), y(pi)=0. "

what I did:

y(out)=Ʃ(ε^n)y(n)
εy''(out)+y'(out)*sinx+y(out)*sin(2x)=0
for order 0: y'(out)*sinx+y(out)*sin(2x)=0
with the B.C y(pi)=0---> y(out)=0.

for the inner region:
x=εX
y(x)=Y(in)(X)
Y(in)=Ʃ(ε^n)Y(n)
Y''(in)+Y'(in)*sin(εX)+Y(in)*sin(2εX)=0
sin(εX)≈εX+ O(ε^3)
for order 0: Y''(in)=0
Y(in)=AX+B
with the B.C Y(0)=pi---> Y(out)=AX+(pi).
the matching:
x=δz, X=x/ε
ε-->0
Y(in)=A*δz/ε+(pi) -->∞
y(out)=0.
can't do the matching and find B!

Where is my mistake?

Thank you very much!

 

[lippyka]

Your mistake is in the matching process. To match the two solutions, you must set the values of A, B, and δz. The way to do this is to find the value of Y(in) at the boundary x=π by setting X=π/ε. This gives you Y(in)=A*π/ε+B so that you can solve for B. You then substitute this back into the inner solution to get Y(in)=A*δz/ε+B. From this you can solve for A and δz.