Maybe someone can come up with any idea about this.
Let us assume nonlinear two-point boundary problem
x′′=f(t,x,x′)
with boundary conditions written as
L1(x(a),x(b),x′(a),x′(b))=0,
L2(x(a),x(b),x′(a),x′(b))=0,
where f∈C([a,b]×R2,R) and L1,L2∈C(R4,R).
Further assume that there exists upper and lower boundary functions α and β,such that
α(t)≤β(t),t∈[a,b] and L1(α(a),α(b),α′(a),α′(b))≥0≥L2(β(a),β(b),β′(a),β′(b)).
There is a theorem that states:
if L1 is non-decreasing in second and third term, and L2 is non-decreasing in first, but non-increasing in fourth argument, then
there is a solution to this problem such that
α(t)≤x(t)≤β(t).
Can anyone help with constructing an example where theorem assumptions are not met and showing that there is no soluton for given problem in such case?
Let us assume nonlinear two-point boundary problem
x′′=f(t,x,x′)
with boundary conditions written as
L1(x(a),x(b),x′(a),x′(b))=0,
L2(x(a),x(b),x′(a),x′(b))=0,
where f∈C([a,b]×R2,R) and L1,L2∈C(R4,R).
Further assume that there exists upper and lower boundary functions α and β,such that
α(t)≤β(t),t∈[a,b] and L1(α(a),α(b),α′(a),α′(b))≥0≥L2(β(a),β(b),β′(a),β′(b)).
There is a theorem that states:
if L1 is non-decreasing in second and third term, and L2 is non-decreasing in first, but non-increasing in fourth argument, then
there is a solution to this problem such that
α(t)≤x(t)≤β(t).
Can anyone help with constructing an example where theorem assumptions are not met and showing that there is no soluton for given problem in such case?