Use the Rayleigh-Ritz method to find two successive approximate solutions to the extremum problem associated with the functional:

J[y]=Integral from 0 to 1 of (y'')^2-y^2 dx, y(0)=0, y'(0)=0, y(1)=1, y'(1)=1,

using the trial function Y_0(x)=a_3 x^3+a_2 x^2+a_1 x+a_0, where a_0, a_1, a_2, and a_3 are chosen so that Y_0(x) satisfies the boundary conditions and Y_1(x)=Y_0(x)+c x^2(1-x)^2.

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For Y_0(x)=a_3x^3+a_2x^2+a_1x+a_0, we find the values of the constants so that Y_0 satisfies the boundary conditions. If
Y_0(0)=0, then a_0=0. If Y_0'(0)=0, then a_1=0. If Y_0(1)=1, then a_3+a_2=1. If Y_0'(1)=1, then 3a_3+2a_2=1.
Solving the system for a_3 and a_2 gives a_2=2, a_3=-1. Then there is only one function of the form Y_0 that satisfies
the boundary conditions, namely Y_0(x)=-x^3+2x^2.

For that Y_0 we have Y_0''(x)=-6x+4. We evaluate J(Y_0) to obtain ...

Solution Summary

We use the Rayleigh-Ritz method to find two successive approximate solutions to the extremum problem associated with a functional. We use as trial functions polynomials of degree 3 and 4.

Use the Rayleigh-Ritzmethod to find three successive approximate solutions to the extremum problem associated with the functional:
J[y]=Integral from 0 to 1 of (y')^2+xy^2+2x^2y dx,
y(0)=0, y(1)=1,
using the trial functions Y_0(x)=x, Y_1(x)=x+c_1 x(1-x) and Y_2(x)=x+c_1 x(1-x)+c_2 x^2(1-x)^2.

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