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Rayleigh Ritz method and the minimum value of a functional

Use the Rayleigh-Ritz method 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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The trial function Y_0(x)=x satisfies the boundary conditions. We have

J(Y_0)=int_0^1(1+x^3+2x^3)dx=frac{7}{4}=1.75.

From here we conclude
inf J(y)leq 1.75<= sup J(y),
where the infimum and supremum range over all functions y satisfying the boundary conditions.

For Y_1(x)=x+c_1x(1-x) we see that Y_1(0)=0 and Y_1(1)=1, so Y_1 satisfies the boundary conditions for all c_1. We
evaluate the ...

Solution Summary

We use the Rayleigh-Ritz method to find three successive approximate solutions to the extremum problem associated with a functional. We use three polynomial trial functions.

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