Explore BrainMass
Share

Rayleigh Ritz method and the minimum value of a functional

This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!

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.

© BrainMass Inc. brainmass.com October 17, 2018, 10:53 am ad1c9bdddf
https://brainmass.com/math/integrals/rayleigh-ritz-method-and-the-minimum-value-of-a-functional-516734

Attachments

Solution Preview

See the attached pdf for the formatted version of your solution.

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.

$2.19
Similar Posting

Rayleigh Ritz method

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.

View Full Posting Details