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# Rayleigh-Ritz Method watch

1. Ok I'm doing a question where I try to estimate the first eigenvalue of a Sturm-Liouville equation via the Rayleigh-Ritz method. It first asks me to use the trial function y1=a-x^2 and then y2=cos(pi*x/2) and then compare the two trial functions (for boundary condition y(+-1)=0). The SL equation is: -((1-x^2)y')'=lamba*y

For y1 I added some adjustable parameters so that it looked like a-bx^2. After integrating appropriately I obtained F[y] and G[y], since I wanted to maximise F subject to the constraint G, I found the stationary value of F-lambaG. Eventually I got a value of 4 for the eigenvalue. I understand that this gives an upper bound of the eigenvalue, but I was under the impression that the estimates should be very accurate? Since the SL equation is Legendre's equation I expect the eigenvalues to be given by n(n+1) so the first eigenvalue =2.

For the second trial function I carried out the same steps and got the eigenvalue estimate as 0.66(2dp) which is obv below the true eigenvalue.

I've checked the integration and it seems fine, so if you can spot something that I'm doing wrong then that would be awesome.

Thanks.
2. After trying a few more of these questions I can see where I'm having problems- its with the additional adjustable parameter(s) that must be added to the trial function. So I kinda need help with that.

For example, when the question asks me to use the trial function y1=1-x^2, where would I add the parameter? I'm guessing I should go with: ytrial= 1-ax^2 ? also would this parameter only be added to the trial function and not for example q(x), one of the non-constant coefficients of y?

Another example is when y2=cos(pi*x/2), would I simply add an adjustable parameter such that ytrial=cos(a*pi*x/2)?

I can understand the rest of the method and can do problems where I'm already given trial functions with parameters added in- its just doing that part myself that I'm finding tricky, even though its probably easy.

Thanks.
3. Anyone?

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Updated: February 5, 2010
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