Anharmonic Oscillation | Calculate Period w/ Gaussian Quadrature

[Robben]

Homework Statement

Assume that the potential is symmetric with respect to zero and the system has amplitude $a$ suppose that $V(x)=x^4$ and the mass of the particle is $m=1$. Write a java function that calculates the period of the oscillator for given amplitude $a$ using Gaussian quadrature with $N=20$ points.

Homework Equations

$E = \frac12 m(\frac{dx}{dt})^2+V(x)$
$T=\sqrt{8m}\int^a_0\frac{dx}{\sqrt{V(a)-V(x)}}.$

The Attempt at a Solution

I don't have much experience with numerical methods so I am having difficulty understanding the question and how to proceed.

Is the question asking instead of using $T$, i.e the period given by : $T=\sqrt{8m}\int^a_0\frac{dx}{\sqrt{V(a)-V(x)}}$, we instead must use the Gaussian quadrature of $T$? Also, will $T=\sqrt{8m}\int^a_0\frac{dx}{\sqrt{V(a)-V(x)}}$ work for a nonquadratic function $V(x)$?

 

[stevendaryl]

The method of Gaussian quadratures is (in my opinion) a really difficult way to efficiently compute integrals numerically. I think it's too complicated to explain in a message. The basic idea is simple enough: You can approximate an integral

\int_{a}^b f(x) dx = \sum_{i=1}^N w_i f(x_i) where x_1, x_2, ... are points in the interval [a,b] and w_i is a weighting function. The simplest approach is the "rectangle rule", which uses w_i = \frac{b-a}{N} and x_i = a + i \cdot \frac{b-a}{N}. Gaussian quadrature is a way to get a much more accurate approximation by choosing w_i and x_i in a more sophisticated way, but it's beyond me to explain how to do it in a post.

As to the second question, if a particle goes from a to b and back, then the time taken for a full period is twice the time to go from a to b. To compute that time, you use:

T = 2 \int dt = \int_a^b \frac{dt}{dx} dx = \int_a^b \frac{1}{v} dx where v = \frac{dx}{dt}. To compute v, you use conservation of energy:

E = \frac{1}{2} m v^2 + V
v = \sqrt{\frac{2}{m} (E - V)}

So T = 2 \int_a^b \sqrt{\frac{m}{2(E-V)}} dx

If the potential is symmetric about x=0, then you can just do half the integral and double it:

T = 4 \int_0^b \sqrt{\frac{m}{2(E-V)}} dx

So it works no matter what the potential V is--it doesn't have to be quadratic.

 

[Robben]

The method of Gaussian quadratures is (in my opinion) a really difficult way to efficiently compute integrals numerically. I think it's too complicated to explain in a message. The basic idea is simple enough: You can approximate an integral

\int_{a}^b f(x) dx = \sum_{i=1}^N w_i f(x_i) where x_1, x_2, ... are points in the interval [a,b] and w_i is a weighting function. The simplest approach is the "rectangle rule", which uses w_i = \frac{b-a}{N} and x_i = a + i \cdot \frac{b-a}{N}. Gaussian quadrature is a way to get a much more accurate approximation by choosing w_i and x_i in a more sophisticated way, but it's beyond me to explain how to do it in a post.

As to the second question, if a particle goes from a to b and back, then the time taken for a full period is twice the time to go from a to b. To compute that time, you use:

T = 2 \int dt = \int_a^b \frac{dt}{dx} dx = \int_a^b \frac{1}{v} dx where v = \frac{dx}{dt}. To compute v, you use conservation of energy:

E = \frac{1}{2} m v^2 + V
v = \sqrt{\frac{2}{m} (E - V)}

So T = 2 \int_a^b \sqrt{\frac{m}{2(E-V)}} dx

If the potential is symmetric about x=0, then you can just do half the integral and double it:

T = 4 \int_0^b \sqrt{\frac{m}{2(E-V)}} dx

So it works no matter what the potential V is--it doesn't have to be quadratic.

I see, thank you very much!

So our weighted function for this particular question will be $w_i = \frac{b-a}{N} = \frac{a-0}{20}$ which implies $\int_{a}^b f(x) dx = \sum_{i=1}^N w_i f(x_i) = \sum_{i=1}^{20} \frac{a-0}{20}\sqrt{8 \dot\ (1)}\frac{1}{\sqrt{V(a)-V(x_i = 0 +i\frac{a-0}{N})}}?$

 

[BvU]

Wouldn't count as a gaussian quadrature in my book. (a/20 is equal weights) But I wouldn't know how to do it more sophisticated, sorry.

 

[stevendaryl]

I see, thank you very much!

So our weighted function for this particular question will be $w_i = \frac{b-a}{N} = \frac{a-0}{20}$ which implies $\int_{a}^b f(x) dx = \sum_{i=1}^N w_i f(x_i) = \sum_{i=1}^{20} \frac{a-0}{20}\sqrt{8 \dot\ (1)}\frac{1}{\sqrt{V(a)-V(x_i = 0 +i\frac{a-0}{N})}}?$

That's NOT using quadratures. That's using the much simpler "rectangle rule", also known as Newton's method (I think).