Rayleigh-Ritz Approximation (Obtaining the constants)

[roldy]

Homework Statement

Consider the differential equation

$$-\frac{d^2u}{dx^2}=cos\pi x$$ for 0<x<1

subject to the boundary condition

u(0)=0, u(1)=0

Determine a three-parameter solution, with trigonometric functions the least-squares method

Given: $$u=\phi_0 + c_1\phi_1 + c_2\phi_2 + c_3\phi_3$$

$$\phi_0=0, \phi_1=sin\pi x, \phi_2=sin2\pi x, \phi_3=sin3\pi x$$

Homework Equations

1st and 2nd derivatives of u with respect to x
Derivative of R with respect to ci

$$0=\int_0^1R\frac{\partial R}{\partial c_i}dx$$

The Attempt at a Solution

$$u=c_1sin\pi x + c_2sin2\pi x + c_3sin3\pi x=\Sigma_{i=1}^n c_isin(i\pi x)$$

$$\frac{d^2u}{dx^2}=-\Sigma_{i=1}^n c_i(i\pi)^2sin(i\pi x)$$

and

$$\frac{\partial R}{\partial c_i}=(i\pi)^2sin(i\pi x)$$

so then,

$$R=-\frac{d^2u}{dx^2}-cos(\pi x)=\Sigma_{i=1}^n c_i(i\pi)^2sin(i\pi x)-cos(\pi x)$$
$$0=\int_0^1R\frac{\partial R}{\partial c_i}dx=\int_0^1\left[\Sigma_{i=1}^n c_i(i\pi)^2sin(i\pi x)-cos(\pi x)\right]\left[(i\pi)^2sin(i\pi x)\right]dx$$

The term,
$$(i\pi)^2$$ is dropped out because it's just a multiplicative factor.

So then I arrive here, at which point I'm stuck
$$0=\int_0^1\left[\Sigma_{i=1}^n c_i(i\pi)^2sin(i\pi x)-cos(\pi x)\right]sin(i\pi x)dx$$

I know that I need to somehow get $$c_1, c_2, c_3$$
If I integrate the above integral across the summation, I'll end up with one equation with the three constants.

I would think that I need three equations. Do I perform the summation from i=1,2,3 and then integrate, which would give me one equation, or do I integrate and then do the summation? If I remember correctly I can take the summation out before the integral. Is this correct?

 

[mighty2000]

Thank you for your post. I understand that you are trying to solve the given differential equation using the least-squares method and trigonometric functions. Your approach so far seems to be on the right track. However, there are a few things that I would like to clarify and suggest for your solution.

Firstly, the given differential equation is a second-order one, so we would expect to have a solution with two arbitrary constants. However, your solution includes three arbitrary constants. This may be because you have included the constant term \phi_0 in your solution, which is not necessary. The general solution for the given differential equation should be u=c_1sin\pi x+c_2sin2\pi x, where c_1 and c_2 are arbitrary constants. Therefore, we need to find two equations to determine c_1 and c_2 using the least-squares method.

Secondly, in your attempt at a solution, you have not taken into account the given boundary conditions. These conditions should be satisfied by the solution u=c_1sin\pi x+c_2sin2\pi x. Therefore, we can write the following equations using the given boundary conditions:

u(0)=0=c_1sin(0)+c_2sin(0)=0
u(1)=0=c_1sin(\pi)+c_2sin(2\pi)=0

These two equations will help us determine the values of c_1 and c_2. Now, let's move on to the least-squares method. The idea behind this method is to minimize the sum of the squared errors between the given differential equation and the solution. In other words, we want to minimize the integral of the squared residual (R) over the given domain. This can be written as:

min\int_0^1R^2dx=\int_0^1\left(-\frac{d^2u}{dx^2}-cos(\pi x)\right)^2dx

To minimize this integral, we can take the derivative of the integrand with respect to c_1 and c_2 and set them equal to 0. This will give us two equations that we can solve to find the values of c_1 and c_2. The resulting equations will be:

\frac{\partial}{\partial c_1}\int_0^1R^2dx=0
\