Second order differential equation

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SUMMARY

The discussion focuses on finding an approximate solution to the second-order differential equation $$\lambda \frac{d^2u}{dx^2} + q = 0$$, defined over the interval $$x \in [-L/2, L/2]$$, where $$q = a + bx$$. The proposed solution is represented as a Fourier series: $$u_N(x) = \frac{a_0}{2} + \sum_{n=1}^N a_n \cos nx + b_n \sin nx$$. The boundary conditions lead to a system of equations that must be solved using orthogonality principles to construct a coefficient matrix, specifically $$M_{mn} = \int_{-L/2}^{L/2} \phi_n''(x) \phi_m(x)\,dx$$, where $$\phi_n(x)$$ satisfies the boundary conditions.

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Linder88
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Homework Statement
I need help with finding the approximate solution of a second order differential equation.
Relevant Equations
I am given the differential equation as in
$$
\lambda \frac{d^2u}{dx^2} + q = 0
$$
which is defined over $$x \in [-L/2,L/2]$$ and where $$q = a+bx$$. I am supposed to find an approximative solution using a spectral method, i.e. using cosine and sine terms that fulfills the essential boundary conditions given by
$$
u \bigg ( -\frac{L}{2} \bigg ) = 0, u \bigg (\frac{L}{2} \bigg ) = 0
$$
The appoximative solution should be represented by a Fourier series and by noticing that the terms in the integral are orthogonal, the matrix of the system of equations should be diagonal.
We choose an approximative solution given by
$$
u_N(x) = \frac{a_0}{2} + \sum_{n=1}^N a_n \cos nx + b_n \sin nx
$$
Comparing this approximative solution with the differential equation yields that
$$
\frac{a_0}{2} = a
$$
and the boundary conditions yields the equation system
$$
a + \sum_{n=1}^N a_n \cos \bigg ( \frac{nL}{2} \bigg ) + b_n \sin \bigg ( \frac{nL}{2} \bigg ) = 0 \\
a + \sum_{n=1}^N a_n \cos \bigg ( -\frac{nL}{2} \bigg ) + b_n \sin \bigg ( -\frac{nL}{2} \bigg ) = 0
$$
So at least I got a system of two equations but I do not know where to go from here. How should I use the orthogonality to set up a matrix from these two equations?
 
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Linder88 said:
Comparing this approximative solution with the differential equation yields that
##{a_0\over 2}=a##
How ?

and the boundary conditions yields the equation system
I miss the ## +bx## from
$$
\lambda \frac{d^2u}{dx^2} + a + bx = 0
$$

(and I find the use of ##a_n## and ##b_n## in your notation confusing, since ##a## and ##b## are already used).
 
Linder88 said:
Homework Statement:: I need help with finding the approximate solution of a second order differential equation.
Relevant Equations:: I am given the differential equation as in
$$
\lambda \frac{d^2u}{dx^2} + q = 0
$$
which is defined over $$x \in [-L/2,L/2]$$ and where $$q = a+bx$$.

I am supposed to find an approximative solution using a spectral method, i.e. using cosine and sine terms that fulfills the essential boundary conditions given by
$$
u \bigg ( -\frac{L}{2} \bigg ) = 0, u \bigg (\frac{L}{2} \bigg ) = 0
$$
The appoximative solution should be represented by a Fourier series and by noticing that the terms in the integral are orthogonal, the matrix of the system of equations should be diagonal.

We choose an approximative solution given by
$$
u_N(x) = \frac{a_0}{2} + \sum_{n=1}^N a_n \cos nx + b_n \sin nx
$$
Comparing this approximative solution with the differential equation yields that
$$
\frac{a_0}{2} = a
$$

No. It's the second derivative of u that is equal to \frac{-q(x)}{\lambda}, not u itself. When you differentiate the constant term you get zero, and in any case a non-zero constant does not satisfy the boundary condition.

and the boundary conditions yields the equation system
$$
a + \sum_{n=1}^N a_n \cos \bigg ( \frac{nL}{2} \bigg ) + b_n \sin \bigg ( \frac{nL}{2} \bigg ) = 0 \\
a + \sum_{n=1}^N a_n \cos \bigg ( -\frac{nL}{2} \bigg ) + b_n \sin \bigg ( -\frac{nL}{2} \bigg ) = 0
$$
So at least I got a system of two equations but I do not know where to go from here. How should I use the orthogonality to set up a matrix from these two equations?

You have u(x) \approx \sum_n u_n\phi_n(x) where <br /> \phi_n(-L/2) = \phi_n(L/2) = 0.<br /> Therefore <br /> \lambda \sum_n u_n \phi_n&#039;&#039;(x) \approx -q(x) and we determine the u_n by requiring that the inner product with \phi_m should be exact, ie. <br /> \lambda \sum_n u_n \int_{-L/2}^{L/2} \phi_n&#039;&#039;(x) \phi_m(x)\,dx = -\int_{-L/2}^{L/2} q(x)\phi_m(x)\,dx. The matrix you are looking for is the matrix of coefficients <br /> M_{mn} = \int_{-L/2}^{L/2} \phi_n&#039;&#039;(x) \phi_m(x)\,dx.

Now the choice of a Fourier basis means that \phi_n(x) = A_n\cos(k_n x) + B_n\sin(k_n x) and k_n is such that the boundary conditions are satisfied by something other than A_n = B_n = 0.
 
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