University of Florida/Egm4313/s12.team11.imponenti/R6.4

R6.4

solved by Luca Imponenti

Problem Statement

Consider the L2-ODE-CC (5) p.7b-7 with the window function f(x) p.9-8 as excitation:

$y^{″}-3y^{\prime }+2y=f(x)$

and the initial conditions

$y(0)=1,y^{\prime }(0)=0$

1. Find $y_n(x)$ such that:

${y_n}^{″}+a{y_n}^{\prime }+b{y}_n=r_n(x)$

with the same initial conditions as above.

Plot $y_n(x)$ for $n=2,4,8$ for x in $[0,10]$

2. Use the matlab command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution.

Level 1: $n=0,1$

Fourier Series

One period of the window function p9.8 is described as follows

$f(x)=\{\begin{array}{cc}0& -1.75<x<l0.25\\ A& 0.25<x<2.25\end{array}$

From the above intervals one can see that the period, $p=4$ and therefore $L=2$ Applying the Euler formulas from $-1.75$ to $2.25$ the Fourier coefficients are computed:

$a_0=\frac{1}{2L}\underset{-1.75}{\overset{2.25}{\int }}f(x)dx$

$a_0=\frac{1}{4}(\underset{-1.75}{\overset{0.25}{\int }}0dx+\underset{0.25}{\overset{2.25}{\int }}Adx)$

$a_0=+\frac{1}{4}(0+\underset{0.25}{\overset{2.25}{\int }}Adx)$

$a_0=\frac{1}{4}(2.25A-0.25A)$

$a_0=\frac{A}{2}$

The integral from $-1.75$ to $0.25$ can be omitted from this point on since it is always zero.

$a_n=\frac{1}{L}\underset{0.25}{\overset{2.25}{\int }}f(x)cos(\frac{n\pi x}{L})dx$

$a_n=\frac{1}{2}\underset{0.25}{\overset{2.25}{\int }}Acos(\frac{n\pi x}{2})dx$

$a_n=\frac{2A}{2n\pi }(sin(\frac{2.25n\pi }{2})-sin(\frac{0.25n\pi }{2})$

$a_n=\frac{A}{n\pi }(sin(\frac{9n\pi }{8})-sin(\frac{n\pi }{8})$

and

$b_n=\frac{1}{L}\underset{0.25}{\overset{2.25}{\int }}f(x)sin(\frac{n\pi x}{L})dx$

$b_n=\frac{1}{2}\underset{0.25}{\overset{2.25}{\int }}Asin(\frac{n\pi x}{2})dx$

$b_n=\frac{2A}{2n\pi }(cos(\frac{2.25n\pi }{2})-cos(\frac{0.25n\pi }{2})$

$b_n=\frac{A}{n\pi }(Acos(\frac{n\pi }{8})-Acos(\frac{9n\pi }{8})$

The coefficients give the Fourier series:

$f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}[a_{n}cos(\frac{n\pi x}{L})+b_{n}sin(\frac{n\pi x}{L})]$

$f(x)=\frac{A}{2}+\sum _{n=1}^{\mathrm{\infty }}[\frac{A}{n\pi }(sin(\frac{9n\pi }{8})-sin(\frac{n\pi }{8}))cos(\frac{n\pi x}{2})$

$+\frac{A}{n\pi }(cos(\frac{n\pi }{8})-cos(\frac{9n\pi }{8}))sin(\frac{n\pi x}{2})]$

Homogeneous Solution

Considering the homogeneous case of our ODE:

$y^{″}-3y^{\prime }+2y=0$

The characteristic equation is

${\lambda }^2-3\lambda +2=0$

$(\lambda -1)(\lambda -2)=0$

${\lambda }_1=1,{\lambda }_1=2$

Therefore our homogeneous solution is of the form

$y_h=c_1{e}^x+c_2{e}^{2x}$

Particular Solution

Considering the case with f(x) as excitation

$y^{″}-3y^{\prime }+2y=\frac{A}{2}+\sum _{k=1}^n\frac{A}{k\pi }[(sin(\frac{9k\pi }{8})-sin(\frac{k\pi }{8}))cos(\frac{k\pi x}{2})$

$+(cos(\frac{k\pi }{8})-cos(\frac{9k\pi }{8}))sin(\frac{k\pi x}{2})]$

The solution will be of the form

$y_n=A_0+\sum _{k=1}^n{A}_{k}cos(\frac{k\pi x}{2})+\sum _{k=1}^n{B}_{k}sin(\frac{k\pi x}{2})$

Taking the derivatives

${y_n}^{\prime }=-A_n\frac{\pi }{2}\sum _{k=2}^{n}ksin(\frac{k\pi x}{2})+B_n\frac{\pi }{2}\sum _{k=2}^{n}kcos(\frac{k\pi x}{2})$

${y_n}^{″}=-A_n\frac{{\pi }^2}{4}\sum _{k=3}^n{k}^{2}cos(\frac{k\pi x}{2})-B_n\frac{{\pi }^2}{4}\sum _{k=3}^n{k}^{2}sin(\frac{k\pi x}{2})$

Plugging these back into the ODE:

$-\frac{{\pi }^2}{4}\sum _{k=3}^n{A}_k{k}^{2}cos(\frac{k\pi x}{2})-\frac{{\pi }^2}{4}\sum _{k=3}^n{B}_k{k}^{2}sin(\frac{k\pi x}{2})-3[-\frac{\pi }{2}\sum _{k=2}^n{A}_{k}ksin(\frac{k\pi x}{2})$

$+\frac{\pi }{2}\sum _{k=2}^n{B}_{k}kcos(\frac{k\pi x}{2})]+2[A_0+\sum _{k=1}^n{A}_{k}cos(\frac{k\pi x}{2})+\sum _{k=1}^n{B}_{k}sin(\frac{k\pi x}{2})]$

$=\frac{A}{2}+\sum _{k=1}^n\frac{A}{k\pi }[(sin(\frac{9k\pi }{8})-sin(\frac{k\pi }{8}))cos(\frac{k\pi x}{2})+(cos(\frac{k\pi }{8})-cos(\frac{9k\pi }{8}))sin(\frac{k\pi x}{2})]$

Setting the two constants equal

$2A_0=\frac{A}{2}$

$A_0=\frac{A}{4}$

This is valid for all values of n. Since the coefficients of the excitation $sin(\frac{9k\pi }{8})-sin(\frac{k\pi }{8})$ and $cos(\frac{k\pi }{8})-cos(\frac{9k\pi }{8})$ are zero for all even n, then the coefficients $A_k$ and $B_k$ will also be zero, so we must only find these coefficients for odd n's. Now carrying out the sum to $n=1$ and comparing like terms yields the following sets of equations. Written in matrix form:

$\left[\begin{array}{cc}2& 0\\ \\ 0& 2\end{array}\right]*\left[\begin{array}{c}{A}_1\\ B_1\end{array}\right]=\left[\begin{array}{c}\frac{A}{\pi }(sin(\frac{9\pi }{8})-sin(\frac{\pi }{8}))\\ \frac{A}{\pi }(cos(\frac{\pi }{8})-cos(\frac{9\pi }{8}))\end{array}\right]$

Assuming $A=1$ this matrix can be solved to obtain

$A_1=-0.1218,B_1=0.2941$

For the remaining coefficients to be solved all sums will be used so a more general equation may be written:

$\left[\begin{array}{cc}2-\frac{(\pi k{)}^2}{4}& \frac{-3\pi k}{2}\\ \frac{-3\pi k}{2}& 2-\frac{(\pi k{)}^2}{4}\end{array}\right]*\left[\begin{array}{c}{A}_k\\ B_k\end{array}\right]=\left[\begin{array}{c}\frac{A}{\pi k}(sin(\frac{9\pi k}{8})-sin(\frac{\pi k}{8}))\\ \frac{A}{\pi k}(cos(\frac{\pi k}{8})-cos(\frac{9\pi }{8}))\end{array}\right]$

Results of these calculations are shown below:

$A=\left[\begin{array}{c}{A}_1\\ A_2\\ .\\ .\\ A_7\\ A_8\end{array}\right]=\left[\begin{array}{c}-0.1218\\ 0\\ 0.0084\\ 0\\ 0.0014\\ 0\\ 0.0001\\ 0\end{array}\right],B=\left[\begin{array}{c}{B}_1\\ B_2\\ .\\ .\\ B_7\\ B_8\end{array}\right]=\left[\begin{array}{c}0.2941\\ 0\\ 0.0019\\ 0\\ 0.0014\\ 0\\ 0.0007\\ 0\end{array}\right]$

The solution to the particular case can be written for all n (assuming A=1):

      $y_n=\frac{1}{4}+\sum _{k=1}^n{A}_{k}cos(\frac{k\pi x}{2})+\sum _{k=1}^n{B}_{k}sin(\frac{k\pi x}{2})$

General Solution

The general solution is

$y=y_h+y_p$

where

$y_h=c_1{e}^x+c_2{e}^{2x}$

Different coefficients $c_1,c_2$ will be calculated for each n. These coefficients are easily solved for by applying the given initial conditions. Below are the calculations for n=2.

$y=c_1{e}^x+c_2{e}^{2x}+\frac{1}{4}+\sum _{k=1}^2{A}_{k}cos(\frac{k\pi x}{2})+\sum _{k=1}^2{B}_{k}sin(\frac{k\pi x}{2})$

Applying the first initial condition $y(0)=1$

$y(0)=c_1+c_2+A_1+A_2=1$

Taking the derivative

$y^{\prime }=c_1{e}^x+2c_2{e}^{2x}-\sum _{k=2}^2\frac{k\pi }{2}{A}_{k}sin(\frac{k\pi x}{2})+\sum _{k=2}^2\frac{k\pi }{2}{B}_{k}cos(\frac{k\pi x}{2})$

Applying the second initial condition $y^{\prime }(0)=0$

$y^{\prime }(0)=c_1+2c_2+\frac{2\pi }{2}{B}_2=0$

Solving the two equations for two unknowns yields:

$c_1=2.2436,c_2=-1.1218$

So the general solution for n=2 is:

$y=2.2436e^x-1.1218e^{2x}+\frac{1}{4}+\sum _{k=1}^2{A}_{k}cos(\frac{k\pi x}{2})+\sum _{k=1}^2{B}_{k}sin(\frac{k\pi x}{2})$

Below is a plot showing the general solutions for n=2,4,8:

File:Plot64.jpg

File:Plot64zoom.jpg

Matlab Plots

Using ode45 the following graph was generated for n=0:

File:Matlab0.jpg

and for n=1

File:Matlab1.jpg