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:

$y''-3y'+2y=f(x)\!$

and the initial conditions

$y(0)=1\ ,\ y'(0)=0\!$

1. Find $y_{n}(x)\!$ such that:

$y_{n}''+ay_{n}'+by_{n}=r_{n}(x)\!$

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

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

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:

$a_{0}={\frac {1}{2L}}\int _{-1.75}^{2.25}f(x)\ dx\!$

$a_{0}={\frac {1}{4}}(\int _{-1.75}^{0.25}0\ dx+\int _{0.25}^{2.25}A\ dx)\!$

$a_{0}=+{\frac {1}{4}}(0+\int _{0.25}^{2.25}A\ dx)\!$

$a_{0}={\frac {1}{4}}(2.25A-0.25A)\!$

$a_{0}={\frac {A}{2}}\!$

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

$a_{n}={\frac {1}{L}}\int _{0.25}^{2.25}f(x)cos({\frac {n\pi x}{L}})\ dx\!$

$a_{n}={\frac {1}{2}}\int _{0.25}^{2.25}Acos({\frac {n\pi x}{2}})\ dx\!$

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

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

and

$b_{n}={\frac {1}{L}}\int _{0.25}^{2.25}f(x)sin({\frac {n\pi x}{L}})\ dx\!$

$b_{n}={\frac {1}{2}}\int _{0.25}^{2.25}Asin({\frac {n\pi x}{2}})\ dx\!$

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

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

The coefficients give the Fourier series:

$f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos({\frac {n\pi x}{L}})+b_{n}sin({\frac {n\pi x}{L}})]\!$

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

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

Homogeneous Solution

Considering the homogeneous case of our ODE:

$y''-3y'+2y=0\!$

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

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

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

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

$y_{n}'=-A_{n}{\frac {\pi }{2}}\sum _{k=2}^{n}ksin({\frac {k\pi x}{2}})+B_{n}{\frac {\pi }{2}}\sum _{k=2}^{n}kcos({\frac {k\pi x}{2}})\!$

$y_{n}''=-A_{n}{\frac {\pi ^{2}}{4}}\sum _{k=3}^{n}k^{2}cos({\frac {k\pi x}{2}})-B_{n}{\frac {\pi ^{2}}{4}}\sum _{k=3}^{n}k^{2}sin({\frac {k\pi x}{2}})\!$

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:

${\begin{bmatrix}2&0\\\\0&2\end{bmatrix}}*{\begin{bmatrix}A_{1}\\B_{1}\end{bmatrix}}={\begin{bmatrix}\ {\frac {A}{\pi }}(sin({\frac {9\pi }{8}})-sin({\frac {\pi }{8}}))\\\ {\frac {A}{\pi }}(cos({\frac {\pi }{8}})-cos({\frac {9\pi }{8}}))\end{bmatrix}}\!$

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:

${\begin{bmatrix}2-{\frac {(\pi k)^{2}}{4}}&{\frac {-3\pi k}{2}}\\{\frac {-3\pi k}{2}}&2-{\frac {(\pi k)^{2}}{4}}\end{bmatrix}}*{\begin{bmatrix}A_{k}\\B_{k}\end{bmatrix}}={\begin{bmatrix}\ {\frac {A}{\pi k}}(sin({\frac {9\pi k}{8}})-sin({\frac {\pi k}{8}}))\\\ {\frac {A}{\pi k}}(cos({\frac {\pi k}{8}})-cos({\frac {9\pi }{8}}))\end{bmatrix}}\!$

Results of these calculations are shown below:

$A={\begin{bmatrix}A_{1}\\A_{2}\\.\\.\\A_{7}\\A_{8}\end{bmatrix}}={\begin{bmatrix}-0.1218\\0\\0.0084\\0\\0.0014\\0\\0.0001\\0\end{bmatrix}},\ B={\begin{bmatrix}B_{1}\\B_{2}\\.\\.\\B_{7}\\B_{8}\end{bmatrix}}={\begin{bmatrix}0.2941\\0\\0.0019\\0\\0.0014\\0\\0.0007\\0\end{bmatrix}}\!$

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

$y'=c_{1}e^{x}+2c_{2}e^{2x}-\sum _{k=2}^{2}{\frac {k\pi }{2}}A_{k}sin({\frac {k\pi x}{2}})+\sum _{k=2}^{2}{\frac {k\pi }{2}}B_{k}cos({\frac {k\pi x}{2}})\!$

Applying the second initial condition $y'(0)=0\!$

$y'(0)=c_{1}+2c_{2}+{\frac {2\pi }{2}}B_{2}=0\!$

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:

Matlab Plots

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

and for n=1