Boundary Value Problems/Lesson 4.1

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Sturm Liouville and Orthogonal Functions

The solutions in this BVP course will ALL be expressed as series built on orthogonal functions. Understanding that the simple problem $X^{″}+{\lambda }^{2}X=0$ with the boundary conditions ${\alpha }_{1}X(a)+{\alpha }_2{X}^{\prime }(a)=0$ and ${\beta }_{1}X(b)+{\beta }_2{X}^{\prime }(b)=0$ leads to solutions $X(x)$ that are orthogonal functions is crucial. Once this concept is grasped the majority of the work in this course is repetitive.
In the following notes think of the function $\mathrm{\Phi }(x)$ as a substitution for $X(x)$.

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File:Sturmliouville.pdf

Fourier Series

From the above work, solving the problem:
$X^{″}+{\lambda }^{2}X=0$ with the boundary conditions $X(0)=0$ and $X(L)=0$ leads to an infinite number of solutions $X_n(x)={\mathrm{\Phi }}_n(x)=sin\left(\frac{n\pi }{L}x\right)$
. These are eigenfunctions with eigenvalues ${\lambda }_n=\frac{n\pi }{L}$

Homework Assignment from Powell's sixth edition Boundary Value Problems page 71.

Project 1.2

This is a fourier series application problem.
You are given the piecewise defined function $f(t)$ shown in the following graph.

Plot of f(t)

The positive unit pulse is 150 μs in duration and is followed by a 100 μs interval where f(t) =0. Then f(t) is a negative unit pulse for 150 μs once again returning to zero. This pattern is repeated every 2860 μs. We will attempt to represent f(t) as a Fourier series,

1. Determine the value of the period: Ans. Period is 2860 μs. The time for a complete repetition of the waveform.
2. Find the Fourier Series representation: $f(t)=a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cos(n\pi t/a)+b_{n}sin(n\pi t/a)$ .The video provides an explanation of the determining the coefficients $a_0,a_n,b_n$
   

File:Proj1 2 0003.ogv

This is the first image.

. The results are:

$a_0=0$ $a_n=\frac{\sin \left(\frac{15}{143}n\pi \right)+\sin \left(\frac{25}{143}n\pi \right)-\sin \left(\frac{40}{143}n\pi \right)}{n\pi }$ $b_n=\frac{-(-1+\cos \left(\frac{15}{143}n\pi \right)+\cos \left(\frac{25}{143}n\pi \right)-\cos \left(\frac{40}{143}n\pi \right))}{n\pi }$

1. Using 100 terms an approximation is;File:Plotproj1 2 approx.jpg
2. Shift $f(t)$ right or left by an amount $b$ such that the resulting periodic function is an odd function. Here is a plot of shifting it to the left half way between the +1 and -1 pulses. This is a shiift of b= 200 μs. The new funnction is $f(t+200)$. A plot follows: File:Proj1 2shiftleft.jpg. It could also be shifted to the right by 1230 μs, that is $f(t-1230)$ is the new function.