# 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 ${\displaystyle X''+{\lambda }^{2}X=0}$ with the boundary conditions ${\displaystyle \alpha _{1}X(a)+\alpha _{2}X'(a)=0}$ and ${\displaystyle \beta _{1}X(b)+\beta _{2}X'(b)=0}$ leads to solutions ${\displaystyle X(x)}$ 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 ${\displaystyle \Phi (x)}$ as a substitution for ${\displaystyle X(x)}$.

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#### Fourier Series

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

## 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 ${\displaystyle f(t)}$ shown in the following graph.

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: ${\displaystyle f(t)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos(n\pi t/a)+b_{n}sin(n\pi t/a)}$ .The video provides an explanation of the determining the coefficients ${\displaystyle a_{0},a_{n},b_{n}}$
This is the first image.

. The results are:${\displaystyle a_{0}=0}$ ${\displaystyle a_{n}={\frac {\sin \left({\frac {15}{143}}\,n\,\pi \right)+\sin \left({\frac {25}{143}}\,n\,\pi \right)-\sin \left({\frac {40}{143}}\,n\,\pi \right)}{n\pi }}}$ ${\displaystyle b_{n}={\frac {-\left(-1+\cos \left({\frac {15}{143}}\,n\,\pi \right)+\cos \left({\frac {25}{143}}\,n\,\pi \right)-\cos \left({\frac {40}{143}}\,n\,\pi \right)\right)}{n\pi }}}$

1. Using 100 terms an approximation is;
2. Shift ${\displaystyle f(t)}$ right or left by an amount ${\displaystyle b}$ 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 ${\displaystyle f(t+200)}$. A plot follows: . It could also be shifted to the right by 1230 μs, that is ${\displaystyle f(t-1230)}$ is the new function.