University of Florida/Egm4313/s12.team11.perez.gp/R5.9

R5.9

Problem Statement

Consider the L2-ODE-CC (5) p.7b-7 with $log(1+x)\!$ as excitation:

$y''-3y'+2y=r(x)\!$ (5) p.7b-7

$r(x)=log(1+x)\!$ (1) p.7c-28

and the initial conditions

$y({\frac {-3}{4}})=1,y'({\frac {-3}{4}})=0\!$.

Part 1

Project the excitation $r(x)\!$ on the polynomial basis

${b_{j}(x)=x^{j},j=0,1,...,n}\!$ (1)

i.e., find $d_{j}\!$ such that

$r(x)\approx r_{n}(x)=\sum _{j=0}^{n}d_{j}x^{j}\!$ (2)

for x in $[{\frac {-3}{4}},3]\!$, and for n = 3, 6, 9.

Plot $r(x)\!$ and $r_{n}(x)\!$ to show uniform approximation and convergence.

Note that:

$\left\langle x^{i},r\right\rangle =\int _{a}^{b}x^{i}log(1+x)dx\!$ (3)

Solution

Using Matlab, this is the code that was used to produce the results:

Part 2

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

$y_{n}''+ay_{n}'+by_{n}=r_{n}(x)\!$ (1) p.7c-27

with the same initial conditions as in (2) p.7c-28.

Plot $y_{n}(x)\!$ for n = 3, 6, 9, for x in $[{\frac {-3}{4}},3]\!$.

In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.

Solution

---

---

---

Using integration by parts, and then with the help of of

General Binomial Theorem

$(x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{n-k}y^{k}\!$

Solution

For $n=0\!$:

$\int x^{0}log(1+x)dx=\int log(1+x)dx\!$

For substitution by parts, $u=log(1+x),du={\frac {1}{1+x}},dv=dx,v=x\!$

$\int log(1+x)dx=xlog(1+x)-\int {\frac {x}{1+x}}dx\!$

$\int log(1+x)dx=xlog(1+x)-\int (1-{\frac {1}{1+x}})dx\!$

$\int log(1+x)dx=xlog(1+x)-x+log(1+x)+C\!$

Therefore:

                                     $\int log(1+x)dx=(x+1)log(1+x)-x+C\!$

Using the General Binomial Theorem:

$(x+y)^{0}=\sum _{k=0}^{0}{\binom {0}{k}}x^{0-k}y^{k}=1\!$

Therefore: $\int (1)log(1+x)dx=\int log(1+x)dx\!$

Which we have previously found that answer as:

                                     $\int log(1+x)dx=(x+1)log(1+x)-x+C\!$

For $n=1\!$:

$\int x^{1}log(1+x)dx=\int xlog(1+x)dx\!$

Initially we use the following substitutions: $t=1+x,x=t-1,dt=dx\!$

$\int xlog(1+x)dx=\int (t-1)log(t)dt=\int (tlog(t)-\log(t))dt\!$

First let us consider the first term: $\int tlog(t)dt\!$

Next, we use the integration by parts: $u=\log {t},du={\frac {1}{t}}dt,dv=tdt,v={\frac {1}{2}}t^{2}\!$
$\int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-\int {\frac {1}{2}}t^{2}({\frac {1}{t}}dt)\!$

$\int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-\int {\frac {1}{2}}tdt)\!$

$\int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}\!$

Next let us consider the second term: $\int log(t)dt\!$

Again, we will use integration by parts: $u=\log {t},du={\frac {1}{t}}dt,dv=dt,v=t\!$
$\int tlog(t)dt=tlog(t)-\int t({\frac {1}{t}}dt)\!$

$\int tlog(t)dt=tlog(t)-\int dt\!$

$\int tlog(t)dt=tlog(t)-t\!$

Therefore:

$\int (tlog(t)-\log(t))dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}-(tlog(t)-t)\!$

$\int (tlog(t)-\log(t))dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}-tlog(t)+t\!$

Re-substituting for t:

$\int xlog(1+x)dx={\frac {1}{2}}(1+x)^{2}log(1+x)-{\frac {1}{4}}(1+x)^{2}-(1+x)log(1+x)+(1+x)+C\!$

$\int xlog(1+x)dx=(1+x)({\frac {1}{2}}(1+x)log(1+x)-{\frac {1}{4}}(1+x)-log(1+x)+1)+C\!$

$\int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!$

Therefore:

                                     $\int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!$

Using the General Binomial Theorem for the integral with t substitution $\int (t-1)log(t)dt\!$:

$(x+y)^{1}=(t+(-1))^{1}=(t+(-1))=\sum _{k=0}^{1}{\binom {1}{k}}x^{1-k}y^{k}=\sum _{k=0}^{1}{\binom {1}{k}}t^{1-k}(-1)^{k}=t-1\!$

Therefore: $\int (t-1)log(t)dt=\int xlog(1+x)dx\!$

Which we have previously found that answer as:

                                     $\int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!$