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:

$y^{″}-3y^{\prime }+2y=r(x)$ (5) p.7b-7

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

and the initial conditions

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

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:

$⟨x^i,r⟩=\underset{a}{\overset{b}{\int }}{x}^{i}log(1+x)dx$ (3)

Solution

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

Part 2

Find $y_n(x)$ such that:

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

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

General Binomial Theorem

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

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:

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

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 $\int (t-1)log(t)dt$:

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

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

Which we have previously found that answer as:

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