University of Florida/Egm4313/IEA-f13-team10/R6

Report 6

Problem 1

Problem Statement

ODE: $y^{″}-3y^{\prime }-10y=3cos7x$

Part 1: show that cos7x and sin7x are linearly independent using the Wronskian and Gramian.

Part 2: Find 2 equations for the two unknowns M, N, and solve for M, N.

Part 3: Find the overall solution y(x) that corresponds to the initial conditions: $y(0)=1,y^{\prime }(0)=0$

Plot the solution over 3 periods

Solution

Part 1

Wronskian: Function is linearly independent if $W(f,g)\ne 0$
$W(f,g):=\left[\begin{array}{cc}f& g\\ f^{\prime }& g^{\prime }\end{array}\right]=f{g}^{\prime }-g{f}^{\prime }$

$f=cos7x,f^{\prime }=-7sin7x$

$g=sin7x,g^{\prime }=7cos7x$

$W(cos7x,sin7x):=\left[\begin{array}{cc}cos7x& sin7x\\ -7sin7x& 7cos7x\end{array}\right]=7co{s}^{2}7x+7si{n}^{2}7x=1$

g(x) and f(x) are linearly independent

Gramian: Function is linearly independent if $\mathrm{\Gamma }(f,g)\ne 0$
$\mathrm{\Gamma }(f,g):=\left[\begin{array}{cc}<f,f>& <f,g>\\ <g,f>& <g,g>\end{array}\right]$

$f(x)=cos7x,g(x)=sin7x$
$<f,f>=\underset{-1}{\overset{1}{\int }}cos7x*cos7xdx=1.07$
$<f,g>=\underset{-1}{\overset{1}{\int }}cos7x*sin7xdx=0$
$<g,f>=\underset{-1}{\overset{1}{\int }}sin7x*cos7xdx=0$
$<g,g>=\underset{-1}{\overset{1}{\int }}sin7x*sin7xdx=0.93$

$\mathrm{\Gamma }(f,g):=\left[\begin{array}{cc}1.07& 0\\ 0& 0.93\end{array}\right]=0.9951$

g(x) and f(x) are linearly independent

Part 2

The particular solution for a $r(x)=3cos7x$ will be:

$y_p(x)=Mcos7x+Nsin7x$

Differentiate to get:

${y_p}^{\prime }(x)=-M7sin7x+N7cos7x$

${y_p}^{″}(x)=-M{7}^{2}cos7x-N{7}^{2}sin7x$

Plug the derivatives into the equation:

${y_p}^{″}(x)-3{y_p}^{\prime }(x)-10y_p(x)=3cos7x$

$(-M{7}^{2}cos7x-N{7}^{2}sin7x)-3(-M7sin7x+N7cos7x)-10(Mcos7x+Nsin7x)$

Separate the sin and cos terms to get 2 equations in order to solve for M and N

$-M{7}^{2}cos7x-3N7cos7x-10Mcos7x=3cos7x$

$-N{7}^{2}sin7x+3M7sin7x-10Nsin7x=0$

dividing each equation by cos7x and sin7x respectively:

$-49M-21N-10M=3$

$-49N+21M-10N=0$

$M=-0.0454$

$N=-0.0154$

So the particular solution is:

$y_p(x)=-0.0454cos7x-0.0154sin7x$

Part 3

The overall solution can be found by:

$y(x)=y_h(x)+y_p(x)$

The roots given in the problem statement ${\lambda }_1=-2,{\lambda }_2=+5$

Lead to the homogeneous solution of:

$y_h(x)=C_1{e}^{-2x}+C_2{e}^{5x}$

Combining the homogeneous and particular solution gives us:

$y(x)=C_1{e}^{-2x}+C_2{e}^{5x}-0.0454cos7x-0.0154sin7x$

Solving for the constants by using the initial conditions $y(0)=1,y^{\prime }(0)=0$

$y(0)=C_1{e}^0+C_2{e}^0-0.0454cos(0)-0.0154sin(0)=1$

$y^{\prime }(0)=-2C_1{e}^0+5C_2{e}^0+0.3178sin(0)-0.1078cos(0)=0$

$C_1=0.73$

$C_2=0.31$

The overall solution is:

$y(x)=0.73e^{-2x}+0.31e^{5x}-0.0454cos7x-0.0154sin7x$

Plot

Plot $y(x)=0.73e^{-2x}+0.31e^{5x}-0.0454cos7x-0.0154sin7x$

over 3 periods:

File:R6.1.PNG

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 2

Problem Statement

Complete the solution to problem on p.8-6.
Find the overall solution $y(x)$
that corresponds to the initial condition $y(0)=1,y^{\prime }(0)=0$
Plot solution over 3 periods.

Solution

Given:
$y^{″}+4y^{\prime }+13y=2*e^{-2x}*3cos7x$
$y_h(x)=e^{-2x}*(Acos(3x)+BNsin(3x))$
$y_P(x)=x*e^{-2x}*(Mcos(3x)+Nsin(3x))$


${y_P}^{\prime }(x)=e^{-2x}*cos(3x)(x(-2M+N)+M)+e^{-2x}*sin(3x)(x(-2N-M)+N)$
${y_P}^{″}(x)=e^{-2x}(-3sin(3x)(x(-2M+N)+M)$
$+cos(3x)(-2M+N))-2e^{-2x}cos(3x)(x(-2M+N)+M)+e^{-2x}(3cos(3x)(x(-2N-M)$
$+N)+sin(3x)(-2N-M))-2e^{-2x}sin(3x)(x(-2N-M)+N)$

${y_p}^{″}+4{y_p}^{\prime }+13y_p=2*e^{-2x}*3cos7x$


Solve for M and N:
$4N=2,-2M+4N$
$M=1,N=0.5$

$y(x)=e^{-2x}*(Acos(3x)+BNsin(3x))+x*e^{-2x}*(cos(3x)+.5*sin(3x))$
Using initial conditions given find A and B

After applying initial conditions, we get
$A=1,-2A+3B+1=0$
$A=1,B=1/3$

$y(x)=e^{-2x}*(cos(3x)+sin(3x)/3)+x*e^{-2x}*(cos(3x)+.5*sin(3x))$

File:Plot 1.PNG
File:Plot 2.PNG

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 3

Problem Statement

Is the given function even or odd or neither even nor odd? Find its Fourier Series.

Solution

$f(x)=x^2,(-1<x<1),p=2$

$f(-x)=f(x)$ so $f(x)=x^2$ is an even function.

The Fourier series is $f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}[a_{n}cos(nwx)+b_{n}sin(nwx)]$.

$a_0=\frac{1}{2L}\underset{-L}{\overset{L}{\int }}f(x)dx$

For $n=1,2,...$
$a_n=\frac{1}{L}\underset{-L}{\overset{L}{\int }}f(x)cos(nwx)dx$
$b_n=\frac{1}{L}\underset{-L}{\overset{L}{\int }}f(x)sin(nwx)dx$

For $f(x)=x^2$
$a_0=\frac{1}{2(1)}\underset{-1}{\overset{1}{\int }}{x}^{2}dx=\frac{1}{3}$
$a_n=\frac{1}{1}\underset{-1}{\overset{1}{\int }}{x}^{2}cos(n\pi x)dx$
The above integral requires two iterations of integration by parts. Which gives
$a_n=\frac{1}{n\pi }[sin(n\pi )-sin(-n\pi )]+\frac{2}{n^2{\pi }^2}[cos(n\pi )+cos(-n\pi )]-\frac{2}{n^3{\pi }^3}[sin(n\pi )-sin(-n\pi )]$
Similarly, integration by parts needs to be used twice to solve the following integral.
$b_n=\frac{1}{1}\underset{-1}{\overset{1}{\int }}{x}^{2}sin(n\pi x)dx$
$b_n=\frac{-1}{n\pi }[cos(n\pi )-cos(-n\pi )]+\frac{2}{n^2{\pi }^2}[sin(n\pi )+sin(-n\pi )]+\frac{2}{n^3{\pi }^3}[cos(n\pi )-cos(-n\pi )]$
So the Fourier series for $f(x)=x^2$ is
$\frac{1}{3}+\sum _{n=1}^{\mathrm{\infty }}cos(nwx)[\frac{1}{n\pi }[sin(n\pi )-sin(-n\pi )]+\frac{2}{n^2{\pi }^2}[cos(n\pi )+cos(-n\pi )]-\frac{2}{n^3{\pi }^3}[sin(n\pi )-sin(-n\pi )]]$
$+sin(nwx)[-\frac{1}{n\pi }[cos(n\pi )-cos(-n\pi )]+\frac{2}{n^2{\pi }^2}[sin(n\pi )+sin(-n\pi )]+\frac{2}{n^3{\pi }^3}[cos(n\pi )-cos(-n\pi )]]$

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 4

Problem Statement

1) Develop the Fourier series of$f(\overline{x})$. Plot$f(\overline{x})$and develop the truncated Fourier series$f_n(\overline{x})$.
$f_n(\overline{x}):={\overline{a}}_0+\sum _{k=1}^n[{\overline{a}}_k\cos k\omega \overline{x}+{\overline{b}}_k\sin k\omega \overline{x}]$
for n = 0,1,2,4,8. Observe the values of at the points of discontinuities, and the Gibbs phenomenon. Transform the variable so to obtain the Fourier series expansion of . Level 1: n=0,1.

2)Do the same as above, but using$f(\tilde{x})$to obtain the Fourier series expansion of$f(x)$; compare to the result obtained above. Level 1: n=0,1.

Solution

Part 1

To begin, the function $f(\overline{x})$ was determined to be even. Even functions reduce to a cosine Fourier series. Because $f(\overline{x})$, has a period of 4, the length is 2.

$a_0=\frac{1}{L}\underset{0}{\overset{2}{\int }}f(\overline{x})d\overline{x}=\frac{A}{2}$

$a_k=\frac{2}{L}\underset{0}{\overset{2}{\int }}f(\overline{x})\cos (\frac{k\pi \overline{x}}{2})d\overline{x}$

$a_k=\frac{2A}{k\pi }\sin (\frac{k\pi }{2})$

$f_k(\overline{x})=a_0+\sum _{k=1}^n[a_n\cos \frac{n\pi \overline{x}}{2}]$

$f_k=\frac{A}{2}+\sum _{k=1}^n[\frac{2A}{k\pi }\sin (\frac{k\pi }{2})\cos \frac{k\pi \overline{x}}{2}]$

For n=0,
$f_0(\overline{x})=\frac{A}{2}$

For n=1,
$f_1(\overline{x})=\frac{A}{2}+\frac{2A}{\pi }\cos (\frac{\pi \overline{x}}{2})$

$\overline{x}=x-1.25$

$f_k(x)=\frac{A}{2}+\frac{A}{\pi }\cos (\frac{\pi (x-1.25)}{2})$

Plot (A=1)

File:R6 p4(1).JPG

Part 2

To begin, the function $f(\overline{x})$ was determined to be odd. Even functions reduce to a sine Fourier series. Because $f(\overline{x})$, has a period of 4, the length is 2.

$a_0=\frac{1}{L}\underset{0}{\overset{2}{\int }}f(\overline{x})d\overline{x}=\frac{A}{2}$

$a_0=\frac{1}{2L}\underset{0}{\overset{4}{\int }}f(\tilde{x})d\tilde{x}=\frac{A}{2}$

$a_n=\frac{1}{2}\underset{0}{\overset{4}{\int }}f\tilde{x}\cos (\frac{n\pi \tilde{x}}{2}d\tilde{x}$

$a_n=\frac{1}{2}f(\tilde{x})\cos (\frac{n\pi \tilde{x}}{2}$ from 0 to 4

$a_n=\frac{A}{n\pi }\sin (\pi k)$

$b_n=\frac{1}{2}\underset{0}{\overset{4}{\int }}f\tilde{x}\sin (\frac{n\pi \tilde{x}}{2}d\tilde{x}$

$b_n=\frac{1}{2}f(\tilde{x})\sin (\frac{n\pi \tilde{x}}{2})$ from 0 to 4

$b_n=\frac{A}{n\pi }(1-\cos (\pi k))$

$f_k(\tilde{x})=\frac{A}{2}+\sum _{k=1}^n[\frac{A}{k\pi }(1-\cos (\pi k))(\sin (\frac{k\pi \tilde{x}}{2}))]$

For n=0,
$f_0(\tilde{x})=\frac{A}{2}$

For n=1,
$f_1(\tilde{x})=\frac{A}{2}+\frac{A}{\pi }\sin (\frac{\pi \tilde{x}}{2})$

$\tilde{x}=x-.25$

$f_1(x)=\frac{A}{2}+\frac{A}{\pi }\sin (\frac{\pi (x-.25)}{2})$
Plot (A=1)

File:R6 p4(2).JPG

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 5

Problem Statement

Find the separated ODE's for the Heat Equation:

$\frac{{\partial }^{2}u}{\partial t^2}=k\frac{{\partial }^{2}u}{\partial x^2}$ (1)

$k=$ heat capacity

Solution

Separation of Variables:

Assume: $u(x,t)=F(x)\cdot G(t)$

$\frac{\partial u(x,t)}{\partial x}=F^{\prime }(x)\cdot G(t)$ (2)

$\frac{{\partial }^{2}u(x,t)}{\partial x^2}=F^{″}(x)\cdot G(t)$ (3)

$\frac{\partial u(x,t)}{\partial t}=F(x)\cdot \dot{G}(t)$ (4)

$\frac{{\partial }^{2}u(x,t)}{\partial t^2}=F(x)\cdot \ddot{G}(t)$ (5)

Plug (2) and (3) into Heat Equation (1):

$F(x)\cdot \dot{G}(t)=k{F}^{″}(x)\cdot G(t)$ (6)

Rearrange (6) to combine like terms:

$\frac{\dot{G}(t)}{kG(t)}=\frac{F^{″}(x)}{F(x)}=c\text{ (constant)}$

$\frac{\dot{G}(t)}{kG(t)}=c$

$\dot{G}(t)=kcG(t)$

$\dot{G}(t)-kcG(t)=0$

$\frac{F^{″}(x)}{F(x)}=c$

$F^{″}(x)=cF(x)$

$F^{″}(x)-cF(x)=0$

Solution:

Separated ODE's for Heat Equation:

$\dot{G}(t)-kcG(t)=0$

$F^{″}(x)-cF(x)=0$

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 6

Problem Statement

Verify (4)-(5) p.19-9
(4)$<{\varphi }_i,{\varphi }_j>=0$ for $i\ne j$
(5)$<{\varphi }_i,{\varphi }_j>=L/2$ for $i=j$

Solution

Verification of (4)

Using the integral scalar product calculation,
$\underset{0}{\overset{L}{\int }}{\varphi }_i(x){\varphi }_j(x)dx$

Substituting in sin values,
$\underset{0}{\overset{L}{\int }}sin({\omega }_i(x))sin({\omega }_j(x))dx$

Using $z=\frac{\pi x}{L}$ and $dz=\frac{\pi }{L}dx$

You can substitute z into the integral instead of x.
$\underset{0}{\overset{\pi }{\int }}sin(iz)sin(jz)dz\frac{L}{\pi }$

Integrating,
$\frac{1}{2}[\frac{sin(i-j)z}{i-j}-\frac{sin(i+j)z}{i+j}]$ from $z=0$ to $z=\pi$
Since $i\ne j$, the equation with its sin values turns into 0-0=0

Verification of (5)

You can use the same equation from the verification of (4) from this point:
$\frac{1}{2}[\frac{sin(i-j)z}{i-j}-\frac{sin(i+j)z}{i+j}]$ from $z=0$ to $z=\pi$
Putting those values in and substituting L back in the equation, it turns into $\frac{L}{2}$

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 7

Problem Statement

$u(x,t)=\sum _{j=1}^{\mathrm{\infty }}{a}_j\cos (c{w}_{j}t)\sin (w_{j}x)$
Plot the truncated series for n=5. $t=\alpha p_1=\frac{\alpha 2L}{c}$

$\alpha =.5,1,1.5,2$

Solution

$a_j=\frac{2[((-1{)}^j)-1]}{{\pi }^3{j}^3}$

$w_j=\frac{j\pi }{L}$

C=3 and L=2
Plot $u(x,2/3)$

File:P7.1.JPG

Plot $u(x,4/3)$

File:P7.2.JPG

Plot $u(x,2)$

File:P7.3.JPG

Plot $u(x,8/3)$

File:P7.4.JPG

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.