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

Report 6

Problem 1

Problem Statement

ODE: $y''-3y'-10y=3cos7x$

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: $y(0)=1,y'(0)=0$

Plot the solution over 3 periods

Solution

Part 1

Wronskian: Function is linearly independent if $W(f,g)\neq 0$
$W(f,g):={\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}=fg'-gf'$

$f=cos7x,f'=-7sin7x$

$g=sin7x,g'=7cos7x$

$W(cos7x,sin7x):={\begin{bmatrix}cos7x&sin7x\\-7sin7x&7cos7x\end{bmatrix}}=7cos^{2}7x+7sin^{2}7x=1$

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

Gramian: Function is linearly independent if $\Gamma (f,g)\neq 0$
$\Gamma (f,g):={\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}$

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

$\Gamma (f,g):={\begin{bmatrix}1.07&0\\0&0.93\end{bmatrix}}=0.9951$

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:

$y_{p}'(x)=-M7sin7x+N7cos7x$

$y_{p}''(x)=-M7^{2}cos7x-N7^{2}sin7x$

Plug the derivatives into the equation:

$y_{p}''(x)-3y_{p}'(x)-10y_{p}(x)=3cos7x$

$(-M7^{2}cos7x-N7^{2}sin7x)-3(-M7sin7x+N7cos7x)-10(Mcos7x+Nsin7x)$

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

$-M7^{2}cos7x-3N7cos7x-10Mcos7x=3cos7x$

$-N7^{2}sin7x+3M7sin7x-10Nsin7x=0$

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 $y(0)=1,y'(0)=0$

$y(0)=C_{1}e^{0}+C_{2}e^{0}-0.0454cos(0)-0.0154sin(0)=1$

$y'(0)=-2C_{1}e^{0}+5C_{2}e^{0}+0.3178sin(0)-0.1078cos(0)=0$

$C_{1}=0.73$

$C_{2}=0.31$

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:

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 $y(0)=1,y'(0)=0$
Plot solution over 3 periods.

Solution

Given:
$y''+4y'+13y=2*e^{-2x}*3cos7x$
$y_{h}(x)=e^{-2x}*(Acos(3x)+BNsin(3x))$
$y_{P}(x)=x*e^{-2x}*(Mcos(3x)+Nsin(3x))$

$y_{P}'(x)=e^{-2x}*cos(3x)(x(-2M+N)+M)+e^{-2x}*sin(3x)(x(-2N-M)+N)$
$y_{P}''(x)=e^{-2x}(-3sin(3x)(x(-2M+N)+M)$
$+cos(3x)(-2M+N))-2e^{-2x}cos(3x)(x(-2M+N)+M)+e^{-2x}(3cos(3x)(x(-2N-M)$
$+N)+sin(3x)(-2N-M))-2e^{-2x}sin(3x)(x(-2N-M)+N)$

$y_{p}''+4y_{p}'+13y_{p}=2*e^{-2x}*3cos7x$

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
$A=1,-2A+3B+1=0$
$A=1,B=1/3$

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

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 $f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos(nwx)+b_{n}sin(nwx)]$ .

$a_{0}={\frac {1}{2L}}\int _{-L}^{L}f(x)dx$

For $n=1,2,...$
$a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)cos(nwx)dx$
$b_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)sin(nwx)dx$

For $f(x)=x^{2}$
$a_{0}={\frac {1}{2(1)}}\int _{-1}^{1}x^{2}dx={\frac {1}{3}}$
$a_{n}={\frac {1}{1}}\int _{-1}^{1}x^{2}cos(n\pi x)dx$
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.
$b_{n}={\frac {1}{1}}\int _{-1}^{1}x^{2}sin(n\pi x)dx$
$b_{n}={\frac {-1}{n\pi }}[cos(n\pi )-cos(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[sin(n\pi )+sin(-n\pi )]+{\frac {2}{n^{3}\pi ^{3}}}[cos(n\pi )-cos(-n\pi )]$
So the Fourier series for $f(x)=x^{2}$ is
${\frac {1}{3}}+\sum _{n=1}^{\infty }cos(nwx)[{\frac {1}{n\pi }}[sin(n\pi )-sin(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[cos(n\pi )+cos(-n\pi )]-{\frac {2}{n^{3}\pi ^{3}}}[sin(n\pi )-sin(-n\pi )]]$
$+sin(nwx)[-{\frac {1}{n\pi }}[cos(n\pi )-cos(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[sin(n\pi )+sin(-n\pi )]+{\frac {2}{n^{3}\pi ^{3}}}[cos(n\pi )-cos(-n\pi )]]$

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 $f({\bar {x}})$ . Plot $f({\bar {x}})$ and develop the truncated Fourier series $f_{n}({\bar {x}})$ .
$f_{n}({\bar {x}}):={\bar {a}}_{0}+\sum _{k=1}^{n}[{\bar {a}}_{k}\cos k\omega {\bar {x}}+{\bar {b}}_{k}\sin k\omega {\bar {x}}]$
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 $f({\bar {x}})$ was determined to be even. Even functions reduce to a cosine Fourier series. Because $f({\bar {x}})$ , has a period of 4, the length is 2.

$a_{0}={\frac {1}{L}}\int _{0}^{2}f({\bar {x}})d{\bar {x}}={\frac {A}{2}}$

$a_{k}={\frac {2}{L}}\int _{0}^{2}f({\bar {x}})\cos({\frac {k\pi {\bar {x}}}{2}})d{\bar {x}}$

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

$f_{k}({\bar {x}})=a_{0}+\sum _{k=1}^{n}[a_{n}\cos {\frac {n\pi {\bar {x}}}{2}}]$

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

For n=0,
$f_{0}({\bar {x}})={\frac {A}{2}}$

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

${\bar {x}}=x-1.25$

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

Plot (A=1)

Part 2

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

$a_{0}={\frac {1}{L}}\int _{0}^{2}f({\bar {x}})d{\bar {x}}={\frac {A}{2}}$

$a_{0}={\frac {1}{2L}}\int _{0}^{4}f({\tilde {x}})d{\tilde {x}}={\frac {A}{2}}$

$a_{n}={\frac {1}{2}}\int _{0}^{4}f{\tilde {x}}\cos({\frac {n\pi {\tilde {x}}}{2}}d{\tilde {x}}$

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

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

$b_{n}={\frac {1}{2}}\int _{0}^{4}f{\tilde {x}}\sin({\frac {n\pi {\tilde {x}}}{2}}d{\tilde {x}}$

$b_{n}={\frac {1}{2}}f({\tilde {x}})\sin({\frac {n\pi {\tilde {x}}}{2}})$ 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)

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

${\frac {\partial u(x,t)}{\partial x}}=F'(x)\cdot G(t)$ (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):

$F(x)\cdot {\dot {G}}(t)=kF''(x)\cdot G(t)$ (6)

Rearrange (6) to combine like terms:

${\frac {{\dot {G}}(t)}{kG(t)}}={\frac {F''(x)}{F(x)}}=c{\text{ (constant)}}$

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

${\dot {G}}(t)=k\,c\,G(t)$

${\dot {G}}(t)-k\,c\,G(t)=0$

${\frac {F''(x)}{F(x)}}=c$

$F''(x)=c\,F(x)$

$F''(x)-c\,F(x)=0$

Solution:

Separated ODE's for Heat Equation:

${\dot {G}}(t)-k\,c\,G(t)=0$

$F''(x)-c\,F(x)=0$

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) $<\phi _{i},\phi _{j}>=0$ for $i\neq j$
(5) $<\phi _{i},\phi _{j}>=L/2$ for $i=j$

Solution

Verification of (4)

Using the integral scalar product calculation,
$\int _{0}^{L}\phi _{i}(x)\phi _{j}(x)dx$

Substituting in sin values,
$\int _{0}^{L}sin(\omega _{i}(x))sin(\omega _{j}(x))dx$

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

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

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 $i\neq j$ , 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

$u(x,t)=\sum _{j=1}^{\infty }a_{j}\cos(cw_{j}t)\sin(w_{j}x)$
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 $u(x,2/3)$

Plot $u(x,4/3)$

Plot $u(x,2)$

Plot $u(x,8/3)$

Honor Pledge

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

This article is issued from Wikiversity. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.