Zach

The displacement from the equilibrium of the midpoint of the string at time t=1.3 seconds into the vibration is approximately -0.351.

Heat flow with a constant internal heat source is governed by:

$$u_t = 1u_{xx} + 2, u(0,t) = -3, u(1,t) = 5.$$

When at the steady state temperature distribution, $u_t = 0$ therefore:

$$0 = u_{xx} + 2$$

This can be rearranged to give $u_{xx} = -2$. Next this function was integrated with respect to $x$ twice giving $$u(x) = -1x^2 + \alpha*x + \beta$$.

Plugging in the conditions $u(0,t) = -3$ and $u(1,t) = 5$ allowed for finding of $\alpha = 9$ and $\beta = -3$.

Therefore, the steady state temperature distribution is:

$$u(x) = -1x^2 + 9x - 3$$

A metal bar of length 1 lies along the unit interval. Its temperature distribution is given by:

$$g(x) = 5x^2 - 1x .$$

At time t = 0, its left end is set to temperature -2 and its right end to -1. Sketch the temperature distribution at times:

t = 0, t = 0.01, t = 0.1, t = 10.

For my problem, the basic heat equation was:

$$u_t = 0.47\Delta u$$

The vertical side of the triangle is insulated

The Dirichlet condition on the circle and the 2 other sides of the triangle is

$$u(x,y,t) = 0.87^2 - (x^2+y^2)$$

The initial condition is:

$$u(x,y,0) = 0.87^2 - (x^2+y^2)$$

I found the temperature of the domain near the midpoint of the insulated edge at time, $t = 0.99$s to be 0.11568

Here's the photo of my house at t=0.6, the temperature just above the door is 0.37670. I had set the diffusivity constant equal to 1.

The full Fourier Series is generally:

$$f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} a_ncos(\frac{n\pi x}{L}) + b_nsin(\frac{n\pi x}{L})$$

First find the value of $a_0$ for $f(x) = x - x^2$ over the interval $[0,L]$

$$a_0 = \frac{2}{L} \int_{0}^{L} f(x)\,dx$$

$$\implies a_0 = 4 \int_{0}^{\frac{1}{2}}(x - x^2)\,dx = 4\int_{0}^{\frac{1}{2}} x\,dx - \int_{0}^{\frac{1}{2}} x^2\,dx$$

$$\implies 4[\frac{1}{2} x^2 \biggr\rvert_{0}^{\frac{1}{2}} - \frac{1}{3} x^3 \biggr\rvert_{0}^{\frac{1}{2}}] = 4[\frac{1}{2} (\frac{1}{2})^2 - \frac{1}{3} (\frac{1}{2})^3]$$

$$\implies 4[\frac{1}{8} - \frac{1}{24}] = 4(\frac{1}{12})$$

Therefore $a_0 = \frac{1}{3}$ and plugging in this value into the general equation gives $\frac{a_0}{2} = \frac{1}{6}$

Next, find the value for $a_n$ for $f(x)$

$$\int_{0}^{L} f(x)cos(\frac{n\pi x}{L})\,dx = 4\int_{0}^{\frac{1}{2}} (x - x^2)cos(2\pi nx)\,dx$$

Can split into integrals of $x$ and $x^2$

$$4\int_{0}^{\frac{1}{2}}xcos(2\pi nx)\,dx$$

Using integration by parts, with 

$u = x, du = dx, v = \frac{sin(2\pi nx)}{2\pi n}, dv = cos(2\pi nx)dx$

$$\implies 4[ \frac{xsin(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} - \frac{1}{2\pi n} \int_{0}^{\frac{1}{2}}sin(2\pi nx)\,dx ]$$

$$\implies \frac{1}{(\pi n)^2} cos(2\pi nx) \bigg\rvert_{0}^{\frac{1}{2}} = \frac{1}{(\pi n)^2} [(-1)^n - 1] = \frac{(-1)^n -1}{\pi^2n^2}$$

Now, plugging in $x^2$ gives

$$4 \int_{0}^{\frac{1}{2}} x^2cos(2\pi nx)\,dx$$

Again using integration by parts, with 

$u = x^2, du = 2xdx, v = \frac{sin(2\pi nx)}{2\pi n}, dv = cos(2\pi nx)dx$

$$\implies 4[ \frac{(x^2)sin(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} - \frac{1}{\pi n} \int_{0}^{\frac{1}{2}} xsin(2\pi nx)\,dx ]$$

The first term is equal to $0$, and integration by parts must be used to solve the integral with

$u = x, du = dx, v = \frac{-cos(2\pi nx)}{2\pi n}, dv = sin(2\pi nx)dx$

$$\implies \frac{-4}{\pi n} \frac{-xcos(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} + \frac{1}{2\pi n} \int_{0}^{\frac{1}{2}} cos(2\pi nx)\,dx$$

$$\implies \frac{4}{\pi n}(\frac{(-1)^n}{4\pi n}) = \frac{(-1)^n}{\pi^2n^2}$$

$$\implies a_n = \frac{(-1)^n - 1}{\pi^2n^2} - \frac{(-1)^n}{\pi^2n^2} = \frac{(-1)}{\pi^2n^2}$$

Now move onto $b_n$ and again split the integrals of $x$ and $x^2$

$$b_n = \frac{2}{L} \int_{0}^{L} f(x)sin(\frac{\pi nx}{L})\,dx $$

$$b_n = 4[\int_{0}^{\frac{1}{2}} xsin(2\pi nx)\,dx - \int_{0}^{\frac{1}{2}} x^2sin(2\pi nx)\,dx ]$$

Use integration by parts for the integral of $x$, with

$u = x, du = dx, v = \frac{-cos(2\pi nx)}{2\pi n}, dv = sin(2\pi nx)dx$

$$\implies 4[\frac{-xcos(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} + \frac{1}{2\pi n} \int_{0}^{\frac{1}{2}} cos(2\pi nx)\,dx ]$$

$$\implies 4[-\frac{(-1)^n}{4\pi n}] = -\frac{(-1)^n}{\pi n}$$

Use integration by parts for the integral of $x^2$, with 

$u = x^2, du = 2xdx, v = \frac{-cos(2\pi nx)}{2\pi n}, dv = sin(2\pi nx)dx$

$$\implies 4[\frac{-x^2cos(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} + \frac{1}{\pi n} \int_{0}^{\frac{1}{2}} xcos(2\pi nx)\,dx ]$$

$$\implies 4[-\frac{(-1)^n}{8\pi n}] = -\frac{(-1)^n}{2\pi n}$$

$$\implies b_n = -\frac{(-1)^n}{\pi n} + \frac{(-1)^n}{2\pi n} = -\frac{(-1)^n}{2\pi n}$$

Therefore, the full Fourier Series for $f(x) = x - x^2$ over $[0,\frac{1}{2}]$ is:

$$\frac{1}{6} + \sum_{n=1}^{\infty} \frac{(-1)}{\pi^2n^2}cos(2\pi nx) - \frac{(-1)^n}{2\pi n}sin(2\pi nx)$$

This differs from the Fourier series we saw in class as it contains both the Fourier Sine series and the Fourier Cosine series.

My matrix was : [

[0,1,2,3,1],

[2,0,4,2,4],

[1,2,0,4,1],

[4,3,2,0,3],

[3,4,3,4,0] ]

Team 5: rating = 0.5552749621639953
Team 2: rating = 0.493080305685833
Team 4: rating = 0.4796470095274982
Team 3: rating = 0.3560934619397378
Team 1: rating = 0.3027832907437918

Team 5 was the best.

For my unique conditions, $\kappa = 0$ and $f = 0$. The three sides on the lower left, sticking out of the bar, are set to temperature zero. The side at the very top of the rectangle on the upper right is set to temperature 1. The remaining sides are insulated. At the steady state for the heat distribution, the temperature for the lower right corner of the bar is $u = 0.80488$.