### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Calvin Arnott

Pages: 1 [2] 3
16
##### Home Assignment 6 / Re: Bonus Web Problem--4
« on: November 10, 2012, 02:26:11 PM »
Does fractional integration make sense in the language of differential forms?

17
##### Home Assignment 6 / Re: Problem 2
« on: November 07, 2012, 11:54:42 PM »
Calvin, if you type your solutions in LaTeX or TeX, you could post just the source fixing some discrepancies on the fly

I changed the .pdfs to png's for now, when I have some time later I'll try to massage the tex code into mathjax/forum format.

18
##### Home Assignment 6 / Re: Bonus Web Problem--4
« on: November 07, 2012, 11:33:30 PM »
[\thicksim \text{For}  f   \text{which decays fast as}   x \rightarrow \infty: \forall n \in \mathbb{N}: (\frac{x^{n-1}_+}{( n-1)!} \ast f(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}f(x_n)dx_n\cdots dx_1 \thicksim]
[\text{We proceed by induction on  } n.  \text{ Let:  }f  \text{be a nice function on } \mathbb{R}   \text{which decays fast as}   x \rightarrow -\infty.  \text{Base case where   }n = 1:]
[(\frac{x^{n-1}_+}{( n-1)!} \ast f(x)) = (\frac{x^{0}_+}{(0)!} \ast f(x)) = \int_{-\infty}^{\infty}(x-x_1)^{0}_+ f(x_1)dx_1 = \int_{-\infty}^{x}(x-x_1)^{0}_+ f(x_1)dx_1 + \int_{x}^{\infty}(x-x_1)^{0}_+ f(x_1)dx_1]
[= \int_{-\infty}^{x}1 f(x_1)dx_1 + \int_{x}^{\infty}(0) f(x_1)dx_1 = \int_{-\infty}^{x} f(x_1)dx_1,    \text{as for:}    x_1 > x,   (x-x_1)_+ = 0]
[\text{So for   } n = 1: (\frac{x^{n-1}_+}{( n-1)!} \ast f(x)) = (\frac{x^{0}_+}{(0)!} \ast f(x)) = \int_{-\infty}^{x} f(x_1)dx_1 = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}f(x_n)dx_n\cdots dx_1]
[\text{This integrals converges because } f(x) \text{ decays fast and vanishes as } x \rightarrow \infty]
[\text{Now suppose for some}    n\in\mathbb{N}: (\frac{x^{n-1}_+}{( n-1)!} \ast f(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}f(x_n)dx_n\cdots dx_1.    \text{Then for}    n + 1 \in \mathbb{N} \text{ we show:}]
[(\frac{x^{(n+1)-1}_+}{( (n+1)-1)!} \ast f(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-2}} \int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}f(x_n)dx_n\cdots dx_2 dx_1]
[\text{Recall the previous property that: } (f' \ast g) = (f \ast g)' = (f \ast g') ]
[\text{Let: } h(x) = \frac{x^{n}_+}{(n)!},    h'(x) = \frac{x^{n-1}_+}{(n-1)!}]
[\text{And let: } g(x) = \int_{-\infty}^{x}f(y)dy,    g'(x) = f(x)]
[\text{Then we have: } (h \ast g')(x) = (h' \ast g)(x) ]
[ \text{Because } f(x) \text{ is nice and decays fast, we have that } g(x) \text{is nice and decays fast as } x \rightarrow -\infty]
[\text{... and we can use our induction hypothesis on } g(x) \text{, namely:}]
[(h' \ast g)(x) = (\frac{x^{n-1}_+}{( n-1)!} \ast g(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}(\int_{-\infty}^{x_{n}}f(y)dy)dx_{n}\cdots dx_1]
[= \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}} \int_{-\infty}^{x_{n}}}_{n+1\text{ integrals}}f(x_{n+1})dx_{n+1}\cdots dx_2 dx_1 = (h \ast g')(x) = (\frac{x^{(n+1)-1}_+}{( (n+1)-1)!} \ast f(x))]
[\text{And for } (n+1) \text{ we have } (\frac{x^{(n+1)-1}_+}{( (n+1)-1)!} \ast f(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}} \int_{-\infty}^{x_{n}}}_{n+1\text{ integrals}}f(x_{n+1})dx_{n+1}\cdots dx_2 dx_1]
[\text{By induction then: } \forall n \in \mathbb{N}: (\frac{x^{n-1}_+}{( n-1)!} \ast f(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}f(x_n)dx_n\cdots dx_1 \blacksquare]

19
##### Home Assignment 6 / Re: Problem 2
« on: November 07, 2012, 09:36:34 PM »
Part 2

20
##### Home Assignment 6 / Re: Problem 2
« on: November 07, 2012, 09:36:22 PM »
Problem 2

Let: $\alpha > 0, \beta > 0, n\in\mathbb{N}$. Compute the Fourier transform for:

a. $\left(x^2 + \alpha^2\right)^{-1}$

Let $f\left(z\right)$ be holomorphic on a domain $D$. Cauchy's formula states that for any positively oriented piecewise smooth simple closed curve $\gamma \in D$ whose inside $\Omega$ lies in $D$:

$$\forall z \in \Omega: f\left(z\right) = \frac{1}{2 \pi i}\int_\gamma \frac{f\left(\zeta\right)}{\left(\zeta - z\right)} d\zeta$$

$$\text{Claim: } F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{\left(x - i\alpha\right)\left(x + i\alpha\right)}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha }$$

Proof: Let: $\gamma_r^+$ be the positively oriented semi-circle on the upper half-plane with radius $r$, and $\gamma_r^-$ be the positively oriented semi-circle on the lower half-plane with radius $r$. Then both $\{\gamma_r^+,\gamma_r^-\}$ are positively oriented piecewise smooth simple closed curves. Moreover, $f^+\left(z\right)= \frac{e^{-i k z}}{\left(z + i\alpha\right)}$ and $f^-\left(z\right) = \frac{e^{-i k z}}{\left(z - i\alpha\right)}$ are holomorphic on $\gamma_r^+$ and $\gamma_r^-$ respectively, as $e^{z}$ is entire and we avoid the poles of the rational functions on their respective domains. We take cases and apply Cauchy's formula:

$$\text{Suppose: } k < 0. \text{ Then: } \int_{\gamma_r^+}\frac{\frac{e^{-i k x}}{\left(x + i\alpha\right)}}{\left(x - i\alpha\right)}dx = 2 \pi i f^+\left(i\alpha\right) = 2\pi i \frac{e^{-i k \left(i \alpha\right)}}{\left(\left(i \alpha\right) + i\alpha\right)} = \frac{ \pi e^{\alpha k} }{\alpha }$$

$$\text{Suppose: } k > 0. \text{ Then: } \int_{\gamma_r^-}\frac{\frac{e^{-i k x}}{\left(x - i\alpha\right)}}{\left(x + i\alpha\right)}dx = 2 \pi i f^-\left(i\alpha\right) = 2\pi i \frac{e^{-i k \left(i \alpha\right)}}{\left(\left(-i \alpha\right) - i\alpha\right)} = -\frac{ \pi e^{- \alpha k} }{\alpha }$$

$$\text{Now, } \int_{\gamma_r}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx \text{ is the sum of two integrals:}$$

$$\int_{\gamma_r^+}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = \int_{-r}^{r}\frac{e^{-i k x}}{x^2 + \alpha^2}dx + \int_{C_r^+}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = \frac{ \pi e^{\alpha k} }{\alpha }$$

$$\int_{\gamma_r^-}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = \int_{r}^{-r}\frac{e^{-i k x}}{x^2 + \alpha^2}dx + \int_{C_r^-}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = -\frac{ \pi e^{- \alpha k} }{\alpha }$$

Where $C_r$ is the upper or lower half of the half circle. We show that the $C_r$ portion of the integral vanishes as $r \rightarrow \infty$. Without loss of generality we set $\alpha$ to $1$ to simplify the inequalities:

$$\mid \int_{C_r^+}\frac{e^{-i k x}}{x^2 + 1}dx \mid \le \int_{C_r^+} \mid \frac{e^{-i k x}}{x^2 + 1} \mid \mid dx \mid \le \int_{C_r^+}\frac{1}{r^2 - 1} \mid dx \mid = \frac{\pi r}{r^2 - 1} \rightarrow 0 \text{ as } r \rightarrow \infty$$

$$\text{Then: } \lim_{r \to +\infty} \int_{\gamma_r^+}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx \rightarrow \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx + 0 = \frac{ \pi e^{\alpha k} }{\alpha }$$

$$\implies \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{\alpha k} }{\alpha } \text{ when } k < 0$$

And similarly for our $\gamma_r^-$ integral, changing the bounds of integration and the sign. We have then: $\frac{ \pi e^{\alpha k} }{\alpha }$ for $k < 0$ and $\frac{ \pi e^{-\alpha k} }{\alpha }$ for $k > 0$. Thus:

$$F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha } \blacksquare$$

b. $x\left(x^2 + \alpha^2\right)^{-1}$

We have for for any function $f\left(x\right)$ with Fourier transform $F\left(k\right),$ the transform of $g\left(x\right) = x f\left(x\right)$  is given by: $G\left(k\right) = i\frac{dF}{dk}$. Then:
$$g\left(x\right) = x\left(x^2 + \alpha^2\right)^{-1} = x f\left(x\right) \implies G\left(k\right) = i\frac{dF}{dk} = i \partial_k \left(\frac{ \pi e^{-\alpha |k|} }{\alpha }\right) = - i \pi sgn\left(k\right) e^{-\alpha |k|} \blacksquare$$

Where $f\left(x\right) = \left(x^2 + \alpha^2\right)^{-1}$ and $F\left(k\right) = \frac{ \pi e^{-\alpha |k|} }{\alpha }$ as found in part a.

c. i) $\left(x^2 + \alpha^2\right)^{-1} \cos\left(\beta x\right)$

Two properties of the Fourier transform are that for: $g\left(x\right) = e^{i a x} f\left(x\right)$ the Fourier transform of $g\left(x\right)$ is given by $G\left(k\right) = F\left(k-a\right)$, and that the transform is linear: $h\left(x\right) = a f\left(x\right) + b g\left(x\right) \implies H\left(k\right) = a F\left(k\right) + b G\left(k\right)$

$$\text{Now, } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:}$$

$$\left(x^2 + \alpha^2\right)^{-1}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} \left(x^2 + \alpha^2\right)^{-1} + e^{- i \beta x} \left(x^2 + \alpha^2\right)^{-1}\right)$$

$$\text{Then our transform is given by: } G\left(k\right) = \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: }$$

$$F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha } \text{ is our transform from part a.}$$

$$\implies G\left(k\right) = \frac{1}{2} \left( \frac{ \pi e^{-\alpha |k - \beta|} }{\alpha } + \frac{ \pi e^{-\alpha |k + \beta|} }{\alpha }\right) \blacksquare$$

ii) $\left(x^2 + \alpha^2\right)^{-1} \sin\left(\beta x\right)$

$$\text{Proceeding as in part c. i): } \sin{\beta x} = \frac{1}{2i}\left(e^{i \beta x} - e^{- i \beta x}\right) \text{ so we write:}$$

$$\left(x^2 + \alpha^2\right)^{-1}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} \left(x^2 + \alpha^2\right)^{-1} - e^{- i \beta x} \left(x^2 + \alpha^2\right)^{-1}\right)$$

$$\text{Then our transform is given by: } G\left(k\right) = \frac{1}{2i}\left(F\left(k-\beta\right) - F\left(k+\beta\right)\right) \text{ where: }$$

$$F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha } \text{ is our transform from part a.}$$

$$\implies G\left(k\right) = \frac{1}{2i} \left( \frac{ \pi e^{-\alpha |k - \beta|} }{\alpha } - \frac{ \pi e^{-\alpha |k + \beta|} }{\alpha }\right) \blacksquare$$

d. i) $x \left(x^2 + \alpha^2\right)^{-1} \cos\left(\beta x\right)$

$$\text{Proceeding as in part c. i): } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:}$$

$$x \left(x^2 + \alpha^2\right)^{-1}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} x \left(x^2 + \alpha^2\right)^{-1} + e^{- i \beta x} x \left(x^2 + \alpha^2\right)^{-1}\right)$$

$$\text{Then our transform is given by: } G\left(k\right) = \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: }$$

$$F\left(k\right) = \int_{-\infty}^{\infty}\frac{x e^{-i k x}}{x^2 + \alpha^2}dx = - i \pi sgn\left(k\right) e^{-\alpha |k|} \text{ is our transform from part b.}$$

$$\implies G\left(k\right) = \frac{1}{2} \left( - i \pi sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} - i \pi sgn\left(k + \beta\right) e^{-\alpha |k + \beta|}\right)$$

$$= - \frac{i \pi}{2} \left(sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} + sgn\left(k + \beta\right) e^{-\alpha |k + \beta|}\right) \blacksquare$$

ii) $x \left(x^2 + \alpha^2\right)^{-1} \sin\left(\beta x\right)$

$$\text{Proceeding as in part c. ii): } \sin{\beta x} = \frac{1}{2i}\left(e^{i \beta x} - e^{- i \beta x}\right) \text{ so we write:}$$

$$x \left(x^2 + \alpha^2\right)^{-1}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} x \left(x^2 + \alpha^2\right)^{-1} - e^{- i \beta x} x \left(x^2 + \alpha^2\right)^{-1}\right)$$

$$\text{Then our transform is given by: } G\left(k\right) = \frac{1}{2i}\left(F\left(k-\beta\right) - F\left(k+\beta\right)\right) \text{ where: }$$

$$F\left(k\right) = \int_{-\infty}^{\infty}\frac{x e^{-i k x}}{x^2 + \alpha^2}dx = - i \pi sgn\left(k\right) e^{-\alpha |k|} \text{ is our transform from part b.}$$

$$\implies G\left(k\right) = \frac{1}{2i} \left( - i \pi sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} {\alpha } + i \pi sgn\left(k + \beta\right) e^{-\alpha |k + \beta|} {\alpha }\right)$$

$$= - \frac{\pi}{2} \left(sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} - sgn\left(k + \beta\right) e^{-\alpha |k + \beta|}\right) \blacksquare$$

21
##### Home Assignment 6 / Re: Problem3
« on: November 07, 2012, 09:34:57 PM »
Problem 3

Let: $\alpha > 0, \beta > 0$. Using the Fourier transform of $e^{-\frac{\alpha x^2}{2}}$ compute the Fourier transform for:

a. i) $e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right)$

We have that the Fourier transform for $f\left(x\right) = e^{-\frac{x^2}{2}}$ is given by $F\left(k\right) = \sqrt{2 \pi} e^{-\frac{k^2}{2}}$, and that a property of the Fourier transform is $f\left(a x\right) \rightarrow \frac{1}{|a|}F\left(\frac{k}{a}\right)$. So for $f\left(x\right) = e^{-\frac{x^2}{2}}, f\left(\sqrt{\alpha} x\right) = e^{-\frac{\alpha x^2}{2}} = g\left(x\right) \implies G\left(k\right) = \frac{1}{\sqrt{\alpha}}F\left(\frac{k}{\sqrt{\alpha}}\right) = \sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha k^2}{2}}$ is the transform for $e^{-\frac{\alpha x^2}{2}}$, using that $\alpha > 0 \implies | \sqrt{\alpha} | = \sqrt{\alpha}$.

$$\text{ Again; } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:}$$

$$e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} e^{-\frac{\alpha x^2}{2}} + e^{- i \beta x} e^{-\frac{\alpha x^2}{2}}\right)$$

Using that for: $g\left(x\right) = e^{i a x} f\left(x\right)$ the Fourier transform of $g\left(x\right)$ is given by $G\left(k\right) = F\left(k-a\right)$, and that the transform is linear: $h\left(x\right) = a f\left(x\right) + b g\left(x\right) \implies H\left(k\right) = a F\left(k\right) + b G\left(k\right)$ our Fourier transform is then given by $\frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right)$ where $F\left(k\right) = \sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha k^2}{2}}$

$$\implies \frac{1}{2}\left(\sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} + \sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)$$

$$= \sqrt{\frac{\pi}{2\alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} + e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right) \text{ is our transform } \blacksquare$$

ii) $e^{-\frac{\alpha x^2}{2}}\sin\left(\beta x\right)$

As in part i) $\cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right)$ so we write:

$$e^{-\frac{\alpha x^2}{2}}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} e^{-\frac{\alpha x^2}{2}} - e^{- i \beta x} e^{-\frac{\alpha x^2}{2}}\right)$$

$$\text{and our transform is given by: } \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: } F\left(k\right) = \sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha k^2}{2}}$$

$$\implies \frac{1}{2}\left(\sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} - \sqrt{\frac{2 \pi}{\alpha}} e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)$$

$$= \sqrt{\frac{\pi}{2 i\alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} - e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right) \text{ is our transform } \blacksquare$$

b. i) $x e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right)$

We have for for any function $f\left(x\right)$ with Fourier transform $F\left(k\right),$ the transform of $g\left(x\right) = x f\left(x\right)$  is given by: $G\left(k\right) = i\frac{dF}{dk}$. Then for $f\left(x\right) = e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right)$, $F\left(k\right) = \sqrt{\frac{\pi}{2\alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} + e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)$, $g\left(x\right) = x e^{-\frac{\alpha x^2}{2}}\cos\left(\beta x\right) = x f\left(x\right)$, $G\left(k\right) = i\frac{dF}{dk} = i \partial_k \left(\sqrt{\frac{\pi}{2\alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} + e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)\right)$

$$\implies G\left(k\right) = i \left(\left(k-\beta\right)\sqrt{\frac{\pi}{2\alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} + \left(k+\beta\right)e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)\right) \text{ is our transform} \blacksquare$$

ii) $x e^{-\frac{\alpha x^2}{2}}\sin\left(\beta x\right)$

Again, for for any function $f\left(x\right)$ with Fourier transform $F\left(k\right),$ the transform of $g\left(x\right) = x f\left(x\right)$  is given by: $G\left(k\right) = i\frac{dF}{dk}$. Then for $f\left(x\right) = e^{-\frac{\alpha x^2}{2}}\sin\left(\beta x\right)$, $F\left(k\right) = \sqrt{\frac{\pi}{2 i \alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} - e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)$, $g\left(x\right) = x e^{-\frac{\alpha x^2}{2}}\sin\left(\beta x\right) = x f\left(x\right)$, $G\left(k\right) = i\frac{dF}{dk} = i \partial_k \left(\sqrt{\frac{\pi}{2 i \alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} - e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)\right)$

$$\implies G\left(k\right) = i \left(\left(k-\beta\right)\sqrt{\frac{\pi}{2 i \alpha}}\left(e^{-\frac{\alpha \left(k-\beta\right)^2}{2}} - \left(k+\beta\right)e^{-\frac{\alpha \left(k+\beta\right)^2}{2}}\right)\right) \text{ is our transform} \blacksquare$$

22
##### Home Assignment 6 / Re: Problem 4
« on: November 07, 2012, 09:34:32 PM »
Problem 4
Compute the Fourier transform:

Problem a. f(x) = \left\{\begin{aligned} &1 &&: |x| \le a\\ &0 &&: |x| > a \end{aligned} \right.

$$F\left(k\right) = \int_{-\infty}^{\infty}f\left(x\right) e^{-i k x}dx = \int_{-a}^{a} e^{-i k x}dx \text{ when } a > 0 \text{ and } 0 \text{ otherwise.}$$

$$= -\frac{e^{-i k x}}{i k}\Bigr|_{x=-a}^{a} = \frac{1}{i k}\left(e^{i k a} - e^{-i k a}\right) = \frac{2}{k}\sin\left(a k\right) \blacksquare$$

Problem b.  f(x) = \left\{\begin{aligned} &x &&: |x| \le 1\\ &0 &&: |x| > 1 \end{aligned} \right.

$g\left(x\right) = x f\left(x\right)$ where $f\left(x\right)$ is defined as in part a. Then we have the Fourier transform of $g\left(x\right)$, is given by $G\left(k\right) = i\frac{dF}{dk}$ where $F\left(k\right) = \frac{2}{k}\sin\left(a k\right)$ as derived in part a. Then $G\left(k\right) = i \partial_k \left( \frac{2}{k}\sin\left(a k\right)\right) = \frac{2 i}{k^2} \left(a k \cos \left(a k\right) - \sin \left(a k\right)\right)$ is our transform for $g\left(x\right) \blacksquare$

Problem c. Compute $\int_{-\infty}^{\infty} \frac{\sin\left(x\right)}{x} d x$

Let $f\left(x\right)$ be defined as in part a. with $a = 1$

f(x) = \left\{\begin{aligned} &1 &&: |x| \le 1\\ &0 &&: |x| > 1 \end{aligned} \right.

The Fourier transform of $f\left(x\right)$ is given by $F\left(k\right) = \int_{-\infty}^{\infty}f\left(x\right) e^{-i k x}dx = \frac{2 sin\left( k\right)}{k}\$. Then the inverse Fourier transform is given by: $f\left(x\right) = \int_{-\infty}^{\infty}\frac{2 \sin\left(k\right)}{k} e^{i k x} \frac{dk}{2\pi} \implies \pi f\left(x\right) = \int_{-\infty}^{\infty}\frac{\sin\left(k\right)}{k} e^{i k x} dk$. We use the continuity of $f\left(x\right)$ at $x = 0$ and continuity of the exponential inside of the transform to take the limit as $x \rightarrow 0$.

$$\lim_{x \to 0+} \pi f\left(x\right) \rightarrow \pi \cdot 1 = \lim_{x \to 0+} \int_{-\infty}^{\infty}\frac{\sin\left(k\right)}{k} e^{i k x} dk \rightarrow \int_{-\infty}^{\infty}\frac{\sin\left(k\right)}{k} dk$$

$$\implies \pi = \int_{-\infty}^{\infty}\frac{\sin\left(k\right)}{k} dk \blacksquare$$

23
##### Home Assignment 6 / Re: Problem 1--simplified
« on: November 07, 2012, 09:34:13 PM »
Part 2 of solution.

*edited to png

24
##### Home Assignment 6 / Re: Problem 1--simplified
« on: November 07, 2012, 09:33:54 PM »
I ended up solving with $x^n$ in this problem. Notice that the messy integration in part b. could have been avoided by using properties of the Fourier transform as is done later in the problem.

I take the definition of the Fourier transform $\hat{f}$ for a function $f$ to be:  $F\left(k\right) = \hat{f}\left(k\right) = \int_{-\infty}^{\infty}f\left(x\right) e^{-i k x}dx,$ with inverse Fourier transform $f\left(x\right) = \check{F}\left(x\right) = \int_{-\infty}^{\infty}F\left(k\right) e^{i k x}\frac{dk}{2 \pi}$.

Problem

Let: $\alpha > 0, \beta > 0, n\in\mathbb{N}$. Compute the Fourier transform for:

a. $e^{-\alpha |x|}$

$$F\left(k\right) = \int_{-\infty}^{\infty}e^{-\alpha |x|} e^{-i k x}dx = \int_{0}^{\infty}e^{-\alpha \left(x\right)} e^{-i k x}dx + \int_{-\infty}^{0}e^{-\alpha \left(-x\right)} e^{-i k x}dx$$

$$= \int_{0}^{\infty}e^{-\left(\alpha + i k\right)x}dx + \int_{-\infty}^{0}e^{\left(\alpha - i k \right)x}dx = \frac{1}{-\left(\alpha + i k \right)}e^{-\left(\alpha + i k\right)x}\Bigr|_{x=0}^{\infty} + \frac{1}{\left(\alpha- i k\right)}e^{\left(\alpha - i k\right)x}\Bigr|_{x=-\infty}^{0}$$

$$= \lim_{x \to +\infty} \frac{1}{-\left(\alpha + i k \right)}e^{-\left(\alpha + i k\right)x} - \frac{1}{-\left(\alpha + i k \right)}e^{-\left(\alpha + i k\right)\left(0\right)} + \frac{1}{\left(\alpha- i k\right)}e^{\left(\alpha - i k\right)\left(0\right)} - \lim_{x \to -\infty} \frac{1}{\left(\alpha- i k\right)}e^{\left(\alpha - i k\right)x}$$

$$= 0 + \frac{1}{\left(\alpha + i k \right)} + \frac{1}{\left(\alpha- i k\right)} - 0 = \frac{\left(\alpha- i k\right)+\left(\alpha + i k \right)}{\left(\alpha + i k \right)\left(\alpha- i k\right)} = \frac{2\alpha}{\alpha ^2 - k^2}$$

$$\implies \frac{2\alpha}{\alpha ^2 - k^2} = F\left(k\right) \text{ is our Fourier transform }\blacksquare$$

b. i) $e^{-\alpha |x|}\cos\left(\beta x\right)$

$$F\left(k\right) = \int_{-\infty}^{\infty}e^{-\alpha |x|} \cos\left(\beta x\right) e^{-i k x}dx = \int_{0}^{\infty}e^{-\alpha \left(x\right)} e^{-i k x} \cos\left(\beta x\right)dx + \int_{-\infty}^{0}e^{-\alpha \left(-x\right)} e^{-i k x} \cos\left(\beta x\right)dx$$

$$\text{Now, } \forall\{a,b\}\in\mathbb{C}: \int e^{a x} \cos\left(b x\right) dx = \frac{e^{a x}}{a^2 + b^2}\left(b \sin\left(b x\right) + a \cos\left(b x\right)\right) \text{ so out integral is:}$$

$$\int_{0}^{\infty}e^{-\left(\alpha + i k\right) x} \cos\left(\beta x\right)dx + \int_{-\infty}^{0}e^{\left(\alpha -i k\right) x} \cos\left(\beta x\right)dx$$

$$= \frac{e^{\left(-\left(\alpha + i k\right)\right) x}}{\left(-\left(\alpha + i k\right)\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) + \left(-\left(\alpha + i k\right)\right) \cos\left(\beta x\right)\right)\Bigr|_{x=0}^{\infty} + \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) + \left(\alpha -i k\right) \cos\left(\beta x\right)\right)\Bigr|_{x=-\infty}^{0}$$

$$= \lim_{x \to +\infty} \frac{e^{-\left(\alpha + i k\right) x}}{\left(\alpha + i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) - \left(\alpha + i k\right) \cos\left(\beta x\right)\right) - \frac{e^{-\left(\alpha + i k\right) \left(0\right)}}{\left(\alpha + i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta \left(0\right)\right) -\left(\alpha + i k\right) \cos\left(\beta \left(0\right)\right)\right)$$

$$+ \frac{e^{\left(\alpha -i k\right) \left(0\right)}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta \left(0\right)\right) + \left(\alpha -i k\right) \cos\left(\beta \left(0\right)\right)\right) - \lim_{x \to -\infty} \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) + \left(\alpha -i k\right) \cos\left(\beta x\right)\right)$$

$$= 0 - \frac{-\left(\alpha + i k\right)}{\left(\alpha + i k\right)^2 + \beta^2} + \frac{\left(\alpha -i k\right)}{\left(\alpha -i k\right)^2 + \beta^2} - 0$$

$$\implies \frac{\left(\alpha + i k\right)}{\left(\alpha + i k\right)^2 + \beta^2} + \frac{\left(\alpha -i k\right)}{\left(\alpha -i k\right)^2 + \beta^2} = F\left(k\right) \text{ is our Fourier transform } \blacksquare$$

b. ii) $e^{-\alpha |x|}\sin\left(\beta x\right)$

$$F\left(k\right) = \int_{-\infty}^{\infty}e^{-\alpha |x|} \sin\left(\beta x\right) e^{-i k x}dx = \int_{0}^{\infty}e^{-\alpha \left(x\right)} e^{-i k x} \sin\left(\beta x\right)dx + \int_{-\infty}^{0}e^{-\alpha \left(-x\right)} e^{-i k x} \sin\left(\beta x\right)dx$$

$$\text{Now, } \forall\{a,b\}\in\mathbb{C}: \int e^{a x} \sin\left(b x\right) dx = \frac{e^{a x}}{a^2 + b^2}\left(a \sin\left(b x\right) - b \cos\left(b x\right)\right) \text{ so out integral is:}$$

$$\int_{0}^{\infty}e^{-\left(\alpha + i k\right) x} \sin\left(\beta x\right)dx + \int_{-\infty}^{0}e^{\left(\alpha -i k\right) x} \sin\left(\beta x\right)dx$$

$$= \frac{e^{\left(-\left(\alpha + i k\right)\right) x}}{\left(-\left(\alpha + i k\right)\right)^2 + \beta^2}\left(\left(-\left(\alpha + i k\right)\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right)\Bigr|_{x=0}^{\infty}$$
$$+ \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\left(\alpha -i k\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right)\Bigr|_{x=-\infty}^{0}$$

$$= \lim_{x \to +\infty} \frac{e^{-\left(\alpha + i k\right) x}}{\left(\alpha + i k\right)^2 + \beta^2}\left(-\left(\alpha + i k\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right) - \frac{e^{-\left(\alpha + i k\right) \left(0\right)}}{\left(\alpha + i k\right)^2 + \beta^2}\left(-\left(\alpha + i k\right) \sin\left(\beta \left(0\right)\right) - \beta \cos\left(\beta \left(0\right)\right)\right)$$

$$+ \frac{e^{\left(\alpha -i k\right) \left(0\right)}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\left(\alpha -i k\right) \sin\left(\beta \left(0\right)\right) - \beta \cos\left(\beta \left(0\right)\right)\right) - \lim_{x \to -\infty} \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\left(\alpha -i k\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right)$$

$$= 0 - \frac{-\beta}{\left(\alpha + i k\right)^2 + \beta^2} + \frac{- \beta}{\left(\alpha -i k\right)^2 + \beta^2} - 0$$

$$\implies \frac{\beta}{\left(\alpha + i k\right)^2 + \beta^2} - \frac{\beta}{\left(\alpha -i k\right)^2 + \beta^2} = F\left(k\right) \text{ is our Fourier transform } \blacksquare$$

c. $x^n e^{-\alpha |x|}$

We have for for any function $f\left(x\right)$ with Fourier transform $F\left(k\right),$ the transform of $g\left(x\right) = x f\left(x\right)$  is given by: $G\left(k\right) = i\frac{dF}{dk}$.  Moreover, it's clear that for: $g\left(x\right) = x^n f\left(x\right)$, $G\left(k\right) = i^n\frac{d^n F}{d k^n}$.

Proof:} We proceed by induction. Our base case is obvious: for $n = 1$, $g\left(x\right) = x^n f\left(x\right) = x f\left(x\right) \implies G\left(k\right) = i\frac{dF}{dk} = i^n\frac{d^n F}{d k^n}$. Suppose we have for some $n \in \mathbb{N}$, $h\left(x\right) = x^n f\left(x\right) \implies H\left(k\right) = i^n\frac{d^n F}{d k^n}.$ Then, $g\left(x\right) = x^{n+1} f\left(x\right) = x h\left(x\right) \implies G\left(k\right) = i \cdot i^n\partial_k H\left(k\right) = i^{n+1} \partial_k \frac{d^n F}{d k^n} = i^{n+1}\frac{d^{n+1} F}{d k^{n+1}}$ and $\forall n \in \mathbb{N}: g\left(x\right) = x^n f\left(x\right) \implies i^n\frac{d^n F}{d k^n}$, as needed $\square$

Now, we found in part a. that the Fourier transform for $f\left(x\right) = e^{-\alpha |x|}$ is given by: $F\left(k\right) = \frac{2\alpha}{\alpha ^2 - k^2}$. Examining the derivatives of $F\left(k\right)$:

$$F\left(k\right) = \frac {2\alpha} {\alpha^2-k^2}, F'\left(k\right) = \frac{4 k \alpha }{\left(\alpha ^2-k^2\right)^2}, F''\left(k\right) = 2 \left(\frac{-1}{\left(\alpha-k \right)^3}+\frac{1}{\left(\alpha+k \right)^3}\right)$$

$$F'''\left(k\right) = 6 \left( \frac{1}{\left(\alpha-k\right)^4}-\frac{1}{\left(\alpha+k \right)^4}\right), F^{\left(4\right)}\left(k\right) = 24 \left(\frac{-1}{\left(\alpha-k \right)^5}+\frac{1}{ \left(\alpha+k \right)^5}\right)$$

$$\text{So for: } n \in \mathbb{N}: F^n \left(k\right) = n! \left(\frac{\left(-1\right)^{n+1}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^n}{\left(\alpha+k\right)^{n+1}} \right) \text{, and:}$$

$$g\left(x\right) = x^n f\left(x\right) = x^n e^{-\alpha |x|} \implies G\left(k\right) = i^n\frac{d^n F}{d k^n} = i^n n! \left(\frac{\left(-1\right)^{n}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+k\right)^{n+1}} \right) \text{ is our transform } \blacksquare$$

d. i) $x^n e^{-\alpha |x|}\cos\left(\beta x\right)$

Two properties of the Fourier transform are that for: $g\left(x\right) = e^{i a x} f\left(x\right)$ the Fourier transform of $g\left(x\right)$ is given by $G\left(k\right) = F\left(k-a\right)$, and that the transform is linear: $h\left(x\right) = a f\left(x\right) + b g\left(x\right) \implies H\left(k\right) = a F\left(k\right) + b G\left(k\right)$

$$\text{Now, } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:}$$

$$x^n e^{-\alpha |x|}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} x^n e^{-\alpha |x|} + e^{- i \beta x} x^n e^{-\alpha |x|}\right)$$

$$\text{Then our transform is given by: } G\left(k\right) = \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: }$$

$$F\left(k\right) = i^n n! \left(\frac{\left(-1\right)^{n}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+k\right)^{n+1}} \right) \text{ is our transform from part c. for } x^n e^{-\alpha |x|}$$

$$\implies G\left(k\right) = \frac{1}{2} i^n n! \left( \frac{\left(-1\right)^{n}}{\left(\alpha+ \beta - k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha -\beta + k\right)^{n+1}} + \frac{\left(-1\right)^{n}}{\left(\alpha -\beta -k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+\beta +k\right)^{n+1}} \right) \blacksquare$$

ii) $x^n e^{-\alpha |x|}\sin\left(\beta x\right)$

Proceeding as in part i) we use that: $\sin{\beta x} = \frac{1}{2 i}\left(e^{i \beta x} - e^{- i \beta x}\right)$ to write:

$$g\left(x\right) = x^n e^{-\alpha |x|}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} x^n e^{-\alpha |x|} - e^{- i \beta x} x^n e^{-\alpha |x|}\right)$$

$$\text{Which yields the transform: } G\left(k\right) = \frac{1}{2i}\left(F\left(k-\beta\right) - F\left(k+\beta\right)\right) \text{ with: } F\left(k\right) = i^n n! \left(\frac{\left(-1\right)^{n}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+k\right)^{n+1}} \right) \text{ and so:}$$

$$G\left(k\right) = \frac{1}{2 i} i^n n! \left( \frac{\left(-1\right)^{n}}{\left(\alpha+ \beta - k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha -\beta + k\right)^{n+1}} - \frac{\left(-1\right)^{n}}{\left(\alpha -\beta -k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+\beta +k\right)^{n+1}} \right) \blacksquare$$

25
##### Home Assignment 6 / Re: Bonus Web Problem--4
« on: November 07, 2012, 09:31:19 PM »
[\thicksim \left(f \ast g\right) = \left(g \ast f\right) \thicksim]
[\text{Let:  }t=\left(x-y\right) \implies y=\left(x-t\right),    dy=\left(-dt\right).   \text{Then:}]
[\left(f \ast g\right)\left(x\right) = \int_{-\infty}^{\infty}f\left(x-y\right)g\left(y\right)dy = \int_{\infty}^{-\infty}f\left(t\right)g\left(x-t\right)\left(-dt\right) = -\int_{\infty}^{-\infty}g\left(x-t\right)f\left(t\right)dt = \int_{-\infty}^{\infty}g\left(x-t\right)f\left(t\right)dt = \left(g \ast f\right)\left(x\right)    \blacksquare]

[\thicksim \left(f' \ast g\right) = \left(f \ast g\right)' = \left(f \ast g'\right) \thicksim]
[\text{We have:    }\left(f \ast g\right) = \left(g \ast f\right),   \text{so it suffices to show:    }\left(f \ast g\right)' = \left(f' \ast g\right)    \text{as:    }\left(f \ast g\right)' = \left(g \ast f\right)' = \left(g' \ast f\right)   \text{since both    }\{f,g\}   \text{arbitrary.}]
[\text{Let:  }\{f,g\}\left(x\right)   \text{be nice enough functions that we can differentiate them under the sign. Then:}]
[\left(f \ast g\right)'\left(x\right) = \partial_x\int_{-\infty}^{\infty}f\left(x-y\right)g\left(y\right)dy = \int_{-\infty}^{\infty}\partial_x \left(f\left(x-y\right)g\left(y\right)\right)dy = \int_{-\infty}^{\infty}f'\left(x-y\right)g\left(y\right)dy = \left(f' \ast g\right)\left(x\right)    \blacksquare]

[\thicksim \left(f\ast\left(g \ast h\right)\right) = \left(\left(f \ast g\right)\ast h\right) \thicksim]
[\text{Let:  }\{f,g,h\}\left(x\right)   \text{be nice enough functions that the order of integration is arbitrary for any combination of them.}]
[\left(f\ast\left(g \ast h\right)\right)\left(x\right) = \left(f\left(x\right)\ast\int_{-\infty}^{\infty}g\left(x-s\right)h\left(s\right)ds\right) = \int_{-\infty}^{\infty}f\left(x-t\right)\int_{-\infty}^{\infty}g\left(t-s\right)h\left(s\right)dsdt = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x-t\right)g\left(t-s\right)h\left(s\right)dsdt]
[\text{Let:  }u=\left(t-s\right) \implies t=\left(u+s\right),    dt=du.    \text{Then:  }]
[\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x-\left(u+s\right)\right)g\left(u\right)h\left(s\right)dsdu = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(\left(x-s\right)-u\right)g\left(u\right)h\left(s\right)duds = \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f\left(\left(x-s\right)-u\right)g\left(u\right)du\right)h\left(s\right)ds]
[= \left(\int_{-\infty}^{\infty}f\left(x-u\right)g\left(u\right)du \ast h\left(x\right)\right) = \left(\left(f \ast g\right)\ast h\right)\left(x\right)    \blacksquare]

26
##### Home Assignment 6 / Re: Problem 1
« on: November 03, 2012, 08:37:38 PM »
In part d. of this question, the Fourier transform of $x^ne^{-\alpha |x|}\cos(\beta x)$, I can find quite a few representations for $F(k)$ but none of them have a really tractable closed form. Is it OK to write an answer for the transform in the form of operators acting on a function?- this form of the solution is easily the most elegant in appearance.

27
##### Home Assignment 6 / Problem 2
« on: November 03, 2012, 03:36:12 PM »
To be perfectly clear- in problem 2 and problem 3, we have that $\beta > 0$, correct?

28
##### Home Assignment 4 / Re: Re: Web Bonus Problem--3
« on: October 27, 2012, 07:43:30 PM »
Suppose we somehow have several time-like variables. Does it make sense to consider this kind of system? How would we represent such a structure in the language of equations we're developing in this course?- perhaps in this specific case a Laplacian on time as well? What kind of properties do systems of this kind have that don't appear with a single time-like variable?

29
##### Home Assignment 4 / Re: Re: Web Bonus Problem--3
« on: October 27, 2012, 05:10:45 PM »
Hmm, interesting. It seems as if that perspective gives some insight as to why the result was an even and odd eigenfunction for the respective problems. What does the generalization to this problem in several spacial dimensions look like?

30
##### Home Assignment 4 / Re: Web Bonus Problem--3
« on: October 27, 2012, 02:06:11 PM »
In the clamped case, things get a lot easier! Some modifications to the previous post:

Let: [u(x,t) = X(x)T(t)] in [u_{tt}(x,t) + K u_{xxxx} = 0,   K > 0,    u(0,t) = u_{xx}(0,t)=u(l,t)=u_{xx}(l,t)=0]

Then: [u_{tt}(x,t) = X(x)T''(t),  u_{xxxx}(x,t) = X''''(x)T(t)]
[u_{tt}(x,t) + K u_{xxxx} = X(x)T''(t) + K X''''(x)T(t) = 0]
[\frac{X''''(x)}{X(x)} = \frac{-T''(t)}{K T(t)} = \lambda]

For some constant, as both sides are independant of the other respective variable. Now, we have by assumptions that [\lambda = c^4 > 0,   K = k^2 > 0]

So we are left with two ODE's in the form: [X''''(x) = c^4 X(x),  T''(t) = -c^4 k^2 T(t)]

Which yield solutions: [X(x) = A \cosh(c x) + B \sinh(c x) + C \cos(c x) + D \sin(c x)]
[T(t) = A \cos(c^2 k t) + B \sin(c^2 k t)]

Using the two boundary conditions: [u(0,t) = 0 = u_{xx}(0,t)]

[X(0) = A \cosh(c 0) + B \sinh(c 0) + C \cos(c 0) + D \sin(c 0) = A + C = 0,   A = -C]
[X''(0) = A c^2 \cosh(c 0) + B c^2 \sinh(c 0) - C c^2 \cos(c 0) - D c^2 \sin(c 0) = A c^2 + D c^2,   A = C]
[A = C = -C \implies A = C = 0]

As we disregard the case where: [X(0) \ne 0,   X''(x) \ne 0 \implies T(t) â‰¡ 0]

OK, nice. We plug into the 3rd and 4th boundary conditions, using our new function for X:

[X(x) = B \sinh(c x) + D \sin(c x)]

[u(l,t) = 0 = u_{xx}(l,t)]

[X(l) = B \sinh(c l) + D \sin(c l) = 0]
[X''(l) = B c^2 \sinh(c l) - c^2 D\sin(c l) = B \sinh(c l) - D\sin(c l) = 0]

As c != 0. Adding equations gives us that: [B \sinh(c l) + D \sin(c l) + B \sinh(c l) - D\sin(c l) = 2 B \sinh(c l) = 0]

Now, for real x, sinh(x) has a root only at c = 0, which would yield an eigenvalue of 0, which we disregard by assumption. So:

[2 B \sinh(c l) = B \sinh(c l) = 0 \implies B = 0]

Now we cannot have D = 0, or our solution is trivial. So, resubstituting for B, we're left with only:

[B \sinh(c l) + D \sin(c l) = D \sin(c l) = \sin(c l) = 0]

And our eigenvalues are those c which satisfy [\sin(c l) = 0 \blacksquare]

Which we can provide a closed form for in this case since our solution is so nice:
[\sin(c l) = 0 \implies c = \frac{\pi n}{l} \implies \lambda = \frac{\pi ^4 n^4}{l^4}]

These next two sections need basically no modification, as only which term vanishes in our boundary case during integration by parts changes.

[\int_0^l X_n(x)X_m(x) dx,   n \ne m,   \lambda_n \ne \lambda_m]

Then [(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = \int_0^l \lambda_n X_n(x)X_m(x) -  \lambda_m X_n(x)X_m(x) dx]
[\int_0^l X''''_n(x)X_m(x) - X_n(x)X''''_m(x) dx = (X'''_n(x)X_m(x) - X_n(x)X'''_m(x))_{x=(0,l)} - \int_0^l X'''_n(x)X'_m(x) - X'_n(x)X'''_m(x) dx]
[= 0 - (X''_n(x)X'_m(x) - X_n'(x)X''_m(x))_{x=(0,l)} + \int_0^l X''_n(x)X''_m(x) - X''_n(x)X''_m(x) dx = 0 + 0 = 0]

Using integration by parts, and the fact that our boundary conditions vanish at:

[X(0) = 0, X''(0) = 0, X(l) = 0, X''(l) = 0]

Then we have [(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = 0,     \lambda_n \ne \lambda_m \ne 0 \implies \int_0^l X_n(x)X_m(x) dx = 0]

And our eigenfunctions are orthogonal â– .

Let our differential operator be [\mathcal{I}, st.  \mathcal{I} X = \lambda X,   \mathcal{I} Y = \lambda Y]

We show that: [\langle \mathcal{I}X,Y\rangle = \langle X,\mathcal{I}^*Y\rangle]

[\langle \mathcal{I}X,Y\rangle = \int_0^l \mathcal{I}X(x)Y(x)^* dx = \int_0^l \lambda X(x)Y(x)^* dx = \int_0^l X(x)\lambda^* Y(x)^* dx = \int_0^l X(x)\mathcal{I}^*Y(x)^* dx = \langle X,\mathcal{I}^*Y\rangle]

As we had by assumption our eigenvalues were real. Then our operator is hermetian, so the eigenfuncitons of different eigenvalues are linearly independant, and the eigenfunctions of the same eigenvalue are linearly dependant.â–

Pages: 1 [2] 3