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### Topics - Calvin Arnott

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1
##### Technical Questions / Typesetting piecewise functions
« on: November 18, 2012, 05:09:34 PM »
I ran into a problem typesetting the piecewise functions of problem 4 when converting my solutions for homework 6 to mathjax here: http://forum.math.toronto.edu/index.php?topic=118.msg626#msg626

Namely, I can't seem to get everything to line up nicely without cheating and using \phantom{}, which is just a really clumsy solution. Here's the code I'm using:

Code: [Select]
   f\left(x\right) = \left\{     \begin{array}{lr}       1 & : |x| \le 1\\       0 & : |x| > 1     \end{array}   \right.\\
Which looks like this:

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

Is there a better way of typesetting this?

2
##### Home Assignment Y / Problem 2
« on: November 12, 2012, 12:17:59 PM »
Problem
Consider the Laplace equation in the half-strip:
$$\Delta u = u_{xx} + u_{yy} = 0 : \{x > 0, -1 < y < 1\}$$
with the boundary conditions:
$$u(x,-1) = u(x,1) = 0$$
$$u(0,y) = 1 - |y|, \phantom{\ } \max_{\{x,y\}} |u| < \infty$$

Part  a. Write the associated eigenvalue problem
First we separate variables in $\{x,y\}$. Let: $(x,y) = X(x)Y(y)$.
$$\implies u_{xx} = X''(x)Y(y), \phantom{\ } u_{yy} = X(x)Y''(y)$$
$$u_{xx} + u_{yy} = 0 \rightarrow X''(x)Y(y) + X(x)Y''(y) = 0$$
$$\implies \frac{X''(x)}{X(x)} = - \frac{Y''(y)}{Y(y)} = \lambda$$
For some constant $\lambda$, as each side depends only on one variable and so is constant with repsect to the other. Then we have two ODEs with BCs, and our eigenvalue problem is:
$$X''(x) - \lambda X(x) = 0, \phantom{\ } Y''(y) + \lambda Y(y) = 0$$
$$Y(-1) = Y(1) = 0, \phantom{\ } \max_{\{x,y\}} |X(x)Y(y)| < \infty, \phantom{\ } u(0,y) = 1 - |y| \phantom{\ } \square$$

Part b. Find all eigenvalues and corresponding eigenfunctions
Answer As previously derived, we make the assumption that all eigenvalues $\lambda_n = \beta_n^2 > 0$. Solving first for $Y(y)$ in:
$$\{Y(y) \phantom{\ } | Y''(y) + \lambda Y(y) = 0, \phantom{\ } Y(-1) = Y(1) = 0\}$$
Let: $\lambda = \beta^2$. Then the ODE $Y''(y) + \lambda Y(y) = Y''(y) + \beta^2 Y(y) = 0$ has solution $Y(y) = A \cos(\beta y) + B \sin(\beta y)$ for some constants $\{A,B\} \in \mathbb{R}$. Plugging in our BCs $\{Y(-1) = Y(1) = 0\}$ yields:
$$Y(-1) = (A \cos(\beta y) + B \sin(\beta y))\bigr|_{y=-1} = A \cos(-\beta) + B \sin(-\beta) = 0$$
$$Y(1) = (A \cos(\beta y) + B \sin(\beta y))\bigr|_{y=1} = A \cos(\beta) + B \sin(\beta) = 0$$
Using that $\sin$ is odd, and $\cos$ is even, we have the equations:
$$A \cos(\beta) - B \sin(\beta) = A \cos(\beta) + B \sin(\beta) = 0$$
Adding and subtracting these equations, then dividing by $2$, gives us:
$$A \cos(\beta) = B \sin(\beta) = 0$$
Now, as $\nexists \beta \in \mathbb{R}$ where $\cos(\beta) = \sin(\beta) = 0$, and we discard the trivial solution $X(x) \equiv 0$ where $A = B = 0$, we must have either $A = 0, B \ne 0, \sin(\beta) = 0 \implies \beta = n \pi$, or $A \ne 0, B = 0, \cos(\beta) = 0 \implies \beta = (n \pi - \frac{\pi}{2})$, for $n \in \mathbb{N}$. Because our BC $u(0,y) = 1 - |y|$ is clearly an even function in $y$, we choose the case where $A \ne 0, B = 0, \cos(\beta) = 0 \implies \beta = (n \pi - \frac{\pi}{2})$. Then our eigenfunction for $Y_n(y)$, and eigenvalues $\lambda_n = \beta_n^2 > 0$, are given by:
$$n \in \mathbb{N} : \lambda_n = \beta_n^2 = (n \pi - \frac{\pi}{2})^2, \phantom{\ } Y_n(y) = \cos(\beta_n y) = \cos((n \pi - \frac{\pi}{2}) y)$$
Next, using that $\lambda_n = \beta_n^2 = (n \pi - \frac{\pi}{2})^2$, we solve for $X(x)$ in:
$$\{X(x) \phantom{\ } | X''(x) - (n \pi - \frac{\pi}{2})^2 X(x) = 0, \phantom{\ } \max_{\{x\}} |X(x)| < \infty \}$$
This ODE has solutions in the form $X(x) = C e^{(n \pi - \frac{\pi}{2}) x} + D e^{-(n \pi - \frac{\pi}{2}) x}$. But, because $x > 0$, and $(n \pi - \frac{\pi}{2}) > 0$, we must have that $C = 0$ since we require $|X(x)|$ to be bounded as $x \rightarrow \infty$. Otherwise: $C \ne 0 \implies \lim_{x \to \infty} |X(x)| \ge |C e^{(n \pi - \frac{\pi}{2}) x}| \rightarrow \infty$. So each $X_n(x)$ is given by:
$$n \in \mathbb{N} : X_n(x) = e^{-(n \pi - \frac{\pi}{2}) x}$$

Thus, our eigenvalues are given by:
$$n \in \mathbb{N} : \lambda_n = (n \pi - \frac{\pi}{2})^2$$
With corresponding eigenfunctions:
$$u_n(x,y) = A_n X_n(x) Y_n(y) = A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y), \phantom{\ } A_n \in \mathbb{R} \phantom{\ } \square$$

Part c. Write the solution in the form of a series expansion

We have for $n \in \mathbb{N}, A_n \in \mathbb{R}$ the eigenvalues $\lambda_n = \beta_n^2 = (n \pi - \frac{\pi}{2})^2$ and eigenfunctions $u_n(x,y) = A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y)$. Then the series expansion of our solution $u(x,y)$ is given by:
$$u(x,y) = \sum_{n=1}^{\infty} A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y)$$
Applying our final BC: $u(0,y) = 1 - |y|$:
$$u(0,y) = (\sum_{n=1}^{\infty} A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y))\bigr|_{x=0}$$
$$= \sum_{n=1}^{\infty} A_n \cos(\pi(n - \frac{1}{2}) y) = 1 - |y| : \{ -1 < y < 1 \}$$
Proceeding; recall that the set of half-integer cosines:
$$\{ X_n(x) = \cos(\pi(n - \frac{1}{2}) x) : n \in \mathbb{N} \}$$
$$\text{With inner product: } \langle f,g \rangle = \int_{-1}^{1} f(x)\overline{g(x)} dx$$
forms an orthonormal basis for the space of even functions which converge on $[-1,1]$ in the $L^2$ sense: $\| f \| = \langle f,f \rangle ^{\frac{1}{2}} < \infty$. Then the projection of $f(y) = 1 - |y|$ onto the function space is given by:
$$f(y) = 1 - |y| = \sum_{n=1}^{\infty} A_n X_n(y) = \sum_{n=1}^{\infty} \langle f,X_n \rangle \frac{X_n(y)}{\| X_n \| ^2}$$
$$\implies A_n = \langle f,X_n \rangle \cdot \frac{1}{\| X_n \|^2} = \int_{-1}^{1} f(x)\overline{X_n(x)}dx \cdot 1$$
$$= \int_{-1}^{1} (1-|x|)\cos(\pi(n - \frac{1}{2}) x)dx = 2 \int_{0}^{1} (1-x)\cos(\pi(n - \frac{1}{2}) x)dx$$
$$= \frac{8 - \sin(n \pi)}{(2 n \pi - \pi)^2} = \frac{8}{\pi^2(2 n - 1)^2} = A_n$$
Finally then, our solution $u(x,y)$ is given by the series:
$$u(x,y) = \sum_{n=1}^{\infty} A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y) = \sum_{n=1}^{\infty} \frac{8}{\pi^2(2 n - 1)^2} e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y) \phantom{\ } \blacksquare$$ \\ \\

3
##### Home Assignment Y / Problem 1
« on: November 11, 2012, 12:49:08 PM »
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: Find a solution using the Fourier transform for the Laplace equation in the half-plane (it was a misprint in the problem. V.I.):
$$\Delta u = u_{xx} + u_{yy} = 0 : \{x > 0, -\infty < y < \infty\}$$
with the boundary conditions:
$$u(0,y) = e^{-|y|}, \phantom{\ } \max_{\{x,y\}} |u| < \infty$$
Answer: Now, $x > 0$ so we cannot perform a Fourier transform over $x$. We proceed instead by partially transforming with respect to $y \mapsto k$, $u(x,y) \mapsto \mathcal{F}_y (u)(x,k)$:
$$\text{Let: } U(x,k) = \mathcal{F}_y (u)(x,k) = \int_{-\infty}^{\infty} u(x,y) e^{-i k y} d y$$
We know that this transformation converges and exists as we have that $|u|$ is bounded by some constant $c < \infty$. Recall that the Fourier transform $\mathcal{F}$ is an operator, and that for a function $f(x)$ with transform $\mathcal{F}(f)(k)$, a property of the transform $\mathcal{F}$ is that for $g(x) = \partial_x f(x)$, $\mathcal{F}(g)(k) = i k \mathcal{F}(f)(k)$. Applying the partial transformation $\mathcal{F}_y$ to our PDE and BC then yields us:
$$\mathcal{F}_y((u_{xx} + u_{yy})(x,y) = (0)(x,y)) \mapsto \mathcal{F}_y(u_{xx})(x,k) + \mathcal{F}_y(u_{yy})(x,k) = \mathcal{F}_y(0)(x,k) = (0)(x,k)$$
$$= \mathcal{F}(\partial_{x}^2 u)(x,k) + \mathcal{F}(\partial_{y}^2 u)(x,k) = \partial_{x}^2 \mathcal{F}(u)(x,k) + (i k)^2 \mathcal{F}(u)(x,k) = 0$$
$$\text{And BC: } \mathcal{F}_y(u(0,y) = e^{-|y|}) \mapsto \mathcal{F}_y(u)(0,k) = \mathcal{F}_y (e^{-|y|})(k) = \frac{2}{1+k^2}$$
Where we differentiated $\mathcal{F}_y(\partial_{x}^2 u)(x,k)$ under the sign, and used that because $\mathcal{F}$ is an operator, we must have $\mathcal{F}(0) = 0$. Using $U(x,k) = \mathcal{F}_y (u)(x,k)$, we then have an ODE in $x$ with BC:
$$U_{xx} + (i k)^2 U = U_{xx} -k^2 U = 0$$
$$U(0,k) = \mathcal{F}_y (e^{-|y|})(k) = \frac{2}{1+k^2}$$
Using our previously derived transformation for $e^{-|y|}$. This has solution $U(x,k) = \Psi(k) e^{- |k| x} + \Phi(k) e^{+ |k| x}$, with $\{ \Psi(k),\Phi(k)\}$ arbitrary functions of $k$, where we take $|k|$ in place of $k$ to control the asymptotic behavior as $\{|k|,x\} \rightarrow \infty$. Since $x > 0$ we must exclude $e^{|k| x}$ in our solution since it grows without bound as $|k| \rightarrow \infty$ and so doesn't allow for the convergence of the inverse Fourier transformation. Then $U(x,k) = \Psi(k) e^{- |k| x}$. Plugging in our BC:
$$U(0,k) = \frac{2}{1+k^2} = (\Psi(k) e^{- |k| x})\bigr|_{x=0} = \Psi(k)$$
$$\implies \Psi(k) = \frac{2}{1+k^2} \implies U(x,k) = \frac{2}{1+k^2} e^{- |k| x}$$
Finally, we apply the inverse Fourier Transformation $\mathcal{F}^{-1}$ with respect to $k$ on $U(x,k)$ to solve in terms of $u(x,y)$:
$$\mathcal{F}^{-1}_k (U(x,k)) = \mathcal{F}_k^{-1}(\mathcal{F}_y(u(x,y))) \mapsto u(x,y)$$
$$\implies u(x,y) = \mathcal{F}^{-1}_k (\frac{2}{1+k^2} e^{- |k| x}) = \int_{-\infty}^{\infty}(\frac{2}{1+k^2} e^{- |k| x}) e^{i k y} \frac{dk}{2 \pi} \phantom{\ } \blacksquare$$ \\ \\

4
##### 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?

5
##### Home Assignment 4 / Problem 1
« on: October 18, 2012, 04:39:28 PM »
It seems to me that question 1 part c) doesn't ask any question and instead makes a statement. Is there anything I'm missing there?

6
##### Home Assignment 4 / Problem 2
« on: October 17, 2012, 04:46:26 PM »
In problem 2, do we have: [K > 0] for [u_{tt} + K u_{xxxx} = 0]?

7
##### Home Assignment X / Problem 4
« on: October 13, 2012, 10:02:10 PM »
In problem 4, we have the functions:

[ ÆŸ(x) = 0  \{x < 0\}, \space ÆŸ(x) = 1  \{x > 0\}]            [ f(x) = 1   \{|x| > 1\},    f(x) = 0   \{|x| < 1\}]

and an idea to use the functional relation [f(x) = ÆŸ(x+1) - ÆŸ(x-1)]

But this identity does not hold. For instance, [f(0) = ÆŸ(1) - ÆŸ(-1) = 1 - 0 = 1 \ne 0]

An identity which I found to work instead for the integral is: [f(x) = ÆŸ(x - 1) + ÆŸ(-x - 1) ]