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\title{37335 - Differential Equations Assignment 2}
\author{Zachary Zerafa - 24557656}

\begin{document}

\maketitle

\vspace{0.5in}

\section{Problem 1}

The Laplace transform can be used to solve a wide range of problems. Use the Laplace transform to solve the equation
\[ y'(t) - 2 \int^{t}_{0} y(u) \cos (t-u) du= 1 , \]
with $y(0)=-1$. Hint : Convolution.


\subsection{Laplace transform}


\[ \mathcal{L} \{ y'(t) - 2 \int^{t}_{0} y(u) \cos (t-u) du\}(s)= \mathcal{L}\{1\}(s) , \]

\[ \mathcal{L} \{ y' \}(s) - 2 \mathcal{L} \{ \int^{t}_{0} y(u) \cos (t-u) du\}(s)= \mathcal{L}\{1\}(s) \]

\[ \mathcal{L} \{ y' \}(s) - 2 \mathcal{L} \{ \int^{t}_{0} y(u) \cos (t-u) du\}(s)= \frac{1}{s} \]

\[ \mathcal{L} \{ y' \}(s) - 2 \mathcal{L} \{  y * \cos t\}(s)= \frac{1}{s} \]


\subsubsection{Convolution theorem}


\[ \mathcal{L} \{ y' \}(s) - 2 \mathcal{L} \{ y \}(s) \mathcal{L} \{ \cos t\}(s)= \frac{1}{s} \]


\[ \mathcal{L} \{ y' \}(s) - 2 \mathcal{L} \{  y \}(s) \frac{1}{s^2 + 1} = \frac{1}{s} \]




\subsubsection{Differentiation proposition}



\[ s\mathcal{L}\{y\}(s) - y(0) - 2 \mathcal{L} \{ y \} (s) \frac{1}{s^2 + 1} = \frac{1}{s} \]

\[ s\mathcal{L}\{y\}(s) + 1 - 2 \mathcal{L} \{ y \}(s) \frac{1}{s^2 + 1} = \frac{1}{s} \]



\[ su +1 -\frac{2su}{s^2 + 1} = \frac{1}{s} \]
\[ u(s -\frac{2s}{s^2 + 1}) = \frac{1-s}{s} \]
\[ u(s -\frac{2s}{s^2 + 1}) = \frac{-(s-1)}{s} \]
\[ u(s(s^2 +1) -2s) = \frac{-(s-1)(s^2 +1)}{s} \]
\[ u(s^3 -s) = \frac{-(s-1)(s^2 +1)}{s} \]
\[ us(s-1)(s+1) = \frac{-(s-1)(s^2 +1)}{s} \]
\[ u = \frac{-(s^2 +1)}{s^2 (s+1)} \]




\subsection{Inverse Laplace transform}



The natural way to begin an inverse Laplace transform is to exploit any linearity of terms as such.
\[ y(t) = \mathcal{L}^{-1} \{ \frac{-(s^2 +1)}{s^2 (s+1)} \}(t) \]
\[ = \mathcal{L}^{-1} \{ \frac{-(s^2 +1)}{s^2 (s+1)} \}(t) \]
\[ = \mathcal{L}^{-1} \{ \frac{-s^2 -1)}{s^2 (s+1)} \}(t) \]
\[ = \mathcal{L}^{-1} \{ \frac{-s^2}{s^2 (s+1)} + \frac{-1}{s^2 (s+1)} \}(t) \]
\[ = -\mathcal{L}^{-1} \{ \frac{s^2}{s^2 (s+1)} \}(t) - \mathcal{L}^{-1} \{ \frac{1}{s^2 (s+1)} \}(t) \]
\[ = -\mathcal{L}^{-1} \{ \frac{1}{s+1} \}(t) - \mathcal{L}^{-1} \{ \frac{1}{s^2 (s+1)} \}(t) \]

By reversing the Laplace transform $\mathcal{L}\{\frac{1}{s+n}\} = e^{-nt}$ one deduces the following. 
\[ = -e^{-t} - \mathcal{L}^{-1} \{ \frac{1}{s^2 (s+1)} \}(t) \]
By reversing the convolution theorem one inverses the product of two Laplace transforms as the convolution of their respective inverses. Further noting that $\mathcal{L}\{\frac{\Gamma (n+1)}{s^{n+1}}\} = t^n$, one is lead to the following.
\[ = -e^{-t} - (e^{-t} * t)(t) \]

The convolution can be represented in terms of elementary functions by the reverse product rule.
\[  (e^{-t} * t)(t) = \int^{t}_{0} (t-u)e^{-u}du \]
\[  = [-e^{-u}(t-u)]^{t}_{0} - \int^{t}_{0} -e^{-u}du\]
\[  = t - \int^{t}_{0} -e^{-u}du\]
\[  = t - [e^{-u}]^{t}_{0}\]
\[  = t - (e^{-t} + 1) \]
\[  = t - e^{-t} - 1\]

Substituting this into the original expression gives the following.
\[ y(t)= -e^{-t} - (e^{-t} + t - 1) \]
\[ = 1-2e^{-t} -t \]

Hence the ODE is satisfied by the following function.
\[ y(t)= 1-2e^{-t} -t \]


\section{Problem 2}

It is possible to expand periodic functions in temrs of functions other than sines and cosines. A differentiable function $f$ on $(0,1)$ can be written as
\[ f(x) = \sum^{\infty}_{k=1} A_k J_n (\lambda_{k} x), x \in (0,1) , \]
where $J_n$ are order $n$ Bessel functions and $\{ \lambda_k : k \in \mathbb{N} \setminus \{0\} \}$ is the set of zeroes of $J_n$. That is $J_n (\lambda_k) = 0 , k \in \mathbb{N}\setminus \{0\}$. The problem is to find $A_k$. We have two useful facts which you can assume

Look carefully at how the formula for the Fourier coefficients are derived. Mimic this procedure to show that

\[ A_k = \frac{2}{(J_{n+1}(\lambda_k))^2}  \int^{1}_{0} x f(x) J_n (\lambda_k x) dx\]

\subsection{Taking $x$-weighted inner product}

As the problem suggests, the set \( \{J_n (\lambda_k x) : k \in \mathbb{N} \setminus \{1\} \}  \) forms the basis for the function space $L^1 (0,1)$, however interestingly this function space uses a weighted inner product, with weight $x$. In a similar fashion to the Fourier series, one equates the inner product of both sides of the equation, taken with the second argument being an arbitrary basis element $J_n (\lambda_{m} x)$.



\[ f(x) = \sum^{\infty}_{k=1} A_k J_n (\lambda_{k} x) \]
\[ xf(x) J_n (\lambda_m x) = \sum^{\infty}_{k=1} A_k J_n (\lambda_{k} x) J_n (\lambda_m x) x \]
\[ \int^{1}_{0} xf(x) J_n (\lambda_m x)dx =  \int^{1}_{0} \sum^{\infty}_{k=1} A_k J_n (\lambda_{k} x) J_n (\lambda_m x) x dx \]


\subsection{Evaluating integrals}



Ideally one would like to pass the integral though the limit. The dominated convergence theorem (DCT) ensures the ability to swap limits if on $(0,1)$ there is an $L^1 (0,1) $ function $g$ (note also that $f$ should be in $L^1(0,1)$ if we are integrating over it at all, which we are) that dominates each sequence term (in this case, the partial sums). Since such a function can be found, limit swapping is justified.



\[ \int^{1}_{0} xf(x) J_n (\lambda_m x)dx =   \sum^{\infty}_{k=1} A_k \int^{1}_{0} J_n (\lambda_{k} x) J_n (\lambda_m x) x dx \]
\[ \scalebox{0.8}{$\displaystyle \int^{1}_{0} xf(x) J_n (\lambda_m x)dx = \frac{A_m}{2}[(J'_{n}(\lambda_{m}))^2 + (1 - \frac{n^2}{\lambda_{m}^2})(J_n (\lambda_{m}))^2] + \sum_{k \in \mathbb{N}\setminus \{m,0\} } A_k \frac{ \lambda_{k} J_{n}(\lambda_{m}) J'_{n}(\lambda_{k}) -  \lambda_{m} J_{n}(\lambda_{k}) J'_{n}(\lambda_{m}) }{\lambda_{m}^2 - \lambda_{k}^2} $ } \]

Noting that $\forall k \in \mathbb{N} \setminus \{0\} [ J_{n}(\lambda_{k})=0 ]$, one has the following.

\[ \int^{1}_{0} xf(x) J_n (\lambda_m x)dx = \frac{A_m}{2}[(J'_{n}(\lambda_{m}))^2] \]
\[ A_m = \frac{2}{(J'_{n}(\lambda_m))^2}\int^{1}_{0} xf(x) J_n (\lambda_m x)dx \]


\subsection{Expressing $J'_n$ as a Bessel function}

A lemma relating $n$ order Bessel function derivatives to Bessel functions of other orders will now be proven. Note the following manipulation of the $n$ order Bessel function.

\[ J_{n}(x) =  \sum^{\infty}_{m=0} \frac{(-1)^m}{m! (m+n)!} (\frac{x}{2})^{2m+n} \]
\[ J'_{n}(x) =  \sum^{\infty}_{m=0} \frac{(-1)^m (2m+n) x^{2m+n-1}}{m! (m+n)! 2^{2m+n}} \]
\[ J'_{n}(x) =  \sum^{\infty}_{m=0} 2m \frac{(-1)^m x^{2m+n-1}}{m! (m+n)! 2^{2m+n}}  + n \sum^{\infty}_{m=0}   \frac{(-1)^m x^{2m+n-1}}{m! (m+n)! 2^{2m+n}}\]
\[ J'_{n}(x) =  \sum^{\infty}_{m=0} 2m \frac{(-1)^m x^{2m+n-1}}{m! (m+n)! 2^{2m+n}}  + \frac{n}{x} \sum^{\infty}_{m=0} \frac{(-1)^m}{m! (m+n)!} (\frac{x}{2})^{2m+n}\]
\[ J'_{n}(x) =  \sum^{\infty}_{m=0} 2m \frac{(-1)^m x^{2m+n-1}}{m! (m+n)! 2^{2m+n}}  + \frac{n}{x} J_n (x) \]
\[ J'_{n}(x) =  \sum^{\infty}_{m=-1} 2(m+1) \frac{(-1)^{m+1} x^{2(m+1)+n-1}}{(m+1)! (m+n+1)! 2^{2(m+1)+n}}  + \frac{n}{x} J_n (x) \]
\[ J'_{n}(x) =  \sum^{\infty}_{m=0} 2(m+1) \frac{(-1)^{m+1} x^{2(m+1)+n-1}}{(m+1)! (m+n+1)! 2^{2(m+1)+n}}  + \frac{n}{x} J_n (x) \]
\[ J'_{n}(x) =  -\sum^{\infty}_{m=0} \frac{(-1)^{m} x^{2(m+1)+n-1}}{m! (m+n+1)! 2^{2(m+1)+n-1}}  + \frac{n}{x} J_n (x) \]
\[ J'_{n}(x) =  -\sum^{\infty}_{m=0} \frac{(-1)^{m} }{m! (m+n+1)! } (\frac{x}{2})^{2m+n+1}  + \frac{n}{x} J_n (x) \]
\[ J'_{n}(x) = \frac{n}{x} J_{n}(x) - J_{n+1}(x) \]



\[ J'_{n}(x) = \frac{n}{x} J_{n}(x) - J_{n+1}(x) \]

Recalling that $J_n ( \lambda_m ) = 0$ and applying this lemma completes the proof.
\[ A_m = \frac{2}{(\frac{n}{x} J_{n}(\lambda_m) - J_{n+1}(\lambda_m))^2}\int^{1}_{0} xf(x) J_n (\lambda_m x)dx \]
\[ A_m = \frac{2}{(- J_{n+1}(\lambda_m))^2}\int^{1}_{0} xf(x) J_n (\lambda_m x)dx \]
\[ A_m = \frac{2}{(J_{n+1}(\lambda_m))^2}\int^{1}_{0} xf(x) J_n (\lambda_m x)dx \]



\section{Problem 3}

This uses question 2 to solve an important engineering problem. We have a circular playe, such as a stove top. Take its radius to be 1. The temperature of the plate depends only on the distance $r$ from the origin. The initial temperature is $f(r)$ and the temperature at $r=1$ is kept equal to zero. The temperature $u$ at time $t$ will satisfy the partial differential equation
\[ \frac{1}{k} \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r}, \]
\[ u(1,t)=0, u(r,0)=f(r) , |u(r,t)| < M ,\]
for some finite $M > 0$. This last condition just says that the temperature is finite. $k$ is a constant depending on the plate. Use separatin of variables to solve this problem. That is, let $u(r,t)=R(r)T(t)$ and follow the procedure we used in the lectures for the heat equation. You will need the fact that $ \lim_{x \to 0^{+}} |Y_{0}(x)| = \infty$.

\subsection{Applying separation of variables}

One makes the ansatz $u(r,t)=R(r)T(t)$

\[ \frac{1}{k} R T' = R'' T + \frac{1}{r} R' T\]
\[ \frac{1}{k} R T' = T ( R''  + \frac{1}{r} R' ) \]
\[ \frac{T'}{T} = \frac{kR'' + \frac{k}{r}R' }{R}\]

Since this relation holds for any setting of $r,t$, this implies both sides of the equation equate to a constant. Let such a constant be denoted as $\mu$.

\[ \frac{T'}{T} = \mu , \frac{kR'' + \frac{k}{r}R' }{R} = \mu\]

Now one seeks to solve these two resultant ODEs.

\[ T' = \mu T \]
This is a first order linear ODE with the following general solution.
\[ T(t)=c_t e^{\mu t} \]

One now turns to the other ODE.
\[ r^2 R'' + r R' -r^2 \frac{\mu}{k}R=0\]

This second order linear ODE can be mapped to a Bessel equation! The following lemma is used to map this ODE to a Bessel equation. 

\[ x^2 y'' + (1-2s)xy' + [(s^2 -r^2 \alpha^2 ) + a^2 r^2 x^2]y = 0 \]
\[ \implies y(x)=c_1 x^s J_{\alpha}(ax^r) + c_2 x^s Y_{\alpha}(ax^r)\]

Letting $\mu < 0$ and considering that the solution must be bounded on $[0,1]$, one must omit the Bessel function of the second kind from the solution, hence one has the following.


\[ R(r)=c_r J_{0}(\sqrt{\frac{-\mu}{k}}r) \]

The Dirichlet condition $u(1,t)=0$ allows the deduction of $\mu$.
\[ 0=c_r J_{0}(\sqrt{\frac{-\mu}{k}}) \]
One avoids the possibility that $c_r=0$ since this leads to the trivial solution, instead one considers $ 0=J_{0}(\sqrt{\frac{-\mu}{k}})$, hence the argument within $J_{0}$ must be a zero of $J_{0}$; this article denotes the sequence of ascending real zeroes of $J_0$ as $ ( \lambda_n )_{n \in \mathbb{N} \setminus \{0\}}$. This implies the following.
\[ \sqrt{\frac{-\mu}{k}} = \lambda_n \]
\[ \mu = -k \lambda_{n}^2 \]

Considering the linearity of the PDE, infinite sets of solutions, and condensing the coefficients of $R,T$ into one $c_n$, the solution now turns toward the following Bessel-Fourier expansion.


\[u(r,t) = \sum^{\infty}_{n=1} c_n e^{-k \lambda_{n}^2} J_{0}( \lambda_n r) \]

\subsection{Calculating Bessel-Fourier coefficients}

Now one considers the condition $u(r,0)=f(r)$. Applying this to the unrefined solution leads to the following.

\[f(r) = \sum^{\infty}_{n=1} c_n J_{0}( \lambda_n r) \]

By problem 2, the Bessel-Fourier coefficients for a $L^1 ([0,1])$ function is the following.
\[ c_n = \frac{2}{(J_{1}(\lambda_n))^2}  \int^{1}_{0} r f(r) J_0 (\lambda_n r) dr\]

Therefore given $k,f(r)$, one has the following solution to the PDE.

\[u(r,t) = \sum^{\infty}_{n=1} \frac{2}{(J_{1}(\lambda_n))^2}   e^{-k \lambda_{n}^2} J_{0}( \lambda_n r) \int^{1}_{0} r f(r) J_0 (\lambda_n r) dr \]

\end{document}
