\chapter{Possible Future Work}
\label{chap:future}

The work in this thesis is pertinent to emerging methods in machine learning as well as current problems in structural bioinformatics.
As such, extensions of this work can be made in several distict veins, some focused on model formulation and training procedure, others on possible applications outside of interface prediction.
Several of these extensions are documented below, categorized as either extensions of method or extensions of application.

\section{Extensions of Method}

\subsection{Double Coupling and Ensemble Approaches}

As discussed in Chapter \ref{chap:experiments}, Sum Coupling outperforms Product Coupling on rigid and medium difficulty complexes, whereas Product Coupling is better for difficult complexes.
Therefore it would be reasonable to incorporate both sum and Product Coupling into a single convolution operation:

\begin{equation}
h_i(x | W^\textsc{c}, W^\textsc{n}, W^\textsc{e}, b) = \sigma \bigg( W^{\textsc{c}} x_i + \frac{1}{|\mathcal{N}_i|}\sum_{j \in \mathcal{N}_i} (W^{\textsc{n}} x_j) \odot (W^{\textsc{e}} A_{ij}) + W^{\textsc{n}} x_j + W^{\textsc{e}} A_{ij} + b \bigg),
\label{eq:double_coupling}
\end{equation}

\noindent
where here the same weight matrices are used in both the Sum Coupling and Product Coupling components to help prevent overfitting.
Alternately, separate weight matrices could be used to increase the model's expressive power.

Recall that concerning the RFPP metric, the optimal network depth increased with complex difficulty, where more layers were needed for complexes of greater difficulty.
This suggests that an ensemble model with multiple networks of varying depth could perform  better than each individual network.
Alternative ensemble approaches like \emph{boosting} or \emph{bagging} may also prove beneficial.
Boosting trains a sequence of "weak" models, with each added model being trained by giving more importance (loss weight) to examples misclassified by the existing set of models.
For interface prediction, this weighting can be performed on a per-complex basis, using RFPP or AUC as an indication of performance on each complex, or on a per-residue-pair basis, using the cross entropy loss of that specific training example.
The final prediction is a weighted average of the weak models' predictions, with the weights determined by the validation error of each respective model (with higher weight given to models with lower error).
Bagging creates multiple datasets by sampling with replacement from the original data set, and trains a different model on each sampled dataset.
This sampling can be performed at the complex level or the residue pair level.
Again, the final prediction is a weighted average of the weak models' predictions, weighted according to validation error.

\subsection{RFPP Optimization}

Predicting an entire interface is a more difficult task than predicting only a few interacting residues in that problem.
It could be that training a classifier in the former may hinder its performance in the latter.
An alternative approach would be to just focus on generating a few predictions of high quality (i.e. confidence), which means optimizing performance on just the top positive predictions.
The existing model uses a loss function which incorporates every training example, not just the top positive prediction, hence the model is optimized for all examples.
The new approach implies treating RFPP as the primary metric and optimizing it directly.
An alternate loss function which includes the loss from only the highest scoring positive example could be used.
Hence when the loss is minimized, the performance of the highest performing positive example will be maximized.
Here a sum over all examples has been replaced by a max function, which is not differentiable, but differentiable variants of the loss function could be constructed to enable gradient descent based learning.
This method is currently being investigated by another student in Professor Ben-Hur's research group.

\subsection{Additional Data Sources}

The success of deep learning methods has been attributed to large volumes of data and deep architectures~\cite{krizhevsky2012}.
For interface prediction, despite the large volume of recorded protein structures, precious few complexes have been labeled in bound and unbound forms for use in model training and testing.
This dependence on small curated subsets of the available proteins has potentially limited the full leveraging of deep learning methods, however some opportunities exist for enlarging the training and testing data.

The Docking Benchmark Dataset was conceived as a method to evaluate docking methods, and correspondingly was carefully constructed to be non-redundant with respect to SCOP families, in order to give a fair evaluation.
However for training purposes, it may be useful to include redundant proteins simply to provide the model with more training data.
The Critical Assessment of PRediction of Interactions (CAPRI) is an annual competition aimed at evaluating protein-protein docking methods~\cite{janin2003}, and could also be used to help train or evaluate partner-specific protein interface prediction methods.

\subsection{Unsupervised Pretraining}

Another approach to the problem of limited labeled complexes is to utilize the vast quantity (>125,000) of recorded protein structures via unsupervised pre-training.
Hinton and Salakhutdinov (2006)~\cite{hinton2006b} proposed a greedy layer-wise training algorithm for a particular form of neural network called an \emph{autoencoder}.
Generally speaking, an autoencoder is simply a network trained to reproduce the input at the output under certain constraints~\cite{goodfellow2016}.
These constraints, such as undercompleteness, regularization, and sparsity, prevent the network from simply learning an identity mapping of the input, and instead promote the learning of a compressed representation of the input (i.e. an "encoding").
The portion of the network which compresses the input is called the encoder, while the portion which reconstructs the input from the encoding is called the decoder.
The encoding half of the autoencoder can be used independently of the decoding half to create representations of the input that are often useful in supervised tasks like classification~\cite{hinton2006b, bengio2007}.

Masci et al. (2011)~\cite{masci2011} proposed convolutional autoencoders (CAEs) which perform convolution to encode an image and \emph{deconvolution} to decode the image.
There is also a corresponding \emph{unpooling} operation which upsamples images which were pooled during the encoding step.
Deconvolution is simply a convolution operation performed on the result of an encoding convolution, with weights being tied between convolution and deconvolution layers.
For $k$ filters of size $m \times m$ and $c$ channels, the corresponding deconvolution would consist of $c$ filters of size $m \times m$ and $k$ channels, where the weights have been reflected in both spatial dimensions (e.g. the weights for the lower left pixel of the receptive field are now applied to the upper right pixel).
The spatial reflection allows a particular weight to carry a specific meaning, namely it characterizes the relationship between a particular pixel in an image and a pixel in its encoding.
To illustrate this, Figure \ref{fig:deconv} shows a deconvolution for a single filter and single channel.
CAEs are trained in the same greedy, layerwise fashion as conventional autoencoders.

\begin{figure}
	\includegraphics[width=0.8\textwidth]{conv_ae.png}
	\caption{Application of convolution, then deconvolution on a single channel input image. The relationship between the purple pixel in the input/reconstruction and the red pixel in the encoding is captured by the appropriate weight in the filters (lower left for convolution and upper right for deconvolution). By reflecting the convolution filter horizontally and vertically for use in deconvolution, the same weight ($W_{31}$) is used for this relationship during both encoding and decoding.
		\label{fig:deconv}}
\end{figure}

The concept of deconvolution can be applied to graph convolutions as well.
Note that when reflecting image convolution filters, the center weight remains in the same position.
In the same way, graph deconvolutions retain the same weights for the central vertex.
Furthermore, since all neighbors use the same weights, there is no spatial reflection necessary.
Deconvolution simply consists of transposing the center and neighbor weight matrices so that channels become filters and filters become channels.
For Sum Coupling, deconvolution becomes:

\begin{equation}
\hat{x_i}(h_i | W^\textsc{c}, W^\textsc{n}, W^{\textsc{e}*}, b^*) = \sigma \bigg( (W^{\textsc{c}})' h_i + \frac{1}{|\mathcal{N}_i|}\sum_{j \in \mathcal{N}_i} \big[ (W^{\textsc{n}})' h_j + W^{\textsc{e}*} A_{ij} \big] + b^* \bigg),
\label{eq:sum_deconv}
\end{equation}

\noindent
where $\hat{x}$ denotes the reconstruction of $x$ after deconvolving, $h_i$ is the convolved representation at vertex $i$, $(W^\textsc{c})'$ is the transpose of $W^\textsc{c}$, likewise for $(W^\textsc{n})'$ and $W^\textsc{n}$, and both $W^{\textsc{e}*}$ and $b^*$ are weight matrices with unshared weights.
Note that graph convolutions generate representations on vertices of the graph, not the edges, so the non-encoded edge information ($A_{ij}$) must be used when deconvolving, and the weight matrix $W^{\textsc{e}*}$ cannot be tied to the encoding matrix $W^{\textsc{e}}$.

Alternately, edge representations could be created in a separate convolution operation, where an edge's receptive field is simply the incident vertices:

\begin{equation}
h_{ij}(A_{ij} |W^{\textsc{ee}}, W^{\textsc{ev}}, b_e) = \sigma\bigg( W^{\textsc{ee}} A_{ij} + \frac{1}{2}\big(
W^{\textsc{ev}} x_i + W^{\textsc{ev}} x_j \big) + b_\textsc{e} \bigg),
\label{eq:edge_conv}
\end{equation}

\noindent
where $h_{ij}$ is the representation of edge $(i, j)$ , $W^{\textsc{ee}}$ is the weight matrix associated with the edge, $W^{\textsc{ev}}$ is the weight matrix associated with the incident vertices, and $b_\textsc{e}$ is the bias.
In this case, vertex deconvolution can use the encoded edge representations and associated weights.
Equation (\ref{eq:sum_deconv}) then becomes:

\begin{equation}
\hat{x_i}(h_i | W^\textsc{c}, W^\textsc{n}, W^\textsc{ev}, b) = \sigma \bigg( (W^{\textsc{c}})' h_i + \frac{1}{|\mathcal{N}_i|}\sum_{j \in \mathcal{N}_i} \big[ (W^{\textsc{n}})' h_j + (W^{\textsc{ev}})' h_{ij} \big] + b^* \bigg),
\label{eq:sum_deconv2}
\end{equation}

\noindent
where $(W^{\textsc{ev}})'$ is the transpose of $W^{\textsc{ef}}$.
Edges can also be deconvolved using the edge and vertex representations:

\begin{equation}
\hat{A_{ij}}(h_{ij} | W^{\textsc{ee}}, W^{\textsc{e}}, b) = \sigma \bigg( (W^{\textsc{ee}})' h_{ij} + \frac{1}{2}\big[
(W^{\textsc{e}})' h_i + (W^{\textsc{e}})' h_j \big] + b^{*}_\textsc{e} \bigg).
\label{eq:edge_deconv}
\end{equation}

\noindent
This added weight sharing and symmetry between convolution and deconvolution operations may allow training of deeper networks which recognize more sophisticated structures.

\subsection{Simplicial Complex Convolution}

Graphs are only one way to model structural relationships between entities.
A simplicial complex is a generalization of a graph in which higher order simplices (triangles, tetrahedra, etc), in addition to edges, describe relationships between groups of vertices.
For example, just as an edge indicates a relationship between two vertices, a triangle (or 2-simplex) indicates the relationship between three vertices.
For example, both the \v{C}ech and Vietoris-Rips definitions can be used to construct a simplicial complex from a set of points in an underlying metric space:

\begin{definition}
	Given a point set $X$ in some metric space and a number $\epsilon>0$, the \v{C}ech complex $C^{\textsc{\v{C}}}_\epsilon$ is the simplicial complex whose simplices are formed as follows. For each subset $S \subset X$ of points, form an ($\epsilon/2$)-ball around each point in $S$, and include $S$ as a simplex (of dimension $|S|$) if there is a common point contained in all of the balls in $S$.
\end{definition}

\begin{definition}
	Given a point set $X$ in some metric space and a number $\epsilon>0$, the Vietoris-Rips complex $C^{\textsc{VR}}_\epsilon$ is the simplicial complex whose simplices are formed as follows. For each subset $S \subset X$ of points, form an ($\epsilon/2$)-ball around each point in $S$, and include $S$ as a simplex (of dimension $|S|$) if for each pair of balls in $S$ there is a common point contained in both balls.
\end{definition}

In both definitions, simplices are included in the simplicial complex based on some notion of proximity.
This means simplicial complexes are naturally equipped to identify clusters of vertices within a neighborhood, something which can be useful for interface prediction, where local regions of residues may constitute portions of an interface.
Though the number of possible convolution operations on simplicial complexes are many, one example is provided here as an illustration.
In this formulation, rather than sum the signal from all neighbors in the receptive field, each simplex $S_k, k \in \{1, 2,... K\}$ (not counting vertices and edges, which are the lowest order simplexes) in the neighborhood is summed individually and the maximum simplex signal is added to the central signal:

\begin{equation}
\begin{split}
h_i(x |  & W^{\textsc{c}}, W^{\textsc{n}}, W^{\textsc{e1}}, W^{\textsc{e2}}, b) = \\
&\sigma \bigg( W^{\textsc{c}} x_i + \max_{k} \Big[ \frac{1}{|S_k|}  \sum_{j \in \mathcal{S}_k} ( W^{\textsc{n}} x_j + W^{\textsc{e1}} A_{ij} ) + \frac{1}{|S_k|^{2}}\sum_{j, l \in \mathcal{S}_k} (W^{\textsc{e2}} A_{jl}) \Big] + b \bigg).
\end{split}
\label{eq:simplicial_complex_conv}
\end{equation}

\noindent
This formulation also includes \emph{secondary} edges ($A_{jl}$), those between different neighbors in the same simplex.
It is possible that neighbors and edges occur in more than one simplex in the neighborhood, but that doesn't pose a problem because of the normalization and max function.
The use of simplicial complexes in defining convolution operations is currently being investigated by a student in Professor Ben-Hur's research group.

\section{Extensions of Application}

Interface prediction is not the only active area of protein research, and so the modeling of proteins as graphs and use of graph convolution could potentially be used for other protein problems as well.
A common characteristic of docking methods is that they create a high number of putative 3D bound structures~\cite{janin2013}.
Graph convolutional neural networks could also be used to evaluate putative docking solutions and rank them in order of likelihood.
For example, the interface derived from a docking solution could be compared to the interface predicted by a trained pairwise graph convolutional neural network and ranked according to similarity with the network prediction.
Similarly, much research has been performed in protein modeling, which seeks to predict a protein's secondary and tertiary structures from sequence alone~\cite{schwede2013}.
Though graph convolutions cannot be used to create protein folds \emph{de novo}, they can be used to evaluate a set of putative folds and likewise rank them.
In both of these applications, scoring is being performed not at the individual vertex level, but at the graph level.
These could take advantage of a global pooling method like taking the maximum or average across all vertices, or developing a fingerprint similar to work by Duvenaud \& Maclaurin, et al.~\cite{duvenaud2015}.

Any application involving structured data that do not fit naturally into a grid is a potential candidate for graph convolutional approaches.
Aside from proteins, other biological and chemical data often meet this criterion.
In particular, Quantitative Structure-Activity Relationship (QSAR) and Quantitative Structure-Property Relationship (QSPR) both attempt to predict chemical behavior and properties of proposed chemical structures in order to identify structures with certain desired properties.
Identified structures become candidates for laboratory synthesis and experimentation, a much more laborious and time consuming process.
This is the focus of the Fingerprint convolution from Duvenaud \& Maclaurin et al.~\cite{duvenaud2015}.
Studies in QSAR and QSPR have long been considered "unquestionably of great importance in modern chemistry and biochemistry"~\cite{karelson1996}.
Like proteins, chemical structures can be modeled as a graph and used to train graph convolutional networks to predict the properties of interest.
This is made possible by the vast and growing databases of chemical structures and their corresponding structures~\cite{olah2008, judson2008}.

Beyond the biochemical realm, graph structured data abound.
Knowledge bases are collections of structured data in the form of triples indicating \emph{(subject, predicate, object)}, where \emph{subject} and \emph{object} are named entities and \emph{predicate} indicates the relationship between them, and pairs indicating \emph{(subject, attribute)}, where \emph{attribute} is a type or property label for \emph{subject}.
These data can easily be thought of graphs, where entities are vertices, each type of predicate is one of many different types of edge features, and each attribute is one of many different types of vertex features.
Knowledge bases emerge in online information repositories such as YAGO~\cite{mahdisoltani2014}, DBPedia~\cite{auer2007}, and Wikidata~\cite{vrandevcic2014}, and are used in the Google search engine \cite{singhal2012}.
Due to their large size, knowledge graphs are often under-annotated, where vertex relationships and properties are missing.
A graph convolutional neural network can be used to predict properties or relationships on graphs, essentially performing automatic inference on the graph.
This is the chief problem of interest for Schichtkrull \& Kipf~\cite{schlichtkrull2017}.


\section{Option valuation using Fourier inversion method}
\label{chap:fourier}

Now we show how to value options using the Fourier inversion method introduced by \textcite{heston1993closed}. Our approach is based on \textcite[pp. 222--233]{nawalkabeliaevasoto2007dynamic}.

We make the following assumptions. The interest rate model is affine $A_M(N)$ model. Thus
	\begin{align}
		r(t) = \delta(t) + \sum_{m=1}^M X_m(t) + \sum_{i=1}^{N-M} Y_i(t) .
	\end{align}
The correlated gaussian process are
\begin{align}
\dx Y_i(t) &= - k_i Y_i(t) \dx t + \nu_i \dx W_i (t),
\end{align}
where $W_i$ is a Wiener process and
\begin{align}
\dx W_i (t) \dx W_j (t) &= \rho_{ij} \dx t
\end{align}
for all $i,j = 1,2, \ldots, N-M$. Here $-1 < \rho_{ij} = \rho_{ji} < 1$ and $\rho_{ii} = 1$. The $M$ square-root processes are
\begin{align}
\dx X_m(t) = \alpha_m ( \theta_m - X_m(t) ) \dx t + \sigma_m \sqrt{ X_m(t) } \dx Z_m (t)
\end{align}
where $Z_m$ are independent Wiener process and
\begin{align}
\dx W_i (t) \dx Z_m (t) &= 0
\end{align}
for all $i = 1,2, \ldots, N-M$ and $m = 1,2, \ldots, M$.

We explicitly assume that $\delta(t) = 0$ for all $t \geq 0$ to keep notation simple and we do the dynamic extension of Section \ref{sec:dynamicextension} to add the $\delta(t)$ afterwards. Now zero-coupon bond prices are given by
	\begin{align}
		\label{fourierbondaffineassumption}
		\Bond(t,T) = \e^{A(\tau) - B^{\top} (\tau) X(t) - C^{\top} (\tau) Y(t) } ,
	\end{align}
where $\tau = T-t$. This is a simplification from \textcite[p. 433--435]{nawalkabeliaevasoto2007dynamic}.

The price of a call option expiring on $S$ written on $T$-bond with strike price $K$ is
	\begin{align}
		c(t) &= \E_{\Pm_0} \left( \e^{ - \int_t^S r(s) \dx s } \left( \Bond(S,T)) - K \right)_+ \ | \ \F_t \right) \\
			&= \E_{\Pm_0} \left( \e^{ - \int_t^S r(s) \dx s } \left( \Bond(S,T)) - K \right) \1_{ \Bond(S,T) - K \geq 0 } \ | \ \F_t \right) \\
			&= \Bond(t,T) \Pi_1 - K \Bond(t,S) \Pi_2 ,
	\end{align}
where
	\begin{align}
		\Pi_1 &= \E_{\Pm_0} \left( \frac{\e^{ - \int_t^S r(s) \dx s } \Bond(S,T) \1_{ \Bond(S,T) \geq K }}{\Bond(t,T)} \ | \ \F_t \right) \\
		\Pi_2 &= \E_{\Pm_0} \left( \frac{\e^{ - \int_t^S r(s) \dx s } \1_{ \Bond(S,T) \geq K }}{\Bond(t,T)} \ | \ \F_t \right) .
	\end{align}
Here all the expectations are taken under the risk-free measure. We now write $\Pi_1$ under a different measure. As $\Bond(t,T) \geq K$ is equivalent to $\ln \Bond(t,T) \geq  \ln K$, we change the variable by $y = \ln \Bond(t,T)$ and get
	\begin{align}
		\label{fourierchangeofvariable}
		\Pi_1 &= \int_{\ln K}^{\infty} \left( \frac{\e^{ - \int_t^S r(s) \dx s } \Bond(S,T) }{\Bond(t,T)} f(y) \right) \ \dx y .
	\end{align}
We notice that
	\begin{align}
		\xi_1 (t) &= \frac{\Bond(S,T) \Bank(t)}{\Bond(t,T) \Bank(S)} \\
		&= \frac{ \Bond(S,T) }{\Bond(t,T)} \e^{ - \int_t^S r(s) \dx s } .
\end{align}
is the Radon-Nikod\'{y}m derivative of $T$-forward measure with respect to risk-free measure. Thus
	\begin{align}
		\Pi_1 &= \int_{\ln K}^{\infty} \xi_1(t) f(y) \ \dx y \\
			&= \int_{\ln K}^{\infty} f_1(y) \ \dx y
\end{align}
where $f_1$ is the probability density function under the $T$-forward measure. Let $g_1$ be the characteristic function of $T$-forward measure, hence
	\begin{align}
		g_1 (\omega) &= \int_{-\infty}^{\infty} \e^{i\omega y} f_1(y) \ \dx y \\
			&= \int_{-\infty}^{\infty} \left( \e^{i\omega y} \xi_1(t) f(y) \right) \ \dx y \\
			&= \E_{\Pm_0} \left( \e^{i\omega \ln \Bond(S,T)} \xi_1(t)  \ | \ \F_t \right) \text{ and } \\
		g_1 (\omega) \Bond(S,T) &= \E_{\Pm_0} \left( \e^{(1+i\omega) \ln \Bond(S,T)} \e^{ - \int_t^S r(s) \dx s } \ | \ \F_t \right)
	\end{align}
as, by Equation \ref{fourierbondaffineassumption},
	\begin{align}
		y = \ln \Bond(S,T) = A(\tau) - B^{\top} (\tau) X(t) - C^{\top} (\tau) Y(t) .
	\end{align}
Feynman-Kac theorem this expected value may be presented as a $N$-dimensional stochastic partial differential equation. \textcite{nawalkabeliaevasoto2007dynamic} shows that a solution is
\begin{align}
\exp \left( A_1^*(s) - \sum_{m=1}^M B_{1m}^*(s) X_m(t) - \sum_{i=1}^{N-M} C_{1m}^*(s) Y_i(t) \right) ,
\end{align}
where
\begin{align}
A_1^*(0) &= a_1 = A(U) ( 1+\boldsymbol{i} \omega ) \\
B_{1m}^*(0) &= b_{1m} = B_m(U) ( 1+\boldsymbol{i} \omega ) \\
C_{1i}^*(0) &= c_{1i} = C_i(U) ( 1+\boldsymbol{i} \omega )
\end{align}
for $i= 1,2, \ldots, N-M$ and $m=1,2, \ldots , M$. Here
\begin{align}
A_1^*(z) &= a_1  +  \frac{1}{2} \sum_{i=1}^{N-M} \sum_{j=1}^{N-M} \frac{\nu_i \nu_j \rho_{ij}}{k_ik_j} ( z - q_i C_i(z) - q_j C_j(z) \\ &+ q_iq_j \frac{1 - \exp^{ - (k_i+k_j)z } }{k_i+k_j} ) \\
&- 2 \sum_{m=1}^M \frac{\alpha_m \theta_m}{\sigma_m^2} \left( \beta_{3m} z + \log ( \frac{1 - \beta_ {4m} \exp^{ \beta_{1m} z } }{1 - \beta_{4m}} )  \right) \\
B_{1m}^*(z) &= \frac{2}{\sigma_m^2} \left( \frac{ \beta_{2m} \beta_{4m} \exp^{\beta_{1m} z} - \beta_{3m} }{ \beta_{4m} \exp^{\beta_{1m} z} - 1 } \right) \\
C_{1i}^* (z) &= \frac{1-q_i \exp^{-k_iz}}{k_i},
\end{align}
where
\begin{align}
q_i &= 1 - k c_{1j} \\
\beta_{1m} &= \sqrt{\alpha_m^2 + 2 \sigma_m^2} \\
\beta_{2m} &= \frac{- \alpha_m + \beta_{1m}}{2} \\
\beta_{3m} &= \frac{- \alpha_m - \beta_{1m}}{2} \\
\beta_{4m} &= \frac{ - \alpha_m - \beta_{1m} - b_{1m} \sigma_m^2 }{ - \alpha_m + \beta_{1m} - b_{1m} \sigma_m^2 }
\end{align}
for $i= 1,2, \ldots, N-M$ and $m=1,2, \ldots , M$. It should be noted that $a_1, b_{1m}$ and $c_{1i}$ are actually functions of $\omega$ and therefore $A_1^*, B_{1m}^*$ and $C_{1i}^*$ also depend on $\omega$.

Thus the characteristic function is
\begin{align}
g_1 (\omega) &= \frac{ \exp \left( A_1^*(s) - \sum_{m=1}^M B_{1m}^*(s) X_m(t) - \sum_{i=1}^{N-M} C_{1m}^*(s) Y_i(t)  \right) }{\Bond(t,T)}
\end{align}
which allows us to calculate the values of characteristic functions. Now
	\begin{align}
		\Pi_1 &= \int_{\ln K}^{\infty} f_1(y) \ \dx y \\
			&= \frac{1}{2} + \frac{1}{\pi} \int_0^{\infty} \Re \left( \frac{\exp^{-\boldsymbol{i}\omega \log K} g_1(\omega)}{\boldsymbol{i} \omega} \right) \dx \omega
	\end{align}
which can be calculated numerically. \textcite{nawalkabeliaevasoto2007dynamic} note that this computation only requires that the model has analytical bond pricing formulas, so it can be utilized in variety of models.

We can solve $\Pi_2$ similarly. Instead of Equation \ref{fourierchangeofvariable}, we have
	\begin{align}
		\Pi_2 &= \int_{\ln K}^{\infty} \left( \frac{\e^{ - \int_t^S r(s) \dx s } \Bond(S,T) }{\Bond(t,T)} f(y) \right) \ \dx y .
	\end{align}
and similar reasoning shows that
	\begin{align}
		\Pi_2 = \frac{1}{2} + \frac{1}{\pi} \int_0^{\infty} \Re \left( \frac{\exp^{-\boldsymbol{i}\omega \log K} g_2(\omega)}{\boldsymbol{i} \omega} \right) \dx \omega .
	\end{align}
Now
	\begin{align}
		g_2 (\omega) &= \frac{ \exp \left( A_2^*(s) - \sum_{m=1}^M B_{2m}^*(s) X_m(t) - \sum_{i=1}^{N-M} C_{1m}^*(s) Y_i(t)  \right) }{\Bond(t,S)}
	\end{align}
where
\begin{align}
A_2^*(0) &= a_2 = A(U) ( \boldsymbol{i} \omega ) \\
B_{2m}^*(0) &= b_{2m} = B_m(U) ( \boldsymbol{i} \omega ) \\
C_{2i}^*(0) &= c_{2i} = C_i(U) ( \boldsymbol{i} \omega )
\end{align}
for $i= 1,2, \ldots, N-M$ and $m=1,2, \ldots , M$. Similarly as before
\begin{align}
A_2^*(z) &= a_2  +  \frac{1}{2} \sum_{i=1}^{N-M} \sum_{j=1}^{N-M} \frac{\nu_i \nu_j \rho_{ij}}{k_ik_j} ( z - q_i C_i(z) - q_j C_j(z) \\ &+ q_iq_j \frac{1 - \exp^{ - (k_i+k_j)z } }{k_i+k_j} ) \\
&- 2 \sum_{m=1}^M \frac{\alpha_m \theta_m}{\sigma_m^2} \left( \beta_{3m} z + \log ( \frac{1 - \beta_ {4m} \exp^{ \beta_{1m} z } }{1 - \beta_{4m}} )  \right) \\
B_{2m}^*(z) &= \frac{2}{\sigma_m^2} \left( \frac{ \beta_{2m} \beta_{4m} \exp^{\beta_{1m} z} - \beta_{3m} }{ \beta_{4m} \exp^{\beta_{1m} z} - 1 } \right) \\
C_{2i}^* (z) &= \frac{1-q_i \exp^{-k_iz}}{k_i},
\end{align}
where
\begin{align}
q_i &= 1 - k c_{2j} \\
\beta_{1m} &= \sqrt{\alpha_m^2 + 2 \sigma_m^2} \\
\beta_{2m} &= \frac{- \alpha_m + \beta_{2m}}{2} \\
\beta_{3m} &= \frac{- \alpha_m - \beta_{2m}}{2} \\
\beta_{4m} &= \frac{ - \alpha_m - \beta_{2m} - b_{2m} \sigma_m^2 }{ - \alpha_m + \beta_{2m} - b_{2m} \sigma_m^2 }
\end{align}
for $i= 1,2, \ldots, N-M$ and $m=1,2, \ldots , M$. Again, it should be noted that $a_2, b_{2m}$ and $c_{2i}$ are actually functions of $\omega$ and therefore $A_2^*, B_{2m}^*$ and $C_{2i}^*$ also depend on $\omega$.

Since the computations above involves numerical integration, it is computationally costly. However \textcite{carrmadan1999optionvaluation} showed that Fast Fourier Transform (FFT) can utilized in the computation, which can reduce the complexity significantly. Sadly, this was not attempted in this thesis.

As we explicitly assumed that $\delta = 0$, we can add the shift to the model by the extension method introduced in Section \ref{sec:dynamicextension}.


\section{Cellular Complexes with GraphBLAS }\label{graphblas-implementation}
%=================================================================

We have implemented in Julia~\cite{BEKS14}---the novel language for scientific computing---our topological operations over cell complexes, using the package \href{https://github.com/abhinavmehndiratta/GraphBLAS.jl}{\texttt{SuiteSparseGraphBLAS.jl}}, which is a Julia wrapper~\cite{Mehndiratta:2019} for \texttt{SuiteSparse:GraphBLAS}, i.e., the GraphBLAS standard~\cite{osti_1208646,DBLP:journals/corr/KepnerABBFGHKLM16,Buluc:7965104} provided within the \emph{SuiteSparse} library~\cite{Davis:2018}  of sparse matrix software.

\subsection{Geometric / topological sparse matrices}
%------------------------------------------------

Within a typical computational pipeline in geometric applications, we may distinguish at least three types of sparse matrices: (a) characteristic matrices of cells as subsets of vertices; (b) boundary representations of edges, faces and solid cells; (c) matrix representation of binary incidence/adjacency relations between cells of different dimension.

\paragraph{Characteristic matrices} provide the simplest representation of the independent elements (i.e., those that cannot be generated by linear combination of other elements) of a $p$-chain space $C_p$ ($0\leq p \leq d$) from a cellular $d$-complex ($2\leq d \leq 3$). They are built from arrays of arrays of vertex indices using the sol-called ``coords'' method, i.e., starting from $(i,j,x)$ triples.

\paragraph{Boundary operators} give a simple mathematical representation of the so-called ``B-reps'', normally used for solid models, in our case extended to cells of every dimension. In particular, every column of the $[\partial_p]$ matrix gives the signed representation of a basis $p$-cell (i.e., an independent $p$-chain) as a ($p$-1)-cycle (i.e., a ($p$-1)-chain without-boundary). They are built by multiplication of two characteristic matrices, followed by suitable ``filtering'' of  values.

\paragraph{Incidence relations} are generated by multiplication of the characteristic matrices of the two types of cells (of dimension $p$ and $q$, say) under consideration. The non-zero $(i,j)$ element of the product matrix provides the ``strength'' of the elementary incidence, i.e., the number of vertices shared between the $i$-th $p$-cell and the $j$-th $q$-cell.  Their building is done by multiplication of two appropriate instances of (co)boundary matrices, as shown in Table~\ref{tab:relations}b.

\subsection{Matrix computation of boundary chain}
%------------------------------------------------

The mathematics used to compute the graded chain complex $C_\bullet = (C_p, \partial_p)$ starting from sparse binary characteristic matrices $M_p$, with $p$-cells indexing the rows and $0$-cells indexing the columns is given below.
The boundary matrices $\partial_p$ ($1\leq p\leq 3$) between non-oriented chain spaces are computed by \emph{sparse matrix multiplication} of characteristic matrices, followed by \emph{matrix filtering},  produced in Julia by broadcasting vectorized integer division, i.e., ``$\texttt{.}\!\div$'', as follows:

{\footnotesize\begin{lstlisting}
$\partial_1 \texttt{ = } \texttt{M}_0 * \texttt{M}_1' \texttt{ = } \texttt{M}_1'$
$\partial_2 \texttt{ = } (\texttt{M}_1 * \texttt{M}_2')\ \texttt{.}\!\div\ \texttt{sum(}\texttt{M}_1,\texttt{dims=}2)$
$\partial_3 \texttt{ = } (\texttt{M}_2 * \texttt{M}_3')\ \texttt{.}\!\div\ \texttt{sum(}\texttt{M}_2,\texttt{dims=}2)$
\end{lstlisting}}

\subsection{GraphBLAS computation of boundary chain}
%------------------------------------------------

Our current open-source implementation of cellular and chain complexes using sparse matrices and \href{https://github.com/abhinavmehndiratta/GraphBLAS.jl}{\texttt{SuiteSparseGraphBLAS.jl}} is mantained in \href{https://github.com/gmgigi96/SparseMM}{\texttt{https://github.com/gmgigi96/SparseMM}}, where it provides fast and easy-to-use matrix tools for geometric and topological computing, including the input of cellular complexes, the computation of (unsigned) boundary operators, the answer to single and multiple queries about incidence or adjacency of cells.


\documentclass{article}

%\documentclass[aps,pra,notitlepage,amsmath,amssymb,letterpaper,12pt]{revtex4-1}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{titling}
\usepackage{amsthm}
\usepackage{graphicx}
\usepackage{epstopdf}

\epstopdfDeclareGraphicsRule{.gif}{png}{.png}{convert gif:#1 png:\OutputFile}
\AppendGraphicsExtensions{.gif}

%  Below define helpful commands to set up problem environments easily
\newenvironment{problem}[2][Problem]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}\hskip \labelsep {\bfseries #2.}]}{\end{trivlist}}
\newenvironment{solution}{\begin{proof}[Solution]}{\end{proof}}

% --------------------------------------------------------------
%                   Document Begins Here
% --------------------------------------------------------------

\begin{document}

\title{definition of derivative of a function }
\author{Jianhua Li}
\date{\today}

\maketitle

\begin{abstract}
Create a LaTeX file testlatex.tex (using the template in the info repository). In this file write an explanation of what the definition of the derivative $f'(x)$ of a function $f(x)$ means. Include both inline and numbered equations, as well as a proper title, abstract, and section headings. Find a suitable image to illustrate your definition online, and include it as a figure, with proper citation of the source.
\end{abstract}

%\section{Section Title Here} % Specify main sections this way

\section{Definition}
%The derivative of f(x) is defined as the limit:
%\lim_{x\to\infty} f(x) $$= 2x
%https://apcalcnotebookarjw.wikispaces.com/Definition+of+the+derivative+of+a+function

\begin{align}
  %f(x) = x^{2}
  f'(x_0) = \lim_{x\to x_0} \frac{f(x_0)-f(x)}{x_0-x}\\
  f'(x) = \lim_{h\to\infty} \frac{f(x+h)-f(x)}{h} %it worked
  %f'(x_0) = \lim_{x\to\x_0}\frac{f(x_0)-f(x)}{x_0-x}
\end{align}

%\begin{align}
	%f'(x_0) &=\lim_{x\to\x_0}\frac{f(x_0)-f(x)}{x_0-x}
   %$$f'(x) &=\lim_{h\to\infty}f(x)\frac{f(x+h)-f(x)}{h}
%\[
%    \binom{n}{k} = \frac{n!}{k!(n-k)!}
%\]
%f(x) &= 2x^2-16x+35\\
%f'(x) &= 2x+2\\
%\[ \lim_{x \to +\infty} \frac{3x^2 +7x^3}{x^2 +5x^4} = 3.\]

%\end{align}
% Use align environments for equations. The \\ is a newline character. The & is the alignment character.
% Using align* or \nonumber on each line removes equation numbers

Example Illustration:

\begin{figure}[h!] % h forces the figure to be placed here, in the text
  \includegraphics{250px-Lim-secant.png}  % if pdflatex is used, jpg, pdf, and png are permitted
  \caption{A secant approaches a tangent when h approach to 0,Source https://en.wikipedia.org/wiki/Derivative}
  \label{fig:tangent}
\end{figure}

This text should be below the figure unless \LaTeX  decides that a different layout works better.

% Repeat as needed

\end{document}


\chapter{Schr\"{o}dinger Wave Equation}\label{c4}
Bohr's model explained the spectrum of the Hydrogen atom quite well. But not
too well. When a Hydrogen atom is subjected to an electric field, the spectral
lines split. This is called the `Stark effect' after the name of its discoverer.
Similarly, `Zeeman effect' is the splitting of spectral lines when the atom is
subjected to a magnetic field. By splitting we mean that in the place of a
single line in the absence of a field we now see more than one. A spectral line
indicates the energy level of an electron. A splitting of a spectral line tells
us that electrons which used to have the same energies in the absence of a
field have different energies when the field is turned on. The Bohr model
could not explain why this happens. Arnold Sommerfeld tried to extend the Bohr
model by allowing the orbits to be elliptical and by using certain concepts in
advanced classical mechanics. However, it was clear that the theory was getting
\emph{ad hoc} and that a better explanation was needed.

Werner Heisenberg, a German physicist, first came up with a new theory to
explain quantum phenomena \cite{heisenberg1925quantum}. His paper is quite
hard to follow. People of Steven Weinberg's \cite{aitchison2004understanding}
stature have admitted that while they understand what Heisenberg wrote they
cannot fathom what inspired him to write it. He also found out that the
equations of quantum mechanics are best written in terms of matrices. As a
result, his formulation of quantum mechanics is called `matrix mechanics'.

A year later, the Austrian physicist Erwin Schr\"{o}dinger proposed an
alternative formalism to solve atomic problems. His `equation of motion' was
conceived using certain optical analogies and is called the wave equation. A
function satisfying the equation is called the wave function and it describes
the matter waves we studied in chapter \ref{c2}. Schr\"{o}dinger's paper
\cite{schrodinger1926undulatory} is remarkably lucid and it offers an insight
into how he came up with his equation. The Schr\"{o}dinger wave equation
is a linear partial differential equation and the physicists of
Schr\"{o}dinger' time had a expertise and comfort in dealing with them. As a
result, a first introduction to quantum mechanics is almost always using the
Schr\"{o}dinger equation. Schr\"{o}dinger showed \cite{schrodinger1926relation}
that his approach is equivalent to the one proposed by Heisenberg and developed
by Born and Jordan. The equivalence was put on a firm mathematical footing by
John von Neumann \cite{von2018mathematical}.

In this chapter, we will state Schr\"{o}dinger's equation, interpret its
solution and show how to use it in a few very simple situations. Schr\"{o}dinger
derived \cite{schrodinger1926undulatory} his equation using variational
principles. However, we shall take it as a starting point of our discussion.
The way we accepted Newton's laws as `correct' because they explained a few
natural phenomena, we will also accept Schr\"{o}dinger's equation because it
explains a lot many phenomena in the atomic world.

\section{Schr\"{o}dinger equation}\label{c4s1}
Before stating and understanding Schr\"{o}dinger's equation that describes
the unfamiliar matter waves, let us consider start in the more familiar world
of sound waves. We know that sound waves travel through the air as a succession
of compressions and rarefactions. We can detect and measure the compression and
rarefaction using a pressure gauge. If $p(x, t)$ is the air pressure at a point
$x$ and at time $t$ then the equation of sound waves is
\begin{equation}\label{c4s1e1}
\frac{\partial^2 p}{\partial x^2} = \frac{1}{c_s^2}
\frac{\partial^2 p}{\partial t^2}
\end{equation}
where $c_s$ is the speed of sound. It is easy to check that any function of
the form $p(x \pm c_st)$ is a solution of \eqref{c4s1e1}. In particular,
\begin{equation}\label{c4s1e2}
p(x, t) = \cos(x - c_st)
\end{equation}
is a solution. A solution of a wave equation tells us how pressure varys with
position and time.

Equation \eqref{c4s1e1} described waves in one dimension. If the pressure varies
in all three dimensions then the equation becomes
\begin{equation}\label{c4s1e3}
\frac{\partial^2 p}{\partial x^2} + \frac{\partial^2 p}{\partial y^2} +
\frac{\partial^2 p}{\partial z^2} =
\frac{1}{c_s^2} \frac{\partial^2 p}{\partial t^2}
\end{equation}
We can check that
\begin{equation}\label{c4s1e4}
p(x, y, z, t) = \cos(x + y + z - \sqrt{3} c_st)
\end{equation}
is a solution of \eqref{c4s1e4}.

\subsection{Problem set 1}
\begin{enumerate}
\item Verify that \eqref{c4s1e2} is a solution of \eqref{c4s1e1}.
\item Verify that \eqref{c4s1e4} is a solution of \eqref{c4s1e3}.
\item If $p_1(x, t)$ and $p_2(x, t)$ are two solutions of \eqref{c4s1e1} then
show that $\alpha_1 p_1 + \alpha_2 p_2$, where $\alpha_1, \alpha_2$ are real
numbers is also a solution of \eqref{c4s1e1}.
This fact is called the principle of linear superposition. The expression,
$\alpha_1 p_1 + \alpha_2 p_2$ is called a \emph{linear combination} of $p_1$
and $p_2$.
\end{enumerate}

A sound wave is described by the pressure fluctations it causes. de Broglie
waves are described by the variations of a wave function $\Psi$. It could be a
complex-valued function and therefore it does not have a physical interpretation
by itself. However, its squared modulus $|\Psi|^2 = \Psi\Psi^\ast$ is the
probability density of finding at a particle at a point $(x, y, z)$ at time $t$.
In particular, $|\Psi|^2 dV$ is the probability of finding a particle in a
small volume element $dV$ around the point $(x,y,z)$ at time $t$. The way the
air pressure $p$ satisfies the equation \eqref{c4s1e3}, the wave function
$\Psi$ satisfies
\begin{equation}\label{c4s1e5}
-\frac{\hslash^2}{2m}\left(\frac{\partial^2\Psi}{\partial x^2} +
\frac{\partial^2\Psi}{\partial y^2} + \frac{\partial^2\Psi}{\partial z^2}
\right) + V(x,y,z,t)\Psi = i\hslash\frac{\partial\Psi}{\partial t}.
\end{equation}
This is called the Schr\"{o}dinger equation. In this equation,
\begin{enumerate}
\item $\hslash = h/(2\pi)$; we first introduced this quantity when we wrote
\eqref{c3s3e2}.
\item $m$ is the mass of the particle.
\item $V(x, y, z, t)$ is the potential function. It describes the force acting
on the particle. We will describe it more a little later.
\item $i$ is the imaginary number.
\end{enumerate}
A few remarks are in order.
\begin{enumerate}
\item A wave equation always has a second order partial derivative in time.
Equation \eqref{c4s1e5} has just the first order partial derivative in time.
Therefore, strictly speaking, \eqref{c4s1e5} is not a wave equation although
the name `wave function' persists for its solution $\Psi$. It is closer to the
heat equation than it is to a wave equation.
\item The coefficient on the right hand side is an imaginary number. Therefore,
it is not surprising that its solution $\Psi$ is complex valued.
\item The only term in \eqref{c4s1e5} that is specific to a problem is the
potential function $V$. The other terms never change.
\item We remarked that $|\Psi|^2$ has an interpretation of a probability
density function. Therefore, if we want to find the probability of finding
a particle a region of volume $V$ then it is
\begin{equation}\label{c4s1e6}
P(V) = \int_V |\Psi|^2 dV.
\end{equation}
This statement is accurate only when
\begin{equation}\label{c4s1e7}
\int_{-\infty}^\infty |\Psi|^2dV = 1.
\end{equation}
Such a function is said to be `normalised'. If a wave function is not normalised
, that is, if
\begin{equation}\label{c4s1e8}
\int_{-\infty}^\infty |\Psi|^2dV = N,
\end{equation}
then we can alway consider the function
\begin{equation}\label{c4s1e9}
\Psi_1 = \frac{1}{\sqrt{N}}\Psi.
\end{equation}
We will show in the next problem set that $\Psi_1$ is also a solution of
\eqref{c4s1e5}. $\Psi_1$ is called the \emph{normalised} wave function and
the number $1/\sqrt{N}$ is called the \emph{normalisation constant}.
\end{enumerate}

\subsection{Problem set 2}
\begin{enumerate}
\item Show that is $\Psi_1$ and $\Psi_2$ are two solutions of \eqref{c4s1e5}
then $\alpha_1\Psi_1 + \alpha_2\Psi_2$, where $\alpha_1$ and $\alpha_2$ are
complex numbers, is also a solution. Therefore, show that \eqref{c4s1e9} is
also a solution of \eqref{c4s1e5}.
\end{enumerate}

\section{Some solutions of Schr\"{o}dinger equation}\label{c4s2}
Partial differential equations are, in general, much more difficult to solve
than ordinary differential equations. Schr\"{o}dinger equation is no exception.
We will consider a few situations where an explicit, closed-form solution is
available. In the previous section we remarked that the physics of a problem
is in the choice of the potential function $V$. The situations we alluded to
are really the different choices of $V$.

\subsection{The free particle}
A free particle is the one not subject to any forces. Therefore, a convenient
choice of $V$ is to set it to zero. Schr\"{o}dinger equation now becomes
\begin{equation}\label{c4s2e1}
-\frac{\hslash^2}{2m}\left(\frac{\partial^2\Psi}{\partial x^2} +
\frac{\partial^2\Psi}{\partial y^2} + \frac{\partial^2\Psi}{\partial z^2}
\right) = i\hslash\frac{\partial\Psi}{\partial t}.
\end{equation}
Since the particle has no force acting on it, its energy does not change. If
$E$ is its energy then we can write
\begin{equation}\label{c4s2e2}
\Psi(\vec{r}, t) = \psi(\vec{r})e^{-iEt/\hslash}.
\end{equation}
Substituting it in \eqref{c4s2e1} we get
\begin{equation}\label{c4s2e3}
-\frac{\hslash^2}{2m}e^{-iEt/\hslash}\left(\frac{\partial^2\psi}{\partial x^2}
+ \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}
\right) = E\psi(\vec{r})e^{-iEt/\hslash}.
\end{equation}
Cancelling the factor $e^{-iEt/\hslash}$ and rearranging a bit, we get
\begin{equation}\label{c4s2e4}
\frac{\partial^2\psi}{\partial x^2}
+ \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2} =
-\frac{2mE}{\hslash^2}\psi.
\end{equation}
Since we chose $V = 0$, the particle has no potential energy. Its energy is
entirely kinetic. If $E$ is the kinetic energy then we know that the quantity
$2mE$ is the square of momentum. The coefficient on the right hand side is
\begin{equation}\label{c4s2e5}
-\frac{2mE}{\hslash^2} = -\frac{p^2}{\hslash^2} = -k^2,
\end{equation}
where we have use the de Broglie relation,
\begin{equation}\label{c4s2e6}
p = \frac{h}{\lambda} = \frac{\hslash}{k}.
\end{equation}
Thus, the Schr\"{o}dinger equation simplifies to
\begin{equation}\label{c4s2e7}
\frac{\partial^2\psi}{\partial x^2}
+ \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2} =
-k^2\psi.
\end{equation}
If we define the vector
\begin{equation}\label{c4s2e8}
\vec{k} = k_x\hat{i} + k_y\hat{j} + k_z\hat{k},
\end{equation}
and let
\begin{equation}\label{c4s2e9}
\psi(\vec{r}) = e^{i\vec{k}\cdot\vec{r}} = e^{i(k_xx+k_yy+k_zz)},
\end{equation}
then we can readily verify that $\psi$ defined in \eqref{c4s2e9} satisfies
\eqref{c4s2e7}. Thus, the complete solution of equation \eqref{c4s2e1} is
\begin{equation}\label{c4s2e10}
\Psi(\vec{r}, t) = e^{i(\vec{k}\cdot\vec{r} - \omega t)},
\end{equation}
where we have used the relation
\begin{equation}\label{c4s2e11}
E = \hslash\omega.
\end{equation}
The solution of \eqref{c4s2e10} is called a plane wave solution. It can have
any value for its energy $E$.

\subsection{Time-independent potentials}
When $V$ is independent of time we can write
\begin{equation}\label{c4s2e12}
\Psi(\vec{r}, t) = \psi(\vec{r})\tau(t),
\end{equation}
where $\tau$ is a function of time alone. Equation \eqref{c4s1e5} becomes
\begin{equation}\label{c4s2e13}
\tau(t)\left[-\frac{\hslash^2}{2m}\left(\frac{\partial^2\psi}{\partial x^2} +
\frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}
\right) + V(x,y,z)\psi\right] = i\psi(\vec{r})\hslash\frac{d\tau}{dt}
\end{equation}
Divide both sides by $\Psi = \psi\tau$ to get
\begin{equation}\label{c4s2e14}
\frac{1}{\psi(\vec{r})}\left[-\frac{\hslash^2}{2m}\left(
\frac{\partial^2\psi}{\partial x^2} +
\frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}
\right) + V(x,y,z)\psi\right] = \frac{i\hslash}{\tau(t)}\frac{d\tau}{dt}
\end{equation}
The left hand side of this equation depends on $\vec{r}$ alone and the
right hand side on $t$ alone. Therefore, each side must be a constant, say $E$.
Thus, we get two equations
\begin{equation}\label{c4s2e15}
-\frac{\hslash^2}{2m}\left(\frac{\partial^2\psi}{\partial x^2} +
\frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}
\right) + V(x,y,z)\psi = E\psi
\end{equation}
and
\begin{equation}\label{c4s2e16}
\frac{1}{\tau}\frac{d\tau}{dt} = -i\frac{E}{\hslash}.
\end{equation}
Since $E$ is a constant, equation \eqref{c4s2e16} is particularly easy to
solve. We see that
\begin{equation}\label{c4s2e17}
\tau(t) = \alpha e^{-iEt/\hslash},
\end{equation}
where $\alpha$ is a constant of integration that can be chosen while normalising
the wavefunction.

Equation \eqref{c4s2e15} is called  Schr\"{o}dinger time independent equation.
In this course, we will deal only with the situation when $V$ is independent of
$t$ and therefore we will focus on solving \eqref{c4s2e15}.

\subsection{Particle in an infinite potential well}
The free particle is a rather uninteresting. It does not have any features that
are peculiar to a quantum system. We now consider a particles in a box. The box
prevents the particle from escaping and therefore one can consider it be
represented by an infinitely high potential wall. That is, it takes an infinte
energy for the particle to escape it. Let us start with a one-dimensional case
where the potential is defined by
\begin{equation}\label{c4s2e18}
V(x) = \begin{cases}
0 \text{ if } -\frac{L}{2} \le x \le \frac{L}{2} \\
\infty \text{ otherwise.}
\end{cases}
\end{equation}
Within the confines of the box, the particle is free. Without, it cannot even
get. Therefore, we must solve  Schr\"{o}dinger's time independent equation only
in the region
\[
-\frac{L}{2} \le x \le \frac{L}{2}.
\]
Since we are considering a one dimensional box, \eqref{c4s2e15} becomes
\begin{equation}\label{c4s2e19}
-\frac{\hslash^2}{2m}\frac{d^2\psi(x)}{dx^2} = E\psi(x).
\end{equation}
Let
\begin{equation}\label{c4s2e20}
k^2 = \frac{2mE}{\hslash^2}.
\end{equation}
$k$ is called the wave number and it is related to the momentum by the equation
\begin{equation}\label{c4s2e21}
p = \hslash k.
\end{equation}
In terms of $k$, equation \eqref{c4s2e19} becomes
\begin{equation}\label{c4s2e22}
\frac{d^2\psi(x)}{dx^2} + k^2\psi(x) = 0.
\end{equation}
One can check that the general solution of this ordinary differential equation
is
\begin{equation}\label{c4s2e23}
\psi(x) = \alpha e^{-ikx} + \beta e^{ikx},
\end{equation}
where $\alpha$ and $\beta$ are constants of integration. The infinite potential
barrier of the box prevents the particle from escaping. Therefore, at the walls,
the function $\psi$ must be zero. That is
\begin{eqnarray}
\psi\left(-\frac{L}{2}\right) &=& 0 \label{c4s2e24} \\
\psi\left(\frac{L}{2}\right) &=& 0 \label{c4s2e25}
\end{eqnarray}
From \eqref{c4s2e23} and \eqref{c4s2e24} we have
\[
\alpha e^{ikL/2} + \beta e^{-ikL/2} = 0
\]
or that
\begin{equation}\label{c4s2e26}
\alpha = -\beta e^{-ikL}
\end{equation}
and hence equation \eqref{c4s2e23} becomes
\begin{equation}\label{c4s2e27}
\psi(x) = -\beta e^{-ikL}e^{-ikx} + \beta e^{ikx}.
\end{equation}
From \eqref{c4s2e25} and \eqref{c4s2e27},
\[
\psi\left(\frac{L}{2}\right) = -\beta e^{-3ikL/2} + \beta e^{ikL/2} =
-\beta e^{-ikL/2}(e^{-ikL} - e^{ikL}) = 0
\]
or
\begin{equation}\label{c4s2e28}
-2i\beta e^{ikL/2}\sin(kL) = 0
\end{equation}
which can be true only if
\begin{equation}\label{c4s2e29}
kL = n\pi, n = 0, \pm 1, \pm 2, \ldots.
\end{equation}
If $n = 0$ then $k = 0$ and from \eqref{c4s2e23} $\psi(x) = \alpha + \beta$.
Such a solution will not obey the boundary conditions \eqref{c4s2e24} and
\eqref{c4s2e25} unless $\alpha = \beta = 0$. But this is a trivial solution
indicating that there is no particle. Therefore, the only physically interesting
values of $k$ are
\begin{equation}\label{c4s2e30}
k_n = \frac{n\pi}{L}, n = \pm 1, \pm 2, \ldots.
\end{equation}
As a result of this condition, we see that the momentum of the particle is
restricted to
\begin{equation}\label{c4s2e31}
p_n = \frac{n\pi\hslash}{L}, n = \pm 1, \pm 2, \ldots.
\end{equation}
Thus, the particle's momentum is quantised. So is energy, as seen from
\eqref{c4s2e20},
\begin{equation}\label{c4s2e32}
E_n = \frac{n^2\pi^2\hslash^2}{2mL^2}, n = \pm 1, \pm 2, \ldots.
\end{equation}
The different values of $n$ are the energy levels of the particle. If the
particle has to be pushed from a lower energy then it must be supplied with an
energy equal to the difference between the two energy levels. On the other hand,
when a particle makes a transition from a higher energy level to a lower on it
emits a photon of frequency $\nu$ such that $\Delta E = h\nu$.

Equations \eqref{c4s2e31} and \eqref{c4s2e32} indicate that both $p_n$ and
$E_n$ increase with $n$. The lowest energy of the particle is
\begin{equation}\label{c4s2e33}
E_1 = \frac{\pi^2\hslash^2}{2mL^2}.
\end{equation}
The lowest energy of a classical particle is zero. It remains at rest in the
box. However, a quantum particle cannot be at rest because that would violate
the uncertainty principle.

We still have not written the complete solution of the Schr\"{o}dinger equation
\eqref{c4s2e22}. From \eqref{c4s2e27} and \eqref{c4s2e30}, we have
\begin{equation}\label{c4s2e34}
\psi_n(x) = \beta(e^{in\pi x/L} + (-1)^n e^{-in\pi x/L}).
\end{equation}
We can write it in a simpler form
\begin{equation}\label{c4s2e35}
\psi_n(x) = \begin{cases}
2\beta\cos\left(\frac{n\pi x}{L}\right) & \text{ if } n = \pm 2, \pm 4, \ldots\\
2i\beta\sin\left(\frac{n\pi x}{L}\right) & \text{ if } n = \pm 1, \pm 3, \ldots.
\end{cases}
\end{equation}
The constant $\beta$ can be found using the normalisation condition
\begin{equation}\label{c4s2e36}
\int_{-L/2}^{L/2} |\psi_n(x)|^2dx = 1.
\end{equation}
The integration is quite straigtforward and we get
\begin{equation}\label{c4s2e37}
\beta = \frac{1}{\sqrt{2L}}.
\end{equation}
Thus, the complete solution is
\begin{equation}\label{c4s2e38}
\psi_n(x) = \begin{cases}
\sqrt{\frac{2}{L}}\cos\left(\frac{n\pi x}{L}\right) &
   \text{ if } n = \pm 2, \pm 4, \ldots \\
\sqrt{\frac{2}{L}}i\sin\left(\frac{n\pi x}{L}\right) &
   \text{ if } n = \pm 1, \pm 3, \ldots.
\end{cases}
\end{equation}

\subsection{Problem set 3}
\begin{enumerate}
\item Energy is indendepent of the sign of $n$, but momemtum is not. What is
the physical significance of the this observation?
\item Choose any $n$, say $n = 4$, for which
\[
\psi_4(x) = \sqrt{\frac{2}{L}}\cos\left(\frac{4\pi x}{L}\right).
\]
At what $x$ is $\psi_4(x) = 0$? Points at which $\psi$ vanishes are called
nodes.  Thus, there are some points at which the particle just cannot be. This
is another stark difference between the classical and the quantum particles.
\item The normalisation condition is defined by \eqref{c4s1e7}. Why did we
change the limits of integration from $\pm\infty$ to $\pm L/2$?
\end{enumerate}

\section{Operators and eigenvalues}\label{c4s3}
The wavefunction $\Psi$ has the entire information about the quantum particle.
If we know $\Psi$, we can find the particle's position, momentum, angular
momentum and energy, up to a certain extent. We have learnt that
\begin{itemize}
\item At an atomic scale, particles have a wavelike character and therefore
do not have a precise position or momentum.
\item Therefore, they cannot have precise angular momentum.
\item Under some situations, they do have definite values of energy.
\end{itemize}
How to extract this information from the wavefunction $\Psi$? Dynamical
variables are represented by operators in quantum mechanics. An operator is
something that maps one function to another. We shall denote operators with
a hat on top of a letter. From now on, we shall not use the hat notation for
unit vectors. Some examples of operators in
quantum mechanics are
\begin{enumerate}
\item The position operator is just $\hat{r}$. Its action on the wavefunction is
\begin{equation}\label{c4s3e1}
\hat{r}(\Psi) = \vec{r}\Psi.
\end{equation}
\item The momentum operator $\hat{p}$ is
\begin{equation}\label{c4s3e2}
\hat{p}(\Psi) = (i\hslash)\left(\vec{i}\frac{\partial}{\partial x} +
\vec{j}\frac{\partial}{\partial y} + \vec{k}\frac{\partial}{\partial z}
\right)\Psi
= i\hslash\nabla\Psi.
\end{equation}
\item The energy operator $\hat{E}$ is
\begin{equation}\label{c4s3e3}
\hat{E}(\psi) = i\hslash\frac{\partial\Psi}{\partial t}.
\end{equation}
\end{enumerate}
When the potential energy is a function of $\vec{r}$ alone then the potential
energy operator $\hat{V}(\vec{r})$ is
\begin{equation}\label{c4s3e4}
\hat{V}(\vec{r})(\Psi) = V(\vec{r})\Psi.
\end{equation}
The relation between kinetic energy and momentum is $T = p^2/(2m)$. Therefore,
we can infer that the relationship between the operators is
\begin{equation}\label{c4s3e5}
\hat{T}(\Psi) = \frac{\hat{p}^2}{2m}(\Psi).
\end{equation}
Using \eqref{c4s3e2} we can easily infer that
\begin{equation}\label{c4s3e6}
\hat{T}(\Psi) = -\frac{\hslash^2}{2m}\left(\frac{\partial^2}{\partial x^2}
+ \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial x^2}\right)
(\Psi)
\end{equation}
The fact that total energy is the sum of kinetic and potential energies can be
expressed in terms of operators are
\begin{equation}\label{c4s3e7}
\hat{E}(\Psi) = \hat{T}(\Psi) + \hat{V}(\Psi).
\end{equation}
Using equations \eqref{c4s3e3}, \eqref{c4s3e4} and \eqref{c4s3e6} we get
\begin{equation}\label{c4s3e8}
i\hslash\frac{\partial\Psi}{\partial t} = -\frac{\hslash^2}{2m}
\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +
 \frac{\partial^2}{\partial z^2}\right)\Psi + V(\vec{r})\Psi.
\end{equation}
This is Schr\"{o}dinger equation. It is just an expression of the fact that the
total energy is the sum of kinetic and potential energies.

If $\hat{O}$ is an operator then an equation of the form
\begin{equation}\label{c4s3e9}
\hat{O}\Psi = \lambda\Psi,
\end{equation}
where $\lambda$ is a number, is called an `eigenvalue' equation. The number
$\lambda$ is an `eigenvalue` of the operator $\hat{O}$ and the function $\Psi$
is its `eigenfunction'. An example of an eigenvalue equation is
\begin{equation}\label{c4s3e10}
-i\hslash \frac{d}{dx} e^{ikx} = \hslash k e^{ikx}.
\end{equation}

The total energy operator is called the Hamiltonian and is denoted by $\hat{H}$.
It is named after the Irish physicist William Rowan Hamilton who gave an
alternative formulation of classical mechanics. Thus,
\begin{equation}\label{c4s3e11}
\hat{H} = \hat{T} + \hat{V} = -\frac{\hslash^2}{2m}\nabla^2 + V,
\end{equation}
where the symbol $\nabla^2$ stands for the Laplacian and is defined as
\begin{equation}\label{c4s3e12}
\nabla^2 = \frac{\partial^2}{\partial x^2} +
\frac{\partial^2}{\partial y^2} +  \frac{\partial^2}{\partial z^2}.
\end{equation}

\subsection{Problem set 4}
\begin{enumerate}
\item Let $\hat{A}$ and $\hat{B}$ be commute. That is, let $\hat{A}\hat{B} =
\hat{B}\hat{A}$. Show that if $f$ is an eigenfunction of $A$ then it is also an
eigenfunction of $\hat{B}$.
\item Eigenvalues of an operator are the only possible outcomes of an experiment
measuring the dynamical variable that the operator represents. From this fact
and the previous problem conclude that if two operators commute then they can
be measured simultaneously.
\item Find $[\hat{x}, \hat{p}_x]$. Can they be measured simultaneously?
\end{enumerate}

\section{Postulates of quantum mechanics}\label{c4s4}
Non-relativistic quantum mechanics of a single is based on the following
postulates.
\begin{enumerate}
\item The state of a quantum mechanical system of one particle is given by a
wavefunction $\Psi(\vec{r}, t)$.
\item A dynamical variable of classical mechanics is represented by a
(Hermitian) operator in quantum mechanics.
\item The only values of an operator in a measurement are its eigenvalues.
\item The state of a quantum mechanical system can be expressed as a linear
combination of all the eigenfunctions of an operator. If $\hat{O}$ is an
operator with eigenvalue equations
\begin{equation}\label{c4s4e1}
\hat{O}\Psi_i = \lambda_i\Psi, i = 1, \ldots, n,
\end{equation}
and if $\Phi = c_1\Psi_1 + \cdots + c_n\Psi_n$ is an arbitrary normalised state
then the probability of an experiment giving an eigenvalue $\lambda_i$ is
\begin{equation}\label{c4s4e2}
P(\lambda_i) = |c_i|^2.
\end{equation}
\item The wavefunction evolves according to the Schr\"{o}dinger equation.
\end{enumerate}

Note that we have chosen to restrict our attention to systems with just one
particle. For some time to come, we will restrict ourselves only to these
systems.

\subsection{Problem set 5}
\begin{enumerate}
\item Many dynamical variables have a classical analogue. The operator for such
variables is found by expressing these variables in terms of classical position
and momentum and then replacing them by their operator representations. Using
this technique find the operators for the three coordinates of angular momentum.
Recall that the angular momentum is defined as $\vec{L}=\vec{r}\times\vec{p}$.
\item Using the expressions for $\hat{L}_x, \hat{L}_y$ and $\hat{L}_z$ obtained
in the previous problem, show that
\begin{eqnarray}
{[}\hat{L}_x, \hat{L}_y] &=& i\hslash \hat{L}_z \label{c4s4e3} \\
{[}\hat{L}_y, \hat{L}_z] &=& i\hslash \hat{L}_x \label{c4s4e4} \\
{[}\hat{L}_z, \hat{L}_x] &=& i\hslash \hat{L}_y \label{c4s4e5}
\end{eqnarray}
\item The square magnitude of the angular momentum operator is represented by
the operator $\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$. Show that
\begin{equation}\label{c4s4e6}
[\hat{L}^2, \hat{L}_z] = 0.
\end{equation}
\end{enumerate}


\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[margin=1in]{geometry}
\usepackage{setspace}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{graphicx}

\title{Chapter 9\\Partial Differential Equations}
\author{solutions by Hikari}
\date{September 2021}
\begin{document}

\newcommand{\pdv}[2]{\frac{\partial#1}{\partial#2}}
\newcommand{\V}{\mathbf}

\maketitle

\section*{9.2 First-Order Equations}

\paragraph{9.2.1}
Let $s=x+2y$, $t=2x-y$, then
\[
\pdv{\varphi}{x}=\pdv{\varphi}{s}+2\pdv{\varphi}{t}
\]
\[
\pdv{\varphi}{y}=2\pdv{\varphi}{s}-\pdv{\varphi}{t}
\]
so the equation becomes
\[
5\pdv{\varphi}{s}+t\varphi=0
\]
\[
\frac{1}{\varphi}d\varphi=-\frac{t}{5}ds
\]
\[
\ln\varphi=-\frac{ts}{5}+C(t)
\]
\[
\varphi=e^{-\frac{1}{5}(2x^2-2y^2+3xy)}f(2x-y)
\]
Or note that $s+2t=5x$, so the solution can be transformed into
\[
\varphi=e^{-\frac{st}{5}}e^{-\frac{2t^2}{5}}e^{\frac{2t^2}{5}}f(t)
=e^{-\frac{5x\cdot t}{5}}g(t)
\]
\[
=e^{-2x^2+xy}g(2x-y)
\]

\paragraph{9.2.2}
Let $s=x-2y$, $t=2x+y$, then
\[
\pdv{\varphi}{x}=\pdv{\varphi}{s}+2\pdv{\varphi}{t}
\]
\[
\pdv{\varphi}{y}=-2\pdv{\varphi}{s}+\pdv{\varphi}{t}
\]
so the equation becomes
\[
5\pdv{\varphi}{s}+\frac{3t-s}{5}=0
\]
\[
\pdv{\varphi}{s}=\frac{s-3t}{25}
\]
\[
\varphi=\frac{s^2-6st}{50}+f(t)
\]
\[
=\frac{s^2-6st}{50}+\frac{9t^2}{50}-\frac{9t^2}{50}+f(t)
\]
\[
=\frac{(s-3t)^2}{50}+g(t)
\]
\[
=\frac{(x+y)^2}{2}+g(2x+y)
\]

\paragraph{9.2.3}
Let $s=x+y-z,\;t=x-y,\;u=x+y+2z$, then
\[
\pdv{\varphi}{x}=\pdv{\varphi}{s}+\pdv{\varphi}{t}+\pdv{\varphi}{u}
\]
\[
\pdv{\varphi}{y}=\pdv{\varphi}{s}-\pdv{\varphi}{t}+\pdv{\varphi}{u}
\]
\[
\pdv{\varphi}{x}=-\pdv{\varphi}{s}+2\pdv{\varphi}{u}
\]
so the equation becomes
\[
3\pdv{\varphi}{s}=0
\]
\[
\varphi=f(t,u)=f(x-y,x+y+2z)\]
or note that $t+u=2(x+z)$, so the equation can be transformed into
\[
\varphi=f(t,2(x+z)-t)=g(t,x+z)=g(x-y,x+z)
\]

\paragraph{9.2.4}
Let $s=x+y+z,\;t=x-y,\;u=x+y-2z$, then
\[
\pdv{\varphi}{x}=\pdv{\varphi}{s}+\pdv{\varphi}{t}+\pdv{\varphi}{u}
\]
\[
\pdv{\varphi}{y}=\pdv{\varphi}{s}-\pdv{\varphi}{t}+\pdv{\varphi}{u}
\]
\[
\pdv{\varphi}{z}=\pdv{\varphi}{s}-2\pdv{\varphi}{u}
\]
so the equation becomes
\[
3\pdv{\varphi}{s}=t
\]
\[
\varphi=\frac{st}{3}+f(t,u)
\]
\[
=\frac{st}{3}-\frac{ut}{3}+\frac{ut}{3}+f(t,u)
\]
\[
=\frac{3z\cdot t}{3}+g(t,u)
\]
\[
=(x-y)z+g(x-y,x+y-2z)
\]

\paragraph{9.2.5}
(a)
\[
\pdv{\varphi}{x}=\pdv{u}{x}\pdv{\varphi}{u}+\pdv{v}{x}\pdv{\varphi}{v}=y\pdv{\varphi}{u}+2x\pdv{\varphi}{v}
\]
\[
\pdv{\varphi}{y}=\pdv{u}{y}\pdv{\varphi}{u}+\pdv{v}{y}\pdv{\varphi}{v}=x\pdv{\varphi}{u}-2y\pdv{\varphi}{v}
\]
so the equation becomes
\[
(x^2+y^2)\pdv{\varphi}{u}=0
\]
\[
\varphi=f(v)=f(x^2-y^2)
\]

(b)
The characteristics are $x^2-y^2=constant$, which are hyperbolas centered at $(0,0)$ and with $x=y,\;x=-y$ as asymptotes.

\paragraph{9.2.6}
Let $u=x^2-y^2,\;v=xy$, then
\[
\pdv{\varphi}{x}=\pdv{u}{x}\pdv{\varphi}{u}+\pdv{v}{x}\pdv{\varphi}{v}=2x\pdv{\varphi}{u}+y\pdv{\varphi}{v}
\]
\[
\pdv{\varphi}{y}=\pdv{u}{y}\pdv{\varphi}{u}+\pdv{v}{y}\pdv{\varphi}{v}=-2y\pdv{\varphi}{u}+x\pdv{\varphi}{v}
\]
so the equation beocomes
\[
2(x^2+y^2)\pdv{\varphi}{u}=0
\]
\[
\varphi=f(v)=f(xy)
\]

\section*{9.3 Second-Order Equations}

\paragraph{9.3.1}
\begin{align*}
    & \pdv{\varphi}{x}=\pdv{\varphi}{\xi}\pdv{\xi}{x}+\pdv{\varphi}{\eta}\pdv{\eta}{x}=c^{\frac{1}{2}}\pdv{\varphi}{\xi}\\
    & \pdv{^2\varphi}{x^2}=c^{\frac{1}{2}}\left(\pdv{^2\varphi}{\xi^2}\pdv{\xi}{x}+\pdv{^2\varphi}{\xi\partial\eta}\pdv{\eta}{x}\right)=c\pdv{^2\varphi}{\xi^2}\\
    & \pdv{^2\varphi}{x\partial y}=c^{\frac{1}{2}}\left(\pdv{^2\varphi}{\xi^2}\pdv{\xi}{y}+\pdv{^2\varphi}{\xi\partial\eta}\pdv{\eta}{y} \right)=-b\pdv{^2\varphi}{\xi^2}+\pdv{^2\varphi}{\xi\partial\eta}\\
    & \pdv{\varphi}{y}=\pdv{\varphi}{\xi}\pdv{\xi}{y}+\pdv{\varphi}{\eta}\pdv{\eta}{y}=-c^{-\frac{1}{2}}b\pdv{\varphi}{\xi}+c^{-\frac{1}{2}}\pdv{\varphi}{\eta}\\
   & \pdv{^2\varphi}{y^2}=-c^{-\frac{1}{2}}b\left(\pdv{^2\varphi}{\xi^2}\pdv{\xi}{y}+\pdv{^2\varphi}{\xi\partial\eta}\pdv{\eta}{y} \right)+c^{-\frac{1}{2}}\left(\pdv{^2\varphi}{\xi\partial\eta}\pdv{\xi}{y}+\pdv{^2\varphi}{\eta^2}\pdv{\eta}{y} \right)=c^{-1}b^2\pdv{^2\varphi}{\xi^2}-2c^{-1}b\pdv{^2\varphi}{\xi\partial\eta}+c^{-1}\pdv{^2\varphi}{\eta^2}
\end{align*}
Substituting, we get
\[
\mathcal{L}=a\pdv{^2}{x^2}+2b\pdv{^2}{x\partial y}+c\pdv{^2}{y^2}
\]
\[
=(ac-b^2)\pdv{^2}{\xi^2}+\pdv{^2}{\eta^2}
\]

\section*{9.4 Separation of Variables}

\paragraph{9.4.1}
\[
(\nabla^2+k^2)(a_1\varphi_1+a_2\varphi_2)=a_1\nabla^2\varphi_1+a_1k^2\varphi_1+a_2\nabla^2\varphi_2+a_2k^2\varphi_2=a_1(\nabla^2+k^2)\varphi_1+a_2(\nabla^2+k^2)\varphi_2
\]

\paragraph{9.4.2}
Let $\varphi(\rho,\varphi,z)=P(\rho)\Phi(\varphi)Z(z)$ and substitute:
\[
\frac{\Phi Z}{\rho}\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho}\right)+\frac{PZ}{\rho^2}\frac{d^2\Phi}{d\varphi^2}+P\Phi\frac{d^2Z}{dz^2}+\left[k^2+f(\rho)+\frac{1}{\rho^2}g(\varphi)+h(z) \right]P\Phi Z=0
\]
The equation can be seperated into
\[
\frac{1}{Z}\frac{d^2Z}{dz^2}+h(z)=l^2
\]
\[
\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2}+g(\varphi)=-m^2
\]
\[
\frac{\rho}{P}\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho} \right)+\left[f(\rho)+l^2+k^2 \right]\rho^2-m^2=0
\]

\paragraph{9.4.3}
Let $\psi(r,\theta,\varphi)=R(r)\Theta(\theta)\Phi(\phi)$ and substitute, we have
\[
\frac{1}{Rr^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\frac{1}{\Theta r^2\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta} \right)+\frac{1}{\Phi r^2\sin^2\theta}\frac{d^2\Phi}{d\varphi^2}=-k^2
\]
Ir can be rearranged into
\[
\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr} \right)+r^2k^2=-\frac{1}{\Theta\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta} \right)-\frac{1}{\Phi\sin^2\theta}\frac{d^2\Phi}{d\varphi^2}
\]
By equating each side to $\lambda$ we can separate $R$. The rest of the equation can be rearranged into
\[
\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2}=-\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta} \right)-\lambda\sin^2\theta
\]
By equating each side to $-m^2$ we can separate $\Phi$ and $\Theta$.

So the equation can be separated into
\[
\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr} \right)+r^2k^2=\lambda
\]
\[
\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2}=-m^2
\]
\[
-\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta} \right)-\lambda\sin^2\theta=-m^2
\]
which are the same as equation (9.74), (9.77), and (9.78).

\paragraph{9.4.4}
Let $\psi(r,\theta,\varphi)=R(r)\Theta(\theta)\Phi(\varphi)$. Substitute and divide by $R\Theta\Phi$, we get
\[
\frac{1}{Rr^2}\frac{d}{dr}\left(r^2\frac{dR}{dr} \right)+f(r)+\frac{1}{r^2}\left[\frac{1}{\Theta\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta} \right)+g(\theta) \right]+\frac{1}{r^2\sin^2\theta}\left[\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2}+h(\varphi) \right]+k^2=0
\]
It can be seperated into
\[
\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2}+h(\varphi)=-m^2
\]
\[
\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr} \right)+r^2f(r)+r^2k^2=\lambda
\]
\[
-\frac{1}{\Theta\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta} \right)-g(\theta)+\frac{m^2}{\sin^2\theta}=\lambda
\]

\paragraph{9.4.5}
Let $\psi(x,y,z)=X(x)Y(y)Z(z)$ and substitute, we have
\[
YZ\frac{d^2X}{dx^2}+XZ\frac{d^2Y}{dy^2}+XY\frac{d^2Z}{dz^2}+\frac{2mE}{\hbar^2}XYZ=0
\]
It can be separated into
\[
\frac{1}{X}\frac{d^2X}{dx^2}=-l^2\qquad \frac{1}{Y}\frac{d^2Y}{dy^2}=-m^2\qquad \frac{1}{Z}\frac{d^2Z}{dz^2}=-n^2
\]
where $l^2+m^2+n^2=\frac{2mE}{\hbar^2}$.
\medskip

The solution of $X$ is $X=A\sin lx+B\cos lx$. When the boundary conditions $X(0)=X(a)=0$ ia applied, we must require $B=0$ and $la=\lambda\pi$, where $\lambda$ is a positive integer. Similarly $mb=\mu\pi$, $nc=\nu\pi$, where $\mu,\nu$ are positive integers. So
\[
E=\frac{\hbar^2}{2m}(l^2+m^2+n^2)=\frac{\pi^2\hbar^2}{2m}\left(\frac{\lambda^2}{a^2}+\frac{\mu^2}{b^2}+\frac{\nu^2}{c^2}\right)
\]
and the minimum of $E$ is
\[
E_{min}=\frac{\pi^2\hbar^2}{2m}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)
\]
in which case $\lambda=\mu=\nu=1$.

\paragraph{9.4.6}
From Exercise 3.10.32 (c), we have
\[
\V{L}^2=-\frac{1}{\sin\theta}\pdv{}{\theta}\left(\sin\theta\pdv{}{\theta}\right)-\frac{1}{\sin^2\theta}\pdv{^2}{\varphi^2}
\]
Let $\psi(r,\theta,\varphi)=R(r)\Theta(\theta)\Phi(\varphi)$. Substitute into the equation and divide by $R\Theta\Phi$, we have
\[
-\frac{1}{\Theta\sin\theta}\pdv{}{\theta}\left(\sin\theta\pdv{\Theta}{\theta} \right)-\frac{1}{\Phi\sin^2\theta}\pdv{^2\Phi}{\varphi^2}=l(l+1)
\]
which can be separated into
\[
\frac{1}{\Phi}\pdv{^2\Phi}{\varphi^2}=-m^2
\]
and
\[
\frac{1}{\sin\theta}\pdv{}{\theta}\left(\sin\theta\pdv{\Theta}{\theta} \right)-\frac{m^2}{\sin^2\theta}\Theta+l(l+1)\Theta=0
\]
Let $t=\cos\theta$ and $\Theta(\theta)=P(\cos\theta)=P(t)$, the $\Theta$ equation becomes
\[
(1-t^2)P''(t)-2tP'(t)-\frac{m^2}{1-t^2}P(t)+l(l+1)P(t)=0
\]
which is the associated Legendre equation.

\paragraph{9.4.7}
(a) Multiply the equation by $-\frac{2}{\hbar}\left(\frac{m}{k} \right)^{1/2}$ and use the definitions of $a$ and $\lambda$, we have
\[
\frac{1}{a^2}\frac{d^2\psi}{dx^2}-a^2x^2\psi+\lambda\psi=0
\]
Note that $\frac{d^2\psi}{dx^2}=\frac{d^2\psi}{d\xi^2}\left(\frac{d\xi}{dx} \right)^2=a^2\frac{d^2\psi}{d\xi^2}$, and $a^2x^2=\xi^2$, we have
\[
\frac{d^2\psi}{d\xi^2}+(\lambda-\xi^2)\psi=0
\]

(b)
\[
\frac{d\psi}{d\xi}=\left[y'(\xi)-\xi y(\xi) \right]e^{-\frac{\xi^2}{2}}
\]
\[
\frac{d^2\psi}{d\xi^2}=\left[y''(\xi)-2\xi y'(\xi)+(\xi^2-1)y(\xi) \right]e^{-\frac{\xi^2}{2}}
\]
Substitute and eliminate $e^{-\frac{\xi^2}{2}}$, we have
\[
y''(\xi)-2\xi y'(\xi)+(\lambda-1)y(\xi)=0
\]
which is the Hermite differential equation.

\section*{9.5 Laplace and Poisson Equations}

\paragraph{9.5.1}
(a) Using Equation (3.158),
\[
\nabla^2\varphi_1=\frac{1}{r^2}\pdv{}{r}\left(r^2\pdv{(\frac{1}{r})}{r} \right)=\frac{1}{r^2}\pdv{}{r}(-1)=0
\]

(b)
Substitute $r\cos\theta$ for $z$, then
\[
\varphi_2=\frac{1}{2r}\ln\frac{1+\cos\theta}{1-\cos\theta}
\]
\[
\pdv{\varphi_2}{r}=-\frac{1}{2r^2}\ln\frac{1+\cos\theta}{1-\cos\theta}
\]
\[
\pdv{\varphi_2}{\theta}=-\frac{1}{r\sin\theta}
\]

Using Equation (3.158),
\[
\nabla^2\varphi_2=\frac{1}{r^2}\pdv{}{r}\left(r^2\pdv{\varphi_2}{r} \right)+\frac{1}{r^2\sin\theta}\pdv{}{\theta}\left(\sin\theta\pdv{\varphi_2}{\theta} \right)
\]
\[
=\frac{1}{r^2}\pdv{}{r}\left(-\frac{1}{2}\ln\frac{1+\cos\theta}{1-\cos\theta} \right)+\frac{1}{r^2\sin\theta}\pdv{}{\theta}\left(-\frac{1}{r} \right)=0
\]

\paragraph{9.5.2}
\[
\nabla^2\Psi=\pdv{^2\Psi}{x^2}+\pdv{^2\Psi}{y^2}+\pdv{^2\Psi}{z^2}=0
\]
so
\[
\nabla^2\left(\pdv{\Psi}{z}\right)=\pdv{^3\Psi}{x^2\partial z}+\pdv{^3\Psi}{y^2\partial z}+\pdv{^3\Psi}{z^3}
\]
\[
=\pdv{}{z}\left(\pdv{^2\Psi}{x^2}+\pdv{^2\Psi}{y^2}+\pdv{^2\Psi}{z^2}\right)=0
\]
which means $\pdv{\Psi}{z}$ is also a solution of Laplace's equation.

\paragraph{9.5.3}
Suppose $\psi_1$ and $\psi_2$ are distinct solutions to the Laplace or Poisson equation for the same Dirichlet boundary conditions, then $\psi=\psi_1-\psi_2$ will also be a solution to the Laplace equation with a zero Dirichlet boundary conditions. From Eq. (9.88),
\[
\int\displaylimits_S\psi\pdv{\psi}{\V{n}}dS=\int\displaylimits_V\psi\nabla^2\psi\,d\tau+\int\displaylimits_V   \nabla\psi\cdot\nabla\psi\,d\tau
\]
$\int\displaylimits_S\psi\pdv{\psi}{\V{n}}dS$ vanishes because $\psi$ vanishes on the boundary. $\int\displaylimits_V\psi\nabla^2\psi\,d\tau$ vanishes because $\psi$ is a solution to the Laplace equation. Therefore $\int\displaylimits_V   \nabla\psi\cdot\nabla\psi\,d\tau$ must vanish, which means $\nabla\psi=0$ everywhere, so $\psi=constant=0$ because it is zero on the boundary. So $\psi_1=\psi_2$, which means the solution is unique.

\section*{9.6 Wave Equation }

\paragraph{9.6.1}
Using the d’Alembert’s solution:
\[
\psi(x,t)=\frac{1}{2}\left[\psi(x+ct,0)+\psi(x-ct,0)\right]+\frac{1}{2c}\int_{x-ct}^{x+ct}\pdv{\psi(x,0)}{t}dx
\]
\[
=\frac{1}{2}\left[\sin(x+ct)+\sin(x-ct)\right]+\frac{1}{2c}\int_{x-ct}^{x+ct}\cos x\,dx
\]
\[
=\sin x\cos ct+\frac{1}{c}\cos x\sin ct
\]

\paragraph{9.6.2}
\[
\psi(x,t)=\frac{1}{2}\left[\psi(x+ct,0)+\psi(x-ct,0)\right]+\frac{1}{2c}\int_{x-ct}^{x+ct}\pdv{\psi(x,0)}{t}dx
\]
\[
=\frac{1}{2}\left[\delta(x+ct)+\delta(x-ct)\right]
\]

\paragraph{9.6.3}
\[
\psi(x,t)=\frac{1}{2}\left[\psi(x+ct,0)+\psi(x-ct,0)\right]+\frac{1}{2c}\int_{x-ct}^{x+ct}\pdv{\psi(x,0)}{t}dx
\]
\[
=\frac{1}{2}\left[\psi_0(x+ct)+\psi_0(x-ct)\right]
\]
where $\psi_0$ is the given square-wave pulse.

\paragraph{9.6.4}
\[
\psi(x,t)=\frac{1}{2}\left[\psi(x+ct,0)+\psi(x-ct,0)\right]+\frac{1}{2c}\int_{x-ct}^{x+ct}\pdv{\psi(x,0)}{t}dx
\]
\[
=\frac{1}{2c}\int_{x-ct}^{x+ct}\sin x\,dx
=\frac{1}{2c}\left[\cos(x-ct)-\cos(x+ct) \right]
=\frac{1}{c}\sin x\cos ct
\]

\section*{9.7 Heat-Flow, or Diffusion PDE}

\paragraph{9.7.1}
Substitute $T(r,t)=R(r)T(t)$ into the equation, we have
\[
R\pdv{T}{t}=KT\nabla^2R=KT\frac{1}{r^2}\pdv{}{r}\left(r^2\pdv{R}{r} \right)
\]
which can be separated into
\[
\frac{1}{KT}\pdv{T}{t}=\frac{1}{Rr^2}\left(r^2\pdv{^2R}{r^2}+2r\pdv{R}{r} \right)=-\alpha^2
\]
so the $R$ equation is
\[
r^2\frac{d^2R}{dr^2}+2r\frac{dR}{dr}+\alpha^2r^2R=0
\]
For $R=\frac{\sin\alpha r}{r}$:
\[
r^2\pdv{^2R}{r^2}+2r\pdv{R}{r}+\alpha^2r^2R
\]
\[
=\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\alpha^2r^2R
\]
\[
=\frac{d}{dr}\left(\alpha r\cos\alpha r-\sin\alpha r\right)+\alpha^2r\sin\alpha r
\]
\[
=\alpha\cos\alpha r-\alpha^2r\sin\alpha r-\alpha\cos\alpha r+\alpha^2 r\sin\alpha r=0
\]
For $R=\frac{\cos\alpha r}{r}$:
\[
r^2\pdv{^2R}{r^2}+2r\pdv{R}{r}+\alpha^2r^2R
\]
\[
=\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\alpha^2r^2R
\]
\[
\frac{d}{dr}\left(-\alpha r\sin\alpha r-\cos\alpha r\right)+\alpha^2r\cos\alpha r
\]
\[
=-\alpha\sin\alpha r-\alpha^2r\cos\alpha r+\alpha\sin\alpha r+\alpha^2r\cos\alpha r=0
\]
so $\frac{\sin\alpha r}{r}$ and $\frac{\cos\alpha r}{r}$ are solutions to the equation.

\paragraph{9.7.2}
Substitute $T(\rho,t)=P(\rho)T(t)$ into the equation, we have
\[
P\pdv{T}{t}=KT\nabla^2P=KT\frac{1}{\rho}\pdv{}{\rho}\left(\rho\pdv{P}{\rho}\right)
\]
which can be separated into
\[
\frac{1}{KT}\pdv{T}{t}=\frac{1}{P\rho}\left(\rho\pdv{^2P}{\rho^2}+\pdv{P}{\rho}\right)=-\alpha^2
\]
so
\[
\frac{dT}{dt}+\alpha^2KT=0
\]
\[
\rho\frac{d^2P}{d\rho^2}+\frac{dP}{d\rho}+\alpha^2\rho P=0
\]

\paragraph{9.7.3}
Use Equation 9.114:
\[
\psi(x,t)=\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty A\delta(x-2a\xi\sqrt{t})e^{-\xi^2}d\xi
\]
\[
=\frac{A}{\sqrt{\pi}}\int_{-\infty}^{\infty}\delta(x-y)e^{-\frac{y^2}{4a^2t}}\frac{dy}{2a\sqrt{t}}
\]
\[
=\frac{A}{2a\sqrt{\pi t}}e^{-\frac{x^2}{4a^2t}}
\]

\paragraph{9.7.4}
From Equation 9.101, the solutions have the form
\[
\psi(x,t)=(A\cos\omega x+B\sin\omega x)e^{-\omega^2a^2t}+C_0'x+C_0
\]
Using the boundary conditions:
\begin{alignat*}{2}
    & \psi(0,\infty)=C_0=1\qquad && C_0=1\\
    & \psi(L,\infty)=C_0'L+C_0=0\qquad && C_0'=-\frac{1}{L}\\
    & \psi(0,t)=Ae^{-\omega^2 a^2t}+1=1\qquad && A=0\\\
    & \psi(L,t)=B\sin(\omega L)\,e^{-\omega^2a^2t}=0\qquad && \omega L=n\pi\qquad\textit{n is a positive integer}
\end{alignat*}
So
\[
\psi(x,t)=\sum_{n=1}^\infty a_n\sin\left(\frac{n\pi x}{L}\right)e^{-\frac{n^2\pi^2a^2}{L^2}t}-\frac{x}{L}+1
\]
To determine $a_n$, use $\psi(x,0)=0$ and the orthogonality $\int_0^L\sin\left(\frac{n\pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right)dx=\frac{L}{2}\delta_{nm}$:
\[
\int_0^L\psi(x,0)\sin\left(\frac{n\pi x}{L}\right)dx=\frac{L}{2}a_n+\frac{L}{n\pi}=0
\]
so $a_n=-\frac{2}{n\pi}$, and the overall solution is
\[
\psi(x,t)=-\sum_{n=1}^\infty \frac{2}{n\pi}\sin\left(\frac{n\pi x}{L}\right)e^{-\frac{n^2\pi^2a^2}{L^2}t}-\frac{x}{L}+1
\]
It can be verified that the solution satisfies the boundary, initial conditions and the PDE.

\end{document}

Mathematical Methods for Physicists/Chapter 09/main.tex