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\begin{document}


\begin{center}
Heat/Diffusion Equation
\end{center}

\bigskip

\begin{align*}
\frac{\partial w}{\partial t}-k\frac{\partial ^{2}w}{\partial x^{2}}&
=0\qquad \text{k constant} \\
w(x,0)& =\varphi (x)\text{ \ \ \ initial condition} \\
w(0,t)& =w(l,t)=0\text{ \ \ \ \ boundary conditions}
\end{align*}

\bigskip

\noindent Energy estimate:

\bigskip

$0=w (w_t-kw_{xx})= (\frac{w^2}{2})_t - (kww_x)_x +k (w_x)^2$

\noindent So

\begin{align*}
0 &= \int\limits_{0}^{l} \left[ (\frac{w^2}{2})_t - (kww_x)_x + k (w_x)^2 %
\right] dx \\
&= \int\limits_{0}^{l} (\frac{w^2}{2})_t dx -kww_x \vert_0^l +
k\int\limits_0^l (w_x)^2 dx
\end{align*}

\noindent or%
\begin{equation*}
\frac{d}{dt}\int\limits_0^l \frac{w^2}{2}dx=-k\int\limits_{0}^{l}(w_{x})^2
dx \leq 0
\end{equation*}

\noindent and therefore $E(t)=\int\limits_{0}^{l}\frac{w^{2}}{2}dx$ is a
non-increasing function of time or equivalently%
\begin{equation*}
0\leq E(t)\leq E_{0}
\end{equation*}

\bigskip

\noindent For the initial value problem if $\varphi (x)=0$ then $E(0)=0$ and 
$E(t)=0$ i.e. uniqueness.

\bigskip

\begin{center}
Stability
\end{center}

\bigskip

\noindent Consider \ \ \ $u_{t}=ku_{xx}$ \ \ \ \ \ \ \ \ \ \ $u(x,0)=\varphi
(x)$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $v_{t}=kv_{xx}$ \ \ \ \ \ \ \ \ \ \ $%
v(x,0)=\psi (x)$

\bigskip

\noindent Let $w(x,t)=u(x,t)-v(x,t)$ \ \ \ \ \ \ \ \ \ \ 

\ \ \ \ \ \ $w(x,0)=u(x,0)-v(x,0)=\varphi (x)-\psi (x)$

\bigskip

\noindent From the energy estimate we get%
\begin{equation*}
\int\limits_{0}^{l}\left[ u(x,t)-v(x,t)\right] ^{2}dx\leq \int\limits_{0}^{l}%
\left[ \varphi (x)-\psi (x,t)\right] ^{2}dx
\end{equation*}

\noindent So small changes in $\varphi $ causes small changes in $v$ in the $%
L^{2}$ norm.

\noindent The maximum principle will give us stability in the maximum norm.

\bigskip

\begin{center}
Maximum Principle
\end{center}

\bigskip

\noindent Let 
\begin{equation*}
Lu=-\frac{\partial u}{\partial t}+\dsum_{i,j}a_{ij}(x,t)\frac{\partial ^{2}u%
}{\partial x_{i}\partial x_{j}}+\dsum_{i}b_{i}(x,t)\frac{\partial u}{%
\partial x_{i}}+c(x,t)u=f
\end{equation*}

\noindent Strong version: Assume $Lu\geq 0$ ,\ \ $M=\max (u)$. Assume $u=M$
at an interior point $(x_{0},t_{0})$ and one of the following is true

\begin{enumerate}
\item $c=0$ and $M$ arbitrary

\item $c\leq 0$ and $M\geq 0$

\item $M=0$ and $c$ arbitrary
\end{enumerate}

\noindent Then $u=M$ everywhere in the domain (i.e. $u$ is constant).

\noindent Weak Version: If $Lu\geq 0$ and $c(x,t)=0$ then the maxiumum
occurs initially or on the boundary.

\noindent So if $|u|<M$ at $t=0,x=0,x=l$ \ then $u<M$ everywhere

\bigskip

\noindent For a minimum principle consider $\frac{-\partial u}{\partial t}-k%
\frac{\partial ^{2}u}{\partial x^{2}}=0$

\noindent Proof \ (weak version in one dimension) 
\begin{equation*}
\frac{\partial u}{\partial t}=k\frac{\partial ^{2}u}{\partial x^{2}}
\end{equation*}

\noindent Intuitively at a maximum the first derivatives are zero and $\frac{%
\partial^{2}u}{\partial x^{2}}<0$ so we have a contradiction. The problem is
that we may have a saddle point with $\frac{\partial^2 u}{\partial x^2}=0$.
So instead we supply a perturbation argument.

\bigskip

\noindent Define: 
\begin{equation*}
v(x,t)=u(x,t)+\epsilon x^{2}\qquad \epsilon >0
\end{equation*}

\noindent Wish: $v(x,t)<M+\epsilon l^2$ \ with $M=\max |u(x,t)|$ \ \ so $%
u(x,t)<M+\epsilon (l^{2}-x^{2})$

\bigskip

\noindent Formally: 
\begin{align*}
v(x,t)& <M+\epsilon l^{2} \\
\text{on }t& =0\text{ and }x=0\text{ and }x=l
\end{align*}%
Also%
\begin{equation*}
(\ast )\quad v_{t}-kv_{xx}=u_{t}-k(u+\epsilon
x^{2})_{xx}=u_{t}-ku_{xx}-2\epsilon k<0
\end{equation*}

\bigskip

\noindent Assume there is a maximum at $(x_{0},t_{0})$ in the interior. 
\newline
At such a maximum $v_{t}=0$ and $v_{xx}\leq 0$ . \newline
Hence, $v_{t}-kv_{xx}\geq 0$ at $(x_{0},t_{0})$. \newline
This contradicts (*). Hence, $v(x,t)<M+\epsilon l^{2}$ . \newline
\noindent Note that low order terms destroy this property. Consider%
\begin{align*}
u_{t} &= u_{xx}+au\quad a>0 \\
u(x,0) &= \sin \pi x\quad u(0,t)=u(1,t)=0
\end{align*}

\noindent has a solution $u(x,t)=e^{(a-\pi ^{2})t}\sin (\pi x)$ which grows
in time if $a>\pi ^{2}$.

\bigskip

Lemma: \ 
\begin{equation*}
I=\int\limits_{-\infty }^{\infty }e^{-x^{2}}dx=\sqrt{\pi }
\end{equation*}

\bigskip

\noindent proof:

\noindent One proof is by a complex integral and Cauchy's Theorem.

\noindent Another proof is%
\begin{align*}
I^{2} &= \int\limits_{-\infty }^{\infty }\int\limits_{-\infty }^{\infty
}e^{-\left( x^{2}+y^{2}\right) }dxdy\qquad r^{2}=x^{2}+y^{2}\quad x=r\cos
\theta \quad y\sin \theta \\
&= \int\limits_{0}^{2\pi }\int\limits_{-\infty }^{\infty} e^{-r^2}rdrd\theta
= 2\pi \int\limits_{-\infty }^{\infty} e^{-r^2}rdr \\
&= -2\pi \int\limits_{-\infty }^{\infty }\frac{1}{2}\frac{d}{dr}
\left(e^{-r^2}\right) dr=-\pi e^{-r^{2}}\vert_0^{\infty }=\pi
\end{align*}

\newpage

\begin{center}
Properties of heat equation
\end{center}

\bigskip

\noindent Let $\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{%
\partial x^2}$

\begin{enumerate}
\item $u(x-y,t)$ is a solution (shift)

\item Any derivative of $u$ is a solution

\item Any linear combinations of solutions is a solution

\item The integral $\int S(x-y,t)g(y)dy$ \ is a solution

\item $u(\sqrt{a}x,at)$ \quad $a>0$ is a solution

\item u=u(p) \ \ \ \ \ \ \ $p=\frac{1}{\sqrt{4k}}\frac{x}{t}$. Then $u(\sqrt{%
a}x,at) =\frac{1}{\sqrt{4k}} \frac{\sqrt{a}x}{\sqrt{at}}=u(x,t)$.
\end{enumerate}

\newpage

\begin{align*}
(\ast )\qquad \frac{\partial u}{\partial t}& =k\frac{\partial ^{2}u}{%
\partial x^{2}} \\
u(x,0)& =\phi (x)
\end{align*}

proof of \#4: If S(x,t) is a solution of \ (*) then by linearity so is%
\begin{equation*}
v(x,t)=\int\limits_{-\infty }^{\infty }S(x-y,t)g(y)dy
\end{equation*}

Then%
\begin{align*}
v_{t} &= \int\limits_{-\infty }^{\infty }S_{t}(x-y,t)g(y)dy \\
v_{xx} &= \int\limits_{-\infty }^{\infty }S_{xx}(x-y,t)g(y)dy \\
v_{t}-kv_{xx} &= \int\limits_{-\infty }^{\infty }\left[
S_{t}(x-y,t)-kS_{xx}(x-y,t)\right] g(y)dy=0
\end{align*}

\bigskip

proof of (5)%
\begin{align*}
v &= u(\sqrt{a}x,at) \\
v_{t} &= au_{t}(\sqrt{a}x,at) \\
v_{xx} &= au_{xx}(\sqrt{a}x,at) \\
v_{t}-kv_{xx} &= a(u_{t}-ku_{xx})=0
\end{align*}

\newpage

\noindent Define%
\begin{align*}
& Q_t =kQ_{xx} \\
& 
\begin{cases}
Q(x,0)=1 & x>0 \\ 
Q(x,0)=0 & x<0%
\end{cases}%
\end{align*}

\bigskip

\noindent Assume $\ \ Q(x,t)=g(p)$ \ \ \ \ \ \ \ $p=\frac{x}{\sqrt{4kt}}$

\noindent then:%
\begin{align*}
Q_{t}& =\frac{dg}{dp}\frac{\partial p}{\partial t}=-\frac{1}{2t}\frac{x}{%
\sqrt{4kt}}g^{\prime }(p)=-\frac{p}{2t}g^{\prime }(p) \\
Q_{x}& =\frac{dg}{dp}\frac{\partial p}{\partial x}=\frac{1}{\sqrt{4kt}}%
g^{\prime }(p) \\
Q_{xx}& =\frac{1}{4kt}g^{\prime \prime }(p)
\end{align*}

\noindent So%
\begin{equation*}
0=Q_{t}-kQ_{xx}=\frac{1}{t}\left[ -\frac{1}{2}pg^{\prime }(p)- \frac{1}{4}
g^{\prime \prime }(p)\right]
\end{equation*}

\noindent Therefore%
\begin{align*}
g^{\prime \prime }(p)+2pg^{\prime }(p) &= 0 \\
g^{\prime }(p) &= C_{1}e^{-p^{2}} \\
Q(x,t) &= g(p)=C_1 \dint\limits_{0}^{p=\frac{x}{\sqrt{4kt}}} e^{-p^2}dp + C_2
\end{align*}

\noindent We now take the limit as $t\downarrow 0$ . \ Then

\noindent Define%
\begin{equation*}
\begin{cases}
x>0 & 1=Q(x,0)=C_{1}\dint\limits_{0}^{\infty }e^{-p^{2}}dp+C_{2}= \frac{C_{1}%
\sqrt{\pi }}{2}+C_{2} \\ 
x<0 & 0=Q(x,0)=C_{1}\dint\limits_{0}^{-\infty }e^{-p^{2}}dp+C_{2}= -\frac{%
C_{1}\sqrt{\pi }}{2}+C_{2}%
\end{cases}%
\end{equation*}

\noindent So%
\begin{equation*}
C_{1}=\frac{1}{\sqrt{\pi }}\qquad C_{2}=\frac{1}{2}
\end{equation*}

\noindent Therefore%
\begin{equation*}
Q(x,t)=\frac{1}{2}+\frac{1}{\sqrt{\pi }}\dint\limits_{0}^{\frac{x}{\sqrt{4kt}%
}}e^{-p^{2}}dp
\end{equation*}

\newpage

\begin{center}
Error Function
\end{center}

\bigskip Remember 
\begin{equation*}
\dint\limits_{0}^{\infty }e^{-p^{2}}dp=\dint\limits_{-\infty
}^{0}e^{-p^{2}}dp=\frac{1}{2}\dint\limits_{-\infty }^{\infty }e^{-p^{2}}dp=%
\frac{\sqrt{\pi }}{2}
\end{equation*}

\noindent Define%
\begin{align*}
\func{erf}(x)& =\frac{2}{\sqrt{\pi }}\dint\limits_{0}^{x}e^{-p^{2}}dp\qquad 
\func{erf}c(x)=1-\func{erf}(x) \\
\func{erf}(0)& =0\qquad \func{erf}(\infty )=1\qquad \func{erf}(-x)=-\func{erf%
}(x) \\
\frac{2}{\sqrt{\pi }}\dint\limits_{-\infty }^{x}e^{-p^{2}}dp& =\frac{2}{%
\sqrt{\pi }}\dint\limits_{-\infty }^{0}e^{-p^{2}}dp+\frac{2}{\sqrt{\pi }}%
\dint\limits_{-0}^{x}e^{-p^{2}}dp=1+\func{erf}(x)
\end{align*}

\bigskip

\noindent Then%
\begin{equation*}
Q(x,t)=\frac{1+\func{erf}\left( \frac{x}{\sqrt{4kt}}\right) }{2}
\end{equation*}

\newpage

\bigskip

\noindent Define 
\begin{equation*}
S(x,t)=\frac{\partial Q}{\partial x}
\end{equation*}%
\noindent Then $S$ is also a solution and

\begin{equation*}
u(x,t)=\dint\limits_{-\infty }^{\infty }S(x-y,t)\varphi (y)dy
\end{equation*}%
is also a solution. \ Does it satisfy the initial condition \ $%
u(x,0)=\varphi (x)$ ?

\begin{align*}
u(x,t)& =\dint\limits_{-\infty }^{\infty }S(x-y,t)\varphi
(y)dy=\dint\limits_{-\infty }^{\infty }\frac{\partial Q}{\partial x}%
(x-y,t)\varphi (y)dy \\
& =-\dint\limits_{-\infty }^{\infty }\frac{\partial Q}{\partial y}%
(x-y,t)\varphi (y)dy\qquad \text{since}\frac{\partial Q}{\partial x}\!=\!-%
\frac{\partial Q}{\partial y} \\
& =\dint\limits_{-\infty }^{\infty }Q(x-y,t)\frac{\partial \varphi }{%
\partial y}(y)dy-Q(x-y,t)\varphi (y)|_{-\infty }^{\infty }
\end{align*}%
Assume $\varphi $ is zero at infinity then

\begin{equation*}
u(x,t)=\dint\limits_{-\infty }^{\infty }Q(x-y,t)\frac{\partial \varphi }{%
\partial y}(y)dy
\end{equation*}

\noindent Using 
\begin{equation*}
\begin{cases}
Q(s,0)=1 & s>0 \\ 
Q(s,0)=0 & s<0%
\end{cases}%
\end{equation*}%
\noindent We have%
\begin{align*}
u(x,0)& =\dint\limits_{-\infty }^{\infty }Q(x-y,0)\frac{\partial \varphi }{%
\partial y}(y)dy \\
& =\dint\limits_{-\infty }^{x}\frac{\partial \varphi }{\partial y}%
(y)dy=\varphi |_{-\infty }^{x}=\varphi (x)
\end{align*}

\bigskip

\newpage

\begin{center}
Conclusion
\end{center}

\begin{align*}
S(x,t) &= \frac{\partial Q}{\partial x}=\frac{1}{\sqrt{4k\pi }t}e^{-\frac{%
x^{2}}{4kt}} \\
u(x,t) &= \frac{1}{\sqrt{4k\pi }t}\dint\limits_{-\infty }^{\infty }e^{-\frac{%
\left( x-y\right) ^{2}}{4kt}}\varphi (y)dy
\end{align*}

\bigskip

\noindent 
\begin{align*}
\dint\limits_{-\infty }^{\infty }S(x,t)dx &= \frac{1}{\sqrt{\pi }}%
\dint\limits_{-\infty }^{\infty }e^{-p^{2}}dp=1 \\
\lim_{t=0}S(x,t) &= \delta (x)
\end{align*}

\newpage

\begin{center}
Importance of Limiting Process
\end{center}

\bigskip

\begin{align*}
\frac{\partial u}{\partial t} &= \frac{\partial^2 u}{\partial x^2} \\
u(x,0) &= 0
\end{align*}

Then a possible solution is

\begin{equation*}
u(x,t)=\frac{1}{4\sqrt{\pi t^{3}}}e^{-\frac{x^{2}}{4t}}
\end{equation*}

\noindent For $x$ fixed and $t\rightarrow 0$ then $u(x,t)\rightarrow 0$.
\medskip

\noindent However, for $\frac{x^{2}}{4t}$ constant and $t\rightarrow 0$ the
solution is not bounded !! Hence, this is {\Large not }a legitimate solution

\newpage

\begin{center}
Main Theorem
\end{center}

\bigskip \noindent Theorem: Assume

\begin{align*}
(\ast )\quad \frac{\partial u}{\partial t}& =k\frac{\partial ^{2}u}{\partial
x^{2}} \\
u(x,0)& =\varphi (x)\qquad |\phi (x)|<\infty
\end{align*}%
\medskip \noindent Then

\begin{equation*}
u(x,t)=\frac{1}{\sqrt{4k\pi t}}\dint\limits_{-\infty }^{\infty }e^{-\frac{%
(x-y)^{2}}{4kt}}\varphi (y)dy
\end{equation*}

\begin{itemize}
\item $u\epsilon C^{\infty }$ \qquad $-\infty <x<\infty \quad ,\qquad
0<t<\infty $

\item each derivative of $u(x,t)$ satisfies $(\ast)$

\item $\lim_{t\downarrow 0} \; u(x,t)=\varphi(x)$
\end{itemize}

\noindent proof

\begin{align*}
u(x,t)& =\dint\limits_{-\infty }^{\infty }S(x-y,t)\varphi
(y)dy=\dint\limits_{-\infty }^{\infty }S(z,t)\varphi (x-z)dz \\
S(z,t)& =\frac{1}{\sqrt{4k\pi t}}e^{-\frac{z^{2}}{4kt}}
\end{align*}%
\medskip \noindent Let $p=\frac{z}{\sqrt{kt}}$ \ $z=\sqrt{kt}p$. Then%
\begin{equation*}
u(x,t)=\frac{1}{\sqrt{4\pi }}\dint\limits_{0}^{x}e^{-\frac{p^{2}}{4}}\varphi
(x-p\sqrt{kt})dp
\end{equation*}%
\medskip \noindent Assume $|\varphi (x)|\leq M$ . Then 
\begin{equation*}
|u(x,t)|\leq \frac{M}{\sqrt{4\pi }}\dint\limits_{-\infty }^{\infty }e^{-%
\frac{p^{2}}{4}}dp=M
\end{equation*}

\noindent Differentiating we get 
\begin{align*}
\frac{\partial u}{\partial x} &= \dint\limits_{-\infty }^{\infty }\frac{%
\partial S}{\partial x}(x-y,t)\varphi (y)dy=-\frac{1}{\sqrt{4\pi kt}}%
\dint\limits_{-\infty }^{\infty }\frac{z}{2kt}e^{-\frac{z^{2}}{4kt}}\varphi
(x-z)dz \\
&= \frac{\text{const}}{\sqrt{t}}\dint\limits_{-\infty }^{\infty }p e^{-\frac{%
p^2}{4}}\varphi (x-p\sqrt{kt})dp \\
&\leq \frac{\text{const}}{\sqrt{t}}M\dint\limits_{-\infty }^{\infty }pe^{-%
\frac{p^{2}}{4}}dp\leq \frac{\text{const}}{\sqrt{t}}M
\end{align*}

\noindent Similarly%
\begin{equation*}
\frac{\partial ^{n}u}{\partial x^{n}}\leq \frac{\text{const}}{\sqrt{t}}%
M\dint\limits_{-\infty }^{\infty }p^{n}e^{-\frac{p^{2}}{4}}dp\leq \frac{%
\text{const}}{\sqrt{t}}M
\end{equation*}

\bigskip

\noindent Remembering that $\dint\limits_{-\infty }^{\infty }S(x,t)dx=1$ we
get%
\begin{align*}
u(x,t)-\varphi (x) &= \dint\limits_{-\infty }^{\infty }S(x-y,t) \left[
\varphi (y)-\varphi (x)\right] dy \\
&= \frac{1}{\sqrt{4\pi }}\dint\limits_{-\infty }^{\infty } e^{-\frac{p^2}{4}%
} \left[ \varphi (x-p\sqrt{kt})-\varphi (x)\right] dp
\end{align*}

\noindent We assume that $\varphi (x)$ is continuous. Hence, \newline
\noindent $|\varphi (y)-\varphi (x)|\leq \varepsilon $ when $|x-y|\leq
\delta $. So%
\begin{align*}
|u(x,t)& -\varphi (x)|\;=\underbrace{\dint\limits_{-\frac{\delta }{\sqrt{kt}}%
}^{\frac{\delta }{\sqrt{kt}}}...dp}+\underbrace{\dint\limits_{|p|>\frac{%
\delta }{\sqrt{kt}}}...dp} \\
\text{bounded by}& \qquad \ \ \ \qquad \qquad \frac{\varepsilon }{2}\text{ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\frac{\varepsilon }{2}\quad \ \ \ \ \ \ \ \  \\
\text{because}& \qquad \ \ \qquad \ \delta \text{ is small \ \ \ \ \ \ \ \ t
is small} \\
\text{So}& \qquad |\varphi (y)-\varphi (x)|\leq \frac{\varepsilon }{2}\quad
p\rightarrow \infty
\end{align*}

\newpage

\begin{center}
Backward heat equation
\end{center}

\bigskip 
\begin{equation*}
\frac{\partial u}{\partial t}=-k\frac{\partial ^{2}u}{\partial x^{2}}
\end{equation*}

This gives growth in time and is not well posed. Equivalently we are trying
to solve the heat equation backward in time, i.e. given distribution of
temperature at time T find original temperature.

\bigskip

Consider%
\begin{eqnarray*}
u_{n}(x,0) &=&\frac{1}{n}\sin (nx)\rightarrow 0\quad \text{as \ \ }%
n\rightarrow \infty \\
u_{n}(x,t) &=&\frac{1}{n}\sin (nx)e^{n^{2}kt}\quad \text{is unbounded}
\end{eqnarray*}

\bigskip

\begin{center}
\textbf{Black Scholes}
\end{center}

\begin{equation*}
\frac{\sigma ^{2}s^{2}}{2}V_{ss}+rsV_{s}-rV+V_{t}=0
\end{equation*}

s=market value of asset being optioned

t=time

$\sigma $=constant volatility

r=constant interest rate

Given the final value we wish to determine how to price it initially. So we
have backward problem ! - but we have additional minus sign \ $%
\Longrightarrow $ well posed

\newpage

\begin{center}
Exercise
\end{center}

\bigskip

\begin{equation*}
\begin{cases}
\varphi (x,0)=1 & |x|<l \\ 
\varphi (x,0)=0 & |x|>l%
\end{cases}%
\end{equation*}

Then

\begin{align*}
u(x,t)& =\frac{1}{\sqrt{4k\pi t}}\dint\limits_{-\infty }^{\infty }e^{-\frac{%
(x-y)^{2}}{4kt}}\varphi (y)dy \\
& =\frac{1}{\sqrt{4k\pi t}}\dint\limits_{-l}^{l}e^{-\frac{(x-y)^{2}}{4kt}}dy
\\
& =\frac{1}{\sqrt{4k\pi t}}\sqrt{4kt}\dint\limits_{\frac{x+l}{\sqrt{4kt}}}^{%
\frac{x-l}{\sqrt{4kt}}}e^{-q^{2}}dq \\
\overset{q\!=\!\frac{x\!-\!y}{\sqrt{4kt}}}{=} &\!-\frac{1}{\sqrt{4\pi }}%
\dint\limits_{\frac{x+l}{\sqrt{4kt}}}^{\frac{x-l}{\sqrt{4kt}}}e^{-q^{2}}dq =%
\frac{1}{\sqrt{4\pi }}\dint\limits_{\frac{x-l}{\sqrt{4kt}}}^{\frac{x+l}{%
\sqrt{4kt}}}e^{-q^2}dq \\
& =\func{erf}\left( \frac{x+l}{\sqrt{4kt}}\right) -\func{erf}\left( \frac{x-l%
}{\sqrt{4kt}}\right)
\end{align*}

\noindent Note: Solution is different from zero for \textbf{all} x when $t>
0 $.

\newpage

\begin{center}
Lower Order Terms
\end{center}

\bigskip

\begin{align*}
u_{t} &= ku_{xx}-\gamma u \\
u(x,0) &= \varphi (x)
\end{align*}

\noindent Let $v(x,t)=e^{\gamma t}u(x,t)$ so $u(x,t)=e^{-\gamma t}v(x,t)$ \
\ \ \ $v(x,0)=\varphi (x)$

\begin{align*}
\frac{\partial u}{\partial t} &= e^{-\gamma t}\frac{\partial v}{\partial t}%
-\gamma e^{-\gamma t}v \\
\frac{\partial ^{2}u}{\partial x^{2}} &= e^{-\gamma t}\frac{\partial ^{2}v}{%
\partial x^{2}}
\end{align*}

\noindent So%
\begin{equation*}
e^{-\gamma t}\frac{\partial v}{\partial t}-\gamma e^{-\gamma t}v=e^{-\gamma
t}\frac{\partial ^{2}v}{\partial x^{2}}-\gamma e^{-\gamma t}v
\end{equation*}

\noindent or%
\begin{align*}
v_{t} &= kv_{xx} \\
v(x,0) &= \varphi (x)
\end{align*}

\noindent Hence,%
\begin{align*}
v(x,t) &= \frac{1}{\sqrt{4k\pi t}}\dint\limits_{-\infty }^{\infty }e^{-\frac{%
(x-y)^{2}}{4kt}}\varphi (y)dy \\
u(x,t) &= \frac{e^{-\gamma t}}{\sqrt{4k\pi t}} \dint\limits_{-\infty}^{%
\infty }e^{-\frac{(x-y)^{2}}{4kt}}\varphi (y)dy
\end{align*}

\newpage

\begin{center}
Convection - Diffusion
\end{center}

\bigskip

\begin{align*}
u_{t}+cu_{x}& =ku_{xx} \\
u(x,0)& =\varphi (x)
\end{align*}

Let $\ v(x,t)=u(x+ct,t)$ \ or \ $u(x,t)=v(\underset{y}{\underbrace{\underset{%
}{x-ct}}},t)=v(y,\tau )$

Then using the chain rule

\begin{eqnarray*}
\frac{\partial u}{\partial t} &=&\frac{\partial v}{\partial \tau }\frac{%
\partial \tau }{\partial t}+\frac{\partial v}{\partial y}\frac{\partial y}{%
\partial t}=\frac{\partial v}{\partial \tau }-c\frac{\partial v}{\partial y}
\\
\frac{\partial u}{\partial x} &=&\frac{\partial v}{\partial \tau }\frac{%
\partial \tau }{\partial x}+\frac{\partial v}{\partial y}\frac{\partial y}{%
\partial x}=\frac{\partial v}{\partial y}
\end{eqnarray*}

So%
\begin{eqnarray*}
\frac{\partial v}{\partial \tau }-c\frac{\partial v}{\partial y}+c\frac{%
\partial v}{\partial y} &=&k\frac{\partial ^{2}v}{\partial y^{2}} \\
\frac{\partial v}{\partial \tau } &=&k\frac{\partial ^{2}v}{\partial y^{2}}
\end{eqnarray*}

Therefore%
\begin{eqnarray*}
v(x,t) &=&\frac{1}{\sqrt{4k\pi t}}\dint\limits_{-\infty }^{\infty }e^{-\frac{%
(x-y)^{2}}{4kt}}\varphi (y)dy \\
u(x,t) &=&\frac{1}{\sqrt{4k\pi t}}\dint\limits_{-\infty }^{\infty }e^{-\frac{%
(x-ct-y)^{2}}{4kt}}\varphi (y)dy
\end{eqnarray*}

For example choose%
\begin{equation*}
\varphi (x)=%
\begin{cases}
\alpha & x<0 \\ 
\beta & |x|>0%
\end{cases}%
\end{equation*}

Then%
\begin{equation*}
u(x,t)=\frac{1}{\sqrt{4k\pi t}}\left\{ \alpha \dint\limits_{-\infty }^{0}e^{-%
\frac{(x-ct-y)^{2}}{4kt}}\varphi (y)dy+\beta \dint\limits_{0}^{\infty }e^{-%
\frac{(x-ct-y)^{2}}{4kt}}\varphi (y)dy\right\}
\end{equation*}

Define \ $p=\frac{x-ct-y}{\sqrt{4kt}}$ . Then \ $y=\sqrt{4kt}p+x-ct$.%
\begin{eqnarray*}
u(x,t) &=&\frac{1}{\sqrt{\pi }}\left\{ \alpha \dint\limits_{-\infty }^{\frac{%
x-ct}{\sqrt{4kt}}}e^{-p^{2}}dy+\beta \dint\limits_{\frac{x-ct}{\sqrt{4kt}}%
}^{0}e^{-p^{2}}dy+\right\} \\
&=&\frac{1}{\sqrt{\pi }}\left\{ \alpha \left[ \left( \dint\limits_{-\infty
}^{0}+\dint\limits_{0}^{\frac{x-ct}{\sqrt{4kt}}}\right) e^{-p^{2}}dy\right]
+\beta \left[ \left( \dint\limits_{0}^{\infty }-\dint\limits_{0}^{\frac{x-ct%
}{\sqrt{4kt}}}\right) e^{-p^{2}}dy\right] \right\} \\
&=&\frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2}\func{erf}(\frac{x-ct}{%
\sqrt{4kt}})
\end{eqnarray*}

\newpage

\begin{center}
Comparison: Wave Equation -- Diffusion Equation

\bigskip
\end{center}

$%
\begin{array}{ccc}
\text{Property} & \text{Wave} & \text{Diffusion} \\ 
\text{speed} & \text{finite } & \text{infinite} \\ 
\text{singularity} & \text{along characteristics} & \text{disappear} \\ 
\text{well posed }t>0 & \checkmark & \checkmark \\ 
\text{well posed }t<0 & \checkmark & X \\ 
\text{maximum principle} & X & \checkmark \\ 
\text{Energy} & \text{conserved} & \text{decays} \\ 
\text{information} & \text{transported} & \text{decays}%
\end{array}%
$

\bigskip \newpage

\end{document}
