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\begin{document}

\begin{center}
\textbf{Wronskian}
\end{center}

Given two functions $u(x)$ and $v(x)$

\begin{equation*}
W[u,v]=%
\begin{vmatrix}
u(x) & v(x) \\ 
u^{^{\prime }}(x) & v^{^{\prime }}(x)%
\end{vmatrix}%
=uv^{^{\prime }}-u^{^{\prime }}v
\end{equation*}

Theorem

\bigskip

\begin{itemize}
\item $u$ and $v$ are linearly independent iff $W[u,v] \neq 0$ for all x

\item If $W[u,v]=0$ then $v(x)=cu(x)$
\end{itemize}

\newpage

\begin{center}
\textbf{Jacobian}
\end{center}

Given two functions $u(x,y)$ and $v(x,y)$%
\begin{equation*}
J(u,v)=%
\begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ 
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}%
\end{vmatrix}%
=\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u%
}{\partial y}\frac{\partial v}{\partial x}
\end{equation*}

Then $dudv=J(u,v)dxdy$

\bigskip

Theorem:

Given $u(x,y)$ and $v(x,y)$

\begin{itemize}
\item If $J(u,v)\ne 0$ in $D$ then this can be inverted in D%
\begin{align*}
x &=x(u,v) \\
y &=y(u,v)
\end{align*}

\item If $J(u,v) \equiv =0$ in $D$ then%
\begin{equation*}
v=F(u)
\end{equation*}
\end{itemize}

\begin{center}
\textit{Examples}
\end{center}

\bigskip

\begin{enumerate}
\item 
\begin{align*}
x &=s+t \\
y &=\left( s+t\right) ^{2}
\end{align*}%
then $\frac{\partial (x,y)}{\partial (s,t)}=0$

\item 
\begin{align*}
x& =s+t \\
y& =s-t
\end{align*}%
then $\frac{\partial (x,y)}{\partial (s,t)}=-2$ so%
\begin{align*}
s& =\frac{x+y}{2} \\
t& =\frac{x-y}{2}
\end{align*}

\item 
\begin{align*}
x& =s+t \\
y& =s^{2}-t^{2}
\end{align*}

then $\frac{\partial (x,y)}{\partial (s,t)}=-2(s+t)$. So $J$ vanishes on the
line $s=-t$. A global inverse may not exist. However, in this case%
\begin{align*}
s& =\frac{x^{2}+y}{2x} \\
t& =\frac{x^{2}-y}{2x}
\end{align*}

\item 
\begin{align*}
x& =s+t \\
y& =s^{2}+t^{2}
\end{align*}%
then $\frac{\partial (x,y)}{\partial (s,t)}=2(t-s)$ so $J$ vanishes on the
line $s=t$. Only a double-valued inverse exists in any domain containing $%
s=t $.
\end{enumerate}

\newpage

\begin{center}
\textbf{Inhomogenous equations - Duhamel}
\end{center}

\begin{align*}
\frac{du}{dt}+Au(t) &= f(t) \qquad u(0)=\phi \\
\text{multiply by}e^{At} \qquad e^{At}\frac{du}{dt}+e^{At}Au(t) &= e^{At}f(t)
\\
\frac{d}{dt}(e^{At}u) &= e^{At}f(t) \\
\text{integrate} \qquad \qquad \quad \qquad \qquad \quad e^{At}u(t) &=
\int_0^t e^{As}f(s)ds+C \\
\text{set t=0} \qquad \qquad \quad \qquad \qquad \qquad \quad u(0) &=C=\phi
\\
\text{So} \qquad \qquad \quad \qquad \qquad \qquad \qquad \quad u(t)
&=e^{-At}\int_{0}^{t}e^{As}f(s)ds+\phi e^{-At} \\
u(t) &=\phi e^{-At}+\int_0^{t}e^{A(s-t)}f(s)ds
\end{align*}

\newpage

\begin{center}
General case
\end{center}

\begin{equation}
{cal L}(u) = F(x,y,u,u_x,u_y,u_{xx},u_{xy},u{yy}) = 0  \label{eqf}
\end{equation}

\medskip

Notation: \newline
d = number of dimesnions \newline
p=order of equation, i.e. highest derivative

We shall concentrate on first and center order equations

\begin{center}
Strong Solution
\end{center}

A strong solution satisfies \eqref{eqf} in some region subject to other
constraints (initial \& boundary conditions).

\bigskip

\begin{center}
\textit{Linear Equations}
\end{center}

\noindent An equation is linear if when $u$ and $v$ are solutions so is $a \
u(x,y) + b \ v(x,y)$. i.e. 
\begin{equation*}
{cal L}(au+bv) = a \ {cal L}(u) + b \ {cal L}(v)
\end{equation*}

An equation is homogenous if ${cal L}(u)=0$.

\newpage

\begin{center}
\textbf{Fredholm Alternative}
\end{center}

Given $A \epsilon R^{n \times n}$ 
\newline
Then exactly one of the following alternatives is true

\begin{itemize}
\item The null space of A is trivial. So for every $b \epsilon R^n$ there
exists a unique solution $x \epsilon R^n$.

\item The null space of A is non-trivial. Then $Ax=b$ has a solution iff 
whenever $y$ has the property that
\begin{equation*}
y \epsilon R^n \qquad A^{T} y = 0
\end{equation*}
So 
\begin{equation*}
y \epsilon N(A^T) \qquad (y,b) = 0
\end{equation*}
If this is true then there are infinitely many solutions of the form 
\begin{equation*}
x = x_0 + y \qquad y \epsilon N(A^T)
\end{equation*}
Note: If $y^TA=0$ then 
\begin{equation*}
(y,b) = y^T b = y^T Ax = (y^TA) x =0
\end{equation*}
so the condition is necessary. 
\end{itemize}

\newpage

\begin{center}
\textit{example}
\end{center}

Consider 
\begin{align*}
-\frac{d^{2}u}{dx^{2}} &=f\qquad 0\leq x\leq l \\
\frac{du}{dx}(0) &=0 \qquad \frac{du}{dx}(l)=0
\end{align*}
Then $u$ = constant is a solution of the homogenous equation. Hence, the
second alternatve exists.

\bigskip

\begin{center}
\textit{Compatibility Condition}
\end{center}

\begin{equation*}
\int_{0}^{l}f(x)dx=-\int_{0}^{l}\frac{d^{2}u}{dx^{2}}dx=
-\frac{du}{dx} \vert_0^l=\frac{du}{dx}(0)-\frac{du}{dx}(l) = 0
\end{equation*}%
So a solution exists only if \quad  $\int_{0}^{l}f(x)dx=0$.
\newline
If this is true then u + constant is also a solution.

\end{document}
