Lecture 1
First-Order Differential Equation:
Lecture 2
First-order differential equation
1. Case 1: y'=g(x)
This case is trivial using the
fundamental theorem calculus, which gives us
$y=\int\limits_{x_0}^xg(x)dx+c$
2. Case 2: Linear equations
A) Homogeneous: $y’=-p(x)y$
For this
case we will have $y=e^{-\int p(t)dt}$. (Say $P(x)=-\int\$)
This
brings us to the initial value theorem which wants $y(x)$ which satisfies the
DE and $y_0=y(x_0)$
It is
worth noting that for a $p:I\to X$, there exists a unique function $y=e^{-\int\limits_{x_0}^x}p(t)dt$
such that $y(x_0)=y_0$ and satisfies the DE.
(Proof
later)
B) Non Homogeneous: $y’=-p(x)y-q(x)$
In this
case we will have $\dfrac{d}{dx}ye^{P(x)}=-q(x)e^{P(x)}$ as it will give us
$-ye^{P(x)}p(x)+y’e^{P(x)}=-q(x)e^{P(x)}$
Now from
the above equation, we can say
$ye^{P(x)}=-\int\limits_{x_0}^x
q(t)e^{P(t)}dt$
3. Case 3: Non linear DE
A) Separable: y’=f(x,y)=g(x)h(y)
Now $\dfrac{dy}{dx}=g(x)h(y)\Rightarrow
\int\dfrac{dy}{h(y)}=\int\dfrac{g(x)}{dx}$ and $h(y)$ must be non vanishing.
Theorem: Suppose $I$ is a closed interval
and $U$ an open interval and we have $g:I\to \mathbb{R}$ and $h:U\to \mathbb{R}$.
Then there exists $J\subseteq I$ such that $x\in J$ and there exists a unique $y(x)$
such that it satisfies the differential equation and $y(x_0)=y_0$.
$G(x)=\int\limits_{x_0}^x
g(t)dt$ and $H(y)=\int\limits_{y_0}^y\dfrac{1}{h(t)}dt$ then we will have $y=H^{-1}(G(x))$
Proof: (Later
XP)
B) Exact DE
Suppose
we have a differential equation of the form
$P(x,y)dx+Q(x,y)dy=0$
We may
consider the above equation as $\phi(x,y)=c$ which means $\dfrac{d\phi(x,y)}{dx}=\dfrac{\partial
\phi}{\partial x}+\dfrac{\partial \phi}{\partial y}\dfrac{dy}{dx}$
On comparison we can see that $\dfrac{\partial \phi}{\partial x}=P(x,y)$ and $\dfrac{\partial \phi}{\partial y}=Q(x,y)$.
Lecture 3
A system of linear differential equations
Say $\overline{x}=(x_1,x_2,\ldots x_n)$ and we are given the system of
equations $\dot{\overline{x}}=\overline{v}(t,\overline{x})$ where $\overline{v}(t,
\overline{x})=(v_1(t,\overline{x}),\ldots v_n(t, \overline{x}))$. This would
like the following
$\dfrac{dx_1}{dt}=v_1(t, \overline{x})$
$\dfrac{dx_2}{dt}=v_2(t, \overline{x})$
$\dfrac{dx_3}{dt}=v_3(t, \overline{x})$
$\ldots$
$\dfrac{dx_n}{dt}=v_n(t, \overline{x})$
A solution
to these family of equations, say is $\overline{\phi}$,
and it satisfies $\dot{\overline{\phi}}=\overline{v}(t,\overline{\phi})$
Theorem: Any $n$th order differential
equation can be converted into a system of 1st order DEs.
$x^{(n)}=F(t,x,x’,x’’\ldots
x^{(n-1)})$
Consider
$\overline{y}=\overline{v}(t,\overline{y})$ where $\overline{y}=(y_1,y_2\ldots
y_n)$. Here we will consider $y_i=\dfrac{dy_{i-1}}{dt}$ where $1\leq i \leq n$
and $y^{(n)}=F(t,y’,y’’,\ldots y^{n-1})$. Now this gives us $\overline{v}(t,\overline{y})=\dot{\overline{y}}=(y_2,\ldots
F(t,\overline{y}))$.
For a
differential equation which is of the form
$\dot{\overline{x}}=\overline{v}(t,\overline{x})$, we call it Non-autonomous and for types of the form $\dot{\overline{x}}=\overline{y(\overline{x})}$, we
call it Autonomous
Geometric interpretation
Let’s focus on the Autonomous equations $\dot{\overline{x}}=\overline{v}(\overline{x})$.
The set of all possible $x$s forms the phase space or the state phase, $U\subseteq
\mathbb{R}^n$.
Now let’s consider a deterministic, finite dimensionafinite-dimensionalle process. Say the phase space is $M$. Therefore
the process is $M\to M$. Say the state $\overline{x}\to g^t(\overline{x})$
under this process. Here the variable $t$ represents the time elapsed to reach
that state. Note that
$g^{t+s}(\overline{x})=g^{t}(g^s(\overline{x}))$. It
is interesting to note that $\{g^t:M\to M | t\in \mathbb{R} \}$ can be treated
as a group. Say $g^{t}(\overline{x})=f(t)$. We will have $f(s+t)=f(s)\oplus f(t)$.
This brings us to the idea of Phase curves $\{M,\{g^t\}\}$
which connects all the different states of $\overline{x}$, forming a curve. (Note
that $g^t$ is differentiable).
Say $\overline{v}(\overline{x})=\dfrac{d}{dt}g^t(\overline{x})\vert_{t=0}$.
The right-hand side gives us a tangent vector at every state. We get a vector
field from the family of differential equations, $\dot{\overline{x}}=\overline{v}(\overline{x})$. Below is a representation of a vector field with the phase curve.
Lecture 4
Consider the phase space $U\subseteq
\mathbb{R}^n$. Recall that a vector field is a system of
differential equations $\dot{\overline{x}}=\overline{v}(\overline{x})$. Now if the
map $\overline{v}$ is continuous/differentiable/continuously differentiable, then
the vector field is continuous/differentiable/continuously
differentiable.
Aside: The vector field is a
section of tangent bundles
Using the phase field, we can get
solutions for $\dot{\overline{x}}=\overline{w}(\overline{x})$.
Have a look at the picture below. (The phase space is in yellow)
For given $U,\{g^t\}$ where $g^t:U\to U$ is a diffeomorphism, we say that the phase velocity is $\dfrac{d}{dt}g^t(x)\vert_{t=0}:=\overline{v}(\overline{x})$
Definition:
We
say a point, an equilibrium in the phase space if $g^t(x)=x$. It simply does
not change with time ($t$).
Theorem:
There
exists one and only one phase curve passing through a point on the phase space.
(Proof?)
Extended
Phase Space: We call the space $\mathbb{R}\times U$,
where $U$ is the phase space. We will get to know soon on its importance.
Integral
curves: Here we will consider the product space $\mathbb{R}\times
U$ (which we have defines as the extended phases space). The curve $(t,\phi (t))$
where $\phi=g^t$, is called the Integral curve. (They are just normal freaking
graphs!)
Now under which conditions on $\overline{v}$ and $t_0,x_0$,
can we guarantee a n unique integral curve through $(t_0,x_0)$. This brings us
to the following theorem.
Theorem:
Before
stating the main statement of the theorem, have a look on the 2 equations below
1. $\dot{x}=v(x)$
2. $x=\phi(t)$ where $t\in \mathbb{R}$ and $x\in U$
(Here $U$ is the phase space).
The Statement: For given $v:U\to \mathbb{R}$, and
given $t_0\in \mathbb{R}$ and $x_0\in U$, we will have a unique solution $\phi$
to equation 1 which satisfies equation 2.
Also, if solutions $\phi_1$ and $\phi_2$ satisfy 2, then they will coincide in a common neighborhood around $t=t_0$.
We will then have
$t-t_0=\displaystyle{\int\limits_{x_0}^{x=\phi(t)}\dfrac{1}{v(\xi)}d \xi}$
Proof:
Take the function $\psi(x)=t_0+\int\limits_{x_0}^{x}\dfrac{d\xi}{v(\xi)}$.
We will then have $\dfrac{d\psi (x)}{dx}\vert_{x=x_0}=\dfrac{1}{\dfrac{1}{v(x_0)}}=v(x_0)\neq
0$.
As $\psi(x)$ is differentiable in a neighborhood of
$x_0$ as $\dfrac{1}{v(\xi)}$ is continuous for $v(\psi)\neq 0$, by inverse
function theorem, we can find $\phi$ such that $\psi(\phi(t))=t$ and $\phi(\psi(x))=x$.
Now we will have
$\dfrac{d\psi (x)}{dx}\vert_{x=\phi(t)}\cdot\dfrac{d\phi
(t)}{t}=1\Rightarrow \dfrac{d\phi }{dt}=\dfrac{1}{\dfrac{d\psi}{dx}}=v(x)$
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