alexandria/documents/by-course/math-4b/course-notes/main.typ

#import "@youwen/zen:0.1.0": *

#show: zen.with(
  title: "Math 4B Course Notes",
  author: "Youwen Wu",
  date: "Winter 2025",
  subtitle: [Taught by Guofang Wei],
)

#outline()

= Course logistics

The textbook is Elementary Differential Equations, 11th edition, 2017. Chapters
1-4, 6, 7, and 9 will be covered.

Attendance to discussion sections is mandatory.

= Lecture #datetime(day: 7, month: 1, year: 2025).display()

== Trivial preliminaries

#definition[
  An ODE involves an unknown function of a single variable and its derivatives
  up to some fixed order. The order of an ODE is the order of the highest
  derivative that appears.
]

#example[First order ODE][
  $ (dif y) / (dif x) = y^2 $
]

#example[Second order ODE][
  $
    y'' &= x \
    integral.double y'' dif x &= integral.double x dif x \
    y &= 1 / 6 x^3 + C
  $
]

#definition[
  A function $y(x)$ defined on $(a,b)$ is a *solution* of the ODE
  $ y' = F(x,y) $
  if and only if
  $ y'(x) = F(x, y(x)), forall x in (a,b) $
]

#problem[
  Check that $y(x) = 20 + 10e^(-x / 2)$ is a solution to the ODE
  $ y' = -1 / 2 y + 10 $
]

#definition[
  A first order ODE
  $ y' = F(x,y) $
  is called *linear* if there are functions $A(x)$ and $B(x)$ such that
  $ F(x,y) = A(x) y + B(x) $
]

#example("Linear ODEs")[
  - $y' = x$
  - $y' = y$
  - $y' = x^2$
]

#example("Nonlinear ODEs")[
  - $y' = y^2$
]

#definition[
  In general, a differential equation is called linear if and only if it can be
  written in the form
  $
    a_n (t) (dif^n y) / (dif t^n) + a_(n-1) (t) (d^(n-1) y) / (d t^(n-1)) + dots + a_1 (t) (dif y) / (dif t) + a_0 (t) y = g(t)
  $
  where $a_k (t)$ and $g(t)$ are single variable functions of $t$.
]

#definition[
  *Equilibrium solutions* for the ODE
  $ y' = F(x,y) $
  are solutions $y(x)$ such that $y'(x) = 0$, that is, $y(x)$ is constant.
]

#example[
  The equation
  $ y' = y(y +2) $
  has two equilibria
  $
    y(x) &= 0 \
    y(x) &= -2
  $
]

#problem[
  What are the equilibria of the equation
  $ y' = y(y - x) $
]

== General solution of a first order linear ODE

We start with the differential equation in standard form

$ (dif y) / (dif t) + p(t) y = g(t) $

where $p(t)$ and $g(t)$ are continuous single variable functions of $t$.

Then let us assume the existence of an *integrating factor* $mu(t)$, such that

$ mu(t) p(t) = mu'(t) $

and then multiplying each term by $mu(t)$ to obtain

$ mu(t) (dif y) / (dif t) + mu(t) p(t) y = mu(t) g(t) $

Then

$ mu(t) (dif y) / (dif t) + mu'(t) y = mu(t) g(t) $

Then recognize that the left side of the equation is the product rule to obtain

$
  (mu(t) y(t))' &= mu(t) g(t) \
  integral (mu(t) y(t))' dif t &= integral mu(t) g(t) dif t \
  mu(t) y(t) + C &= integral mu(t) g(t) dif t \
  y(t) &= (integral mu(t) g(t) dif t + C) / (mu(t))
$

Now we have a general solution but we need to determine $mu(t)$.

$
  mu(t) p(t) &= mu'(t) \
  (mu'(t)) / (mu(t)) &= p(t) \
  (ln mu(t))' &= p(t)
$

So now

$
  integral mu(t) + k &= integral p(t) dif t \
  ln mu(t) &= integral p(t) dif t + k \
  mu(t) = e^(integral p(t) dif t + k) &= k e^(integral p(t) dif t)
$

Now substitute

$
  y(t) &= (integral k e^(integral p(t) dif t) g(t) dif t + C) / (k e^(integral p(t))) \
  &= (k integral e^(integral p(t) dif t) g(t) dif t + C) / (k e^(integral p(t)))
$

Now do some cursed constant manipulation to obtain a final solution with only one arbitrary constant

$
  y(t) = (integral e^(integral p(t) dif t) g(t) dif t + C) / (e^(integral p(t)))
$

#remark[
  The most useful result to us is
  $
    y(t) &= (integral mu(t) g(t) dif t + C) / (mu(t)) \
    mu(t) &= e^(integral p(t) dif t)
  $
  We can easily obtain a solution form for any first order linear ODE simply by
  identifying $p(t)$ and $g(t)$.
]

= Lecture #datetime(day: 22, year: 2025, month: 1).display() - existence and uniqueness of solutions

Midterm 1 is #datetime(month: 1, day: 28, year: 2025).display() in class, covering Chapter 2. Five problems:

- Solving separable equations (10pts)
- Solving Linear ODE (10pts)
- Modeling (Newton's law of cooling) (10pts)
- Eulers's method (5 pts)
- True or false problem (if ODE linear, existence, etc) (5 pts)

== Existence and uniqueness of solutions to the initial value problem

Before we looked at particular ODEs. Now we turn our attention to general ODEs.
In general ODEs do not have solutions, so let us discuss methods to identify
when solutions exist.

Given a first order initial value problem ODE

$ y' = F(t,y), underbrace(y(t_0) = y_0, "initial value") $

When does a solution exist? Is it unique?

Warm up:

$ y' = -x / y $

It's separable and we can solve it.

$
  1 / 2 y^2 &= -1 / 2 x^2 + C \
  y^2 &= 2C - x^2 \
  y &= plus.minus sqrt(2C - x^2)
$

Say $y(1) = 1$. Then $C = 1$.

Since $y(1) = 1$, we choose the positive version of solution.

$ y = sqrt(2 - x^2) $

The slope field shows that the solutions are semicircles above and below the
$x$-axis.

Consider this initial value

$
  y(3) &= 0 \
  2C &= 9 \
  y &= plus.minus sqrt(9 - x^2)
$

Now should we choose $+$ or $-$? It seems that either can work, but it turns
out that it is not a solution!

Recall the equation

$ y' = -x / y $

With our initial condition, this equation was not defined! So there is no
solution for this initial condition.

Now consider

$ y' = sqrt(y) $

Again it is separable

$
  integral dif x &= integral sqrt(y) dif y \
  2y^(1 / 2) &= x + C \
$

With initial condition $y(0) = 0$,

$
  y = (x / 2)^2
$

However we lost a solution, since we divided by $sqrt(y)$, which could've been
0!

When $F(t,y)$ (RHS) is not nice (differentiable) at the initial value $(t_0,
y_0)$, existence and uniqueness (E/U) could fail. We do have E/U when $F(t,y)$
is nice.

== Existence of solutions to IVP - linear case

The first order linear ODE is

$
  y'(t) = p(t) y(t) = g(t) \
$

Use integrating factor

$ mu(t) = e^(integral p(t) dif t) $

We know a general solution using the method of integrating factors.

$ y(t) = 1 / mu(t) [C + integral mu(t) g(t) dif t] $

We are assuming that $p(t)$ and $g(t)$ are given continuous functions defined
on some interval $I$. The solution $y(t)$ is defined on the same interval
$I$.

#remark[
  The solution of the initial value problem for a linear equation exists, is
  unique, and is defined as long as the coefficients in the equation are defined.
]

#remark[
  Existence and uniqueness are important. It guarantees there is one and only one
  solution curve passing through an initial point $(t_0, y_0)$.
]

#example[
  Consider
  $ y'(t) - 1 / t y(t) = 1 / (t^2), y(1) = 0 $

  + $(-infinity, 0)$
  + $(0, infinity)$
  + $(-infinity, infinity)$
  + $(0, 2)$

  $0$ is a bad value. So we choose either $(0, infinity)$ or $(-infinity, 0)$.
  But the initial value exists on $(0, infinity)$, so we choose (1).
]

Now let's consider a general first order ODE

$ y' = F(t,y), y(t_0) = y_0 $

Now "nice" is not so clear cut.
Assume $F(t,y)$ is a "nice" function ($F$, $(diff F)/(diff y)$ are continuous)
defined on some rectangle in $(t,y)$-plane.

#theorem("Existence and uniqueness theorem for ODE")[
  If the initial point $(t_0, y_0)$ belongs to the rectangle where $F(t,y)$ is
  defined, then this initial value problem has a *unique solution* $y(t)$ defined
  on some time interval $I$.
]<eutheorem>

#example[
  Find the solution of the initial value problem
  $
    y' = y^2, y(0) = 1 \
    y(t) = 1 / (1-t)
  $

  The function on the right side is
  $ F(t,y) = y^2 $
  It's defined for all $t$ and $y$. Nevertheless the largest interval on which
  the solution is defined is $-infinity < t < 1$.
]

Consider IVP

$
  y' = sqrt(9 - y^2), y(1) = 0
$

Any E/U problems?

== Numerical approximations with Euler's method

We can determine if solutions exist using @eutheorem but in general we cannot
find an explicit solution. Instead we approximate the solution.

Consider the IVP

$ y' = 1 / 2 cos(y) + t, y(0) = -1 $

The righthand side is nice, by @eutheorem, it has a unique solution. We can't
find an explicit solution. So use Euler's method.

= Lecture #datetime(day: 23, month: 1, year: 2025).display()

Second order linear differential equation.

$
  y'' = p(t)y' + q(t)y = g(t)
$

We say it is homogenous if $g(t) = 0$.

== Initial value problem and uniqueness on 2nd order linear ODEs

Suppose $p(t)$, $q(t)$, and $g(t)$ are given continuous functions defined on
$I$.
$
  y'' = p(t)y' + q(t)y = g(t) \
  y(t_0) = y_0, y'(t_0) = y'_0
$

has a unique solution $y(t)$ defined on $I$. Same as first order case. Always write in standard form first before identifying.

#example[
  What is the largest interval on which the IVP
  $
    (t+1)y'' + y = 3, y(0) = 1, y'(0) = 0
  $
  is certain to have a solution?

  $
    y'' + 1 / (t+1) y = 3 / (t+1)
  $

  So the functions are nice except at -1. Our intervals are

  $
    (-infinity, -1), (-1, infinity)
  $

  But we have initial values at 0, so we know for certain that a solution (but not for certain it's the only solution) exists

  $
    (-1, infinity)
  $
]

== 2nd order linear homogenous ODEs

How to find solutions to 2nd order linear ODE? Recall

$
  y'' = p(t)y' + q(t)y = 0
$

Consider 2nd order linear homogenous ODEs with constant coefficients $a,b,c, a!= 0$.

$
  a y'' + b y' + c y = 0
$

Judging from this equation it seems $y$ should be an exponential since we want
a function whose derivatives can cancel each other out (with constants
applied).

Trying $y=e^(r t)$,

$
  y' = r e^(r t) \
  y'' = r^2 e^(r t) \
  a r^2 e^(r t) + b r e^(r t) + c e^(r t) = 0
$

$e^(r t)$ is never 0 (in $RR$) so we can divide without losing or gaining solutions.

$
  e^(r t) (a r^2 + b r + c) = 0
$

Then we simply just need to solve a quadratic for $r$.

$
  a r^2 + b r + c = 0
$

We may have two distinct real solutions, one repeated solution, or complex
solutions (no solutions in $RR$). For now let us consider the distinct real
solutions.

#fact[
  We conclude $y(t) = e^(r t)$ is a solution of
  $
    a y'' + b y' + c y = 0
  $
  provided $r$ satisfies
  $
    a r^2 + b r + c = 0
  $
  This is the _characteristic equation_.
]

#example[
  $
    y'' + 5y' + 6y = 0
  $
  It is homogenous, so we find the characteristic equation.

  $
    r^2 + 5 r + 6 = 0 \
    r = -2, -3 \
    y_1 = e^(-2t), y_2 = e^(-3t)
  $
]<particular-solutions>

#example[
  $ y'' - 2y' + y = 0 $
  The characteristic equation is
  $
    r^2 - 2r + 1 = 0 \
    r = 1 \
    y_1 = e^t
  $
]

== Superposition principle for 2nd order linear homogenous ODE

We can find a few particular solutions to our ODE, but how can we find all of the solutions?

#fact[
  Suppose $y_1(t), y_2(t)$ are a pair of solutions of
  $
    y'' + p(t) y' + q(t) y = 0
  $
  Then the linear combination
  $
    c_1 y_1(t) + c_2 y_2(t)
  $
  is also a solution.
]<superposition-principle>

#proof[
  $
    (c_1 y_1(t) + c_2 y_2(t))'' + p(t) (c_1 y_1(t) + c_2 y_2(t))' + q(t)(c_1 y_1(t) + c_2 y_2(t)) \
    = c_1 y_1'' + c_2 y_2'' + p(t)(c_1 y_1' + c_2 y_2') + g(t)(c_1 y_1 + c_2 y_2) \
    = c_1 (y''_1 + p(t)y'_1 + q(t) y_1) + c_2 (y''_2 + p(t) y'_2 + q(t) y_2) \
    = c_1 (0) + c_2(0) = 0
  $
]

#example[Revisiting @particular-solutions][
  $
    y'' + 5y' + 6y = 0
  $
  It is homogenous, so we find the characteristic equation.

  $
    r^2 + 5 r + 6 = 0 \
    r = -2, -3 \
    y_1 = e^(-2t), y_2 = e^(-3t)
  $

  So by @superposition-principle,
  $
    y(t) = c_1 e^(-2t) + c_2 e^(-3t)
  $
  is a general solution.
]

#example[
  Find the general solution of
  $
    y'' + 2y' - 3y = 0 \
    y(1) = 1, y'(1) = -1
  $
  and then find the solution satisfying the initial conditions
  $
    y(1) = 1, y'(1) = -1
  $
  General solution is
  $
    y(t) &= c_1 e^(-3t) + c_2 e^t \
    y'(t) &= -3 c_1 e^(-3t) + c_2 e^t
  $
  Now we need the particular solution.
  $
    1 &= c_1 e^(-3) + c_2 e \
    -1 &= -3 c_1 e^(-3) + c_2 e \
  $

  Now we have a system and we can solve it using standard linear algebra
  techniques.
]

= Principle of superposition, Wronskian complex roots

== Review

Recall:

Second order ODE:

$
  y'' = F(t,y,y')
$

Linear:

$
  y'' + p(t) y' + q(t) y = g(t)
$

Homogenous:

$
  y'' + p(t) y' + q(t) y = 0
$

Constant coefficients: $a y'' + b y' + c y = 0$, with characteristic equation

$
  a r^2 + b r + c = 0
$

From this characteristic equation we determine either distinct real roots,
complex real roots, or repeated real roots. We already know what to do with
distinct real roots.

$
  y'' + p(t) y' + q(t) y = 0
$

The linear combination

$
  y(t) = c_1 y_1 (t) + c_2 y_2 (t)
$

is a solution for any constants $c_1, c_2$. The solutions from a vector space!
Key: equation is linear and homogenous.

#example[
  Consider the linear, homogenous equation

  $
    y'' + 5y' + 6y = 0
  $

  We have two solutions:

  $
    y_1 (t) e^(-2t) + c_2 e^(-3t)
  $

  is a general solution. Are these all the solutions?
]

== The Wronskian

Given any initial values

$
  y(t_0) y_0, y' (t_0) = y'_0
$

Substitute in:

$
  c_1 y_1(t_0) + c_2 y_2(t_0) = y_0 \
  c_1 y'_1 (t_0) + c_2 y'_2 (t_0) = y'_0 \
$

We have a linear system for $c_1, c_2$:

$
  mat(c_1 y_1(t_0), c_2 y_2(t_0); c_1 y'_1 (t_0), c_2 y'_2 (t_0)) vec(c_1, c_2) = vec(y_0, y'_0)
$

The linear system has a unique solution provided the determinant of the
coefficient matrix is nonzero:

$
  mat(y_1(t_0), y_2(t_0); y'_1(t_0), y'_2(t_0)) != 0
$

$ W = y_1(t_0) y'_2(t_0) - y_2(t_0) y'_1(t_0) != 0 $

#definition[
  This determinant $W$ is called the *Wronskian* of $y_1(t)$ and $y_2(t)$ at
  the point $t_0$.
]

#fact[
  If $W != 0$ at some point $t_0$, then it is nonzero throughout the interval $I$ where the solution is defined.
]

To summarize our description of the solution space, if $y_1(t)$, $y_2(t)$ are two solutions of the linear homogenous ODE

$
  y'' + p(t) y' + q(t) y = 0
$

such that $W(y_1, y_2)$, then the constants $c_1$, $c_2$ can be uniquely determined so that

$
  y(t) = c_1 y_1(t) + c_2 y_2 (t)
$

satisfies any initial condition

$
  y(t_0) = y_0, y'(t_0) = y'_0
$

$y(t)$ is the general solution.

== Solution space with constant coefficients, distinct real roots

Consider the 2nd order homogenous linear diffeq with constant coefficients.

$
  a y'' + b y ' + c y = 0
$

Suppose the characteristic equation

$
  a r^2 + b r + c = 0
$

has a pair of *distinct real roots* $r_1$, $r_2$. Then we have a pair of solutions

$
  y_1(t) = e^(r_1 t), y_2(t) = e^(r_2 t)
$

Question: is this a fundamental set of solutions?

Check Wronskian.

$
  W = det mat(y_1(t_0), y_2(t_0); y'_1(t_0), y'_2(t_0)) = det mat(e^(r_1 t_0), e^(r_2 t_0); r_1 e^(r_1 t_0), r_2 e^(r_2 t_0)) = e^(r_1 t_0) e^(r_2 t_0) (r_2 - r_1) != 0
$

since $r_1 != r_2$.

The Wronskian of our two solutions is $!= 0$ so the general solution is indeed

$
  y(t) = c_1 e^(r_1 t) + c_2 e^(r_2 t)
$

== 2nd order linear homogenous ODE, complex roots

As usual consider
$ a y'' + b y' + c y = 0 $

with characteristic equation

$
  a r^2 + b r + c = 0
$

If $b^2 - 4a c < 0$, then solutions are complex numbers:

$
  r_1 = lambda + i_mu, r_2 = lambda - i_mu
$

with

$
  lambda = -b / (2a), mu = sqrt(4 a c - b^2) / (2a) != 0
$

Complex solutions

$
  z_1(t) = e^(r_1 t) = e^((lambda + i_mu) t) = e^(lambda t) e^(i_mu t), z_2(t) = e^(r_2 t)
$

What is $e^(i_mu t)$?

Euler's formula:

$
  e^(i theta) = cos theta + i sin theta
$

Using Euler's formula we can write

$
  z_1(t) = e^(r_1 t) = e^((lambda + i_mu) t) = e^(lambda t) e^(i_mu t) = e^(lambda t) [cos mu t + i sin mu t] \
  z_2(t) = e^(r_2 t) = e^((lambda - i_mu) t) = e^(lambda t) e^(-i_mu t) = e^(lambda t) [cos mu t - i sin mu t]
$

Define

$
  y_1(t) = 1 / 2 [z_1(t) + z_2(t)] = e^(lambda t) cos mu t, "real part of" z_1(t) \
  y_2(t) = 1 / (2i) [z_1(t) - z_2(t)] = e^(lambda t) sin mu t, "imaginary part of" z_1(t) \
$

By the superposition principle, they are solutions. Are they a fundamental set
of solutions? Are they a basis for the solution space?

Check the Wronskian:

We see that it is nonzero, therefore, when $b^2 - 4 a c < 0$, the equation

$
  a y'' + b y' + c y = 0
$

has two real solutions

$
  y_1(t) = e^(lambda t) cos mu t, y_2(t) = e^(lambda t) sin mu t
$

where

$
  lambda = -b / (2a), mu = sqrt(4a c - b^2) / (2a) != 0
$

The Wronskian of these solutions is nonzero, so $y_1$ and $y_2$ are a fundamental set of solutions. The general solution of the equation is

$
  y(t) = c_1 e^(lambda t) cos mu t + c_2 e^(lambda t) sin mu t
$

== Amplitude and phase angle

Given $c_1 cos(omega_0 t) + c_2 sin(omega_0 t)$, express it as a single cosine
function so we can graph it. Recall this formula:

$
  cos(alpha - beta) = cos(alpha) cos(beta) + sin(alpha) sin(beta)
$

Write

$
  c_1 cos(omega_0 t) + c_2 sin(omega_0 t) \
  = sqrt(c_1 ^2 + c_2 ^2) cos(omega_0 t - theta) = A cos(omega_0 t - theta)
$

where $A$ is the amplitude, $theta$ is the phase angle with $cos theta = (c_1)/sqrt(c_1^2 + c_2^2)$

== Some trig review (yuck)

#fact[
  The following trig identity is useful

  $
    A cos(omega t) + B sin(omega t) = C cos(omega t - gamma)
  $

  where $C = sqrt(A^2 + B^2)$, $sin(gamma) = B/C$, and $cos(gamma) = A/C$.
]<sus-identity>

To find gamma, we can simply use inverse trig functions.

#example[
  Write $x(t) = -2cos(5t) - sin(5t)$ using only one cosine function:

  First note that $A = -2$, $B - -1$, $omega = 5$. Then by @sus-identity,

  $
    x(t) = -2cos(5t) - sin(5t) \
    = sqrt(5) cos(5t - gamma)
  $

  Now note that
  $
    tan(gamma) &= sin(gamma) / cos(gamma) \
    &= 1 / 2
  $

  Therefore we have the following:

  $
    gamma = arctan(1/2)
  $

  But this is not exactly right.

  Recall that $arctan$ produces outputs only in $[-pi/2, pi/2]$. Therefore if
  we want $cos$ and $sin$ to be negative, we need to add an additional $pi$
  term. So in fact

  $
    gamma = arctan(1/2) - pi
  $

  Which makes $cos(5t + pi - arctan(1/2))$ negative.
]