auto-update(nvim): 2025-03-01 01:08:05

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

@@ -79,7 +79,7 @@ Attendance to discussion sections is mandatory.
 #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.
+  are solutions $y(x)$ such that $y'(x) = 0$, that is, $y(x)$ is *constant*.
 ]
 
 #example[
@@ -531,7 +531,7 @@ We can find a few particular solutions to our ODE, but how can we find all of th
   techniques.
 ]
 
-= Principle of superposition, Wronskian complex roots
+= Principle of superposition, the Wronskian, complex roots
 
 == Review
 
@@ -740,6 +740,8 @@ $
   y_2(t) = 1 / (2i) [z_1(t) - z_2(t)] = e^(lambda t) sin mu t, "imaginary part of" z_1(t) \
 $
 
+(in fact this is a variant of the Laplace transform.)
+
 By the superposition principle, they are solutions. Are they a fundamental set
 of solutions? Are they a basis for the solution space?
 
@@ -840,12 +842,59 @@ To find gamma, we can simply use inverse trig functions.
 
 == Linear systems of differential equations
 
+Consider the following *linear system* of ODEs:
+
+$
+  x' = x + 2y \
+  y' = 2x - 2y
+$
+
+We want function $x(t)$ and $y(t)$ that together solve this system. For instance,
+$
+  x(t) = 2e^(2t), y(t) = e^(2t)
+$
+
+Which we write as a vector
+$
+  arrow(x)(t) = vec(2e^2t, e^(2t))
+$
+
+We express our system above in matrix form
+
+$
+  arrow(x)'(t) = mat(1,2;2,-2) arrow(x)(t)
+$
+
+The solution $arrow(x)(t)$ is a *vector valued function* because it takes you from $t :: RR$ to $RR^2$, so $arrow(x) : RR -> RR^2$.
+
+We may consider $arrow(x)$, $arrow(y)$ as populations of competing species or
+concentrations of two compounds in a mixture.
+
+== General first order system
+
+#fact[
+  The general first order linear system:
+
+  $
+    arrow(x)'(t) = A(t) arrow(x)(t) + arrow(g)(t)
+  $
+
+  where $A(t)$ is an $n times n$ matrix and $arrow(g)$ and $arrow(x)$ are
+  vectors of length $n$.
+  $
+    arrow(g)(t) = vec(g_1 (t), g_2 (t), dots.v, g_n (t)) \
+    arrow(x)(t) = vec(mu_1 (t), mu_2 (t), dots.v, mu_n (t)) \
+  $
+]
+
+== Superposition principle for linear system
+
 Consider a matrix $A$ and solution vector $x$.
 $
   x'(t) = A(t) x(t)
 $
 
-#fact[Superposition principle for linear system][
+#fact[
   If $x^((1)) (t)$ and $x^((2)) (t)$ are solutions, then
   $
     c_1 x^((1)) (t) + c_2 x^((2)) (t)
@@ -853,14 +902,95 @@ $
   are also solutions.
 ]
 
-=== Homogenous case
+== Homogenous case
 
-This is when $g(t) = 0$. I want to solve for $arrow(x)' = A arrow(x)$, where
+This is when $g(t) = 0$. Consider
+$
+  arrow(x)'(t) = A arrow(x)(t)
+$
+Then there exists $n$ solutions
+$
+  arrow(x)^((1)) (t), dots, arrow(x)^((n)) (t)
+$
+such that any solution $arrow(x) (t)$ is a (unique) linear combination
+$
+  arrow(x)(t) = c_1 arrow(x)^((1)) (t) + dots + c_2 arrow(x)^((n)) (t)
+$
+
+=== The Wronskian, back back again!
+
+Alternatively form an $n times n$ square matrix $X(t)$ by arranging the fundamental solutions in columns
+
+#definition[
+  This is called a *fundamental matrix*.
+
+  $
+    X(t) = mat(x^((1)) (t), dots.c, x^((n)) (t))
+  $
+]
+
+#definition[
+  $det X(t)$ is called the *Wronskian* of $x^((1)) (t), dots.c, x^((n)) (t)$.
+]
+
+=== Finding solutions
+
+I want to solve for $arrow(x)' = A arrow(x)$, where
 $
   arrow(x) = vec(mu_1 (t), mu_2 (t), dots.v, mu_n (t))
 $
+
 Similar to the scalar case, we look for solutions in terms of exponential
-functions. Guessing
+functions. Substitute
+
+$
+  arrow(x)(t) = e^(bold(r) t) arrow(v)
+$
+where $bold(r)$ is a constant, and $arrow(v)$ is a column vector in $RR^n$.
+
+Then we have
+
+$
+  arrow(x)'(t) = r e^(r t) arrow(v) \
+  A arrow(x)(t) = A e^(r t) arrow(v) = e^(r t) A arrow(v)
+$
+
+Then $arrow(x)(t)$ is a solution if
+
+$
+  r arrow(v) = A arrow(v)
+$
+
+Recall that when we have a linear transformation from $RR^m -> RR^m$, an
+*eigenvector* is a vector whose only transformation is that it gets scaled.
+That is, applying the linear transformation is equivalent to multiplying the
+vector by a scalar. An *eigenvalue* for an eigenvector is the scalar that
+scales the vector after the linear transformation. That is, for a linear
+transformation $A : RR^m -> RR^m$, vector $arrow(v)$ (in $RR^m$), and scalar
+$lambda$, if $A arrow(v) = lambda arrow(v)$, then $lambda$ is an eigenvalue of
+the eigenvector $arrow(v)$. A complex eigenvector somehow corresponds to
+rotation, however we will not discuss geometric interpretation of it here.
+
+To solve for an eigenvector from an eigenvalue, one only needs to write the
+equation $(A - lambda I) arrow(v) = 0$.
+
+Returning our attention to our equation above, we see that $arrow(v)$ is an
+eigenvector of the coefficient matrix $A$ with eigenvalue $r$.
+
+If an $n times n$ coefficient matrix $A$ has $n$ linearly independent
+eigenvectors (eigenbasis)
+
+$
+  arrow(v)^((1)), ..., arrow(v)^((n))
+$
+
+with corresponding eigenvalues
+
+$
+  r_1, ..., r_n
+$
+
+Guessing
 
 $
   arrow(x) (t) = e^(r t) arrow(v) \
@@ -896,4 +1026,179 @@ This is the eigenvalue equation for $A$!
 
   Now we want the eigenvectors for our eigenvalues. Find an eigenvector
   corresponding to $lambda_1 = 1$.
+
+  For $r_1 = 1$:
+  $
+    (A - I)v^((1)) = 0 \
+    mat(1,1;1,1) v^((1)) = 0 \
+    v^((1)) = vec(1,-1)
+  $
+
+  For $r_2 = 3$:
+  $
+    (A - 3I)v^((2)) = 0 \
+    mat(-1,1;1,-1) v^((2)) = 0 \
+    v^((2)) = vec(1,1)
+  $
+
+  Then our fundamental solutions are
+  $
+    x^((1)) (t) &= e^(r_1 t) arrow(v)^((1)) = e^t vec(1,-1) \
+    x^((2)) (t) &= e^(r_2 t) arrow(v)^((2)) = e^(3t) vec(1,1)
+  $
+  Forming a fundamental matrix
+  $
+    X(t) = mat(e^t,e^(3t);-e^t,e^(3t))
+  $
+  Representing a general solution
+  $
+    arrow(x)(t) = c_1 arrow(x)^((1)) (t) + c_2 arrow(x)^((2)) (t) = c_1 e^t vec(1,-1) + c_2 e^(3t) vec(1,1)
+  $
+  or written as a fundamental matrix:
+  $
+    arrow(x)(t) = X(t) c = mat(e^t,e^(3t);-e^t,e^(3t)) vec(c_1,c_2)
+  $
+]<general-solution-system>
+
+#example[Continued @general-solution-system][
+  Now consider the initial value $arrow(x)(0) = arrow(x)_0 = vec(2,-1)$. Now
+  solve the initial value problem.
+
+  General solution
+  $
+    arrow(x)(t) = X(t) c = mat(e^t,e^(3t);-e^t,e^(3t)) vec(c_1,c_2)
+  $
+  $
+    arrow(x)(0) = X(0) arrow(c) = x_0 => arrow(c) = X(0)^(-1) arrow(x)_0 \
+    arrow(c) = mat(1,1;-1,1)^(-1) vec(2,-1) = 1 / 2 mat(1,-1;1,1) vec(2,-1) = vec(3/2,1/2)
+  $
+  Finally giving us
+  $
+    arrow(x)(t) = X(t) c = mat(e^t,e^(3t);-e^t,e^(3t)) vec(3/2,1/2) \
+    vec(3/2 e^t + 1/2 e^(3t), -3/2 e^t + 1/2 e^(3t))
+  $
 ]
+
+== Visualizing solutions
+
+We discuss visualizing the solutions in @general-solution-system. The general
+solution is a vector valued function.
+
+$
+  arrow(x)(t) = c_1 arrow(x)^((1)) (t) = c_2 arrow(x)^((2)) (t) = c_1 e^t vec(1,-1) + c_2 e^(3t) vec(1,1)
+$
+
+As $t$ varies, each solution $arrow(x)(t)$ traces a curve in the plane. When
+$c_1 = c_2 = 0$, $arrow(x) = vec(0,0)$, the *equilibrium solution*.
+
+When $c_1 != 0$ and $c_2 = 0$, the solution is a scalar multiple of $vec(1,-1)$
+and the magnitude tends to $infinity$ as $t -> infinity$. As $t -> -infinity$
+the magnitude tends to 0.
+
+When both $c_1$ and $c_2$ are nonzero then the full solution is a linear
+combination. As $t -> infinity$ the magnitude of $arrow(x)$ tends to
+$infinity$. As $t -> -infinity$ the magnitude of $arrow(x)$ tends to 0.
+
+The equilibrium solution $arrow(x)(t) = vec(0,0)$ is stable (as $t$ moves in
+either direction it tends to $vec(0,0)$ namely because it doesn't depend on
+$t$). It's an example of a *node*.
+
+= Repeated eigenvalues, nonhomogenous systems
+
+== Classification of equilibria $n=2$
+
+We discuss classification of possible equilibria at 0 for a system $arrow(x)' =
+A arrow(x)$ when $n=2$.
+
+If we have real eigenvalues $r_1, r_2 != 0$, then
+- $r_1,r_2 < 0$ means we have an *asymptotically stable* node
+- $r_1,r_2 > 0$ means we have an *unstable* node
+- $r_1,r_2 < 0$ means we have an *unstable* saddle
+
+== Repeated eigenvalues
+
+Consider the system
+$
+  arrow(x)' = mat(1,-2;2,5) arrow(x)
+$
+on $[a,b]$. It has one eigenvalue and eigenvector
+$
+  r_1 = 3, arrow(v)_1 = vec(1,-1)
+$
+So we have one solution
+$
+  arrow(x)_1 (t) = e^(3t) vec(1,-1)
+$
+
+How do we obtain the rest of our fundamental set? We need to try
+$
+  arrow(x)_2 (t) = t e^(3t) arrow(v)_1 + e^(3t) arrow(u)
+$
+and find the right choice of $arrow(u)$.
+
+As long as $arrow(u)$ solves $(A - r_1 I) arrow(u) = arrow(v)_1$, it works.
+
+#definition[
+  We call such a vector $arrow(u)$ a *generalized eigenvector*.
+]
+
+#remark[
+  We can *always* find the rest of our solution space with this method.
+]
+
+== Nonhomogenous linear system
+
+Like before, when we have a set of fundamental solutions to the homogenous
+system, we only need to find any particular solution.
+
+$
+  arrow(x)(t) = c_1 arrow(x)_1 (t) + c_2 arrow(x)_2 (t) + arrow(x)_p (t)
+$
+
+== Methods for finding a particular solution
+
+- When $arrow(g)(t) = arrow(g)$ is a constant vector, there may exist an
+  equilibrium solution which can then be used as a particular solution.
+- Method of undetermined coefficients: can be used in constant coefficient case
+  if $arrow(g)(t)$ has a special form. Very limited.
+- Variation of parameters: more general, but messy integrals
+
+== Equilibrium solution as particular solution
+
+Let $arrow(g) in RR^n$ be a constant vector. Find a particular solution to
+$
+  arrow(x)' + A arrow(x) + arrow(g)
+$
+Solve the linear system to find a constant equilibrium solution
+$
+  A arrow(x) + arrow(g) = 0
+$
+If $A$ is invertible then
+$
+  arrow(x) = -A^(-1) arrow(g)
+$
+is an equilibrium solution. So
+$
+  arrow(x)_p (t) = -A^(-1) arrow(g)
+$
+is a particular solution of the system.
+
+== Undetermined coefficients
+
+Consider
+$
+  arrow(x)' (t) = mat(1,2;2,1) arrow(x)(t) + vec(1,1)
+$
+We want $arrow(x)_p$. Let's try
+$
+  arrow(x)_p = vec(A,B)
+$
+
+The general solution looks like
+$
+  arrow(x)(t) = c_1 e^(-t) vec(1,-2) + c_2 e^(3t) vec(1,2) + arrow(x)_p (t)
+$
+We assume
+$
+  arrow(x)_p = vec(A e^t, B e^t)
+$