auto-update(nvim): 2025-01-21 17:02:55

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

@@ -1,6 +1,6 @@
-#import "./dvd.typ": *
+#import "@youwen/zen:0.1.0": *
 
-#show: dvdtyp.with(
+#show: zen.with(
   title: "Math 4B Course Notes",
   author: "Youwen Wu",
   date: "Winter 2025",
@@ -164,3 +164,178 @@ $
   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 IVT - 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.