From bb0b3da8b7530b3cc3f4d10061c74dede08b123b Mon Sep 17 00:00:00 2001 From: Youwen Wu Date: Sat, 1 Mar 2025 01:08:05 -0800 Subject: [PATCH] auto-update(nvim): 2025-03-01 01:08:05 --- .../by-course/math-4b/course-notes/main.typ | 317 +++++++++++++++++- 1 file changed, 311 insertions(+), 6 deletions(-) diff --git a/documents/by-course/math-4b/course-notes/main.typ b/documents/by-course/math-4b/course-notes/main.typ index 992818f..50c9292 100644 --- a/documents/by-course/math-4b/course-notes/main.typ +++ b/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) + $ +] + +#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) +$