auto-update(nvim): 2025-03-01 01:08:05
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@ -79,7 +79,7 @@ Attendance to discussion sections is mandatory.
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#definition[
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*Equilibrium solutions* for the ODE
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$ y' = F(x,y) $
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are solutions $y(x)$ such that $y'(x) = 0$, that is, $y(x)$ is constant.
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are solutions $y(x)$ such that $y'(x) = 0$, that is, $y(x)$ is *constant*.
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]
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#example[
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@ -531,7 +531,7 @@ We can find a few particular solutions to our ODE, but how can we find all of th
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techniques.
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]
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= Principle of superposition, Wronskian complex roots
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= Principle of superposition, the Wronskian, complex roots
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== Review
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@ -740,6 +740,8 @@ $
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y_2(t) = 1 / (2i) [z_1(t) - z_2(t)] = e^(lambda t) sin mu t, "imaginary part of" z_1(t) \
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$
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(in fact this is a variant of the Laplace transform.)
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By the superposition principle, they are solutions. Are they a fundamental set
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of solutions? Are they a basis for the solution space?
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@ -840,12 +842,59 @@ To find gamma, we can simply use inverse trig functions.
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== Linear systems of differential equations
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Consider the following *linear system* of ODEs:
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$
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x' = x + 2y \
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y' = 2x - 2y
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$
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We want function $x(t)$ and $y(t)$ that together solve this system. For instance,
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$
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x(t) = 2e^(2t), y(t) = e^(2t)
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$
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Which we write as a vector
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$
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arrow(x)(t) = vec(2e^2t, e^(2t))
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$
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We express our system above in matrix form
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$
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arrow(x)'(t) = mat(1,2;2,-2) arrow(x)(t)
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$
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The solution $arrow(x)(t)$ is a *vector valued function* because it takes you from $t :: RR$ to $RR^2$, so $arrow(x) : RR -> RR^2$.
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We may consider $arrow(x)$, $arrow(y)$ as populations of competing species or
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concentrations of two compounds in a mixture.
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== General first order system
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#fact[
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The general first order linear system:
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$
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arrow(x)'(t) = A(t) arrow(x)(t) + arrow(g)(t)
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$
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where $A(t)$ is an $n times n$ matrix and $arrow(g)$ and $arrow(x)$ are
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vectors of length $n$.
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$
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arrow(g)(t) = vec(g_1 (t), g_2 (t), dots.v, g_n (t)) \
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arrow(x)(t) = vec(mu_1 (t), mu_2 (t), dots.v, mu_n (t)) \
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$
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]
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== Superposition principle for linear system
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Consider a matrix $A$ and solution vector $x$.
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$
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x'(t) = A(t) x(t)
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$
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#fact[Superposition principle for linear system][
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#fact[
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If $x^((1)) (t)$ and $x^((2)) (t)$ are solutions, then
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$
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c_1 x^((1)) (t) + c_2 x^((2)) (t)
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@ -853,14 +902,95 @@ $
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are also solutions.
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]
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=== Homogenous case
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== Homogenous case
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This is when $g(t) = 0$. I want to solve for $arrow(x)' = A arrow(x)$, where
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This is when $g(t) = 0$. Consider
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$
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arrow(x)'(t) = A arrow(x)(t)
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$
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Then there exists $n$ solutions
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$
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arrow(x)^((1)) (t), dots, arrow(x)^((n)) (t)
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$
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such that any solution $arrow(x) (t)$ is a (unique) linear combination
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$
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arrow(x)(t) = c_1 arrow(x)^((1)) (t) + dots + c_2 arrow(x)^((n)) (t)
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$
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=== The Wronskian, back back again!
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Alternatively form an $n times n$ square matrix $X(t)$ by arranging the fundamental solutions in columns
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#definition[
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This is called a *fundamental matrix*.
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$
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X(t) = mat(x^((1)) (t), dots.c, x^((n)) (t))
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$
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]
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#definition[
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$det X(t)$ is called the *Wronskian* of $x^((1)) (t), dots.c, x^((n)) (t)$.
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]
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=== Finding solutions
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I want to solve for $arrow(x)' = A arrow(x)$, where
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$
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arrow(x) = vec(mu_1 (t), mu_2 (t), dots.v, mu_n (t))
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$
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Similar to the scalar case, we look for solutions in terms of exponential
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functions. Guessing
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functions. Substitute
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$
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arrow(x)(t) = e^(bold(r) t) arrow(v)
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$
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where $bold(r)$ is a constant, and $arrow(v)$ is a column vector in $RR^n$.
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Then we have
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$
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arrow(x)'(t) = r e^(r t) arrow(v) \
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A arrow(x)(t) = A e^(r t) arrow(v) = e^(r t) A arrow(v)
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$
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Then $arrow(x)(t)$ is a solution if
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$
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r arrow(v) = A arrow(v)
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$
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Recall that when we have a linear transformation from $RR^m -> RR^m$, an
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*eigenvector* is a vector whose only transformation is that it gets scaled.
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That is, applying the linear transformation is equivalent to multiplying the
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vector by a scalar. An *eigenvalue* for an eigenvector is the scalar that
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scales the vector after the linear transformation. That is, for a linear
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transformation $A : RR^m -> RR^m$, vector $arrow(v)$ (in $RR^m$), and scalar
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$lambda$, if $A arrow(v) = lambda arrow(v)$, then $lambda$ is an eigenvalue of
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the eigenvector $arrow(v)$. A complex eigenvector somehow corresponds to
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rotation, however we will not discuss geometric interpretation of it here.
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To solve for an eigenvector from an eigenvalue, one only needs to write the
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equation $(A - lambda I) arrow(v) = 0$.
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Returning our attention to our equation above, we see that $arrow(v)$ is an
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eigenvector of the coefficient matrix $A$ with eigenvalue $r$.
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If an $n times n$ coefficient matrix $A$ has $n$ linearly independent
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eigenvectors (eigenbasis)
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$
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arrow(v)^((1)), ..., arrow(v)^((n))
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$
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with corresponding eigenvalues
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$
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r_1, ..., r_n
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$
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Guessing
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$
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arrow(x) (t) = e^(r t) arrow(v) \
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@ -896,4 +1026,179 @@ This is the eigenvalue equation for $A$!
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Now we want the eigenvectors for our eigenvalues. Find an eigenvector
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corresponding to $lambda_1 = 1$.
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For $r_1 = 1$:
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$
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(A - I)v^((1)) = 0 \
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mat(1,1;1,1) v^((1)) = 0 \
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v^((1)) = vec(1,-1)
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$
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For $r_2 = 3$:
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$
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(A - 3I)v^((2)) = 0 \
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mat(-1,1;1,-1) v^((2)) = 0 \
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v^((2)) = vec(1,1)
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$
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Then our fundamental solutions are
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$
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x^((1)) (t) &= e^(r_1 t) arrow(v)^((1)) = e^t vec(1,-1) \
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x^((2)) (t) &= e^(r_2 t) arrow(v)^((2)) = e^(3t) vec(1,1)
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$
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Forming a fundamental matrix
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$
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X(t) = mat(e^t,e^(3t);-e^t,e^(3t))
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$
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Representing a general solution
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$
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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)
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$
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or written as a fundamental matrix:
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$
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arrow(x)(t) = X(t) c = mat(e^t,e^(3t);-e^t,e^(3t)) vec(c_1,c_2)
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$
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]<general-solution-system>
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#example[Continued @general-solution-system][
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Now consider the initial value $arrow(x)(0) = arrow(x)_0 = vec(2,-1)$. Now
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solve the initial value problem.
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General solution
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$
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arrow(x)(t) = X(t) c = mat(e^t,e^(3t);-e^t,e^(3t)) vec(c_1,c_2)
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$
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$
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arrow(x)(0) = X(0) arrow(c) = x_0 => arrow(c) = X(0)^(-1) arrow(x)_0 \
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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)
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$
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Finally giving us
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$
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arrow(x)(t) = X(t) c = mat(e^t,e^(3t);-e^t,e^(3t)) vec(3/2,1/2) \
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vec(3/2 e^t + 1/2 e^(3t), -3/2 e^t + 1/2 e^(3t))
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$
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]
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== Visualizing solutions
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We discuss visualizing the solutions in @general-solution-system. The general
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solution is a vector valued function.
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$
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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)
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$
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As $t$ varies, each solution $arrow(x)(t)$ traces a curve in the plane. When
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$c_1 = c_2 = 0$, $arrow(x) = vec(0,0)$, the *equilibrium solution*.
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When $c_1 != 0$ and $c_2 = 0$, the solution is a scalar multiple of $vec(1,-1)$
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and the magnitude tends to $infinity$ as $t -> infinity$. As $t -> -infinity$
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the magnitude tends to 0.
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When both $c_1$ and $c_2$ are nonzero then the full solution is a linear
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combination. As $t -> infinity$ the magnitude of $arrow(x)$ tends to
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$infinity$. As $t -> -infinity$ the magnitude of $arrow(x)$ tends to 0.
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The equilibrium solution $arrow(x)(t) = vec(0,0)$ is stable (as $t$ moves in
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either direction it tends to $vec(0,0)$ namely because it doesn't depend on
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$t$). It's an example of a *node*.
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= Repeated eigenvalues, nonhomogenous systems
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== Classification of equilibria $n=2$
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We discuss classification of possible equilibria at 0 for a system $arrow(x)' =
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A arrow(x)$ when $n=2$.
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If we have real eigenvalues $r_1, r_2 != 0$, then
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- $r_1,r_2 < 0$ means we have an *asymptotically stable* node
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- $r_1,r_2 > 0$ means we have an *unstable* node
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- $r_1,r_2 < 0$ means we have an *unstable* saddle
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== Repeated eigenvalues
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Consider the system
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$
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arrow(x)' = mat(1,-2;2,5) arrow(x)
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$
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on $[a,b]$. It has one eigenvalue and eigenvector
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$
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r_1 = 3, arrow(v)_1 = vec(1,-1)
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$
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So we have one solution
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$
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arrow(x)_1 (t) = e^(3t) vec(1,-1)
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$
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How do we obtain the rest of our fundamental set? We need to try
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$
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arrow(x)_2 (t) = t e^(3t) arrow(v)_1 + e^(3t) arrow(u)
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$
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and find the right choice of $arrow(u)$.
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As long as $arrow(u)$ solves $(A - r_1 I) arrow(u) = arrow(v)_1$, it works.
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#definition[
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We call such a vector $arrow(u)$ a *generalized eigenvector*.
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]
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#remark[
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We can *always* find the rest of our solution space with this method.
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]
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== Nonhomogenous linear system
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Like before, when we have a set of fundamental solutions to the homogenous
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system, we only need to find any particular solution.
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$
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arrow(x)(t) = c_1 arrow(x)_1 (t) + c_2 arrow(x)_2 (t) + arrow(x)_p (t)
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$
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== Methods for finding a particular solution
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- When $arrow(g)(t) = arrow(g)$ is a constant vector, there may exist an
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equilibrium solution which can then be used as a particular solution.
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- Method of undetermined coefficients: can be used in constant coefficient case
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if $arrow(g)(t)$ has a special form. Very limited.
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- Variation of parameters: more general, but messy integrals
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== Equilibrium solution as particular solution
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Let $arrow(g) in RR^n$ be a constant vector. Find a particular solution to
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$
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arrow(x)' + A arrow(x) + arrow(g)
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$
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Solve the linear system to find a constant equilibrium solution
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$
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A arrow(x) + arrow(g) = 0
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$
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If $A$ is invertible then
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$
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arrow(x) = -A^(-1) arrow(g)
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$
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is an equilibrium solution. So
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$
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arrow(x)_p (t) = -A^(-1) arrow(g)
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$
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is a particular solution of the system.
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== Undetermined coefficients
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Consider
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$
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arrow(x)' (t) = mat(1,2;2,1) arrow(x)(t) + vec(1,1)
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$
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We want $arrow(x)_p$. Let's try
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$
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arrow(x)_p = vec(A,B)
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$
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The general solution looks like
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$
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arrow(x)(t) = c_1 e^(-t) vec(1,-2) + c_2 e^(3t) vec(1,2) + arrow(x)_p (t)
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$
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We assume
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$
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arrow(x)_p = vec(A e^t, B e^t)
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$
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