129 lines
5.3 KiB
Markdown
129 lines
5.3 KiB
Markdown
---
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title: Frameworks and Review
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allDay: false
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startTime: 12:30
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endTime: 14:30
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date: 2024-09-09
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completed: null
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type: single
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---
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# Introduction
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Where do nonlinearities come from?
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Well, a couple of places...
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1. Geometric nonlinearities (pendulum)
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2. External fields
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3. Material properties
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So we're stuck with them. Bjupyter labut how do we deal with noninearities?
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## A nonlinear equation
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$$ \dot{x} = \frac{dx}{dt} = 1-2\cos x$$
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How do you solve this? You can't use Laplace, you can't separate...
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*insert very long expression that Bajaj wrote.*
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Getting an analytical solution can be a PITA to obtain. For this reason:
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**The general case is that nonlinear equations are unsolvable.**
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This doesn't mean we can't learn things. We can describe these systems *qualitatively*.
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Really our options come down to:
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- Solve exactly (Not likely to happen)
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- Solve numerically
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- Analyze qualitatively (~geometrically)
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- Solve an approximation to the problem
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We mix and match these approaches.
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## Geometric (Qualitative) Methods
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Geometric analysis answers questions like "is this stable?" "what's the response look like?"
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### Linear Systems
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$$ \dot{x} = Ax$$
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This is a simple linear dynamic system.
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How many equilibria does this system have?
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**One.**
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The system is at equilibrium where $\frac{dx}{dt} = 0$. It won't move from this point.
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Is this system stable? Check the eigenvalues of A.
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### Nonlinear Systems
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**Recall: $\dot{x} = 1-2\cos x$**
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We can qualitatively describe systems using the phase plane:
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*Insert graphics from class*
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How is this useful to us engineers?
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We are going to see systems that are nonlinear, and they can give us ideas about where things could blow up. In our second example, we have generally a pretty safe area below x = 2. Anywhere below there, we know we're going to end up at x = -2, but above x =2, all hell breaks loose.
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This is what we care about. We want to know where in our nonlinear system domains things can become dangerous.
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# How do we numerically get a time domain response?
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Numericaly:
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$$ \dot x = f(x) $$
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$$\frac{dx}{dt} = \lim_{\Delta t \rightarrow 0} \frac{f(x(t+\Delta t))-f(x(x))}{\Delta t} $$
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This is the tangent (or the secant while $\Delta t =/ 0$)
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>[!note] Euler's Method
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>Therefore, for finite $\Delta t$:
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> $$ f(x(t+\Delta t)) = \frac{dx}{dt} \Delta t + f(x(t))$$
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> Limitations: innaccurate if time steps are large.
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> There are better methods!
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> ode45() <- Variable Step Runge-Kutta
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We're going to use a lot of odeint in SciPy
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*Insert code here*.
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# Geometric Analyses in Higher Dimensions
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We want to do geometric analysis with n-dimensions, not just one.
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## Linear stability for a nonlinear system
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1. Find the fixed equilibrium points
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$$\dot x = 0 = f(x)$$
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This equation is solved by points $x^\star$
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Define: $$ \epsilon(t) = x(t) - x^\star \rightarrow x(t) = x^\star + \epsilon(t)$$
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$$\frac{dx}{dt} = f(x)$$
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$$\frac{d\epsilon}{dt} = f(x^\star +\epsilon)$$
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$$= f(x^\star) + \epsilon f'(x^\star) + \Theta(\epsilon^2)\rightarrow \epsilon f'(x^\star) + \Theta(\epsilon^2)$$
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$$\frac{d\epsilon}{dt} = \epsilon f'(x^\star) + \Theta(\epsilon^2) \rightarrow \epsilon f'(x^\star)$$
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- If $\epsilon f'(x^\star) > 0 \rightarrow \text{Unstable}$
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- If $\epsilon f'(x^\star) < 0 \rightarrow \text{Stable}$
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- If $\epsilon f'(x^\star) = 0 \rightarrow \text{Inconclusive}$
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## Energy Motivation
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$$\dot{x} = f(x) = -\frac{dV}{dt} \text{V = potential}$$
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$$\frac{dV}{dx} = -\frac{dx}{dt}$$
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$$\frac{dV}{dt} <0 \rightarrow V \text{continuously decreasing in time until } \frac{dV}{dt} = 0$$
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- Stable fixed poitn: $V(x^\star)$ is (a local) minimum of $V(x)$
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- Unstable fixed point: $V(x^\star)$ is (a local) maximum of $V(x)$.
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# A review of Linear Differential Equations
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## Foundation
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$$\dot{\vec{x}} = \frac{dx}{dt} = \bf{A} \vec{x} + \vec{F}$$
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Homogeneous (no input/forcing):
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$$\dot{\vec{x}} = \bf{A} \vec{x}$$
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$$\vec{x}(t) = C e^{\bf{A}t}$$
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This e to the matrix A is kinda gross. We represent this as follows:
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$$ e^{\bf{A}} = \sum_{k=0}^\inf \frac{\bf{A}^k}{k!}$$so then:
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$$ e^{\bf{A}t} = \sum_{k=0}^\inf \frac{\bf{A}^k t^k}{k!}$$
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But since $\frac{d}{dt}e^{\bf{A}t} = \bf{A}e^{\bf{A}t}$
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Some proof that xdot = Ax
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## Diagonalization
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Suppose:
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$$ \dot{\vec{x}} = \bf{A}\vec{x} $$
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with A diagonal. Then
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$$ \dot{x}_n = a_{nn} x_n$$
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This is nice. So we should try and diagonalize things...
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- Find $\bf{P}$ such that $\bf P^{-1} \bf A \bf P = \bf D$
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Then we have new coordinates:
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- $\vec{x} = \bf{P} \vec y$
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- $\vec{\dot{x}} = \bf{P} \vec{\dot{y}}$
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- then $\dot{\vec{y}} = \bf{P^{-1}AP}\vec{y}$
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Choose P to be column wise eigenvectors of A, with diagonal matrix D having the eigenvalues of A.
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This isn't always possible.
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## Fundamental Matrix
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$$ e^{\bf{A}t} = \Psi(t) = \text{Fundamental matrix of } \dot{\vec{x}} = \bf{A} \vec{x}$$
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How do we calculate this thing?
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- series calculation of $A^m$
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- Smarter computer program
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- Laplace Transform
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Consider: $\dot{\vec{x}} = \bf{A}\vec{x} + \vec{F}$
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$$ \vec{x}(t) = \bf{\Psi} \vec{c} \leftarrow \text{autonomous}$$
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or the total solution:
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$$ \vec{x}(t) = \bf{\Psi} \vec{c} + \int_{t_0}^t \bf{\Psi}^{-1}(\bf{I}) \bf{F}(I)dI + \bf{\Psi}(t_0)\vec{c}$$
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$$ \vec{x}(t_0) = \bf{\Psi}(t_0)\vec{c}$$
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Then the solution is
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$$\vec{x}(t) = e^{\bf{A}t}\vec{c} + e^{\bf{A}t} \int_{t_0}^t e^{-\bf{A}I} F(T) \delta T$$
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Then using the Laplace transform:
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$$e^{\bf{A}t} = \mathcal{L}^{-1} \{ (sI-\bf{A})^-1 \} $$
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# Written Notes
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![[How do we deal with nonlinearities?.png]]
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![[Mode Diagram.png]] |