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H1. Consider the scalar differential equation d Define the potential V(x) by Use the chain rule to show that x(t) = f(x). dt dV (x) dx = -f(x). dV(x) dx dV(x(1)) ≤ 0. H2. Let x* be an equilibrium solution of Eq (2). First, show that x* is an extrema of V by calculating (2) -||x=x* = 0. Next, if x* maximizes V, is x* locally stable or unstable? Finally, if x* minimizes V, is x* locally stable or unstable?

H1. Consider the scalar differential equation d Define the potential V(x) by Use the chain rule to show that x(t) = f(x). dt dV (x) dx = -f(x). dV(x) dx dV(x(1)) ≤ 0. H2. Let x* be an equilibrium solution of Eq (2). First, show that x* is an extrema of V by calculating (2) -||x=x* = 0. Next, if x* maximizes V, is x* locally stable or unstable? Finally, if x* minimizes V, is x* locally stable or unstable?

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can you explain the maximising and minimising part of H2 in on more detail please, with a diagram if possible I am still confused.

H1. Consider the scalar differential equation d dtx(t) = f(x). Define the potential V(x) by Use the chain rule to show that dV (x) dx = -f(x). dV (x) dx dV(x(1)) ≤ 0. H2. Let x* be an equilibrium solution of Eq (2). First, show that x* is an extrema of V by calculating (2) -||x=x* = 0. Next, if x* maximizes V, is x* locally stable or unstable? Finally, if x* minimizes V, is x* locally stable or unstable?
Transcribed Image Text:H1. Consider the scalar differential equation d dtx(t) = f(x). Define the potential V(x) by Use the chain rule to show that dV (x) dx = -f(x). dV (x) dx dV(x(1)) ≤ 0. H2. Let x* be an equilibrium solution of Eq (2). First, show that x* is an extrema of V by calculating (2) -||x=x* = 0. Next, if x* maximizes V, is x* locally stable or unstable? Finally, if x* minimizes V, is x* locally stable or unstable?
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