hadaasyj
2022-05-01
Answered

Prove instability using Lyapunov function

${x}^{\prime}={x}^{3}+xy$

$y}^{\prime}=-y+{y}^{2}+xy-{x}^{3$

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Cristian Rosales

Answered 2022-05-02
Author has **26** answers

The dominant term is $x}^{6}+{x}^{3}y+{y}^{2$ . This term is positive (complete the square). All other terms are small when x,y, are small, so they can be controlled. Here are the details.

Write

${V}^{\prime}(x,y)={x}^{6}+{x}^{4}y+{y}^{2}-{y}^{3}-x{y}^{2}+{x}^{3}y$

$={x}^{6}+(1+x){x}^{3}y+(1-x-y){y}^{2}$

$\ge {x}^{6}+(1+x){x}^{3}y+\frac{1}{2}{y}^{2}{\textstyle \phantom{\rule{1em}{0ex}}}\text{if}\left|x\right|+\left|y\right|\le \frac{1}{4}$

$\ge {x}^{6}-\frac{5}{4}{\left|x\right|}^{3}\left|y\right|+\frac{1}{2}{y}^{2}{\textstyle \phantom{\rule{1em}{0ex}}}\text{since}\left|x\right|\le \frac{1}{4}$

$={({\left|x\right|}^{3}-\frac{5}{8}\left|y\right|)}^{2}+\frac{7}{64}{y}^{2}$

$\ge 0$

with equality iff$x=y=0$

Write

with equality iff

asked 2021-01-02

Find

asked 2022-04-08

Finding the extrema of a functional (calculus of variations)

$I\left(y\right)={\int}_{0}^{1}{y}^{\prime 2}-{y}^{2}+2xydx$

Explanation:

$\begin{array}{rl}\frac{\mathrm{\partial}F}{\mathrm{\partial}{y}^{\text{'}}}& =2{y}^{\text{'}}\phantom{\rule{0ex}{0ex}}\u27f9\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\frac{\mathrm{\partial}F}{\mathrm{\partial}{y}^{\text{'}}}\right)=2{y}^{\text{'}\text{'}}\\ \frac{\mathrm{\partial}F}{\mathrm{\partial}y}& =-2y+2x\end{array}$

Which gives the Euler equation $2y{}^{\u2033}+2y-2x=0$ (I think, unless I messed up my math).

Ok now I need to try to find solutions to this differential equation, however, I only know how to solve linear second order DE's. The x term is throwing me off and I am not sure how to solve this.

I know $y\left(x\right)=x$ is a solution by inspection, but inspection is a poor man's approach to solving DE's.

asked 2022-04-29

I try to solve this ode using the variation of parameters theorem.

The characteristic polynomial of the homogenous equation is

Then

I

II

Multiply I by

We get

I don't get why it incorrect, where am I wrong?

asked 2022-04-22

My textbook states we need locally Lipschitz continuous with respect to the second argument and uniformly with respect to the first argument of $f(t,\text{}x)$ . This is said to be equivalent to the result that for all compact subsets $V\subset U$ where $U\subseteq {\mathbb{R}}^{n+1}$ is open we have

$L{\textstyle \phantom{\rule{0.222em}{0ex}}}={\supset}_{(t,x)\ne (t,y)\in V\subset U}\frac{|f(t,x)-f(t,y)|}{|x-y|}<\mathrm{\infty}$

I understand that locally Lipschitz means there is a neighborhood around each point where f is Lipschitz, but what does the uniform part mean with respect to t?

I understand that locally Lipschitz means there is a neighborhood around each point where f is Lipschitz, but what does the uniform part mean with respect to t?

asked 2022-04-23

Nonlinear ODE initial value problem

A student came to me with a problem I couldn't solve. It's the beginning of the semester in his Intro DiffEq class, and so the solution shouldn't be too difficult. But it completely stumped me, and now I can't let it go! Here it is:

Problem. Find all solutions to the IVP:

$\sqrt{{\left(\frac{dy}{dx}\right)}^{2}-4{x}^{2}}=\sqrt{{x}^{4}-{y}^{2}},;;;;y\left(0\right)=0.$

I thought about maybe just doing a sort of guess-and-check method, but the existence/uniqueness theorem doesn't apply, so I'm not sure that would help even if I could find a single solution, which I cannot in any case.

A student came to me with a problem I couldn't solve. It's the beginning of the semester in his Intro DiffEq class, and so the solution shouldn't be too difficult. But it completely stumped me, and now I can't let it go! Here it is:

Problem. Find all solutions to the IVP:

I thought about maybe just doing a sort of guess-and-check method, but the existence/uniqueness theorem doesn't apply, so I'm not sure that would help even if I could find a single solution, which I cannot in any case.

asked 2021-06-11

If ${x}^{2}+xy+{y}^{3}=1$ find the value of y''' at the point where x = 1

asked 2021-06-16

Use the substitution