Slope fields of ordinary differential equations

Classification of critical point of linear systems

Consider the homogeneous linear first-order system differential equations[br][center][i]x[/i]'=[i]ax[/i]+[i]by[/i][br][i]y[/i]'=[i]cx[/i]+[i]dy[/i][/center]The following worksheet is designed to analyse the nature of the critical point of the linear system.

Lotka-Volterra model

Consider the Lotka-Voterra equations of interacting predator and prey systems[br][center][i]x[/i]'=[i]x[/i]([i]a[/i]-c[i]x[/i]-d[i]y[/i])[br][i]y[/i]'=-[i]y[/i]([i]b[/i]-e[i]x[/i])-[i]h[/i][/center][size=100][size=150]where all of [i]x[/i], [i]y[/i], [i]a[/i], [i]b[/i], [i]c[/i], [i]d[/i], [i]k[/i], [i]n [/i] are positive, and[/size][/size][list][*][i]x[/i] represents the number of prey[/*][*][i]y[/i] represents the number of predators[/*][*][i]a[/i] is the growth rate of the prey.[/*][*][i]b[/i] is the death rate of the predators independent of the prey.[/*][*][i]d[/i] is the is the rate of consumption of the prey per predator.[/*][*][i]a[/i] / c is the carrying capacity of the prey independent of the predators.[/*][*][i]e[/i] is the growth rate of the predator per prey consumed,[/*][*][i]h[/i] is prey harvesting.[/*][/list]This equations include the effect of limited resources on the food supply of the prey, and how the prey are culled or harvested. The following simulation demonstrates the solutions to these equations for a=1, b=0.25, c=0.01, d=0.02 and e=0.02.[br][br]Change the initial conditions and the harvesting to analyse the behaviour of the populations.

Simple Harmonic Motion

Lorenz attractor

[b]Note:[/b] Online version is slow. Download file for better performance.

Velocity fields: Potential

This simulation shows the potential surface of the vector field, with particles following the velocity vectors. [br][br]Enter an arbitrary potential function in the box [b]Potential[/b]. Try for example:[br][list][*]-(sqrt(x² + y²) + 1 / sqrt(x² + y²)) cos(atan2(y,x))[br][/*][*]log(sqrt((x + 1)² + y²)) - log(sqrt((x - 1)² + y²))[/*][*]1/sqrt(x^2+y^2)[br][/*][*]sqrt(x^2+y^2)[/*][*]arctan2(y,x)[/*][*]x^2-y^2[/*][*]x-1/2y^2[/*][*]-1/x^2[/*][*]abs(x)[br][/*][*]-x[br][/*][/list]

Fourier series expansion

Description
The following simulation shows the partial sum (up to 20 terms) of the Fourier series for a given function defined on the interval [b][a,b][/b]. [br][br]You can also check your calculations by entering the coefficients a[sub]0[/sub], a[sub]1[/sub] and b[sub]1[/sub].[br]
Instructions:
Change the function and calculate its Fourier series. Then[b] type the correct[/b] values of the terms a[sub]0[/sub], a[sub]1[/sub] and b[sub]1[/sub], rounded to two decimal places.[br][br][b]Remark:[/b] Activate the box [b]Fourier series[/b] and increase, or decrease, the number of terms in the partial sum.

Heat equation: visualisation tool

Consider a thin rod of length [math]L[/math] with an initial temperature [math]f\left(x\right)[/math] throughout and whose ends are held at temperature zero for all time [math]t>0[/math]. The temperature [math]u\left(x,t\right)[/math] in the rod is determined from the boundary-value problem:[br][math]\frac{\partial u}{\partial t}=c^2\frac{\partial^2u}{\partial x^2}[/math] 0<[i]x[/i]<[i]L[/i] and [i]t[/i]>0;[br][math]u\left(0,t\right)=u\left(L,t\right)[/math] [i]t[/i]>0;[br][math]u\left(x,0\right)=f\left(x\right)[/math] 0<[i]x[/i]<[i]L[/i].[br][br]In the following simulation, the temperature [math]u\left(x,t\right)[/math] is graphed as a function of [i]x[/i] for various fixed times.[br][br][b]Things to try: [/b][br][list][*]Change the initial condition u(x,0)=f(x).[br][/*][*]Explore the solutions by clicking on the buttons or type a number to show the graph.[br][/*][/list]

Diffusion equation

Consider the heat equation[br][center][math]\frac{\partial u}{\partial t}=c^2\frac{\partial^2u}{\partial x^2}[/math][/center]A particular solution of the heat equation is given by[br][center][math]u\left(x,t\right)=\frac{1}{\sqrt{4\pi c^2t}}\text{exp}\left(\frac{-x^2}{4c^2t}\right)[/math][/center]The following worksheet shows a representation of this solution.[br][br]Use right click and drag the mouse to rotate the 3D view.

Wave equation: d'Alembert's formula

Consider the initial value problem for the wave equation on the whole number line:[br][center][math]u_{tt}\left(x,t\right)=a^2u_{xx}\left(x,t\right)[/math] with [math]x\in\mathbb{R},t>0[/math],[br][math]u\left(x,0\right)=f\left(x\right)[/math] with [math]x\in\mathbb{R}[/math],[br][math]u_t\left(x,0\right)=g\left(x\right)[/math] with [math]x\in\mathbb{R}[/math].[/center]

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