Items tagged with differential_equation



I want to draw  phase plane of system of three fractional order equations. 


Note that 

Also want the  phase portrait when the values of alpha are not same....






I want to get solutions of this system ,can anyone help me ?

i have solved the coupled equations .... and want to subtract a constant

i want to subtract constant from the result of last equation

A population p(t) governed by the logistic equation with a constant rate of harvesting satisfies the initial value problem diff(p(t), t) = (2/5)*p(t)*(1-(1/100)*p(t))-h, p(0) = a. This model is typically analyzed by setting the derivative equal to zero and finding the two equilibrium solutions p = 50+`&+-`(5*sqrt(100-10*h)). A sketch of solutions p(t) for different values of a suggests that the larger equilibrium is stable; the smaller, unstable.


When a is less that the unstable equilibrium, p(t) becomes zero at a time t[e], and the population becomes extinct. If p(t) is not interpreted as pertaining to a population, its graph exists beyond t[e], and actually has a vertical asymptote between the two branches of its graph.


In the worksheet "Logistic Model with Harvesting", two questions are investigated, namely,


  1. How does the location of this vertical asymptote depend on on a and h?
  2. How does the extinction time t[e], the time at which p(t) = 0, depend on a and h?

To answer the second question, an explicit solution p = p(a, h, t), readily provided by Maple, is set equal to zero and solved for t[e] = t[e](a, h). It turns out to be difficult both to graph the surface t[e](a, h) and to obtain a contour map of the level sets of this function. Instead, we solve for a = a(t[e], h) and obtain a graph of a(h) with t[e] as a slider-controlled parameter.


To answer the first question, the explicit solution, which has the form alpha*tan(phi(a, h, t))*beta(h)+50, exhibits its vertical asymptote when phi(a, h, t) = -(1/2)*Pi. Solving this equation for t[a] = t[a](a, h) gives the time at which the vertical asymptote is located, a function that is as difficult to graph as t[e]. Again the remedy is to solve for, and graph, a = a(h), with t[a] as a slider-controlled parameter.


Download the worksheet:

i don't know much about maple, i need to solve the following odes system... I study a little on the help page of maple about numeric[midrich] that takes bvp and deal singularity as well but dint know how to used in the following system

I use pdsolve to solve this system of equation but the graph I have is different from the author's graph. I think I'm missing out on something. Can anyone help me out with the solution using any Maple command and module.

PDE := {diff(phi(x, t), t) = (diff(phi(x, t), x, x))/S__c-K__r*phi(x, t)+S__r*(diff(theta(x, t), x, x)), diff(u(x, t), t) = diff(u(x, t), x, x)-M^2*(u(x, t)-m*w(x, t))/(m^2+1)-u(x, t)/`ϰ`-2*Omega^2*w(x, t)+Gr*theta(x, t)+Gm*phi(x, t), diff(w(x, t), t) = diff(w(x, t), x, x)+M^2*(m*u(x, t)-w(x, t))/(m^2+1)-w(x, t)/`ϰ`+2*Omega^2*u(x, t), diff(theta(x, t), t) = lambda*(diff(theta(x, t), x, x))/P__r}

With Inittial and boundary condition : {phi(0, t) = 1, phi(9, t) = 0, phi(x, 0) = 0, u(0, t) = t, u(9, t) = 0, u(x, 0) = 0, w(0, t) = 0, w(9, t) = 0, w(x, 0) = 0, theta(0, t) = 1, theta(9, t) = 0, theta(x, 0) = 0}

With the following parameter declared as:

I will appreciate the graph of the solution with time t:0.3, 0.5, 0.7 and 1.0.


i am trying to solve an initial value problem i have applied a numeric code but the progam give me an error but instead of several efforts i couldn't correc the error..rho(r).mw


How do I simulate the follow stochastic differential equations. Thanks very much!




how to find the integration of z(x) form 0 to x with the given condition...

diff(z(x), x) = x*Typesetting:-delayDotProduct(b, 1+3*y(x))/(a^2*(1-(x/a)^2));

diff(z(x), x) = x*(b.(1+3*y(x)))/(a^2*(1-x^2/a^2))


`%%where%`, y(x) = b*((1-(x/a)^2)^(1/2)-(1-(R/a)^2)^(1/2))/(3*(1-(R/a)^2)^(1/2)-(1-(x/a)^2)^(1/2));

`%%where%`, y(x) = b*((1-x^2/a^2)^(1/2)-(1-R^2/a^2)^(1/2))/(3*(1-R^2/a^2)^(1/2)-(1-x^2/a^2)^(1/2))


with*condition; -1; z(R) = ln(1-(R/a)^2)

z(R) = ln(1-R^2/a^2)





I apologise if this information is available on the support pages or on these forums already.

I have found the general solution to a differential equation and now need to plot multiple solution curves that correspond to a different value of a constant.

For example if I had y(x) = x + k, how would I go about plotting this for all the different values of k.

Many Thanks. 

I wrote an equation and intended to iterate P(r) till that value of r when P(r) becomes 0.what is the value of r for which P(r)=0, and also  obtained its convergent decreasing graph which show convergence of P(r) to 0.



G := 6.6743*10^(-8);








sys2 := diff(P(r), r) = -G*(epsilon+P(r))*((8.98*10^14*(1/3))*Pi*r^3+4*Pi*r^3*P(r)/c^2)/(c^2*(r^2-(2*G*r*(8.98*10^14*(1/3))*Pi*(r^3))/c^2)), P(0.1e-9) = 8.5561*10^34

diff(P(r), r) = -0.7426160269e-28*(0.7321400000e36+P(r))*(0.2993333333e15*Pi*r^3+0.4450600224e-20*Pi*r^3*P(r))/(r^2-0.4445794614e-13*r^4*Pi), P(0.1e-9) = 0.8556100000e35






Hello! Hope everything fine with you. I am try to solve the three-point differential by numerical method in attached file but failed. Please see the attachement and solve my problem. I am very thankful your kind effort. Please take care.

With my best regards and sincerely.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China





G := 6.6743*10^(-8); 1; c := 2.99792458*10^10; 1; pi := 3.143; 1; rho := 5.3808*10^14









diff(P(r), r) = -G*(rho*c^2+P(r))*((4*pi*r^3*(1/3))*rho+4*Pi*r^3*P(r)/c^2)/(c^2*(r^2-2*G*r*(4*pi*r^3*(1/3))*rho/c^2)), diff(v(r), r) = 1.485232054*10^(-28)*((4*pi*r^3*(1/3))*rho+4.450600224*10^(-21)*Pi*r^3*P(r))/(r^2-1.485232054*10^(-28)*r*(4*pi*r^3*(1/3))*rho)

diff(P(r), r) = -0.7426160269e-28*(0.4836021866e36+P(r))*(0.2254913920e16*r^3+0.4450600224e-20*Pi*r^3*P(r))/(r^2-0.3349070432e-12*r^4), diff(v(r), r) = 0.1485232054e-27*(0.2254913920e16*r^3+0.4450600224e-20*Pi*r^3*P(r))/(r^2-0.3349070432e-12*r^4)


condition; -1; P(0) = 0, v(1014030) = -.4283

P(0) = 0, v(1014030) = -.4283





I found the solution of P(r) at P(0)=0, but could obtain the result of v(r) at v(1014030)=-0.4283, v(r) may have a graph such that i can goes from -0.4283 to 0.

G := 6.6743*10^(-8);

a := 1.9501*10^24;

b := .3306;

c := 2.99792458*10^10;

d := 2.035;

pi := 3.143;

eps := 3.8220*10^35;

g(r) = 1-s(r)/0.06123;

j(r) = e^(-(1/2)*w(r))*(1-2*G*v(r)/(r*c^2))^.5

sys := diff(v(r), r) = 4*pi*r^2*eps/c^2, ics=v(0)=0

diff(u(r), r) = -G*(eps+u(r))*(v(r)+4*Pi*r^3*u(r)/c^2)/(c^2*(r^2-2*G*r*v(r)/c^2)),u(0)=1.3668*10^34

diff(w(r), r) = 1.485232054*10^(-28)*(v(r)+4.450600224*10^(-21)*pi*r^3*u(r))/(r^2-2*G*r*v(r)/c^2), conditions: w(0)=0,iterate it to find w(688240)=-2.05684, it solve must satistfy the both conditions.

diff(r^4*j(r)*(diff(g(r), r)), r)+4*r^3*g(r)*(diff(j(r), r)) = 0, conditions dg(r)/dr =0  at r=0, g(688240) =0.87214

diff(J(r), r) = (8*pi*(1/3))*(eps/c^2+u(r)/c^2)*(g(r)*j(r).(r^4))/(1-2*G*v(r)/(r*c^2)) condition J(0)=0.

1 2 3 4 5 6 7 Last Page 1 of 10