How I can give condition & plot the solution of this differential equation. . . . . . . Please Guide
2 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
This is the equation for which
boundery condition are
theta(z=0)=0 degree
theta(z=h)=90 degree
where h=6
z=0:h
how to give condition here
e=8.85*10^-12
dele=11
E=1
k11=9
k33=9
k22=11
syms theta(z) z dtheta
dtheta=diff(theta,z)
d2theta=diff(theta,z,2)
eqn=d2theta+((k33-k11)*cos(theta)*sin(theta))*(dtheta)^2*(1/(k11*(cos(theta))^2+k22*(sin(theta))^2))+e*dele*E^2*cos(theta)*sin(theta)*(1/(k11*(cos(theta))^2+k22*(sin(theta))^2))
cond(theta(0)==0, theta(pi/2)==0)
thetaSol = dsolve(eqn,cond)
thetaSol = unique(simplify(thetaSol))
fplot(thetaSol)
7 Commenti
Risposte (3)
Torsten
il 22 Lug 2022
Modificato: Torsten
il 22 Lug 2022
dsolve doesn't succeed. Thus use a numerical solver (bvp4c) to solve your equation.
syms A theta(z)
dtheta=diff(theta,z)
d2theta=diff(theta,z,2)
eqn = d2theta + A/2*sin(2*theta)==0;
cond = [theta(0)==0, theta(6)==pi/2];
thetaSol = dsolve(eqn,cond)
7 Commenti
Torsten
il 28 Lug 2022
Not clear what you mean.
The boundary value for theta at z = 6 can be set by writing it in the variable "bv" of my code above. Experiment with it.
Sam Chak
il 22 Lug 2022
Giiven the parameters, it seems that if you select initial values and , the boundary values are satisfied.
epsilnot = 8.85*10^-12;
dele = 11;
E = 1;
k11 = 9;
k33 = 9;
k22 = 11;
A = sqrt(dele*epsilnot*E^2/k11);
f = @(t, x) [x(2); ...
- (A/2)*sin(2*x(1))];
tspan = [0 6];
initc = [0 pi/12]; % initial condition
[t, x] = ode45(f, tspan, initc);
plot(t, x(:,1), 'linewidth', 1.5), grid on, xlabel('t'), ylabel('\theta')
x(end,1) % π/2 at θ(6)
MOSLI KARIM
il 16 Feb 2023
%%
function answer
clc
clear all
close all
global A
epsilnot = 8.85*10^-12;
dele = 11;
E = 1;
k11 = 9;
k33 = 9;
k22 = 11;
A = sqrt(dele*epsilnot*E^2/k11);
solinit=bvpinit(linspace(0,6),[0;pi/12])
sol=bvp4c(@fct,@bc,solinit)
figure(1)
plot(sol.x,sol.y(1,:))
function dxdy=fct(x,y)
dxdy=[y(2); -(A/2)*sin(2*y(1))];
end
function res=bc(ya,yb)
res=[ya(1);yb(1)-90]
end
end
0 Commenti
Vedere anche
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!