How I can give condition & plot the solution of this differential equation. . . . . . . Please Guide
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![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1071040/image.png)
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 commentaires
Réponses (3)
Torsten
le 22 Juil 2022
Modifié(e) : Torsten
le 22 Juil 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 commentaires
Torsten
le 28 Juil 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
le 22 Juil 2022
Giiven the parameters, it seems that if you select initial values
and
, the boundary values are satisfied.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1074360/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1074365/image.png)
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
le 16 Fév 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
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