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How I can give condition & plot the solution of this differential equation. . . . . . . Please Guide

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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 Kommentare

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Antworten (3)

Torsten
Torsten am 22 Jul. 2022
Bearbeitet: Torsten am 22 Jul. 2022
dsolve doesn't succeed. Thus use a numerical solver (bvp4c) to solve your equation.
syms A theta(z)
dtheta=diff(theta,z)
dtheta(z) = 
d2theta=diff(theta,z,2)
d2theta(z) = 
eqn = d2theta + A/2*sin(2*theta)==0;
cond = [theta(0)==0, theta(6)==pi/2];
thetaSol = dsolve(eqn,cond)
Warning: Unable to find symbolic solution.
thetaSol = [ empty sym ]
  7 Kommentare
DEEPAK KARARWAL
DEEPAK KARARWAL am 28 Jul. 2022
yes sir, but what I want, is to give such a boundery condition at theta(z)=(__) such that I will get varying angle from 0 to 90 degree as we increase the value of E, where E is electric field and contained in the expression of A.
Torsten
Torsten am 28 Jul. 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.

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Sam Chak
Sam Chak am 22 Jul. 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)
ans = 1.5708

MOSLI KARIM
MOSLI KARIM am 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

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