Doubt in MATLAB coding.
8 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Joy Salomi
el 4 de Mzo. de 2020
Editada: Joy Salomi
el 6 de Mzo. de 2020
Hello. I'm trying to plot a differntial equation using MATLAB. When I run my program, I'm getting different graph.
I've attached the differential equation and my MATLAB coding below.
Can anyone tell me how to use the boundary conditions correctly?
function abcd
ex4init=bvpinit(linspace(0.00001,100,21),[0,0]);
sol=bvp4c(@ex4ode,@ex4bc,ex4init)
plot(sol.x,sol.y(1,:),'blue');
end
function f=ex4ode(r,y)
a=0.001391304348;
b=19.31459475;
f=[y(2)
-(2/r)*y(2)-a*(1-exp(b*y(1)))
];
end
function res=ex4bc(ya,yb)
res=[
ya(1)+1
yb(2)
]
end
This is the graph which I got.
But this is the exact graph.
2 comentarios
Robert U
el 4 de Mzo. de 2020
Hi Joy Salomi,
Check your x axis range. I suppose the values should be in nm, i.e. 1e-9 m.
Kind regards,
Robert
Respuesta aceptada
Robert U
el 5 de Mzo. de 2020
Editada: Robert U
el 6 de Mzo. de 2020
Hi Joy Salomi,
Thanks for the paper. The differential equation should be written in spherical coordinates (edit: I think there is a mistake to assume R to be constant). Since R is assumed to be constant there is no singularity. I adjusted the code and tested it in Matlab 2019b. Depending on chosen xmesh results might differ. You have to try a bit to get a stable solution. I attached an example for the D = 200 nm graph.
abcd.m
function abcd(ah,D,phi_s)
ex4init=bvpinit(linspace(0,D/2,3e3),[phi_s 0]);
sol=bvp4c(@ex4ode,@ex4bc,ex4init);
plot(ah,sol.x*1e9,sol.y(1,:));
ah.XLabel.String = 'r [nm]';
ah.YLabel.String = 'phi [V]';
function dphidr=ex4ode(r,phi)
a = -1.602176634e-19*1e17*(1e-2)^(-3)/(8.8541878125e-12*11.5);
b = 1.602176634e-19/(physconst('Boltzmann')*600);
dphidr = [ phi(2)
a * ( 1-exp( b*phi(1) ) ) ];
end
function res=ex4bc(ya,yb)
res=[ ya(1)+ phi_s
yb(2) ];
end
end
Example call:
fh = figure;
ah = axes(fh);
hold(ah,'on')
arrayfun(@(dIn) abcd(ah,200e-9,dIn),[0.1,0.3,0.7,1])
Result:
Kind regards,
Robert
4 comentarios
Robert U
el 6 de Mzo. de 2020
Hi Joy Salomi,
I added the singular term as you proposed according to my first answer:
abcd_wSingularity:
function abcd_wSingularity(ah,D,phi_s)
ex4init=bvpinit(linspace(0,D/2,3e3),[phi_s 0]);
S = [0 0; 0 -2];
options = bvpset('SingularTerm',S);
sol=bvp4c(@ex4ode,@ex4bc,ex4init,options);
plot(ah,sol.x*1e9,sol.y(1,:));
ah.XLabel.String = 'r [nm]';
ah.YLabel.String = 'phi [V]';
function dphidr=ex4ode(r,phi)
a = -1.602176634e-19*1e17*(1e-2)^(-3)/(8.8541878125e-12*11.5);
b = 1.602176634e-19/(physconst('Boltzmann')*600);
dphidr = [ phi(2)
a * ( 1-exp( b*phi(1) ) ) ];
end
function res=ex4bc(ya,yb)
res=[ ya(1)+ phi_s
yb(2) ];
end
end
Example call:
fh = figure;
ah = axes(fh);
hold(ah,'on')
arrayfun(@(dIn) abcd_wSingularity(ah,200e-9,dIn),[0.1,0.3,0.7,1])
Result:
Kind regards,
Robert
Más respuestas (2)
Constantino Carlos Reyes-Aldasoro
el 4 de Mzo. de 2020
This is a bit hard to solve without knowing more of the problem. What I would do is to break down the problem and analyse each step. For example, look at the values you have in the solution "sol". I plot separately:
>> plot(sol.x);
>> plot(sol.y);
And also
>> plot(sol.y');
Is anything here remotely connected to what you are expecting? My guess is that no, sol.y has values between 0 and 12e4 whilst the graph you want is between -1 and 0. So the problem is not really about Matlab but about the equations that you are using.
Hope this helps.
1 comentario
Robert U
el 4 de Mzo. de 2020
Hi Joy Salomi,
the implementation of a boundary value problem with singular term has to be done a bit different than you did.
--> doc: Solve BVP with Singular Term
function abcd
ex4init=bvpinit(linspace(0,100,5),[-1 0]);
S = [0 0; 0 -2];
options = bvpset('SingularTerm',S);
sol=bvp4c(@ex4ode,@ex4bc,ex4init,options);
plot(sol.x,sol.y(1,:),'blue');
end
function f=ex4ode(r,y)
a=0.001391304348;
b=19.31459475;
f = [ y(2)
-a*(1-exp(b*y(1))) ];
end
function res=ex4bc(ya,yb)
res=[ ya(1)+1
yb(2) ];
end
Using the values "a" and "b" you supplied still leads to a singular Jacobian matrix which provoces an error. Changing both these values to one leads to a solution that qualitatively corresponds to the curves drawn.
There are two more points that hinder me to test my solution:
- I don't know how "a" and "b" translate to "phi_s".
- I still assume that r is given in SI units. If r is in unit [m] the interval needs to be adjusted to [nm] as shown in your graph.
Kind regards,
Robert
4 comentarios
Robert U
el 5 de Mzo. de 2020
Hi Joy Salomi,
maybe you can state what physical process is described by that differential equation and what reference you are using (where is your photo from).
I still think that your values for a and b do not correspond to the case that is depicted in the book/article.
Kind regards,
Robert
Ver también
Categorías
Más información sobre Function Creation en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!