4th order ode with eigenvalue
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Harry Lee
el 1 de Jul. de 2017
Comentada: Sahaluddin Mirza
el 24 de Mayo de 2019
The 4th order ode is
y'''' +5*y'''-2*i*y''-(1/A)*y'+(2-lambda)*y=0
b.c.: y(0)=y(1)=y'(0)=y'(1)=0.
In which A is a given real number, i is the imaginary unit (i^2= -1), lambda is the unknown eigenvalue.
I wrote the following code:
function mat4bvp
global A
A = 1;
lambda = -0.5+0.2*i;
solinit = bvpinit(linspace(0,1,1000),@guess,lambda);
sol = bvp4c(@stream,@bc,solinit);
x = linspace(0,1,1000);
y = deval(sol,x);
plot(x,y(1,:));
function v = guess (x)
v = [1-cos(2*pi*x)
2*pi*sin(2*pi*x)
(2*pi)^2*cos(2*pi*x)
-(2*pi)^3*sin(2*pi*x)];
function dxdy = stream(x,y,lambda)
global A
dxdy=[y(2)
y(3)
y(4)
-5*y(4)+2*i*y(3)+(1/A)*y(2)+(lambda-2)*y(1)];
function res = bc(ya,yb)
res=[ya(1)
ya(2)
yb(1)
yb(2)
ya(3)-1]; %%%The last condition is additionally imposed because lambda is to be solved for
After running it, it generated the following error:
Error using trial>bc
Too many input arguments.
3 comentarios
Sahaluddin Mirza
el 24 de Mayo de 2019
Hey T S Singh, I am facing same problem as yours. Can u pls share your code, if u solved it? I need help finding the first 5 modes of vibration and am stuck finding the 5th boundary condition.
Respuesta aceptada
Walter Roberson
el 1 de Jul. de 2017
You are using bvpinit() and passing in a vector of length 1 as the third parameter. That initializes "parameters" to a vector of length 1 in the problem structure, and you see that parameter showing up in the third position for your stream() function. However, when you have parameters, they are also passed to your bc function, but your bc function is not expecting anything for parameters.
function res = bc(ya, yb, lambda)
Whether you need it for the computation or not, it is going to be passed in.
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