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second order nonlinear ode with polynomial terms

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Hi everyone! I would resolve the following nonlinear differential equation:
f(Y) + b(Y) (Y')^2 + g(Y) Y'' = A
where Y is a function of x, i.e. Y = Y(x), and
f(Y) = a1 + a2*Y + a3*Y^2 + a4*Y^3
g(Y) = b1 f(Y)/Y^3
b(Y) = c1 (a1 + a2 Y + a3 Y^2)/Y^3
In this example A = cost, but it could be A = A(x). I have no idea how to solve with matlab.. some suggestions? can I use some usual ode-routines?
Thanks in advance for all your support.
Jan on 30 Apr 2013
@Pinco: Bumping after 2 hours is not useful. When the contributors do not find enough information to create an answer, reading the question again without any additional explanantions, is a waste of time. So please do not bump after 2 hours without showing, what you have done in this time.
Pinco on 30 Apr 2013
@Kye Taylor @Chandrakanth, : You are right, there was a mistake (I fixed it): A = A(x). In this case, the function A is completely known.
' = d/dx.
I have to find Y(x), thus I have only independent variable x.
Is it then sufficient to solve an ODE system?
@Jan Simon: I apologize for refreshing, but I was very worried about this solution

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Answers (1)

Kye Taylor
Kye Taylor on 30 Apr 2013
Edited: Kye Taylor on 30 Apr 2013
You must write the ODE as a system of first-order ODEs. Use the substitution u1 = y and u2 = y'. Then, you'll end up with equations like
u1' = u2
u2' = F(u1,u2)
where F is a function of u1 and u2 (y and y')... Once you have those equations, create a function named F, like
function uPrime = F(u)
uPrime(1) = u(2);
uPrime(2) = % code for your F(u1,u2)
Note that the input u should be a two-dimensional vector where the comoponent u(1) represents u1, and u(2) represents u2. The output is also a two-dimensional vector, one element for each first-order differential equation in the system above. Such an interface to F is dictated by the requirements of the fsolve function.
Pinco on 1 May 2013
I implemented the function in this way:
function xprime = myode(t,x)
global a1 a2 a3 a4 h0 k
A = (12.* x(1).^3)/h0^2;
B = a1 + 2*a2.*x(1) + 3*a3.*x(1).^2 + 4*a4.*x(1).^3;
C = 3*a1 + 4*a2.*x(1).^2 + 3*a3.*x(1).^2;
xprime = [x(1); ...
A.*(1- k./B) + (C./x(1).*B).* x(2).^2];
Now I would impose some boundary conditions, i.e. Y(inf) = L, y(-inf) = l, dY(inf)/dx = dY(-inf)/dx = 0. How can I do this?
Pinco on 1 May 2013
Now I'm trying to use bvp4c for boundary value problems. According to the second example in the Matlab help for bvp4c, I wrote this code:
global a0 a1 a2 a3 a4 h0 k Ja Jb
a0 = 2.0377638272727268 * 10^3;
a1 = -7.105521894545453 * 10^3;
a2 = 9.234000147272726 * 10^3;
a3 = -5.302489919999999 * 10^3;
a4 = 1.1362478399999998 * 10^3;
h0 = 45.5 * 10^-10;
k = 5.92;
Ja = 1.025;
Jb = 1.308;
meshX = linspace(0,16);
Y0 = [Ja, 0]; %[Y(0), Y'(0)]
init = bvpinit(meshX,Y0);
sol = bvp4c(@myode2,@mybc2,init);
function yprime = myode2(x,y)
global a1 a2 a3 a4 h0 k
A = (12.* y(1).^3)/h0^2;
B = a1 + 2*a2.*y(1) + 3*a3.*y(1).^2 + 4*a4.*y(1).^3;
C = 3*a1 + 4*a2.*y(1).^2 + 3*a3.*y(1).^2;
yprime = [y(1); ...
A.*(1- k./B) + (C./y(1).*B).* y(2).^2];
function res = mybc(ya,yb)
global Ja Jb
res =[ya(2)
ya(1) - Ja
yb(1) - Jb];
I would impose the boundary conditions on the first derivative (equals to 0 at the extremes) and the function itself (equals to Ja and Jb, respectively). Matlab returns some errors.. where is the error?

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