How to solve this second order nonlinear differential equation?

Here is the differential equation:

   A*y''=B*y+C*y^3+D*y^5;  
Boundary condition: y(0) = constant(known); y'(0) = 0;
                    y'(end)+E*y(end) = 0.

I wanted to solve it in the numerical way with Matlab, and have tried the finite difference method to discretize the above differential equation and get:

(-3*y(0)+4*y(1)-y(2))/2/h = 0;

A*(y(i+1)-2*y(i)+y(i-1))/h^2 = B*y(i)+C*y(i)^3+D*y(i)^5;

(-3*y(I-2)-4*y(I-1)+3*y(I))/2/h+E*y(I) = 0.

I tried the quasi-Newton method to solve the equation set, but the given initial value (y(0) = constant(known)) could not be used. And the values changed with different pre-set initial values.

Is there another method to solve this problem?

P.S. : A,B,C,D and E have their own exact values and it related to a practical problem, but the values of these coefficients are in different order of magnitudes so I didn't put them on the questions. Also the boundary value y(0) is an exact number and for the same reason I replaced it with constant to make it simple.

[Star Strider]

The bvp4c function may be what you want.

Answer 1

[Torsten]

function mat4bvp

xstart=0.0;
xend=...;
A=...;
B=...;
C=...;
D=...;
E=...;

solinit = bvpinit(linspace(xstart,xend,10),[0 0]);
sol = bvp4c(@(x,y)mat4ode(x,y,A,B,C,D),@(ya,yb)mat4bc(ya,yb,E),solinit);

% ------------------------------------------------------------
function dydx = mat4ode(x,y,A,B,C,D)
dydx = [  y(2)
         (B*y(1)+C*y(1)^3+D*y(1)^5)/A ];
% ------------------------------------------------------------
function res = mat4bc(ya,yb,E)
res = [  ya(1) 
         yb(2)+E*yb(1)];

Best wishes

Torsten.

[Jane]

Thank you very much for your detailed answer!!! I will try it myself.

[Torsten]

At the moment, the boundary condition at x=0 is y(0)=0.

Better use the following code:

function mat4bvp

xstart=0.0;
xend=...;
A=...;
B=...;
C=...;
D=...;
E=...;
constant=...;

solinit = bvpinit(linspace(xstart,xend,10),[0 0]);
sol = bvp4c(@(x,y)mat4ode(x,y,A,B,C,D),@(ya,yb)mat4bc(ya,yb,E,constant),solinit);

% ------------------------------------------------------------
function dydx = mat4ode(x,y,A,B,C,D)
dydx = [  y(2)
         (B*y(1)+C*y(1)^3+D*y(1)^5)/A ];
% ------------------------------------------------------------
function res = mat4bc(ya,yb,E,constant)
res = [  ya(1)-constant 
         yb(2)+E*yb(1)];

Best wishes

Torsten.

Answer 2

[John D'Errico]

No matter what, you can NEVER choose a unique set of coefficients for all four values, A,B,C,D, since if any set of numbers {A,B,C,D} satisfies that equation, then {k*A,k*B,k*C,k*D} also satisfies the equation as well.

So without any loss, you can assume A = 1.

y'' = B*y + C*y^3 + D*y^5

Next, you have completely insufficient information in that single boundary value to choose the three coefficients {B,C,D}. ONE piece of information is not enough.

Sorry, but you are wasting your time trying to solve this.

[Jane]

Thank you very much for your answer. Actually A,B,C,D and E have their own exact values and it related to a practical problem, but the values of these coefficients are in different order of magnitudes so I didn't put them on the questions. Also the boundary value y(0) is an exact number and for the same reason I replaced it with constant to make it simple.

And yes, we can assume A = 1.

The condition in the two boundaries is shown with derivative, therefore the equation contained three conditions: the initial value at the first boundary point, the derivatives of the two boundary points.

I found it difficult for me to deal with the power y^3 and y^5 in the calculation. I hope to find an algorithm to work out the numerical results. So sorry that I did not express it clearly.

[Torsten]

You can only prescribe two boundary conditions, not three, for your differential equation.

That being said, you can use bvp4c to solve.

Best wishes

Torsten.

[Jane]

Torsten, thank you for your answer!

You are right for two boundary conditions, should I use the exact value one or the derivative one?

For a normal problem, this differential equation satisfied the derivative conditions in two boundaries.

Because most cases I have seen are given in two exact values for the boundaries, especially this one: [y'(end)+E*y(end) = 0], now I am not sure how to deal with the derivative conditions in bvp4c, could you offer some hints for the numerical calculation?