# Second nonlinear DE for exciton diffusion

1. Jun 4, 2010

### elfine

Dear Everyone,

I am working on a physic problem of exciton diffusion involved in organic optoelectronics.

It is in the form of

y''+a*y+b*y^2=0.

Is there a general solution to this equation?

Thanks!

elfine

Last edited: Jun 4, 2010
2. Jun 5, 2010

### UndeniablyRex

I don't know how to handsolve the equations, but I sent it through Mathematica, and you don't want to see the answer. Although I posted it anyway.

Code (Text):

in = DSolve[y''[x] + a*y[x] + b*y[x]^2 == 0, y[x], x]
out = Solve[(4 EllipticF[
ArcSin[\[Sqrt]((Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 3] -
y[x])/(-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 2] +
Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &,
3]))], (Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 2] -
Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &,
3])/(Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 1] -
Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 3])]^2 (Root[-3 C[1] +
3 a #1^2 + 2 b #1^3 &, 2] -
Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &,
3]) (-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 1] +
y[x]) (-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 2] +
y[x]) (-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 3] +
y[x]))/((-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 1] +
Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &,
3]) (-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 2] +
Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 3]) (C[1] - a y[x]^2 -
2/3 b y[x]^3)) == (x + C[2])^2, y[x]]

where Solve solves an equation for y[x] and Root finds the roots of an equation.

3. Jun 5, 2010

### elfine

The attached picture is a given method for solving this kind equation, but it is uncompleted.

For I am not good at math and I can't get a good solution from matlab, could anyone help me on this?

Best wishes!

elfine

File size:
18.5 KB
Views:
68
4. Jun 5, 2010

### jackmell

There is suppose to be a general solution to the general elliptic equation:

$$y''=A+By+Cy^2+Dy^3$$

multiplying by y', integrating, and adjusting the constants we obtain the form:

$$\left(y')^2=a+by+cy^2+dy^3+ey^4$$

and by a suitable change of variable $z=z(y)$ we can reduce it to the standard form:

$$\left(\frac{dz}{dx}\right)^2=(1-z^2)(1-k^2 z^2)$$

in which $z(x)=\text{sn}(x,k)$ where sn is the Jacobi elliptic sine function. We then invert the expression $z=z(y)$ to obtain the solution in y. However, the exact details of that procedure is a little unclear to me.

5. Jun 6, 2010

### elfine

Hi jackmell,
Thanks a lot! Your suggestion is very valuable to me.
$$\left(y')^2=a+cy^2+dy^3$$
$$\left y(\infty)=0$$
$$\left(y')^2=cy^2+dy^3$$
$$\left t=\sqrt{c+dy}$$
$$\left \frac{adt}{t^2-b}=dx$$