# Solution to general elliptic equation

1. Jul 16, 2005

### saltydog

An eariler post by Kank concerned a special case of the general elliptic equation:

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

This equation can be solved in terms of elliptic functions but I don't understand the transformation to z below. Consider the simpler case:

$$y^{''}=A+By+Cy^2$$

First multiply the ODE by y' and integrate througout:

$$\int y^{'}y^{''}=\int Ay^{'}+\int Byy^{'}+\int Cy^2y^{'}$$

This reduces to:

$$\frac{1}{2}\left(y^{'}\right)^2=c_1+Ay+\frac{B}{2}y^2+\frac{C}{3}y^3$$

Where $c_1$ is determined by the initial conditions.

By adjusting the constants we obtain the more convenient form:

$$\left(y^{'}\right)^2=a+by+cy^2+dy^3$$

Now, we can write the left side in terms of it's roots (ignore for the moment complex roots which is another problem I have):

$$\left(\frac{dy}{dx}\right)^2=h^2(y-\alpha)(y-\beta)(y-\gamma)=h^2\Delta^2(y)$$

We wish to transform this equation into the standard form:

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

That's because the solution to the standard form is the elliptic sine function, Sn(u,k) described as:

If:

$$u=\int_0^\phi \frac{d\theta}{\sqrt{1-k^2Sin^2\theta}}$$

Then Sn(u,k) is defined as:

$$Sn(u,k)=Sin(\phi)$$

So, the transformation is where I'm having the problem. I'm suppose to use the following:

$$z^2=\frac{\alpha-\gamma}{y-\gamma};\quad k^2=\frac{\beta-\gamma}{\alpha-\gamma};\quad M^2=\frac{\alpha-\gamma}{4}$$

***** this is the part I don't understand *******

Using the transformation above, we obtain:

$$\frac{1}{\Delta(z)}\frac{dz}{dx}=-\frac{M}{\Delta(y)}\frac{dy}{dx}$$

and hence:

$$\frac{1}{h\Delta(y)}\frac{dy}{dx}=\frac{1}{\Delta(z)}\frac{dz}{d(-hMx)}=1$$

From this it follows:

$$z=Sn(-hMx,k)$$

********* end of part I don't understand *********

The expression in z then, can then be inverted by the transform above in z and y, to obtain finally an expression for y.

Can anyone explain the transformation to convert the ODE in y to that standard one in z

Last edited: Jul 16, 2005
2. Jul 16, 2005

### Hurkyl

Staff Emeritus
Hrm.

Well, with the indicated substitution, I can transform the equation for (dy/dx)^2 into:

$$\left( \frac{dz}{dx} \right)^2 = \frac{\alpha - \gamma}{4} h^2 (1 - z^2) (1 - k^2 z^2)$$

(Actually, I first transformed (1 - z^2) (1 - k^2 z^2) into y's to see what they looked like)

Oh, I see... now, if you do a change of variable on x, you can get rid of that constant out of front.

Last edited: Jul 16, 2005
3. Jul 17, 2005

### saltydog

Wonderful . . . and of course you won't show me. That's why you're the mentor I suppose. That's Ok. Best for me if I continue to do so unaided. But I see now it's best to ignore the stuff above between asterisks and just try to do the conversion as you suggested.

Thanks!

4. Jul 17, 2005

### saltydog

Got it!

We have:

$$\left(\frac{dy}{dx}\right)^2=h^2(y-\alpha)(y-\beta)(y-\gamma)$$

and wish to convert it to:

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

using the transformations (I added one for x):

$$z^2=\frac{\alpha-\gamma}{y-\gamma},\quad k^2=\frac{\beta-\gamma}{\alpha-\gamma},\quad M^2=\frac{\alpha-\gamma}{4},\quad x=-\frac{v}{hM}$$

Taking the derivative of the expression for $z^2[/tex] with respect to v we obtain: $$2z\frac{dz}{dv}=\frac{-(\alpha-\gamma)}{(y-\gamma)^2}\frac{dy}{dx}(-\frac{1}{hM})$$ Solving for $\frac{dz}{dv}$ we obtain: $$\left(\frac{dz}{dv}\right)^2=\frac{1}{4z^2}\frac{(\alpha-\gamma)^2}{(y-\gamma)^4)}\left(\frac{dy}{dx}\right)^2\frac{1}{h^2 M^2}$$ Simplifying, we obtain: $$\left(\frac{dz}{dv}\right)^2=\frac{1}{h^2}\frac{1}{(y-\gamma)^3}\left(\frac{dy}{dx}\right)^2$$ Now, substituting the expression for $z^2$ into the first equation and then equating the derivatives we obtain: $$\left(\frac{dy}{dx}\right)^2\frac{1}{h^2}\frac{1}{(y-\gamma)^3}=\left(1-\frac{\alpha-\gamma}{y-\gamma}\right)\left(1-\frac{\beta-\gamma}{y-\gamma}\right)$$ Simplifying: $$\left(\frac{dy}{dx}\right)^2=h^2(y-\gamma)\left[(y-\gamma)^2-(y-\gamma)(\beta-\gamma)-(y-\gamma)(\alpha-\gamma)+(\alpha-\gamma)(\beta-\gamma)\right]$$ Which reduces to: $$\left(\frac{dy}{dx}\right)^2=h^2(y-\alpha)(y-\beta)(y-\gamma)$$ Now, the solution to: $$\left(\frac{dz}{dv}\right)^2=(1-z^2)(1-k^2z^2)$$ is: $$z(v)=Sn(v,k)$ Since: [tex]v=f(-hMx)$$

we have:

$$z(x)=Sn(-hMx,k)$$

Inverting the expression for z in terms of y, we have finally:

$$y(x)=\frac{\alpha-\gamma}{Sn^2[-hMx,k]}+\gamma$$

Now, of course I need to verify all this algebra with real data, real plots, real initial values ...

Thanks Hurkyl!

Last edited: Jul 17, 2005
5. Jul 17, 2005

### Hurkyl

Staff Emeritus
Ah, I see. Instead of transforming the equation for dy/dx into the equation for dz/dv, you did the exact opposite! (I was uneasy about your presentation for a bit) Hurray!

By the way, the work would be simplified if you had noticed that (y - γ) - (α - γ) = y - α.

6. Jul 19, 2005

### kank_39

OK guys. I'm totally dumbfounded by the theorem of elliptic integrals. I've not studied this theory but it seemed that it is a rather powerful and not to mention essential tool in solving many problems, especially non-linear differential equation. Do you know of any useful sites or books where i could find more information.

And if it doesn't bother you, could you help me out with this specific problem of mine?

$$\frac{1}{2}\left(\frac{dy}{dx}\right)^2 = { - Ay - \frac{By^3}{3} + C }$$

Thanks.

Last edited: Jul 19, 2005
7. Jul 19, 2005

### saltydog

Hello Kank. The technique above seems to have a number of limitations: The roots of the RHS should be real for starters. Also, when you factor it into its roots, the coefficient needs to be positive although may be able to adjust the equations for a negative one. Also. I've been working on it several days now and have achieved a rare level of fustration and disappointment to the point, well, you know, when you get to that state when you slam down your books onto the floor, utter some vulgarities, and stomp out of the room?

I tell you what, I have yet to get a single real problem solved using the technique above. It's quite confussing for me to interpret the elliptic integral and it's inverse, the Jacobi Sn() function and then apply it to the theorem and then verify that the solution I get is correct.

And Oh yea, my favorite one: Notice the final form of the solution in terms of the Jacobi Sn() function. Well, the Sn() function is periodically zero and yet its in the denominator! See, I just don't have a good handle on this at all.

Last edited: Jul 19, 2005
8. Jul 19, 2005

### saltydog

Alright, let's approach this methodically:

Consider the second degree elliptic equation WITH real roots:

$$y^{''}=1-y-\frac{3}{2}y^2;\quad y(0)=0;\quad y'(0)=1$$

Using the technique above, I obtain:

$$(y^{'})^2=2y-y^2-y^3+a$$

Using the initial conditions:

$$1=a$$

Thus we have:

$$(y^{'})^2=1+2y-y^2-y^3$$

Solving for the roots of this cubic equation and assigning the values so k and M come out real, I get (approx. values):

$$\alpha=-0.445$$

$$\beta=-1.802$$

$$\gamma=1.247$$

$$k=1.342$$

$$M=3.982$$

Using the format above for the solution of:

$$(y^{'})^2=(y-\alpha)(y-\beta)(y-\gamma)$$

**********************
Edit: See what I mean. h is not one but rather -1 for this case which means I can't use the analysis above since it is based on h>0

**********************

(h is 1 in this case).

I get:

$$y(x)=\frac{\alpha-\gamma}{Sn^2[-hMx,k]}+\gamma$$

The first plot below is a numeric solution of this initial value problem. It makes sense with y(0)=0 and y'(0)=1. The second plot is the equation I obtained above. It's not even close. Can anybody show me where I'm going wrong?

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Last edited: Jul 19, 2005
9. Jul 19, 2005

### Hurkyl

Staff Emeritus
I don't think it assumes anywhere that the roots are real... I think it's the distinct roots that's important. (Though, a repeated root will surely admit a simplification of the problem!)

10. Jul 19, 2005

### saltydog

You know Hurkyl, I'd be very interested in going though this technique with complex roots. However, I feel I need to solve it and understand it first with real roots. I'll continue working on it.

11. Jul 20, 2005

### saltydog

I wish to obtain some form of closure in this matter and after doing some reviewing, I 've come to the initial conclusion that solutions of the elliptic equations obtained using the technique above will in general NOT be real and unlike linear differential equations, the real part and imaginary parts separately will NOT be solutions.