Solution to general elliptic equation

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In summary, the conversation discusses the use of elliptic functions to solve a special case of the general elliptic equation. The conversation also mentions a simpler case and the steps involved in solving it. There is a discussion on the transformation to convert the ODE in y to the standard form in z and the use of the Jacobi Sn() function. The limitations and difficulties of using this technique are also mentioned.
  • #1
saltydog
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An eariler post by Kank concerned a special case of the general elliptic equation:

[tex]y^{''}=A+By+Cy^2+Dy^3[/tex]

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

[tex]y^{''}=A+By+Cy^2[/tex]

First multiply the ODE by y' and integrate througout:

[tex]\int y^{'}y^{''}=\int Ay^{'}+\int Byy^{'}+\int Cy^2y^{'}[/tex]

This reduces to:

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

Where [itex]c_1[/itex] is determined by the initial conditions.

By adjusting the constants we obtain the more convenient form:

[tex]\left(y^{'}\right)^2=a+by+cy^2+dy^3[/tex]

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):

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

We wish to transform this equation into the standard form:

[tex]\left(\frac{dz}{dx}\right)^2=(1-z^2)(1-k^2z^2)=\Delta^2(z)[/tex]

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

If:

[tex]u=\int_0^\phi \frac{d\theta}{\sqrt{1-k^2Sin^2\theta}}[/tex]

Then Sn(u,k) is defined as:

[tex]Sn(u,k)=Sin(\phi)[/tex]

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

[tex]z^2=\frac{\alpha-\gamma}{y-\gamma};\quad k^2=\frac{\beta-\gamma}{\alpha-\gamma};\quad M^2=\frac{\alpha-\gamma}{4}[/tex]

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

Using the transformation above, we obtain:

[tex]\frac{1}{\Delta(z)}\frac{dz}{dx}=-\frac{M}{\Delta(y)}\frac{dy}{dx}[/tex]

and hence:

[tex]\frac{1}{h\Delta(y)}\frac{dy}{dx}=\frac{1}{\Delta(z)}\frac{dz}{d(-hMx)}=1[/tex]

From this it follows:

[tex]z=Sn(-hMx,k)[/tex]

********* 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
 
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  • #2
Hrm.

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

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

(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.
 
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  • #3
Hurkyl said:
Hrm.

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

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

(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.

Wonderful . . . and of course you won't show me. That's why you're the mentor I suppose. :smile: 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
Got it!

We have:

[tex]\left(\frac{dy}{dx}\right)^2=h^2(y-\alpha)(y-\beta)(y-\gamma)[/tex]

and wish to convert it to:

[tex]\left(\frac{dz}{dv}\right)^2=(1-z^2)(1-k^2 z^2)[/tex]

using the transformations (I added one for x):

[tex]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}[/tex]

Taking the derivative of the expression for [itex]z^2[/tex] with respect to v we obtain:

[tex]2z\frac{dz}{dv}=\frac{-(\alpha-\gamma)}{(y-\gamma)^2}\frac{dy}{dx}(-\frac{1}{hM})[/tex]

Solving for [itex]\frac{dz}{dv}[/itex] we obtain:

[tex]\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}[/tex]

Simplifying, we obtain:

[tex]\left(\frac{dz}{dv}\right)^2=\frac{1}{h^2}\frac{1}{(y-\gamma)^3}\left(\frac{dy}{dx}\right)^2[/tex]

Now, substituting the expression for [itex]z^2[/itex] into the first equation and then equating the derivatives we obtain:

[tex]\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)[/tex]

Simplifying:

[tex]\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][/tex]

Which reduces to:

[tex]\left(\frac{dy}{dx}\right)^2=h^2(y-\alpha)(y-\beta)(y-\gamma)[/tex]

Now, the solution to:

[tex]\left(\frac{dz}{dv}\right)^2=(1-z^2)(1-k^2z^2)[/tex]

is:

[tex]z(v)=Sn(v,k)[/itex]

Since:

[tex]v=f(-hMx)[/tex]

we have:

[tex]z(x)=Sn(-hMx,k)[/tex]

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

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

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

Thanks Hurkyl! :smile:
 
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  • #5
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. :smile:
 
  • #6
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?

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

Thanks.
 
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  • #7
kank_39 said:
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?

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

Thanks.

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.
 
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  • #8
Alright, let's approach this methodically:

Consider the second degree elliptic equation WITH real roots:

[tex]y^{''}=1-y-\frac{3}{2}y^2;\quad y(0)=0;\quad y'(0)=1[/tex]

Using the technique above, I obtain:

[tex](y^{'})^2=2y-y^2-y^3+a[/tex]

Using the initial conditions:

[tex]1=a[/tex]

Thus we have:

[tex](y^{'})^2=1+2y-y^2-y^3[/tex]


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


[tex]\alpha=-0.445[/tex]

[tex]\beta=-1.802[/tex]

[tex]\gamma=1.247[/tex]

[tex]k=1.342[/tex]

[tex]M=3.982[/tex]

Using the format above for the solution of:

[tex](y^{'})^2=(y-\alpha)(y-\beta)(y-\gamma)[/tex]

**********************
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:

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

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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  • #9
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
Hurkyl said:
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!)

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. :smile:
 
  • #11
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.
 

What is an elliptic equation?

An elliptic equation is a type of partial differential equation (PDE) that involves second-order derivatives and has constant coefficients. It is often used to model physical phenomena in various fields such as physics, economics, and engineering.

What is the general form of an elliptic equation?

The general form of an elliptic equation is given by:

a11 × uxx + a22 × uyy + a12 × uxy + b1 × ux + b2 × uy + c × u = f(x,y)

where a11, a22, and a12 are constants, b1 and b2 are functions of x and y, c is a constant, and f(x,y) is a given function.

What is the solution to a general elliptic equation?

The solution to a general elliptic equation is a function u(x,y) that satisfies the equation for all values of x and y. This means that when u(x,y) is substituted into the equation, the resulting expression is equal to f(x,y) on both sides.

How do you find the solution to a general elliptic equation?

Finding the solution to a general elliptic equation involves solving for the unknown function u(x,y) using various mathematical techniques such as separation of variables, Fourier series, or numerical methods. The specific method used depends on the complexity of the equation and the boundary conditions given.

What are the applications of the solution to general elliptic equations?

The solution to general elliptic equations has many real-world applications, including modeling heat flow, fluid dynamics, and electromagnetic fields. It is also used in image processing, financial modeling, and other areas of science and engineering.

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