- #1
- 1,590
- 3
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
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:
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.
