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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

[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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