\(\int { \cfrac { 1 }{ \psi } } d\, \psi +\int { \int { \cfrac { 1 }{ \psi ^{ 2 } } } } \left( d\, \psi \right) ^{ 2 }=n^{ 2 }ln(x)+Ax+C\)
where both \(A\) and \(C\) are constants for all \(x\). Simplifying further we have,
\(n^{ 2 }ln(x)+Ax+C=0\) --- (*)
when \(x=1\),
\(C=-A\)
So the constrain on \(x\) is,
\(x=e^{ -\cfrac { Ax+C }{ n^{ 2 } } }\)
\(x=e^{ -\cfrac {A( x-1) }{ n^{ 2 } } }\)
This is wrong because \(x=1\) will then always be a solution irrespective of \(n\). And we have a unit circle onto which we can pack infinite number waves, as \(n\rightarrow\infty\). \(x=1\) is not part of the final solution developed previously.
If we consider,
\(x\rightarrow0\)
\(n\rightarrow0\)
As,
\(\lim\limits_{x\rightarrow0 \, n\rightarrow0}n^{ 2 }ln(x)=0^{ 2 }.\infty=0\)
(*) becomes,
\(0^{ 2 }.\infty+A.0+C=0\)
\(C=0\)
And the constrain on \(x\) is,
\(x=e^{ -\cfrac { Ax}{ n^{ 2 } } }\)
And the constrain on \(f\) is,
\(f=\cfrac{nc}{2\pi}e^{ \cfrac { A }{ 2\pi\,n } \cfrac { c }{ f } }\)
by substituting \(2\pi x=n\lambda\) and \(\lambda=\cfrac{c}{f}\)
A plot of y=n/(2*Pi)*(e^{1/(2*Pi*n*x)}) and y=n/(2*Pi)*(e^{8/(2*Pi*n*x)}) together with the line y=x, are shown below.
The value of \(f\) for \(n=1\) is folded up for larger values of \(A=8\). A plot for high value of \(A=30\), y=n/(2*Pi)*(e^{30/(2*Pi*n*x)}) with y=x is shown below.
More intersections between \(y=x\) and \(y=\cfrac{nc}{2\pi}e^{ \cfrac { A }{ 2\pi\,n } \cfrac { c }{ x } }\) are folded up producing pairs of closely spaced solutions of frequencies, \(f\), when \(A\) is large.
Note: In the plot for \(f\), \(c\) was set to \(1\). The effect is the same as scaling \(f\) by \(\cfrac{1}{c}\).
