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Obtain the first two terms of the asymptotic (large x) solution for each of the 2 real solutions of $$u''+\left ( 1-\frac{\gamma}{x^{2}} \right )u=0$$ in $$x> 4$$

Now, For large x, the above equation reduces to $$u_{0}''+u_{0}=0$$ Solving gives $$u_{0}=C_{1}cos\left ( x \right )+C_{2}Sin\left ( x \right )$$ converting to exponential form yields $$\left\{\left\{u_{0}\to \frac{1}{2} c_1 e^{-i x}+\frac{1}{2} c_1 e^{i x}+\frac{1}{2} i c_2 e^{-i x}-\frac{1}{2} i c_2 e^{i x}\right\}\right\}$$

Dropping the imaginary and the non-decaying terms we get

$$u_{0}=\frac{1}{2} c_1 e^{-i x}+\ $$ Now, the 'real' solution to the original equation is of the form $$u=u_{0}+u_{1} where u_{0}\gg u_{1}$$

Any help is appreciated. What am I missing?

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  • $\begingroup$ $$u_{0}=C_{1}\cos\left ( x \right )+C_{2}\sin\left ( x \right )=ae^{ix}+be^{-ix}$$ $\endgroup$ Commented Nov 12, 2015 at 12:05
  • $\begingroup$ You basically combined terms where a and b are some groups of coefficients. Now, dropping the positive power on exponential, we're left with the negative power on exponential . $\endgroup$
    – stoke's
    Commented Nov 12, 2015 at 12:06
  • $\begingroup$ why shall you drop the positive one? $\endgroup$ Commented Nov 12, 2015 at 12:08
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    $\begingroup$ For a large $x$ your solution will behave as $e^x$ which looks fine to me. However, you do have two linearly independent solutions, though... $\endgroup$ Commented Nov 12, 2015 at 12:12
  • $\begingroup$ Do I leave the solution as it is then? Could you elaborate?@MichaelMedvinsky $\endgroup$
    – stoke's
    Commented Nov 12, 2015 at 12:30

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If you let $u=z \sqrt x$, the equation becomes $$x^2 z''+x z'+\left(x^2-\frac{4\gamma+1}4\right)z=0$$ which is a Bessel equation.

Let $k=\frac{1}{2} \sqrt{4 \gamma+1}$ to obtain $$z=c_1 \,J_k(x)+c_2\, Y_k(x)$$

Have a look at Hankel’s expansions

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