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--- math121b:03-20 [2020/03/19 22:50]
pzhou
+++ math121b:03-20 [2020/03/19 23:42] (current)
pzhou
@@ Line 18: / Line 18: @@
 And to leading order, we can replace $f(z)$ by its value at the critical point.
+** Example **
+Physicists ususally use $1/\hbar$ as $\lambda$. Here are some sample computation.
+$$ \int_\R e^{-\frac{1}{\hbar} x^2/2} dx = \sqrt{2\pi \hbar} $$
+$$ \int_\R x e^{-\frac{1}{\hbar} x^2/2} dx = 0 $$
+$$ \int_\R x^2 e^{-\frac{1}{\hbar} x^2/2} dx = \hbar^{3/2} \int_\R u^2 e^{- u^2/2} du = \hbar^{3/2} \sqrt{2\pi} $$
+The point is that, having $x^2$ in the pre-factor will make the integral much smaller (indeed, the $x^2$ factor is killing the peak of $e^{-x^2/(2\hbar)}$ at $x=0$.)
+Another trick to compute the last equality is that
+$$ \int_\R x^2 e^{- \lambda x^2/2} dx = -2 \frac{d}{d\lambda} \int_\R  e^{- \lambda x^2/2} dx = -2 \frac{d}{d\lambda} \sqrt{2\pi / \lambda} = \sqrt{2\pi} \lambda^{-3/2} $$
 ----
 Now, back to our Bessel function. We let $x/2$ be the large parameter $\lambda$. Then we have phase function
 $$ S(u) = u - 1/u. $$
-  - Critical points of $S(u)$. This is the place where $S'(u)$ vanishes. Solve the equation that $$ 0 = S'(u) = 1 + 1/u^2, $$ we get $u = \pm i$.
+  - Critical points of $S(u)$. This is the place where $S'(u)$ vanishes. Compute $$ S'(u) = 1 + 1/u^2, \quad S"(u) = -2/u^3. $$  We get $$S'(u) = 0 \leadsto u = \pm i. $$
   - Massage our contour so that they pass through the critical point. Yes, it does, since our contour is the unit circle $|u|=1$.
   - Make sure it passes the critical point in the 'right direction'. Now, we need to do Taylor expansion of the phase function $S(u)$ near $\pm i$.
-Consider the critical point $u_0=i$ first. For $$u = u_0 + v,$$ we have $$ S(u_0 + v) \approx S(u_0) + (1/2) S''(u_0) v^2 = (i - 1/i) + (1/2) (-1/i^3) v^2 = (2i) - (i/2) v^2 $$ The constant term is $2i$, we can do nothing about it, just leave it there. The quadratic term is $ - (i/2) v^2$. Recall $v = u - u_0$, as $u$ passes through $u_0$, $v$ passes through $0$. We want to choose the direction in which $v$ passes through $0$, so that $-(i/2) v^2$ remains real, and has a local maximum at $v=0$. So we parametrize $$ v = s e^{- i \pi /4}, $$ for $s$ real, that way $$  -(i/2) v^2 = - (1/2) s^2. $$ So the contour $C$ should passes through $i$, in the direction $3\pi/4$( or $-\pi/4$), depending on which way you look).
+Consider the critical point $u_0=i$ first. For $$u = u_0 + v,$$ we have $$ S(u_0 + v) \approx S(u_0) + (1/2) S''(u_0) v^2 = (i - 1/i) + (1/2) (-2/i^3) v^2 = (2i) - i v^2 $$ The constant term is $2i$, we can do nothing about it, just leave it there. The quadratic term is $ - i v^2$. Recall $v = u - u_0$, as $u$ passes through $u_0$, $v$ passes through $0$. We want to choose the direction in which $v$ passes through $0$, so that $- i v^2$ remains real, and has a local maximum at $v=0$. So we parametrize $$ v = s e^{- i \pi /4}, $$ for $s$ real, that way $$  - i v^2 = - s^2. $$ So the contour $C$ should passes through $i$, in the direction $3\pi/4$( or $-\pi/4$), depending on which way you look).
 Thus, the contribution for $u_0 = i$ is
 $$ \begin{aligned}
-I_1 &\approx \int_{+\infty}^{-\infty} (i)^{-1-n} e^{(x/2) [(2i) - (i/2) v^2]} dv \cr
+I_1 &\approx (2\pi i)^{-1}  \int_{+\infty}^{-\infty} (i)^{-1-n} e^{(x/2) [(2i) - i v^2]} dv \cr
-& \approx (i)^{-1-n} e^{i x} \int_{+\infty}^{-\infty} e^{-(1/2) s^2} d [s e^{- i \pi /4}] \cr
+& =(2\pi i)^{-1} (i)^{-1-n} e^{i x} \int_{+\infty}^{-\infty} e^{-(x/2) s^2} d [s e^{- i \pi /4}] \cr
-& \approx (-1) (i)^{-1-n} e^{i x - i \pi /4} \int_{-\infty}^{+\infty} e^{-(1/2) s^2} ds \cr
+& = (2\pi i)^{-1} (-1) (i)^{-1-n} e^{i x - i \pi /4} \int_{-\infty}^{+\infty} e^{-(x/2) s^2} ds \cr
-& \approx (-1) (i)^{-1-n} e^{i x - i \pi /4} \sqrt{2\pi}
+& = (2\pi i)^{-1} (-1)  e^{i x - i \pi /4 - (n+1) i \pi /2 } \sqrt{2\pi/x}
 \end{aligned} $$
 Similarly, we can compute the contribution at $u_0 = -i$, we get
 $$ \begin{aligned}
-I_2 &\approx \int_{-\infty}^\infty (-i)^{-1-n} e^{(x/2) [(-2i) + (i/2) v^2]} dv \cr
+I_2 &\approx (2\pi i)^{-1} \int_{-\infty}^\infty (-i)^{-1-n} e^{(x/2) [(-2i) + i v^2]} dv \cr
-& \approx (-i)^{-1-n} e^{-i x} \int_{-\infty}^{+\infty} e^{-(1/2) s^2} d [s e^{ i \pi /4}] \cr
+& \approx (2\pi i)^{-1} (-i)^{-1-n} e^{-i x} \int_{-\infty}^{+\infty} e^{-(x/2)s^2} d [s e^{ i \pi /4}] \cr
-& \approx (-i)^{-1-n} e^{-i x + i \pi /4} \int_{-\infty}^{+\infty} e^{-(1/2) s^2} ds \cr
+& \approx (2\pi i)^{-1} (-i)^{-1-n} e^{-i x + i \pi /4} \int_{-\infty}^{+\infty} e^{-(x/2) s^2} ds \cr
-& \approx (-i)^{-1-n} e^{-i x + i \pi /4} \sqrt{2\pi}
+& \approx  (2\pi i)^{-1}  e^{-i x + i \pi /4 + (n+1) i \pi /2} \sqrt{2\pi/x}
 \end{aligned} $$
+Finally, we can add up the two contributions to get
+$$ J_n(x) \approx I_1 + I_2 =  \sqrt{2/\pi x} \sin(- x +  \pi /4 + (n+1)  \pi /2) =  \sqrt{2/\pi x} \cos(x - (2n+1)  \pi /4)$$
+where the last step uses trig identity $\sin(A) = \cos(\pi/2 - A)$, so that we agree with Boas.
+Here is a picture of the 'ideal contour', where the phase $S(u)$ has 'constant phase'.
+{{ :math121b:steepest_descent_method.png?nolink&400 |}}
+It is made with [[https://www.wolframalpha.com/input/?i=contourplot%5B+Im+%5B%28x%2BI+y%29+-+1%2F%28x%2BI+y%29%5D+%2C+%7Bx%2C+-2%2C2%7D%2C+%7By%2C+-2%2C2%7D%5D | Wolfram Alpha.]]
+------
+This finishes Chapter 12. I did not talk about the Airy function, Hermite function, Laguerre functions. I will leave those
+as reading materials, and post homework questions about them.

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