Lecture 42
Matched Inner Tearing-Mode Solution for \(y''=-x(1-xy)\)
Consider the inner-layer model equation
\[\frac {d^2 y}{dx^2} = -x\left (1-xy\right ), \tag{H.1}\]
or equivalently \[y'' - x^2 y = -x. \tag{H.2}\]
We want the odd solution that is regular at the origin and matches to the outer ideal solution,
\[y(0)=0, \qquad xy \to 1 \quad \text {as}\quad x\to +\infty . \tag{H.3}\]
The second condition is the usual tearing-mode matching condition: the inner solution must approach the
odd continuation of the outer ideal behavior \(y\sim 1/x\).
Homogeneous solutions.
The homogeneous equation associated with Eq. (H.2) is
\[y_h'' - x^2 y_h = 0. \tag{H.4}\]
Introduce \[z = \frac {x^2}{2}, \qquad y_h(x)=x^{1/2}W(z). \tag{H.5}\]
Then Eq. (H.4) becomes \[z^2 W_{zz} + z W_z - \left (z^2 + \frac {1}{16}\right )W = 0, \tag{H.6}\]
which is the modified-Bessel equation of order \(1/4\). A convenient basis is therefore \[u(x)=x^{1/2} I_{1/4}\!\left (\frac {x^2}{2}\right ), \qquad v(x)=x^{1/2} K_{1/4}\!\left (\frac {x^2}{2}\right ). \tag{H.7}\]
Using the standard modified-Bessel Wronskian \[I_\nu (z)K_\nu '(z)-I_\nu '(z)K_\nu (z) = -\frac {1}{z}, \tag{H.8}\]
one finds \[W[u,v] \equiv u v' - u' v = -2. \tag{H.9}\]
Green-function representation.
Because the regular solution at the origin is built from \(u\) and the decaying solution at large \(x\) is built from \(v\),
the matched odd solution for \(x>0\) is
\[y(x) = \frac {1}{2}v(x)\int _0^x s\,u(s)\,ds \;+\; \frac {1}{2}u(x)\int _x^\infty s\,v(s)\,ds. \tag{H.10}\]
The odd extension to \(x<0\) is then \[y(-x)=-y(x). \tag{H.11}\]
Small-\(x\) behavior and the exact slope.
The small-argument expansions of the modified Bessel functions give
\[I_{1/4}(z)\sim \frac {(z/2)^{1/4}}{\Gamma (5/4)}, \qquad K_{1/4}(z)\sim \frac {1}{2}\Gamma (1/4)\left (\frac {z}{2}\right )^{-1/4}, \qquad z\to 0, \tag{H.12}\]
so that \[u(x)\sim \frac {2\sqrt {2}}{\Gamma (1/4)}x, \qquad v(x)\sim \frac {\Gamma (1/4)}{\sqrt {2}}, \qquad x\to 0. \tag{H.13}\]
Inserting Eq. (H.13) into Eq. (H.10) shows that the first term is \(O(x^2)\) near the origin, while the second term is
linear: \[y(x)\sim \frac {1}{2}u(x)\int _0^\infty s\,v(s)\,ds, \qquad x\to 0. \tag{H.14}\]
Hence \[y'(0) = \frac {u'(0)}{2}\int _0^\infty s\,v(s)\,ds = \frac {\sqrt {2}}{\Gamma (1/4)} \int _0^\infty s^{3/2}K_{1/4}\!\left (\frac {s^2}{2}\right )ds. \tag{H.15}\]
Now substitute \(t=s^2/2\). Then \[\int _0^\infty s^{3/2}K_{1/4}\!\left (\frac {s^2}{2}\right )ds = 2^{1/4}\int _0^\infty t^{1/4}K_{1/4}(t)\,dt. \tag{H.16}\]
Using the standard Mellin integral \[\int _0^\infty t^{\mu -1}K_\nu (t)\,dt = 2^{\mu -2} \Gamma \!\left (\frac {\mu -\nu }{2}\right ) \Gamma \!\left (\frac {\mu +\nu }{2}\right ), \qquad \Re (\mu ) > |\Re (\nu )|, \tag{H.17}\]
with \[\mu =\frac {5}{4}, \qquad \nu =\frac {1}{4}, \tag{H.18}\]
we obtain \[\int _0^\infty t^{1/4}K_{1/4}(t)\,dt = 2^{-3/4}\Gamma \!\left (\frac {1}{2}\right )\Gamma \!\left (\frac {3}{4}\right ) = 2^{-3/4}\sqrt {\pi }\,\Gamma \!\left (\frac {3}{4}\right ). \tag{H.19}\]
Therefore \[y'(0) = \frac {\sqrt {\pi }\,\Gamma (3/4)}{\Gamma (1/4)} = 0.599070117367796\ldots \tag{H.20}\]
Large-\(x\) asymptotics.
For matching to the outer ideal solution it is useful to construct the large-\(x\) expansion directly. Write
\[y(x)\sim \sum _{n=0}^\infty a_n x^{-(4n+1)}, \qquad x\to +\infty . \tag{H.21}\]
Substituting Eq. (H.21) into Eq. (H.2) gives \[a_0=1, \qquad a_{n+1}=(4n+1)(4n+2)a_n. \tag{H.22}\]
The first few coefficients are \[a_0=1, \qquad a_1=2, \qquad a_2=60, \qquad a_3=5400, \tag{H.23}\]
so that \[y(x)\sim \frac {1}{x} \frac {2}{x^5} \frac {60}{x^9} \frac {5400}{x^{13}} \cdots , \tag{H.24}\]
and therefore \[1-xy \sim -\frac {2}{x^4} -\frac {60}{x^8} -\frac {5400}{x^{12}} \cdots . \tag{H.25}\]
This makes the matching integral absolutely convergent.
Exact matching integral.
Define
\[\mathcal {I} \equiv 2\int _0^\infty \left (1-xy\right )\,dx. \tag{H.26}\]
Integrating the Green representation once more gives \[\int _0^\infty \left (1-xy\right )\,dx = \sqrt {\pi }\,y'(0). \tag{H.27}\]
Combining Eq. (H.27) with Eq. (H.20) yields \[\mathcal {I} = 2\pi \frac {\Gamma (3/4)}{\Gamma (1/4)} = 2.123648272981939\ldots \tag{H.28}\]
Takeaways
This appendix reduces the inner tearing-layer matching problem to one exact number.
The odd solution of \(y''-x^2y=-x\) is regular at the origin, tends to \(y\sim 1/x\) at large \(x\), and has the exact central
slope (H.20). That slope fixes the convergent matching integral \(\mathcal I\) in Eq. (H.28), which
is the constant that later appears in the inner-layer jump condition and in the classical
tearing-mode scaling laws.