§32.3 Graphics

§32.3(i) First Painlevé Equation

Plots of solutions $w_{k}(x)$ of $\mbox{P}_{\mbox{\scriptsize I}}$ with $w_{k}(0)=0$ and $w_{k}^{\prime}(0)=k$ for various values of $k$, and the parabola $6w^{2}+x=0$. For analytical explanation see §32.11(i). Figure 32.3.1: wk⁡(x) for -12≤x≤1.33 and k=0.5, 0.75, 1, 1.25, and the parabola 6⁢w2+x=0, shown in black. Magnify Figure 32.3.3: wk⁡(x) for -12≤x≤0.73 and k=1.85185 3, 1.85185 5. The two graphs are indistinguishable when x exceeds -5.2, approximately. The parabola 6⁢w2+x=0 is shown in black. Magnify

§32.3(ii) Second Painlevé Equation with $\alpha=0$

Here $w_{k}(x)$ is the solution of $\mbox{P}_{\mbox{\scriptsize II}}$ with $\alpha=0$ and such that

 32.3.1 $w_{k}(x)\sim k\mathrm{Ai}\left(x\right),$ $x\to+\infty$; ⓘ Symbols: $\mathrm{Ai}\left(\NVar{z}\right)$: Airy function, $\sim$: asymptotic equality, $x$: real and $k$: real Permalink: http://dlmf.nist.gov/32.3.E1 Encodings: TeX, pMML, png See also: Annotations for §32.3(ii), §32.3 and Ch.32

compare §32.11(ii). Figure 32.3.5: wk⁡(x) and k⁢Ai⁡(x) for -10≤x≤4 with k=0.5. The two graphs are indistinguishable when x exceeds -0.4, approximately. Magnify

§32.3(iii) Fourth Painlevé Equation with $\beta=0$

Here $u=u_{k}(x;\nu)$ is the solution of

 32.3.2 $\frac{{\mathrm{d}}^{2}u}{{\mathrm{d}x}^{2}}=3u^{5}+2xu^{3}+\left(\tfrac{1}{4}x% ^{2}-\nu-\tfrac{1}{2}\right)u,$

such that

 32.3.3 $u\sim kU\left(-\nu-\tfrac{1}{2},x\right),$ $x\to+\infty$.

The corresponding solution of $\mbox{P}_{\mbox{\scriptsize IV}}$ is given by

 32.3.4 $w(x)=2\sqrt{2}u_{k}^{2}(\sqrt{2}x,\nu),$ ⓘ Symbols: $x$: real, $k$: real, $\nu$: parameter and $u_{k}(x;\nu)$: solution Permalink: http://dlmf.nist.gov/32.3.E4 Encodings: TeX, pMML, png See also: Annotations for §32.3(iii), §32.3 and Ch.32

with $\beta=0$, $\alpha=2\nu+1$, and

 32.3.5 $w(x)\sim 2\sqrt{2}k^{2}{U^{2}}\left(-\nu-\tfrac{1}{2},\sqrt{2}x\right),$ $x\to+\infty$;

compare (32.2.11) and §32.11(v). If we set $\ifrac{{\mathrm{d}}^{2}u}{{\mathrm{d}x}^{2}}=0$ in (32.3.2) and solve for $u$, then

 32.3.6 $u^{2}=-\tfrac{1}{3}x\pm\tfrac{1}{6}\sqrt{x^{2}+12\nu+6}.$ ⓘ Symbols: $x$: real, $\nu$: parameter and $u_{k}(x;\nu)$: solution Permalink: http://dlmf.nist.gov/32.3.E6 Encodings: TeX, pMML, png See also: Annotations for §32.3(iii), §32.3 and Ch.32 Figure 32.3.7: uk⁡(x;-12) for -12≤x≤4 with k=0.33554 691, 0.33554 692. The two graphs are indistinguishable when x exceeds -5.0, approximately. The parabolas u2+12⁢x=0, u2+16⁢x=0 are shown in black and green, respectively. Magnify Figure 32.3.9: uk⁡(x;32) for -12≤x≤4 with k=0.38736, 0.38737. The two graphs are indistinguishable when x exceeds -1.0, approximately. The curves u2+13⁢x±16⁢x2+24=0 are shown in green and black, respectively. Magnify