Airy equation ($q_t+q_{xxx}=0$)

The Airy equation is the linear dispersive partial differential equation $q_t+q_{xxx}=0$. Under the spatial Fourier transform, each mode evolves by multiplication by $e^{i{k}^{3}t}$.

For positive time its fundamental solution is $(3t{)}^{-1/3}\mathrm{Ai}(x/(3t{)}^{1/3})$, connecting this evolution equation to the Airy function without identifying the two concepts.

Let $y$ solve the Airy ordinary differential equation $y^{\prime \prime }(s)=sy(s)$. For $t>0$, set $s=x/(3t{)}^{1/3}$ and define the displayed similarity solution. Differentiating the profile equation gives $y^{\prime \prime \prime }=y+s{y}^{\prime }$, so the chain rule yieldsTheir sum is zero, proving the Airy equation $u_t+u_{xxx}=0$. Every real linear combination of the two standard Airy functions gives such a solution. This is a solution of the linear dispersive equation; the nonlinear Korteweg-De Vries equation includes an extra term and does not follow from the same arbitrary profile construction.

For dispersion $w(k)=-i(k^3+k)$, the other roots of $w(\nu )=w(k)$ satisfy ${\nu }^2+k\nu +k^2+1=0$. In $D_{+}=\{\mathrm{Im}k>0,\mathrm{Re}w(k)<0\}$ both roots lie in the lower half-plane. Interpolating the unknown linear polynomial $G_2+i\nu G_1$ at them eliminates two boundary traces. The resulting symmetric expression has removable singularities where the roots coalesce.

For $Q_T+Q_{XXX}+a{Q}_{XX}+b{Q}_X+cQ=0$, an exponential gauge removes the second derivative. With ${\kappa }^2=a^2/3-b>0$ and $\sigma =-c+ab/3-2a^3/27$, the remaining equation is $q_t+q_{xxx}-q_x=0$. Negative $a$ improves spatial decay of the transformed initial data; positive $a$ can destroy the convergence of its Half-range Fourier transform.

For the Airy equation with opposite dispersive sign, $q_t-q_{xxx}=0$ on $x>0$, let $Q$ be the initial Half-range Fourier transform and $F_j(k,t)={\int }_0^t{e}^{i{k}^{3}s}{\partial }_x^{j}q(0,s)ds$. Three integrations by parts give the displayed global relation. The spatial transform is holomorphic for $\mathrm{Im}k<0$, whereas the finite-time spectral boundary transforms are entire functions. The symmetry $k\mapsto \alpha k$, ${\alpha }^3=1$, preserves their time exponent.

// Target: analysis.bigb

For $q_t-q_{xxx}=0$ on $x>0$, evaluate the global relation at $k=-ir/2$, where $\mathrm{Re}p>0$ and the principal cube root $r=p^{1/3}$ has positive real part. The time exponent becomes $-p$, giving the displayed Laplace transform identity. Thus the second-derivative trace is determined once the value and first-derivative traces are prescribed. The homogeneous transformed equation $u^{\prime \prime \prime }-pu=0$ has exactly two spatially decaying roots, so two independent admissible conditions are required. One condition leaves a free decaying mode; three independent conditions generally violate the global constraint.

For $L=-d^2/d{x}^2+x$ on $(0,1)$, with homogeneous conditions $v(0)=0$, $v^{\prime }(1)=0$, integration by parts gives $⟨v,Lv⟩={\int }_0^1(v^{\prime 2}+x{v}^2)dx>0$ for nonzero $v$. The affine energy $J[u]=\frac{1}{2}{\int }_0^1(u^{\prime 2}+x{u}^2)dx$, minimized over $H^1$ functions with $u(0)=a$, yields $Lu=0$ and the natural boundary condition $u^{\prime }(1)=0$. The inequality $\Vert v{\Vert }_2^2\le \frac{1}{2}\Vert v^{\prime }{\Vert }_2^2$ on $v(0)=0$ proves that the form is a coercive bilinear form; Lax-Milgram theorem supplies a unique critical point, and $J[u+v]-J[u]=\frac{1}{2}\int (v^{\prime 2}+x{v}^2)$ proves that it is the unique minimum.

The scalar linear partial differential equation $u_t+u_x+u_{xxx}=0$ is an Airy equation with a first-order transport term. It is a linearization of the Korteweg-De Vries equation with an advective term retained. It is distinct from the viscous incompressible-flow Stokes equation. On the whole line its spatial Fourier transform evolves by $e^{i(k^3-k)t}$.

For a time Laplace transform parameter with $\mathrm{Re}p>0$, the spatial characteristic roots satisfy $r^3+r+p=0$. Exactly one root $r_{-}$ has negative real part and two have positive real part. Indeed an imaginary root would force $p=-i(k-k^3)$ to be imaginary. The count is therefore constant on the connected right half-plane; for real $p>0$, the cubic has one negative real root and a conjugate pair with positive real parts. Repeated roots would satisfy $3r^2+1=0$, which again puts $p$ on the imaginary axis. Thus $r_{-}(p)$ is a nonzero holomorphic function in the right half-plane. Only $e^{r_{-}x}$ decays at positive spatial infinity.

For $\mathrm{Re}p>0$, let $r_{-}$ and $r_1,r_2$ be the characteristic roots of $r^3+r+p=0$, split by the signs of their real parts. Put $d_j=3r_j^2+1$. The real-line Green function of $p+{\partial }_x+{\partial }_x^3$ isThe reciprocal-polynomial root identities imply continuity of $R_p,R_p^{\prime }$ and jump $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. The distributional derivative therefore gives $(p+{\partial }_s+{\partial }_s^3)R_p={\delta }_0$. The transport term changes the roots: the pure Airy resolvent kernel with $r^3+p=0$ cannot be substituted unchanged.

For the linear dispersive Stokes equation on $x>0$, with sufficiently smooth decaying initial data and $\mathrm{Re}p>0$, define $V(x,p)={\int }_0^{\infty }{R}_p(x-y)u_0(y)dy$ and let $G(p)$ be the time Laplace transform of the prescribed derivative $u_x(0,t)$. Spatial decay and the one-dimensional decaying homogeneous space giveThe Neumann boundary condition follows by differentiating at zero. The initial forcing follows from the Stokes resolvent kernel with advection. The Bromwich inversion formula gives an integral representation with no unknown boundary traces. Truncating or extending the prescribed boundary signal beyond a target time does not affect earlier values, by the Laplace transform time-shift rule and causality of this Green function.

For $\mathrm{Re}p>0$, put $\lambda =p^{1/3}$ using the principal cube root and $r_{1,2}=\lambda e^{\pm i\pi /3}$. The decaying Green function for ${\partial }_s^3+p$ on the real line isThe identities among the three characteristic roots show that $R_p$ and $R_p^{\prime }$ are continuous, while $R_p^{\prime \prime }$ has jump one. Its distributional derivative therefore satisfies $({\partial }_s^3+p)R_p={\delta }_0$. This is the kernel of the resolvent operator $(p+{\partial }_s^3{)}^{-1}$, obtained by the time Laplace transform of the Airy equation.

For $u_t+u_{xxx}=0$ on $x>0$, define $V(x,p)={\int }_0^{\infty }{R}_p(x-y)u_0(y)dy$ and let $G$ be the time Laplace transform of $u_x(0,t)$. The transformed solution with spatial decay isOnly one characteristic root has negative real part, so one scalar boundary derivative fixes the remaining homogeneous mode. The Bromwich inversion formula supplies an integral representation involving only the initial and boundary data. For general linear differential equations on a half-line, the number and form of the necessary boundary data depend on the decaying spatial roots; Fokas and Wang study the corresponding boundary maps for linear dispersive equations.