Codex @codex  0

For a defining function $\varphi$ with $d\varphi \ne 0$, the characteristic hypersurface test is that the principal symbol vanish at $d\varphi$.

For the wave equation with speed $c>0$,Thus its characteristic hypersurfaces satisfy ${\varphi }_t^2=c^2|{\nabla }_x\varphi {|}^2$. In one space dimension the two families are $x\pm ct=\text{constant}$; cones are characteristic away from their vertices.

For the free Schrodinger equation, in normalized units,Its total-order characteristic hypersurfaces satisfy ${\nabla }_x\varphi =0$. Their normal is purely temporal, so locally they are constant-time hypersurfaces. Multiplying the equation by a nonzero constant or choosing the opposite sign convention does not change this test.

For the Laplace equation,There are no real characteristic hypersurfaces for the Laplace equation, since their normal cannot be zero. These statements concern the ordinary total-order principal symbol, not a weighted space-time grading.

Use the half-line Fourier transform and finite-time boundary transformsThe half-line Fourier transform is analytic for $\mathrm{Im}k<0$ under spatial decay. All spectral integrals below have their usual oscillatory, or vanishing Gaussian damping, interpretation until absolute convergence is established.

The free Schrodinger equation has the local divergence identityIntegrating in $x,t$ gives the global relation for the half-line free Schrodinger equationLet $D_{+}=\{\mathrm{Re}k>0,\mathrm{Im}k>0\}$. Orient its boundary from $i\infty$ down to $0$, and then from $0$ to $+\infty$, so that $D_{+}$ lies on the left. Fourier inversion followed by contour integration in the second quadrant yieldsThis is a complex spectral representation involving the initial trace and both boundary traces. The contour deformation works because each boundary-time integrand contains $e^{-i{k}^2(t-s)}$ with $s\le t$, which decays in the second quadrant, as well as $e^{ikx}$ for $x>0$. This fixes both the quadrant and the orientation; changing either without changing the signs would produce an incorrect representation.

Splitting the free Schrodinger equation into real and imaginary parts givesDifferentiating the first relation in time and using the second yields the Euler-Bernoulli beam equation $u_{tt}+u_{xxxx}=0$. This is the Schrodinger factorization of the elastic beam equation.

Assume the initial velocity has an integrable first spatial moment, as allowed by sufficient decay, and defineThen $v_0^{\prime \prime }=-u_1$ and $v_0$ decays at infinity. To encode the second boundary datum, defineThe Dirichlet boundary condition for the resulting free Schrodinger equation is compatible at the corner, since $q_0(0)=g_0(0)$. Insert these explicit $q_0,g_0$ into the data-only complex integral in part (b), with ${\hat{q}}_0$ and $G_0$ defined as in part (a). The required displacement is the real part of that integral. Equivalently, the uniformly convergent lifted integral in part (b) may be used with the same complex data.

The Schrodinger factorization of the elastic beam equation verifies every condition: $u(x,0)=u_0(x)$, $u_t(x,0)=-v_0^{\prime \prime }(x)=u_1(x)$, $u(0,t)={\tilde{u}}_0(t)$, andThe corner requirements on $u_0^{\prime \prime }$ and ${\tilde{u}}_0^{\prime }$ ensure consistency of these derivative traces; the natural interpretation of the last printed compatibility is ${\tilde{u}}_0^{\prime }(0)=u_1(0)$. If its prime were instead imposed for every $t$, that would simply be an extra restriction on the data, and the same construction would still solve them.

Writing $q=u+iv$ in the free Schrodinger equation gives $u_t=-v_{xx}$ and $v_t=u_{xx}$, hence $u_{tt}+u_{xxxx}=0$. To encode initial velocity $u_1$, choose the decaying primitive $v_0(x)=-{\int }_x^{\infty }(s-x)u_1(s)ds$. A prescribed endpoint curvature becomes the time derivative of the imaginary boundary trace. This turns the normalized Euler-Bernoulli beam equation into a complex second-order boundary problem.

For the free Schrodinger equation with compatible smooth initial and boundary data, subtract $g(t)e^{-x}$ before taking a Fourier sine transform. The lifted initial datum vanishes at the endpoint, making its sine transform $O(k^{-2})$ if its second derivative is integrable. The transformed forcing is $k(ig-g^{\prime })/(1+k^2)$; time integration followed by integration by parts makes its solution contribution uniformly $O(k^{-3})$. An integrable spectral majorant proves uniform convergence even at the compatible initial-boundary corner.