Calculus of separated phasor fields: real-part derivative identities and amplitude injectivity

[Reduction of the phase and of the components] Since $\beta =p+iq$ and $i^2=-1$, $i\theta =i(\beta x_3-\omega t)=i(p{x}_3-\omega t)+i(iq{x}_3)=-q{x}_3+i(p{x}_3-\omega t)$; with $\phi :=p{x}_3-\omega t$ the cartesian form [F3] gives $e^{i\theta }=e^{-q{x}_3}(\cos \phi +i\sin \phi )$. By [F2] the component of $\tilde{F}{e}^{i\theta }$ with index $l$ is $e^{-q{x}_3}[(a_l\cos \phi -b_l\sin \phi )+i(a_l\sin \phi +b_l\cos \phi )]$; taking real parts, $F_l=e^{-q{x}_3}(a_l\cos \phi -b_l\sin \phi )$.

[The two-phase system] For clause (d), write $C=a+ib$ with $a,b\in {\mathbb{R}}^3$. By [F2], $\mathrm{Re}(C{e}^{i{\phi }_j})=a\cos {\phi }_j-b\sin {\phi }_j$ componentwise, so $\mathrm{Re}(C{e}^{i{\phi }_1})=\mathrm{Re}(C{e}^{i{\phi }_2})=0$ is the system $a_l\cos {\phi }_1=b_l\sin {\phi }_1$ and $a_l\cos {\phi }_2=b_l\sin {\phi }_2$ for every $l$. Multiplying the two equations of index $l$ by $\cos {\phi }_2$ and $\cos {\phi }_1$ and subtracting gives $b_l(\sin {\phi }_1\cos {\phi }_2-\cos {\phi }_1\sin {\phi }_2)=0$, and multiplying them by $\sin {\phi }_2$ and $\sin {\phi }_1$ and subtracting gives $a_l(\cos {\phi }_1\sin {\phi }_2-\cos {\phi }_2\sin {\phi }_1)=0$.

[The phase factor and its derivatives] By step 1.1 the real and imaginary parts of $(x_3,t)\mapsto e^{i\theta }$ are $e^{-q{x}_3}\cos \phi$ and $e^{-q{x}_3}\sin \phi$, which are $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^2$ by [F4] (the chain rule applied to the affine substitutions $x_3\mapsto -q{x}_3$ and $x_3,t\mapsto p{x}_3-\omega t$ composed with the outer exponential, sine and cosine, together with the sum, scalar and product rules), with ${\mathrm{\partial }}_{x_3}{e}^{-q{x}_3}=-q{e}^{-q{x}_3}$, ${\mathrm{\partial }}_t{e}^{-q{x}_3}=0$, ${\mathrm{\partial }}_{x_3}\cos \phi =-p\sin \phi$, ${\mathrm{\partial }}_{x_3}\sin \phi =p\cos \phi$, ${\mathrm{\partial }}_t\cos \phi =\omega \sin \phi$ and ${\mathrm{\partial }}_t\sin \phi =-\omega \cos \phi$. Pairing these, ${\mathrm{\partial }}_{x_3}{e}^{i\theta }=(-q+ip)e^{i\theta }=i\beta e^{i\theta }$ and ${\mathrm{\partial }}_t{e}^{i\theta }=-i\omega e^{i\theta }$ by [F2], while $e^{i\theta }$ does not depend on $x_1,x_2$, so ${\mathrm{\partial }}_{x_j}{e}^{i\theta }=0$ for $j=1,2$.

[Transverse derivatives] Fix $j\in \{1,2\}$ and $l\in \{1,2,3\}$. Along the $x_j$-coordinate line the factors $e^{-q{x}_3}$, $\cos \phi$ and $\sin \phi$ of step 1.1 are constant, so the derivative of the partial application is $e^{-q{x}_3}[({\mathrm{\partial }}_j{a}_l)\cos \phi -({\mathrm{\partial }}_j{b}_l)\sin \phi ]$ by the sum and scalar rules [F4], the derivatives ${\mathrm{\partial }}_j{a}_l,{\mathrm{\partial }}_j{b}_l$ existing by [F1]; by the expansion of step 1.1 applied to ${\mathrm{\partial }}_j{\tilde{F}}_l$ in place of ${\tilde{F}}_l$, this number is $\mathrm{Re}(({\mathrm{\partial }}_j{\tilde{F}}_l)e^{i\theta })$. Hence ${\mathrm{\partial }}_{x_j}F=\mathrm{Re}\left({\mathrm{\partial }}_{x_j}\tilde{F}{e}^{i\theta }\right)$ componentwise for $j=1,2$.

[Injectivity] By the addition and parity formulas [F5], $\sin ({\phi }_1-{\phi }_2)=\sin {\phi }_1\cos {\phi }_2-\cos {\phi }_1\sin {\phi }_2$ and $-\sin ({\phi }_1-{\phi }_2)=\cos {\phi }_1\sin {\phi }_2-\cos {\phi }_2\sin {\phi }_1$. Since ${\phi }_1-{\phi }_2\mathrm{\notin }\pi \mathbb{Z}$, [F5] gives $\sin ({\phi }_1-{\phi }_2)\mathrm{\ne }0$, so the two equations of step 1.2 force $b_l=0$ and $a_l=0$ for every $l$, that is, $C=0$.

[Longitudinal and time derivatives] Fix $l$. Along the $x_3$-line the amplitudes $a_l,b_l$ are constant, so the product and scalar rules [F4] and the derivatives of step 2.1 give ${\mathrm{\partial }}_{x_3}{F}_l=-q{e}^{-q{x}_3}(a_l\cos \phi -b_l\sin \phi )+e^{-q{x}_3}(-p{a}_l\sin \phi -p{b}_l\cos \phi )$; by [F2], $i\beta (a_l+i{b}_l)=(-q{a}_l-p{b}_l)+i(p{a}_l-q{b}_l)$, so $\mathrm{Re}\left(i\beta {\tilde{F}}_l{e}^{i\theta }\right)=e^{-q{x}_3}[(-q{a}_l-p{b}_l)\cos \phi -(p{a}_l-q{b}_l)\sin \phi ]$, which is the displayed expression. Along the $t$-line the amplitudes and $e^{-q{x}_3}$ are constant, so ${\mathrm{\partial }}_t{F}_l=e^{-q{x}_3}(a_l\omega \sin \phi +b_l\omega \cos \phi )$; by [F2], $-i\omega (a_l+i{b}_l)=\omega b_l-i\omega a_l$, so $\mathrm{Re}(-i\omega {\tilde{F}}_l{e}^{i\theta })=e^{-q{x}_3}(\omega b_l\cos \phi +\omega a_l\sin \phi )$, the displayed expression. Assembling the components gives ${\mathrm{\partial }}_{x_3}F=\mathrm{Re}(i\beta \tilde{F}{e}^{i\theta })$ and ${\mathrm{\partial }}_{t}F=\mathrm{Re}(-i\omega \tilde{F}{e}^{i\theta })$.

[Regularity] Every partial derivative displayed in steps 2.2 and 3.1 is continuous on ${\mathrm{\Omega }}_t\times \mathbb{R}\times \mathbb{R}$: the partial derivatives ${\mathrm{\partial }}_j{a}_l,{\mathrm{\partial }}_j{b}_l$ ($j=1,2$) are continuous by [F1], the factors $e^{-q{x}_3}$, $\cos \phi$, $\sin \phi$ and their displayed partial derivatives are smooth by step 2.1, and each displayed formula is a finite sum of products of such continuous functions. Thus all first partial derivatives of $F$ exist and are continuous, so $F$ is $C^1$.

[Divergence] By [F6] and steps 2.2 and 3.1, $\mathrm{div}F={\mathrm{\partial }}_1{F}_1+{\mathrm{\partial }}_2{F}_2+{\mathrm{\partial }}_3{F}_3=\mathrm{Re}(({\mathrm{\partial }}_1{\tilde{F}}_1+{\mathrm{\partial }}_2{\tilde{F}}_2+i\beta {\tilde{F}}_3)e^{i\theta })$: indeed, for complex ${\zeta }_l$ the $\mathbb{R}$-linearity of $\mathrm{Re}$ and distributivity in $\mathbb{C}$ [F2] give $\mathrm{Re}({\zeta }_1{e}^{i\theta })+\mathrm{Re}({\zeta }_2{e}^{i\theta })+\mathrm{Re}({\zeta }_3{e}^{i\theta })=\mathrm{Re}(({\zeta }_1+{\zeta }_2+{\zeta }_3)e^{i\theta })$.

[Curl] Similarly, [F6] together with steps 2.2 and 3.1 gives the three components of $\mathrm{curl}F$ as ${\mathrm{\partial }}_2{F}_3-{\mathrm{\partial }}_3{F}_2=\mathrm{Re}(({\mathrm{\partial }}_2{\tilde{F}}_3-i\beta {\tilde{F}}_2)e^{i\theta })$, ${\mathrm{\partial }}_3{F}_1-{\mathrm{\partial }}_1{F}_3=\mathrm{Re}((i\beta {\tilde{F}}_1-{\mathrm{\partial }}_1{\tilde{F}}_3)e^{i\theta })$ and ${\mathrm{\partial }}_1{F}_2-{\mathrm{\partial }}_2{F}_1=\mathrm{Re}(({\mathrm{\partial }}_1{\tilde{F}}_2-{\mathrm{\partial }}_2{\tilde{F}}_1)e^{i\theta })$, since [F2] gives $\mathrm{Re}(\zeta e^{i\theta })-\mathrm{Re}(\eta e^{i\theta })=\mathrm{Re}((\zeta -\eta )e^{i\theta })$ for complex $\zeta ,\eta$. This is clause (c).

[Open-set consequence] Let $U\subseteq {\mathrm{\Omega }}_t\times \mathbb{R}\times \mathbb{R}$ be nonempty open with $F=G$ on $U$. Its projection $\pi (U)$ to ${\mathrm{\Omega }}_t$ is nonempty and open. Fix $x_t\in \pi (U)$; the fibre $\{(x_3,t):(x_t,x_3,t)\in U\}$ is a nonempty open subset of ${\mathbb{R}}^2$, so there are $x_3\in \mathbb{R}$ and an open interval $J$ of $t$-values on which $(x_t,x_3,t)\in U$; choose $t_1,t_2\in J$ with $0<\omega \mathrm{\mid }{t}_1-t_2\mathrm{\mid }<\pi$. Put $C:=(\tilde{F}(x_t)-\tilde{G}(x_t))e^{i\beta x_3}$. For $j=1,2$, the $\mathbb{R}$-linearity of $\mathrm{Re}$ [F2] and $F=G$ on $U$ give $\mathrm{Re}(C{e}^{-i\omega t_j})=\mathrm{Re}((\tilde{F}-\tilde{G})(x_t)e^{i\theta (x_3,t_j)})=F(x_t,x_3,t_j)-G(x_t,x_3,t_j)=0$. The phases ${\phi }_j:=-\omega t_j$ are real and ${\phi }_1-{\phi }_2=-\omega (t_1-t_2)\mathrm{\notin }\pi \mathbb{Z}$, so step 2.3 gives $C=0$. Finally $e^{i\beta x_3}=e^{-q{x}_3}(\cos (p{x}_3)+i\sin (p{x}_3))$ by step 1.1, where $e^{-q{x}_3}>0$ by [F4] and $\cos (p{x}_3)+i\sin (p{x}_3)\mathrm{\ne }0$ because its real and imaginary parts have ${\cos }^2(p{x}_3)+{\sin }^2(p{x}_3)=1\mathrm{\ne }0$ [F5]; since a product of nonzero elements of the field $\mathbb{C}$ is nonzero [F2], $\tilde{F}(x_t)=\tilde{G}(x_t)$. As $x_t\in \pi (U)$ was arbitrary and $\pi (U)$ is nonempty open, this is the stated consequence.