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Non-characteristic. The first-order principal symbol is ${\nu }_t-i{\nu }_x$, which equals one on $(0,1)$.

Set $z=x+it$. The equation is equivalent to ${\partial }_{\bar{z}}\varphi =({\varphi }_x+i{\varphi }_t)/2=0$. Because $\varphi$ is continuously differentiable, the Cauchy-Riemann equations imply that it is a holomorphic function of $z$. Its convergent complex Taylor series restricts to a real power series along $t=0$, so $g$ is a real analytic function.

For smooth-data instability of the Cauchy-Riemann Cauchy problem, chooseThese are entire functions and satisfy the equation. Their initial derivatives obey ${\sup }_x|{\partial }_x^j{g}_n|\le n^j{e}^{-\sqrt{n}}$, so every finite sum tends to zero. ButThus analytic solutions can exist while continuous dependence on initial data fails for this smooth-data topology.

The Cauchy estimate follows directly from the differentiated Cauchy integral formula:The circle has length $2\pi r$, giving the bound for every integer $j\ge 0$.

Put $w=\varphi -\psi$ and integrate on the straight segment $z=\tau t/|t|$. For fixed $|x|<sR$, a disc centred at $x$ with radius approaching $R(\sigma (\tau )-s)$ stays inside $|\xi |\le \sigma (\tau )R$. Applying the first-derivative Cauchy estimate there givesIntegrating its modulus and taking the supremum over $|x|<sR$ is exactly the asserted integral estimate. The source cancels from $T\varphi -T\psi$.

To obtain a contraction mapping, use the weighted holomorphic norm on a shrinking time domain. Its finite-norm space consists of holomorphic functions on ${\mathcal{C}}_{\alpha }$ with zero initial value; the apparent quotient at $t=0$ is interpreted by a limit. Completeness follows because convergence in this norm implies locally uniform convergence of holomorphic functions, including near $t=0$, and the limiting pointwise bounds give convergence in the norm.

Here is the explicit contraction estimate on a shrinking holomorphic domain. Write $N=\Vert w{\Vert }_{\alpha }$, $r=|t|$, $A=\alpha (1-s)$, and chooseThen $\sigma -s=(A-\tau )/(2\alpha )$ and $\alpha (1-\sigma )-\tau =(A-\tau )/2$. The whole auxiliary polydisc lies inside ${\mathcal{C}}_{\alpha }$, and its supremum of $|w|$ is at most $2N\tau /(A-\tau )$. ConsequentlyThus $\Vert T\varphi -T\psi {\Vert }_{\alpha }\le (4\alpha /R)\Vert \varphi -\psi {\Vert }_{\alpha }$. Choose $0<\alpha <\min (\eta ,R/4)$. The source is bounded on the closed polydisc, say by $M_f$, and $\Vert T0{\Vert }_{\alpha }\le \alpha M_f$. Therefore $T$ maps the Banach space into itself and is a strict contraction mapping. The Banach fixed-point theorem gives a holomorphic fixed point with $\varphi (x,0)=0$. Differentiating the integral identity yields ${\varphi }_t-i{\varphi }_x=f$, establishing this case of the Cauchy-Kovalevskaya theorem.

Finally, near each real initial point, a real analytic function $g$ has a holomorphic extension $G$. Apply the same construction to $\varphi =G(x)+v$ with $v(x,0)=0$ and source $i{G}^{\prime }(x)$, restricting the discs if necessary. This gives a local real analytic function of $(x,t)$ with the prescribed initial value. Equivalently the solution is $\boxed{\varphi (x,t)=G(x+it)}$ wherever the extension is defined.

For the period lattice $\Lambda =⟨\lambda ,\mu ⟩$, define the Weierstrass elliptic function byTo prove Normal convergence of the Weierstrass elliptic-function series, let $K\subset \mathbb{C}\setminus \Lambda$ be compact and choose $R$ with $|z|\le R$ on $K$. For $|\omega |>2R$,The comparison series ${\sum }_{\omega \ne 0}|\omega {|}^{-3}$ converges by the stated lattice-sum criterion. The Weierstrass M-test gives uniform convergence on $K$; since $K$ was arbitrary, the series converges locally uniformly away from $\Lambda$. The locally uniform convergence of holomorphic functions makes its sum holomorphic there, and the displayed principal term gives a double pole at every lattice point.

The function $\wp$ is even and satisfiesThe three nonzero two-torsion points $z_1,z_2,z_3$ are distinct modulo $\Lambda$ and each equals its own negative modulo $\Lambda$. Therefore no two can have the same $\wp$-value, so $e_1,e_2,e_3$ are distinct. Equivalently, $\wp$ is a degree-two branched map and each half-lattice point already occupies its fiber with multiplicity two.

For a transitive example, take the equianharmonic latticeand define $\theta ([z])=[\omega z]$. Multiplication by $\omega$ preserves $\Lambda$, so $\theta :\mathbb{C}/\Lambda \to \mathbb{C}/\Lambda$ is a conformal equivalence. On the nonzero half-lattice points it acts bybecause ${\omega }^2=-1-\omega$. Thus it acts transitively.

For $\sigma =\Re s>1$,On every compact subset of $\Re s>1$, the terms are bounded by a convergent series $\sum n^{-1-\delta }$. The Weierstrass M-test gives locally uniform convergence, and the theorem on locally uniform convergence of holomorphic functions shows that the limit is an analytic function. Hence $D_f$ is analytic throughout $\Re s>1$.

Let $R,\alpha >0$ and ${\mathcal{C}}_{\alpha }=\{(x,t)\in {\mathbb{C}}^2:|x|<R,|t|<\alpha (1-|x|/R)\}$. The displayed norm makes the holomorphic functions with finite norm a Banach space. Finiteness forces $u(x,0)=0$: for fixed $x$, the weight grows like $1/|t|$ near zero. A Cauchy sequence in this norm converges uniformly on every compact subset of ${\mathcal{C}}_{\alpha }$, including at $t=0$. Its limit is holomorphic by locally uniform convergence of holomorphic functions, and the pointwise norm bounds show convergence in the original norm. The shrinking domain compensates for loss of an $x$-derivative in a Cauchy-Kovalevskaya theorem argument.