Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence

[Equivalence with the Volterra equation] Let $u\in C^2([0,L])$. If $u$ solves (IVP), then $(u^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=(V-z)u$; [F3] gives $u^{\mathrm{\prime }}(x)=c_1+{\int }_{x_0}^x(V-z)u$ and, integrating once more and applying [F4] to interchange the order of integration on the triangle $\{x_0\le t\le s\le x\}$ (and its mirror image for $x<x_0$), $u(x)-c_0-c_1(x-x_0)={\int }_{x_0}^x(x-t)(V(t)-z)u(t)dt\mathrm{.}$ Conversely, if $u\in C([0,L])$ satisfies this Volterra equation, then [F3] applied to the integrand $t\mapsto (V(t)-z)u(t)$ shows that $x\mapsto {\int }_{x_0}^x(x-t)(V(t)-z)u(t)dt$ is $C^1$ with derivative ${\int }_{x_0}^x(V(t)-z)u(t)dt$, hence $u\in C^1$ with $u^{\mathrm{\prime }}(x)=c_1+{\int }_{x_0}^x(V-z)u$ and, applying [F3] once more, $u\in C^2$ and $u^{\mathrm{\prime }\mathrm{\prime }}=(V-z)u$; the initial conditions hold at $x=x_0$. Thus (IVP) is equivalent to the Volterra equation for $C^2$ functions.

[Contraction] The space $X$ is complete: a supremum-norm Cauchy sequence converges pointwise in $\mathbb{C}$, its Cauchy bounds imply uniform convergence, and its uniform limit is continuous by the triangle inequality. Fix $z$ and put $\alpha :=M+\mathrm{\mid }z\mathrm{\mid }$. For $\lambda >\alpha L$ define on $X=C([0,L])$ the equivalent norm $\mathrm{\parallel }u{\mathrm{\parallel }}_{\lambda }:={\sup }_x\mathrm{\mid }u(x)\mathrm{\mid }{e}^{-\lambda \mathrm{\mid }x-x_0\mathrm{\mid }}$ and the map $(Tu)(x):=c_0+c_1(x-x_0)+{\int }_{x_0}^x(x-t)(V(t)-z)u(t)dt\mathrm{.}$ For $u,v\in X$ and $x\in [0,L]$, with $r:=\mathrm{\mid }x-x_0\mathrm{\mid }$, substituting $t=x_0\pm s$ in the Volterra integral gives $\mathrm{\mid }Tu(x)-Tv(x)\mathrm{\mid }\le \alpha {\int }_0^r(r-s)\mathrm{\mid }u-v\mathrm{\mid }(x_0\pm s)ds$, and $\mathrm{\mid }(x_0\pm s)-x_0\mathrm{\mid }=s$ gives $\mathrm{\mid }u-v\mathrm{\mid }(x_0\pm s)\le \mathrm{\parallel }u-v{\mathrm{\parallel }}_{\lambda }{e}^{\lambda s}$; hence $\mathrm{\mid }Tu(x)-Tv(x)\mathrm{\mid }\le \alpha \mathrm{\parallel }u-v{\mathrm{\parallel }}_{\lambda }{\int }_0^r(r-s)e^{\lambda s}ds=\alpha \mathrm{\parallel }u-v{\mathrm{\parallel }}_{\lambda }\frac{e^{\lambda r}-1-\lambda r}{{\lambda }^2}\le \frac{\alpha r}{\lambda }{e}^{\lambda r}\mathrm{\parallel }u-v{\mathrm{\parallel }}_{\lambda }\le \frac{\alpha L}{\lambda }{e}^{\lambda r}\mathrm{\parallel }u-v{\mathrm{\parallel }}_{\lambda },$ since $e^{\lambda r}-1-\lambda r\le \lambda r{e}^{\lambda r}$ for $r\ge 0$. Hence $\mathrm{\parallel }Tu-Tv{\mathrm{\parallel }}_{\lambda }\le (\alpha L\mathrm{/}\lambda )\mathrm{\parallel }u-v{\mathrm{\parallel }}_{\lambda }$, a strict contraction because $\lambda >\alpha L$. By [F1] $T$ has exactly one fixed point $u\in X$.

[The fixed point is the solution, and the a priori bound] The fixed point satisfies the Volterra equation, so by step 1.1 it is the unique $C^2$ solution of (IVP): uniqueness follows because any $C^2$ solution satisfies the Volterra equation by step 1.1 and hence is a fixed point of $T$, and [F1] gives at most one fixed point. For the bound, put $m(r):={\sup }_{x\in [0,L],\mathrm{\mid }x-x_0\mathrm{\mid }\le r}\mathrm{\mid }u(x)\mathrm{\mid }$ for $r\in [0,L]$; this is continuous, since uniform continuity of $u$ bounds the change when the interval endpoints move; the Volterra equation gives, for $r=\mathrm{\mid }x-x_0\mathrm{\mid }$, $\mathrm{\mid }u(x)\mathrm{\mid }\le \mathrm{\mid }{c}_0\mathrm{\mid }+\mathrm{\mid }{c}_1\mathrm{\mid }L+\alpha {\int }_0^r(r-s)m(s)ds\le \mathrm{\mid }{c}_0\mathrm{\mid }+\mathrm{\mid }{c}_1\mathrm{\mid }L+\alpha L{\int }_0^{r}m(s)ds,$ so $m(r)\le \mathrm{\mid }{c}_0\mathrm{\mid }+\mathrm{\mid }{c}_1\mathrm{\mid }L+\alpha L{\int }_0^{r}m(s)ds$ by taking suprema, and [F2] gives $m(r)\le (\mathrm{\mid }{c}_0\mathrm{\mid }+\mathrm{\mid }{c}_1\mathrm{\mid }L)e^{\alpha Lr}\le (\mathrm{\mid }{c}_0\mathrm{\mid }+\mathrm{\mid }{c}_1\mathrm{\mid }L)e^{\alpha L^2}$. This proves (a).

[Uniform bounds on a parameter disc] Fix $R>0$ and let $\mathrm{\mid }z\mathrm{\mid }\le R$; put ${\alpha }_R:=M+R$ and $C_R:=(\mathrm{\mid }{c}_0\mathrm{\mid }+\mathrm{\mid }{c}_1\mathrm{\mid }L)e^{{\alpha }_R{L}^2}$. By step 2.1, ${\sup }_{[0,L]}\mathrm{\mid }u(\cdot ;z)\mathrm{\mid }\le C_R$.

[Locally uniform Picard series] Put $p(x):=c_0+c_1(x-x_0)$ and $u_n(\cdot ;z):={\mathcal{K}}_z^{n}p$ for $n\ge 0$. Each $u_n(x;\cdot )$ is a polynomial in $z$ with continuous coefficients in $x$. For $r:=\mathrm{\mid }x-x_0\mathrm{\mid }$ and $\mathrm{\mid }z\mathrm{\mid }\le R$, induction in the Volterra integral, using its reversed orientation when $x<x_0$, gives $\mathrm{\mid }{u}_n(x;z)\mathrm{\mid }\le (\mathrm{\mid }{c}_0\mathrm{\mid }+\mathrm{\mid }{c}_1\mathrm{\mid }L)\frac{{\alpha }_R^n{r}^{2n}}{(2n)!}\mathrm{.}$ For $n\ge 1$, differentiating the Volterra integral gives $\mathrm{\mid }{u}_n^{\mathrm{\prime }}(x;z)\mathrm{\mid }\le (\mathrm{\mid }{c}_0\mathrm{\mid }+\mathrm{\mid }{c}_1\mathrm{\mid }L)\frac{{\alpha }_R^n{r}^{2n-1}}{(2n-1)!}\mathrm{.}$ The two majorant series converge uniformly for $r\le L$, so ${\sum }_{n\ge 0}{u}_n$ and $c_1+{\sum }_{n\ge 1}{u}_n^{\mathrm{\prime }}$ converge uniformly on $[0,L]\times \{\mathrm{\mid }z\mathrm{\mid }\le R\}$. Their sums satisfy the Volterra equation and its first-derivative equation by uniform convergence under the integrals [F3]; step 2.1 identifies the first sum with $u(\cdot ;z)$, and the derivative equation identifies the second with $u^{\mathrm{\prime }}(\cdot ;z)$. Since this holds for every $R>0$, both are locally uniform Picard series in $z$. Uniform convergence on each product compact, whose summands are continuous in $(x,z)$, also gives joint continuity.

[Entire dependence and derivative equation] Fix $z\in \mathbb{C}$ and choose $R>\mathrm{\mid }z\mathrm{\mid }+1$. For sufficiently small complex $h$, both $z$ and $z+h$ lie in $\mathrm{\mid }\zeta \mathrm{\mid }\le R$. Put $e:=u(\cdot ;z+h)-u(\cdot ;z)$. Subtracting the two Volterra equations gives $e={\mathcal{K}}_{z}e-hJu(\cdot ;z+h)$, where $J:C([0,L];\mathbb{C})\to C([0,L];\mathbb{C})$ is defined by $(Jf)(x):={\int }_{x_0}^x(x-t)f(t)dt$, and $e(x_0)=e^{\mathrm{\prime }}(x_0)=0$. By step 3.1 and the kernel bound, for the continuous running supremum $m(r):={\sup }_{x\in [0,L],\mathrm{\mid }x-x_0\mathrm{\mid }\le r}\mathrm{\mid }e(x)\mathrm{\mid }$ we have $m(r)\le \mathrm{\mid }h\mathrm{\mid }{C}_R{L}^2\mathrm{/}2+{\alpha }_{R}L{\int }_0^{r}m(s)ds$. Gronwall [F2] yields $\mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }h\mathrm{\mid }{C}_R{L}^2{e}^{{\alpha }_R{L}^2}\mathrm{/}2$. The differentiated equation $e^{\mathrm{\prime }}(x)={\int }_{x_0}^x[(V(t)-z)e(t)-hu(t;z+h)]dt$ then gives $\mathrm{\parallel }{e}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\alpha }_{R}L\mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\mid }h\mathrm{\mid }L{C}_R$. Let $w$ be the unique continuous solution of $w={\mathcal{K}}_{z}w-Ju(\cdot ;z)$, which exists by the contraction argument of step 1.2; step 1.1 makes it $C^2$ and gives $-w^{\mathrm{\prime }\mathrm{\prime }}+Vw-zw=u(\cdot ;z)$ with zero initial data. The quotient $q_h:=e\mathrm{/}h$ satisfies $q_h={\mathcal{K}}_z{q}_h-Ju(\cdot ;z+h)$. Thus $d_h:=q_h-w$ obeys $d_h={\mathcal{K}}_z{d}_h-Je$. Applying the same kernel estimate and [F2] gives $\mathrm{\parallel }{d}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le (L^2\mathrm{/}2)\mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}{e}^{{\alpha }_R{L}^2}\le \mathrm{\mid }h\mathrm{\mid }{C}_R{L}^4{e}^{2{\alpha }_R{L}^2}\mathrm{/}4\to 0$. Hence the complex derivative ${\mathrm{\partial }}_{z}u(\cdot ;z)$ exists in the supremum norm and equals $w$. Finally, $q_h^{\mathrm{\prime }}(x)={\int }_{x_0}^x[(V(t)-z)q_h(t)-u(t;z+h)]dt,w^{\mathrm{\prime }}(x)={\int }_{x_0}^x[(V(t)-z)w(t)-u(t;z)]dt,$ so $\mathrm{\parallel }{q}_h^{\mathrm{\prime }}-w^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\alpha }_{R}L\mathrm{\parallel }{q}_h-w{\mathrm{\parallel }}_{\mathrm{\infty }}+L\mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Therefore ${\mathrm{\partial }}_z{u}^{\mathrm{\prime }}=w^{\mathrm{\prime }}$ as well; since $z$ was arbitrary, both $u(x;\cdot )$ and $u^{\mathrm{\prime }}(x;\cdot )$ are entire, and the derivative is exactly the stated solution $w$.