For part of a bigger question, I need either a solution or a counterexample to the following 1 dimensional ODE.
Let $h\in L^2(0,\infty)$ (complex valued functions) and $r \in (0,\infty)$. Under boundary conditions $w(0) = c_1a + c_2$ and $w'(0) = c_3a + c_4$ where the $c_i$ are complex constants and $a$ is a complex parameter, is there an $a \in \mathbb{C}$ for which there is a solution in $H^2(0,\infty)$ for:
$$w''(t) = irw(t) – h(t) \ \ \ (t\in (0,\infty))?$$
The strange boundary conditions are just my way of saying that there is one parameter I want to solve for if a solution exists in order to couple this ODE with another one on the negative real line.
If I use Laplace transforms, this is equivalent to asking if there exists $w \in H^2(0,\infty)$ such that
$$\hat{w} = \frac{\hat{h}}{ir-s^2} – \frac{w(0)+sw'(0)}{ir – s^2}.$$
There is a solution to this ODE by taking the inverse Laplace transform of the above equation, namely
$$ w(t) = -\frac{1}{\sqrt{ir}}\int_0^t h(u)\sinh(\sqrt{ir}(t-u))du + \frac{w(0)}{\sqrt{ir}}\sinh(\sqrt{ir}t) + w'(0)\cosh(\sqrt{ir}t).$$
However, this is not obvious to me whether or not there is an $a$ for which it's in $L^2(0,\infty)$. Is there another solution to this ODE that is in $L^2(0,\infty)$ or if not, is there an explicit counterexample? In other words, given $h\in L^2(0,\infty)$, are there nonhomogeneous boundary conditions of the form aforementioned that give me a solutions $w\in H^2(0,\infty)$.
Best Answer
The two linearly independent solutions are $e^{\pm \sqrt{ir} t}$, where we take the principle branch cut of $\sqrt\cdot$. In order to make the solution $L^2[0,\infty)$, you need to discard the exponentially growing solution $e^{\sqrt{ir} t}$ and keep only the decaying one $e^{-\sqrt{ir} t}$. So $\sinh$ and $\cosh$ can not be used alone but only in equal weight pair in each of your terms. Explicitly $$w(t)=w(0)e^{-\sqrt{ir} t},$$ and $w'(0)=-w(0)\sqrt{ir}$. Substitute in the linear form of $w(0)$ and $w'(0)$, we can easily obtain the condition on $a$ which is $a=-\frac{c_1\sqrt{ir}+c_3}{c_2\sqrt{ir}+c_4}$ if the denominator does not vanish, and any $a\in\Bbb C$ if both the denominator and numerator vanish, and does not exist if only the denominator vanishes.
There are two interesting cases for two different Green's functions. The difference of these two Green's function is a homogeneous solution.
2.1 The first Green's function is $$G_1(t)=\frac1{k}\sinh(kt)\Theta(t)$$
where $k=\sqrt{ir}$ and $\Theta$ is the Heaviside step function. $$w(t)=-\int_0^t h(u)G_1(t-u)du.$$ There are many $h$'s (counterexamples saught by the OP) making $w\notin L^2[0,\infty)$ for $G_1$: (1) $h(t)=e^{-at}\in L^2[0,\infty)$ with some positive $a$; (2) any $h\in C[0,\infty)$ compactly supported and positive on $(0,T)$ for some positive $T$.
For $G_1$, to produce $w\in L^2[0,\infty)$ with a nonzero $h\in L^2[0,\infty)\cap C[0,\infty)$, both the real part and imaginary part of $h$ have to alternate their signs over $t\in[0,\infty)$.
2.2 The second Green's function is $$G_2(t)=-\frac1{2k}e^{-k|t|}.$$ Then $$w(t)=-\int_0^\infty h(u)G_2(t-u)du+\int_0^\infty h(u)G_2(-u)du=-h*G_2+(h*G_2)(t=0),$$ where $*$ stands for convolution. The second term is to ensure the null initial condition. $G_2\in L^1[0,\infty)$. By Young's convolution inequality, $$\|h*G\|_2\le \|h\|_2\|G_2\|_1,$$ and $w\in L^2[0,\infty),\, \forall h\in L^2[0,\infty)$.
Note: As indicated before, $G_1-G_2=\frac{e^{kt}}k$ a solution of the homogeneous equation.
Separating out the "explosive" exponentially growing term $e^{kt}$ from $\sinh(kt)$ of $G_1(t)$, we have the following proposition regarding the square integrability of $w$.
Proof: Take the Laplace transformation of $w[h,a]$ \begin{align} \mathscr L[w]&=\mathscr L[h]\mathscr L[e^{kt}]-a\mathscr L[e^{kt}], \tag1\\ \mathscr L[e^{kt}](s)&=\frac1{s-k}. \end{align} $\mathscr L[h]\in H^{2+}$ the Hardy function space on the right half complex plane iff $h\in L^2[0,\infty)$. Let $$\mathscr L[h](s)=\frac{s-k}{(s+\alpha)(s+\beta)},$$ or equivalently $$h(t)=\frac{(\beta+k)e^{-\beta t}-(\alpha+k)e^{-\alpha t}}{\beta-\alpha},\tag2$$ for some $\alpha,\beta\in\Bbb C$ where $\mathbf{Re}(\alpha)>0,\mathbf{Re}(\beta)>0$. $\mathscr L[h]\in H^{2+}$ since it is a proper fractional function and its poles are all in the left half complex plane. By Eq. (1), there is always a simple pole at $k$ in the right half complex plane so long as $a\ne0$. Then $\mathscr L\big[w[h,a]\big]\notin H^{2+}$ and $w[h,a]\notin L^2[0,\infty),\,\forall a\ne0$. This can also be verified by direct computation of $w$ with the chosen $h$ with the explicit expression Eq. (2). $\quad\square$
Proof: Take the Laplace transformation of $w[h,a]$ \begin{align} \mathscr L[w]&=\mathscr L[h]\mathscr L[e^{kt}]-a\mathscr L[e^{kt}], \tag1\\ \mathscr L[e^{kt}](s)&=\frac1{s-k}. \end{align} $\mathscr L[h]\in H^{2+}$ the Hardy function space on the right half complex plane iff $h\in L^2[0,\infty)$. Let $D(\omega;R)$ be the closed disk centered at $\omega$ and radius $0<R<\mathbf{Re}(k)$. For an arbitrary $x\ge 0$, let $\Omega_x:=\{x+iy\}\setminus D(\omega-k,R)$. We consider two cases.
(1)$k$ is not one of the zeros.
$$\frac{\mathscr L[h]}{s-k}=\frac{\mathscr L[h](s)-\mathscr L[h](k)}{s-k}+\frac{\mathscr L[h](k)}{s-k}$$ $\phi(s):=\frac{\mathscr L[h](s)-\mathscr L[h](k)}{s-k}$. $\phi(s)$ is holomorphic on the closed right half complex plane.
We will prove that $\phi(s)$ is square integrable and uniformly bounded along all vertical line on the right hand side complex plane. $|\phi(s)|$ has a maximum on the compact $D(k,R)$.
$$|\phi(s)|^2\le \frac{|\mathscr L[h](s)|^2}{R^2}+\frac{|\mathscr L[h](k)|^2}{|s-k|^2},\ \forall |s-k|\ge R.$$ We have $$\int_{\Omega_x} |\phi(x+iy)|^2 dy\le \frac1{R^2}\int_{\Omega_x} |\mathscr L[h](x+iy)|^2dy+\frac{\pi |\mathscr L[h](k)|^2}{2\,\max(x,R)}.$$ The integral on the right hand side is bounded uniformly over all $x\ge0$ as $\mathscr L[h]\in H^{2+}$. So now too is the left hand side. We conclude $\phi\in H^{2+}$.
Then setting and only setting $a=\mathscr L[h](k)$ leads to the desired result.
(2) $k$ is one of the zeros.
$\phi(s):=\frac{\mathscr L[h](s)}{s-k}$. $\frac{\mathscr L[h]}{s-k}\in H^{2+}$. Again $\phi$ is holomorphic on $B(k;R)$.
We will prove that $\phi(s)$ is square integrable and uniformly bounded along all vertical line on the right hand side complex plane. $|\phi(s)|$ has a maximum on the compact $D(k,R)$. $$\int_{\Omega_x} |\phi(x+iy)|^2 dy\le \frac1{R^2}\int_{\Omega_x} |\mathscr L[h](x+iy)|^2dy.$$ The integral on the right hand side is bounded uniformly over all $x\ge0$ as $\mathscr L[h]\in H^{2+}$. Then so is the left hand side. We conclude $\phi\in H^{2+}$.
Setting and only setting $a=0$ leads to the desired result.
$\quad\square$