In a previous post we studied the problem \begin{equation} i \partial_t \phi(t,x) + \partial^2_{xx}\phi(t,x) = 0 \label{maineq} \end{equation} where the unknown function $\phi: \mathbb{R}\times\mathbb{R}\to \mathbb{C}$, subject to the initial condition \begin{equation} \phi(0,x) = H(x) e^{i k x} \label{initdata} \end{equation} where $H(x)$ is the Heaviside function that vanishes on the negative real line and equals 1 on the positive real line.

There we gave the explicit solution of the equation in terms of the complex complementary error function, and used oscillatory integration to justify the formula, as well as the asymptotic convergence of the solutions to plane waves. In this post, we will do the same but using Laplace transform methods.

## Table of Contents

## Solving the free Schrödinger equation

The equation $\eqref{maineq}$ has constant coefficients, and formally we can treat it by taking its *temporal* Laplace transform.

Denoting by
\[
\hat{\phi}(p,x) = \int_0^\infty e^{-pt} \phi(t,x) ~\mathrm{d}t,
\]
and assuming that the solution $\phi$ grows (point-wise) slower than exponentially, this integral converges.
If we integrate by parts we see that
\[
\widehat{\partial_t\phi}(p,x) = \int_0^\infty e^{-pt} \partial_t\phi(t,x)~\mathrm{d}t = -\phi(0,x) + p \hat{\phi}(p,x).
\]
Here we used the assumption that $\lim_{t\to\infty} e^{-pt} \phi(t,x) = 0$. This means that we can rewrite
Schrödinger's equation as
\begin{equation}
i p \hat{\phi}(p,x) + \partial^2_{xx} \hat{\phi}(p,x) = i \phi(0,x).
\end{equation}
Now, for each fixed $p$ this is an *ordinary* differential equation in the space variable $x$.
This, in general, can be solved by variation of parameters^{1}.
The solution, however, is not unique: one can add or subtract multiples of the homogeneous solutions $\exp( \pm (1-i) \sqrt{p/2}~x)$.
Notice that for generic solutions, the fact that
\[ \sqrt{-ip} = \pm (1-i) \sqrt{\frac{p}{2}} \]
has non-zero real part implies that it will grow exponentially as $x$ grows toward at least one of $\pm\infty$ (which would make it no longer a tempered distribution).
When $\phi(0,x)$ is a tempered distribution (in particular, if it is a bounded continuous function), there exists a unique solution $\hat{\phi}$ for each $p$ in the space $\mathcal{S}'$. This can be represented again as a Fourier multiplication: let $M_p$ denote the multiplication by the function $-\frac{1}{p + i\xi^2}$ (notice that this function is smooth in $\xi$ and bounded when $p$ is real and non-zero); then the operator $\mathcal{F}_{\text{spatial}}^{-1} M_p \mathcal{F}_{\text{spatial}}$ maps $\phi(0,x)$ to the solution $\hat{\phi}(p,x)$. Here $\mathcal{F}_{\text{spatial}}$ refers to the spatial Fourier transform.

When $p = 0$, on the other hand, notice that the solutions to the homogeneous equation is given by linear functions, which are also tempered distributions, so we don't run into the issue with exponential growth at infinity there.
This moreover implies that the solution of the ODE is *non-unique* in $\mathcal{S}'$.

## Recovering the solution, case of Schwartz data

When the data is in $\mathcal{S}$, we note that the function corresponding to the multiplier $M_p$ is certainly such that all its derivatives have at most polynomial growth. Therefore they give continuous linear maps both of $\mathcal{S}$ to itself, and of $\mathcal{S}'$ to itself. It turns out we have good representation formulas for $\hat{\phi}(p,x)$. Given $p > 0$, denote by $\omega = \frac{1}{\sqrt{2}}(1-i)$, so that $\omega^2 = -i$. We can define \begin{equation} g_p(x) = -\frac{1}{2\omega \sqrt{p}} \cdot \begin{cases} \exp( \omega \sqrt{p} x) & x < 0 \newline \exp(-\omega \sqrt{p} x) & x > 0 \end{cases} \end{equation} Notice that $g_p$ is Lipschitz continuous and decays exponentially as $x\to \pm\infty$. And it is a simple exercise to show that, in the sense of distributions, \[ (\partial^2_{xx} + ip) g_p = \delta_0.\] This allows us to write \begin{equation} \hat{\phi}(p,\cdot) = -i g_p * \phi(0,\cdot). \end{equation}

For fixed $x$, notice that this function $g_p$ is **analytic in** $p$, except for the branch cut for $p\mapsto \sqrt{p}$ on on the complex plane, and the singularity from $1/ \sqrt{p}$ at $p = 0$. We shall for convenience take the cut to be along the negative real line.
However, when $p$ is in the left half plane we see that $g_p$ can grow exponentially, especially when $p$ is in the third quadrant.
For convenience we will assume that $\phi(0,x)$ has compact support in this section; then the convolution $g_p * \phi(0,\cdot)$ extends analytically for all $p$ except for the negative real line.

The solution $\phi(t,x)$ can be recovered by the **inverse Laplace transform** which is given in terms of a complex line integral
\begin{equation}\label{invlaptrans}
\phi(t,x) = \frac{1}{2\pi i} \lim_{T\to\infty} \int_{\gamma - i T}^{\gamma + iT} e^{pt} \hat{\phi}(p,x) ~\mathrm{d}p
\end{equation}
for some $\gamma > 0$.
Using contour integration, and that for $p$ with negative real parts, the semi-circle at infinity contributes nothing to the contour, we see that $\phi(t,x)$ is given by the integral along the two sides of the branch cut, $\mathbb{R}_- \pm i\epsilon$.
\begin{equation}
2\pi i \phi(t,x) = \lim_{\epsilon \to 0} \int_{-\infty + i \epsilon}^{0 + i \epsilon} e^{pt} \hat{\phi}(p,x) ~\mathrm{d}p +
\int_{0-i\epsilon}^{-\infty - i\epsilon} e^{pt} \hat{\phi}(p,x) ~\mathrm{d}p.
\end{equation}
With our choice of branch cut, we have that (schematically)
\[
\sqrt{p} = \begin{cases}
i \sqrt{|p|} & p \in \mathbb{R}_- + i\epsilon \newline
-i \sqrt{|p|} & p \in \mathbb{R}_- - i\epsilon
\end{cases}
\]
So if we parameterize $\sqrt{|p|}$ by $\alpha$, the integrals become
\[
2\pi i \phi(t,x) = \int_{\infty}^0 \int_{\mathbb{R}} e^{-\alpha^2 t} \frac{1}{\omega} e^{-\omega i \alpha |x-y|} \phi(0,y) ~\mathrm{d}y ~\mathrm{d}\alpha - \int_0^{\infty} \int_{\mathbb{R}} e^{-\alpha^2 t} \frac{1}{\omega} e^{\omega i \alpha|x-y|} \phi(0,y) ~\mathrm{d}y ~\mathrm{d}\alpha .
\]
Which we can rewrite as
\begin{equation}
\phi(t,x) = \frac{1}{2\pi i \omega} \iint e^{-\alpha^2 t} e^{\omega i \alpha |x-y|} \phi(0,y) ~\mathrm{d}y ~\mathrm{d}\alpha.
\end{equation}
The compact support of $\phi(0,y)$ guarantees that the integral in $y$ converges, and the integral itself grows (for fixed $x$) at most exponentially in $\alpha$. Therefore the integral in $\alpha$ also converges.
By the stationary phase argument to leading order
\begin{equation} \phi(t,x) \propto \frac{1}{\sqrt{t}} \int \phi(0,y) ~\mathrm{d}y \end{equation}
which agree with what we found using oscillatory integration techniques on the fundamental solution during the previous post on this subject.

(Notice that when $\int \phi(0,y) ~\mathrm{d}y = 0$, for small $\alpha$ \[ \int e^{\omega i \alpha |x-y|} \phi(0,y) ~\mathrm{d}y \approx \omega i \alpha \int |x-y| \phi(0,y) ~\mathrm{d}y \] which will give that the leading order decay of the solution is bounded by \[ \frac{1}{\sqrt{t}} \frac{|x|}{t} \] with the constant of proportionality related to the first moment of $\phi(0,y)$. )

## Our special data

Now let's try to extend the analysis to see what happens with our special data, using again the Laplace transform techniques.
The estimates on the fundamental solution may now be unreliable due to convergence issues.
Instead, we will solve explicitly for (the analytic continuation of) the function $\hat{\phi}(p,x)$ and apply to it the inverse Laplace transform.
Our goal is then first to solve
\[ i p \hat{\phi}(p,x) + \partial^2_{xx} \hat{\phi}(p,x) = i H_k(x) \]
where $H_k(x) = H(x) e^{ikx}$. As the inhomogeneity is discontinuous, we cannot expect the solution to be twice continuously differentiable.
We will find a *continuous differentiable* solution that is, hopefully, sufficiently bounded to qualify as a tempered distribution.

### Left side

When $x < 0$, the equation reduces to the homogeneous linear equation $ip \hat{\phi}(p,x) + \partial^2_{xx} \hat{\phi}(p,x) = 0$. Solutions to this equation are spanned by \[ \exp(\pm \omega \sqrt{p} x) \] The case with $-\omega$ coefficient blows up exponentially as $x\searrow -\infty$, and so we rule it out (since we are looking for tempered distribution solutions). We therefore conclude that the admissible solution when $p > 0$ is the solution \begin{equation} \hat{\phi}(p,x) |_{x < 0} = C_- \exp(\omega\sqrt{p} x). \end{equation}

### Right side

When $x > 0$, the equation is the forced oscillator \[ ip \hat{\phi}(p,x) + \partial^2_{xx}\hat{\phi}(p,x) = i e^{ikx}.\] The particular solution is given by $A e^{ikx}$, which a direct calculation shows we need \[ A (ip - k^2) = i \implies A = (p + ik^2)^{-1} .\] We can also add to this a solution to the homogeneous equation, but we require the exponential decaying one which is $\exp(-\omega \sqrt{p} x)$. Putting it all together we see that the solution is given by, when $p > 0$, \begin{equation} \hat{\phi}(p,x)|_{x > 0} = C_+ \exp(-\omega \sqrt{p} x) + \frac{1}{p + ik^2} e^{ikx} . \end{equation}

### Altogether now

The coefficients $C_\pm$ can be solved by requiring the solution be $C^1$. This means \[ C_- = C_+ + \frac{1}{p + ik^2} \] and \[ (C_- + C_+) \omega \sqrt{p} = \frac{ik}{p + ik^2} .\] These can be solved with \begin{equation} C_\pm = \frac12 \frac{i}{ip-k^2} \left( \frac{ik}{\omega \sqrt{p}} \mp 1 \right). \end{equation} Noting that $\omega^2 = -i$, we have that $ip - k^2 = (ik + \omega \sqrt{p})(ik - \omega \sqrt{p})$, and finally arrive at \begin{equation}\label{cpmexpr} C_\pm = \frac{1}{2\omega \sqrt{p}} \frac{i}{ik \pm \omega \sqrt{p}} . \end{equation} Our solution \begin{equation}\label{solnpw} \hat{\phi}(p,x) = \begin{cases} C_- e^{\omega \sqrt{p} x} & x < 0 \newline C_+ e^{-\omega \sqrt{p} x} + \frac{1}{p + ik^2} e^{ikx} & x > 0 \end{cases} \end{equation}

## Asymptotic behavior

### $k < 0$

Using the explicit formula for $\eqref{cpmexpr}$, we can extend the expression $\eqref{solnpw}$ from $p$ being a positive real number to as much of the complex plane as we can manage. We see that the obstacles are

- The branch cut (and associated blow-up at zero) given by the expression $\sqrt{p}$, which as before we will take to happen along the negative real line.
- Poles associated to $ik \pm \omega \sqrt{p} = 0$ and $p + ik^2 = 0$ in the coefficients.

Let us treat the latter first. By our choice of the branch cut, $\sqrt{p}$ always have positive real part. Multiplication by $\omega$ is a clockwise rotation by $\pi/4$ radians.
Therefore we see that $C_+$ has *no poles*, while $C_-$ and $(p+ik^2)^{-1}$ have poles when $p = -i k^2$ on the negative real axis.
We note that at the pole, $\omega\sqrt{p} = i k$.

Performing the contour integration for the inverse Laplace transform $\eqref{invlaptrans}$, we see that when $x < 0$ the pole gives a residue precisely $e^{pt} \hat{\phi}(p,x) |_{p = -ik^2}$, which simplifies to $e^{-ik^2t + ikx}$. The branch cut, having the $1/\sqrt{p}$ coefficient, we can treat exactly as we did in the Schwartz case: by method of stationary phase we conclude for fixed $x$ the branch cut contributes order $O(t^{-1/2})$.

Similarly, for $x > 0$, the term with the $C_+$ coefficient only sees the branch cut, and contributes a term of order $O(t^{-1/2})$. The pole in the remaining term again contributes a lingering term of $e^{-ik^2 t + ikx}$, exactly as expected.

(Here we see a general rule for the Laplace transform method: poles of $\hat{\phi}(p,x)$ in $p$ along the imaginary axis give rise to non-decaying states, while the branch-cut controls the rate of decay toward the stationary states.)

### $k > 0$

When $k > 0$, the analysis is similar, again the pole can only happen when $p = -ik^2$. But now we have $\omega\sqrt{p} = -ik$. We see that $C_-$ is now regular except for the branch cut. Which following the same analysis as above shows that when $x < 0$, the solution decays to $0$ at rate $O(t^{-1/2})$.

What happens when $x > 0$? Here we have to notice that the singularities of $C_+ e^{-\omega \sqrt{p} x}$ and $\frac{1}{p + ik^2} e^{ikx}$ *exactly cancel each other*. Indeed, when $x > 0$ we can rewrite
\[ \hat{\phi}(p,x) = \frac{1}{p + ik^2} \left( e^{ikx} - e^{-\omega \sqrt{p} x}\right) + \frac12 \frac{1}{p + ik^2} \left( \frac{ik}{\omega\sqrt{p}} + 1 \right) e^{-\omega\sqrt{p} x} .\]
In both of the terms, the expressions inside the parentheses are holomorphic in a neighborhood of $p = -ik^2$ and evaluates to zero there. And hence $\hat{\phi}(p,x)$ does not actually have a pole at $p = -ik^2$, and so the solution decays with rate $O(t^{-1/2})$ to the 0 solution, again in agreement with what we've found before.

### $k = 0$

In this case the expression $\eqref{cpmexpr}$ simplifies to read \[ C_\pm = \mp \frac{1}{2p}.\] It is best to rewrite $\hat{\phi}(p,x)$ as follows. When $x < 0$, \[ \hat{\phi}(p,x) = \frac{1}{2p} \left( e^{\omega \sqrt{p} x} - 1 \right) + \frac{1}{2p} .\] And when $x > 0$ \[ \hat{\phi}(p,x) = \frac{1}{2p} + \frac{1}{2p} \left( 1 - e^{-\omega \sqrt{p} x} \right) .\] In both cases the terms within the parentheses vanish at the origin.

When performing the contour integration, the two obvious poles each contribute the value $1/2$ to the integral (one when $x > 0$ and one when $x < 0$), exactly as expected.

Let's explicitly treat the remaining integral to see its decay rate. We will again set $\alpha = \sqrt{|p|}$ and integrate around the branch cut. First consider when $x < 0$. The integral can be written as \[ -\int_{\infty}^0 e^{-\alpha^2 t} \frac{1}{2\alpha^2} \left( e^{\omega i \alpha x} - 1 \right) ~2\alpha ~\mathrm{d}\alpha - \int_0^\infty e^{-\alpha^2 t} \frac{1}{2\alpha^2} \left( e^{-\omega i \alpha x} - 1 \right) ~2\alpha ~\mathrm{d}\alpha .\] This we can rewrite as \[ -\int_{-\infty}^\infty e^{-\alpha^2 t} \frac{1}{\alpha} \left( e^{-\omega i \alpha x} - 1 \right) ~\mathrm{d}\alpha .\] Noting that the function \[ \alpha \mapsto \frac{1}{\alpha} \left( e^{-\omega i \alpha x} - 1 \right) \] is smooth with value $-i\omega x$ at $\alpha = 0$, we see that from stationary phase the integral will decay asymptotically as $O(x / \sqrt{t})$.

The same holds for $x > 0$ via a similar argument.

Note that the $O(x/\sqrt{t})$ here is not necessarily comparable to the $O(t^{-1/2})$ in the previous two cases, *except* when $x = 0$. Here of course we see that $\hat{\phi}(p,0) = \frac{1}{2p}$ identically, and so the solution $\phi(t,0) = \frac12$ always, exactly in agreement with what we saw previously.

- In the PDE context, the method of variation of parameters is also known as Duhamel's principle.
^{^}