The problem that will concern us for the next two parts will be the initial value problem $$\begin{cases} \tag{\#} -y^{\prime\prime} + q(x)y = \lambda y \text{ for $x\in[0,1]$,} \\ y(0) = a, y^{\prime}(0) = b. \end{cases}$$ Here $q(x)$ lies in $L^2_\mathbb{C}([0,1])$, which we will recall to be the space of complex-valued square-integrable functions on $[0,1]$. Also, $\lambda, a, b$ are complex numbers. In the future, we will only concern ourselves with real-valued $q$, but it is very convenient to work with complex-valued $q$ now, as it will let us avail ourselves of complex-analytic tools.
Before we begin, those well-acquainted with measure theory will immediately recognise that $q\in L^2$ is only defined almost everywhere, and need not be continuous. Therefore, we need to clarify what we exactly do we mean by a solution to (#).
Definition 1. A continuously differentiable function $y$ on $[0,1]$ is said to be a solution to (#) if the following holds:
- $y^\prime$ is absolutely continuous on $[0,1]$ (i.e. there exists an integrable function $g$ such that $y^\prime(x) = y^\prime(0) + \int_0^x g(t)\,dt$),
- (#) is satisfied almost everywhere, with $g$ playing the role of $y^{\prime\prime}$.
With that out of the way, let us look into solving this problem. Certainly, $y$ will depend not just on $x$, but also on $\lambda$ and $q$. We will use the celebrated method of power series, expanding $y$ as a power series of $q$ and solving it.
Perhaps it will be instructive to first consider a simpler example. This will serve to illustrate the flavour of the method. Let us consider the initial value problem $$\begin{cases} -u^{\prime\prime} = \lambda u \text{ for $x\in[0,1]$,} \\ u(0) = 1, u^{\prime}(0) = 0. \end{cases}$$ To derive the solution, we will assume that $u$ takes the form $$u(x, \lambda) = \sum_{n=0}^\infty u_n(x)\lambda^n.$$
Setting first $\lambda = 0$, we obtain $$\color{red}\begin{cases} -u_0^{\prime\prime} = 0 \text{ for $x\in[0,1]$,} \\ u_0(0) = 1, u^{\prime}_0(0) = 0. \end{cases}$$
Integrating twice, we obtain $u_0\equiv 1$. Now plugging in the power series, and comparing like powers of $\lambda$ will yield, for $n\geq 1$, $$\color{red}\begin{cases} -u_n^{\prime\prime} = u_{n-1} \text{ for $x\in[0,1]$,} \\ u_n(0) = 0, u^{\prime}_n(0) = 0. \end{cases}$$
For instance, since $-u_1^{\prime\prime}=u_0=1$ and $u_1(0)=u_1^\prime(0)=0$, integrating twice will yield $u_1(x) = -\frac{x^2}{2}$. In fact, a simple induction yields $$u_n(x) = \frac{(-1)^n x^{2n}}{(2n)!},$$ so that $$ u(x, \lambda) = u_0(x) + \sum_{n\geq 1}u_n(x)\lambda^n = 1 + \sum_{n\geq 1}\frac{(-1)^n x^{2n}\lambda^n}{(2n)!} = \cos(\sqrt{\lambda} x). $$ That $\lambda\in \mathbb{C}$ will not hamper us, as any choice of square root for $\lambda$ yields the same function $\cos(\sqrt{\lambda} x)$. We will just fix a branch of square root. Either way, one can check that $u(x,\lambda) = \cos(\sqrt{\lambda} x)$ really solves our ODE, exactly as we desired.
The key step in the power series method is solving the two simpler initial value problems in red. When we solve our actual problem (#), we will also have to deal with analogous problems. To that end, we will use the following lemma:
Lemma 1. Let $f\in L^2_\mathbb{C}$, $a,b\in\mathbb{C}$. Then the unique solution to $$\begin{cases} -y^{\prime\prime} = \lambda y - f(x) \text{ for $x\in[0,1]$,} \\ y(0) = a, y^{\prime}(0) = b \end{cases}$$ is given by $$y(x) = a\cos(\sqrt{\lambda}x) + b\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}} + \int_0^x\frac{\sin(\sqrt{\lambda}(x-t))}{\sqrt{\lambda}}f(t)\,dt.$$
To maintain some sort of sanity for the typist we will write $$c_\lambda(x) = \cos(\sqrt{\lambda}x),\qquad s_{\lambda}(x) = \frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}.$$
Proof. To demonstrate uniqueness, we consider two solutions $y$ and $\tilde{y}$ to the ODE, and note that their difference $v:=y-\tilde{y}$ satisfies $$\begin{cases} -v^{\prime\prime} = \lambda v \text{ for $x\in[0,1]$,} \\ v(0) = 0, v^{\prime}(0) = 0. \end{cases}$$ Then $v=0$, so $y=\tilde{y}$.
To demonstrate existence, we will do high school trigonometry (the addition rule for sine and cosine). Consider the integral
$$ \begin{align*} y_f(x) &= \int_0^x s_\lambda(x-t)f(t)\,dt \\&= s_\lambda(x)\int_0^x c_\lambda(t)f(t)\,dt - c_\lambda(x)\int_0^x s_\lambda(t)f(t)\,dt. \end{align*} $$ Since $f\in L^2_\mathbb{C}$, both integrals are continuous in $x$, and the whole expression is continuous; in fact, $y_f$ is absolutely continuous with derivative $$ \begin{align*} y_f^\prime(x) = c_\lambda(x)\int_0^x c_\lambda(t)f(t)\,dt +\lambda s_\lambda(x)\int_0^x s_\lambda(t)f(t)\,dt. \end{align*} $$ Likewise, $y_f^\prime$ is absolutely continuous with derivative $$ \begin{align*} y_f^{\prime\prime}(x) &= -\lambda s_\lambda(x)\int_0^x c_\lambda(t)f(t)\,dt + c_\lambda(x)^2f(x) \\&\qquad + \lambda c_\lambda(x)\int_0^x s_\lambda(t)f(t)\,dt + \lambda s_\lambda(x)^2f(x) \\&= f(x) - \lambda y_f(x). \end{align*} $$ Together with the fact that $y_f(0) = y_f^\prime(0) = 0$, as well as $$ \begin{cases} -c_\lambda^{\prime\prime}(x) = \lambda c_\lambda(x) \text{ for $x\in[0,1]$,} \\ c_\lambda(0) = 1, c_\lambda^{\prime}(0) = 0, \end{cases} \quad \begin{cases} -s_\lambda^{\prime\prime}(x) = \lambda s_\lambda(x) \text{ for $x\in[0,1]$,} \\ s_\lambda(0) = 0, s_\lambda^{\prime}(0) = 1, \end{cases} $$ our claim follows. $\square$
Now let us attack (#) proper. Actually, we will cheat and look at two special instances: $$\begin{cases} \tag{\#1} -y_1^{\prime\prime} + q(x)y_1 = \lambda y_1 \text{ for $x\in[0,1]$,} \\ y_1(0) = 1, y_1^{\prime}(0) = 0, \end{cases}$$ $$\begin{cases} \tag{\#2} -y_2^{\prime\prime} + q(x)y_2 = \lambda y_2 \text{ for $x\in[0,1]$,} \\ y_2(0) = 0, y_2^{\prime}(0) = 2, \end{cases}$$ Later, we will show that for general initial values we can write the solution in terms of $y_1$ and $y_2$. We will work with $y_1$ first, writing $$y_1(x, \lambda, q) = C_0(x, \lambda) + \sum_{n\geq 1} C_n(x, \lambda, q),$$ where $C_n(x, \lambda, q):= C_n(x, \lambda, q_1, \dots, q_n)|_{q_1=\dots=q_n=q}$ is bounded, symmetric and multilinear in the $q_i$. Setting $q=0$, we obtain $$ \begin{cases} -C_0^{\prime\prime}(x, \lambda) = \lambda C_0(x, \lambda) \text{ for $x\in[0,1]$,} \\ C_0(0, \lambda) = 1, C_0^\prime(0, \lambda) = 0, \end{cases} $$
so $C_0(x,\lambda) = c_\lambda(x)$. Now, plugging the series into (#1) and comparing terms of equal degree yields $$ -C_n^{\prime\prime}(x, \lambda, q) = \lambda C_n(x, \lambda, q) - qC_{n-1}(x, \lambda, q), $$ for each $n\geq 1$. The initial conditions require a little cleverness to determine. Putting the series into the given initial conditions gives us $$ \sum_{n\geq 1} C_n(0, \lambda, q) = \sum_{n\geq 1}C_n^\prime(0, \lambda, q) = 0. $$ Replacing $q$ with $tq$ and using multilinearity, we get $$ \sum_{n\geq 1} C_n(0, \lambda, q)t^n = \sum_{n\geq 1}C_n^\prime(0, \lambda, q)t^n = 0. $$ This series vanishes identically in $t$, so all the $C_n(0, \lambda, q)$ and $C_n^\prime(0, \lambda, q)$ must vanish. Therefore $$ C_n(0, \lambda, q) = C_n^\prime(0, \lambda, q) = 0, $$ for each $n\geq 1$. By Lemma 1, we have $$ C_n(x, \lambda, q) = \int_0^x s_\lambda(x - t)q(t)C_{n-1}(t, \lambda, q)\,dt. $$
For instance, $$ \begin{align*} C_1(x, \lambda, q) &= \int_0^x s_\lambda(x-t_1)q(t_1)C_0(t_1, \lambda)\,dt_1 \\&= \int_0^x c_\lambda(t_1)s_\lambda(x-t_1)q(t_1)\,dt_1, \end{align*} $$ and $$ \begin{align*} C_2(x, \lambda, q) &= \int_0^x s_\lambda(x-t_2)q(t_2)C_1(t_2, \lambda, q)\,dt_2 \\&= \int_0^x s_\lambda(x-t_2)q(t_2)\int_0^{t_2}c_\lambda(t_1)s_\lambda(t_2-t_1)q(t_1)\,dt_1\,dt_2 \\&= \int_{0\leq t_1\leq t_2\leq t_3=x} c_\lambda(t_1)\prod_{i=1}^2\left[s_\lambda(t_{i+1}-t_i)q(t_i)\right]\,dt_1\,dt_2. \end{align*} $$
In general, $$ \begin{align*} C_n(x, \lambda, q) &= \int_{0\leq t_1\leq \dots\leq t_{n+1}=x} c_\lambda(t_1)\prod_{i=1}^n\left[s_\lambda(t_{i+1}-t_i)q(t_i)\right]\,dt_1\dots\,dt_n. \end{align*} $$
Thus, we have the (formal) solution for $y_1$. We will leave it as an exercise for the reader (it’s good practice, trust) to verify that repeating this for $y_2$ yields $$y_2(x, \lambda, q) = S_0(x, \lambda) + \sum_{n\geq 1} S_n(x, \lambda, q),$$ with $$S_0(x, \lambda) = s_\lambda(x),$$ and $$ \begin{align*} S_n(x, \lambda, q) &= \int_{0\leq t_1\leq \dots\leq t_{n+1}=x} s_\lambda(t_1)\prod_{i=1}^n\left[s_\lambda(t_{i+1}-t_i)q(t_i)\right]\,dt_1\dots\,dt_n. \end{align*} $$
So far, all our computations have been formal. Our final task for this part will be to show that the series in fact converges to a solution of (#1) and (#2) respectively.
Theorem 1. The formal power series for $y_1$ and $y_2$, given above, converges uniformly on bounded subsets of $[0,1]\times \mathbb{C}\times L^2_\mathbb{C}$ to the unique solutions of (#1) and (#2) respectively. Moreover, they satisfy the integral equations $$ y_1(x, \lambda, q) = \cos(\sqrt{\lambda}x) + \int_0^x\frac{\sin(\sqrt{\lambda}(x-t))}{\sqrt{\lambda}}q(t)y_1(t, \lambda, q)\,dt, $$ $$ y_2(x, \lambda, q) = \frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}} + \int_0^x\frac{\sin(\sqrt{\lambda}(x-t))}{\sqrt{\lambda}}q(t)y_2(t, \lambda, q)\,dt, $$ respectively, along with the estimates $$ |y_1(x,\lambda, q)|, |y_2(x, \lambda, q)| \leq \exp\left(|\operatorname{Im}\sqrt{\lambda}|x + \|q\|_{L^2}\sqrt{x}\right). $$
Proof. We will only do this for $y_1$ as the proof works exactly the same way for $y_2$. Recall that $$|c_\lambda(x)| = \frac{1}{2}\left|e^{i\sqrt{\lambda x}} + e^{-i\sqrt{\lambda}x}\right| \leq \exp\left(|\operatorname{Im}\sqrt{\lambda}|x\right),$$ and for $0\leq x \leq 1$, $$|s_\lambda(x)| = \left|\int_0^x c_\lambda(t)\,dt\right| \leq x\exp\left(|\operatorname{Im}\sqrt{\lambda}|x\right) \leq \exp\left(|\operatorname{Im}\sqrt{\lambda}|x\right). $$ Hence $$ \begin{align*} &|C_n(x, \lambda, q)| \\&\leq \int_{0\leq t_1\leq \dots\leq t_{n+1}=x}|c_\lambda(t_1)|\prod_{i=1}^n|s_\lambda(t_{i+1}-t_i)| |q(t_i)|\,dt_1\dots\,dt_n \\&\leq \int_{0\leq t_1\leq \dots\leq t_{n+1}=x}e^{|\operatorname{Im}\sqrt{\lambda}|t_1}\prod_{i=1}^n| e^{|\operatorname{Im}\sqrt{\lambda}|(t_{i+1}-t_i)} |q(t_i)|\,dt_1\dots\,dt_n \\&\leq \exp(|\operatorname{Im}\sqrt{\lambda}|x)\int_{0\leq t_1\leq \dots\leq t_{n+1}=x} \prod_{i=1}^n |q(t_i)|\,dt_1\dots\,dt_n. \end{align*} $$ Since the product $\prod_{i=1}^n|q(t_i)|$ is invariant under permuting the $t_i$, and furthermore $$ \bigsqcup_{\sigma\in\operatorname{Perm}(\{1,\dots,n\})} \{0\leq t_{\sigma(1)} \leq \dots \leq t_{\sigma(n)} \leq t_{n+1}=x\} = [0,x]^n, $$ we can continue thus: $$ \begin{align*} |C_n(x, \lambda, q)| &\leq \dots \\&= \exp(|\operatorname{Im}\sqrt{\lambda}|x)\frac{1}{n!}\int_{[0,x]^n} \prod_{i=1}^n |q(t_i)|\,dt_1\dots\,dt_n \\&= \exp(|\operatorname{Im}\sqrt{\lambda}|x)\frac{1}{n!}\left(\int_0^x |q(t)|\,dt\right)^n \\&\leq \exp(|\operatorname{Im}\sqrt{\lambda}|x) \|q\|_{L^2}^n x^{n/2}, \end{align*} $$ the last inequality following from Hölder’s inequality. Summing over $n$, we get
$$ \begin{align*} |y_1(x, \lambda, q)| &\leq \exp(|\operatorname{Im}\sqrt{\lambda}| x) + \sum_{n\geq 1}\exp(|\operatorname{Im}\sqrt{\lambda}|x) \|q\|_{L^2}^n x^{n/2} \\&= \exp(|\operatorname{Im}\sqrt{\lambda}|x + \|q\|_{L^2} \sqrt{x}). \end{align*} $$
This establishes local uniform convergence in $[0,1]\times \mathbb{C}\times L^2_\mathbb{C}$. Local uniform convergence allows us to interchange sums and integrals, so the following integral equation will hold:
$$ \begin{align*} y_1(x, \lambda, q) &= c_\lambda(x) + \sum_{n\geq 1}C_n(x, \lambda, q) \\&= c_\lambda(x) + \sum_{n\geq 1}\int_0^x s_\lambda(x-t)q(t) C_{n-1}(t, \lambda, q),dt \\&= c_\lambda(x) + \int_0^x s_\lambda(x-t)q(t) \sum_{n\geq 1}C_{n-1}(t, \lambda, q),dt \\&= c_\lambda(x) + \int_0^x s_\lambda(x-t)q(t) y_1(t, \lambda, q),dt. \end{align*} $$ (Actually, one can prove this just using Lemma 1 too; see below.)
Finally, to prove uniqueness, we will resort to a Grönwall-like argument. If $y_1$ and $\tilde{y_1}$ both satisfy (#1), then their difference $v:=y_1 - \tilde{y_1}$ will satisfy
$$\begin{cases} -v^{\prime\prime} = \lambda v - qv \text{ for $x\in[0,1]$,} \\ v(0) = 0, v^{\prime}(0) = 0, \end{cases}$$
so that Lemma 1 yields $$ v(x) = \int_0^x s_\lambda(x-t) q(t)v(t)\,dt. $$
By Hölder’s inequality ($1 = \frac{1}{\infty} + \frac{1}{2} + \frac{1}{2}$) we get $$ |v(x)|^2 = \left|\int_0^x s_\lambda(x-t) q(t)v(t)\,dt\right|^2 \leq \max_{t\in [0,1]}|s_\lambda(x-t)|^2 \|q\|_{L^2} \int_0^x|v(t)|^2\,dt. $$ Let $c = \max_{t\in [0,1]}|s_\lambda(x-t)|^2 \|q\|_{L^2}$, so that $|v(x)|^2 \leq c\int_0^x |v(t)|^2\,dt$. We note that $$ e^{-cx}\int_0^x|v(t)|^2\,dt = 0 \text{ when $x=0$}, $$ and for all $0\leq x\leq 1$ we have $$ e^{-cx}\int_0^x|v(t)|^2\,dt \geq 0, $$ and $$ \frac{d}{dx} \left(e^{-cx}\int_0^x|v(t)|^2\,dt\right) =e^{-cx}\left(|v(x)|^2 - c\int_0^x|v(t)|^2\,dt \right) \leq 0, $$ so we must have $e^{-cx}\int_0^x |v(t)|^2,dt\equiv 0$, whence $v\equiv 0$, and therefore $y_1 \equiv \tilde{y}$, as desired. $\square$
I noticed that the blog post says this takes eight minutes to read. Surely this is a lie XD