Well-posedness of IVPs

This is a study note on the IVPs from Kreiss and Lorenz [3].

Introduction

We consider the well-posedness of the initial value problems (IVPs) in the form $$u_t = P(\partial /\partial x)u = \sum_{|\nu|\leq m} A_\nu\frac{\partial ^{|\nu|}}{\partial x_1^{\nu_1}\cdots \partial x_d^{\nu_d}} u,\tag{1}\\ u(x,0) = f(x),$$ where $x \in D := \mathbb{R}^d$ and $u(x,t):D \times (0,\infty )\mapsto \mathbb{R}$ is the unknown. The initial function $f(x)$ is smooth enough and has compact support over the domain. We write $\|\cdot \|=\|\cdot \|_{L^2(D)}$ for short.

We consider the well-posedness of the IVP (1). First, we should clarify what is well-posedness, generally speaking, it states the solution $u$ should continuously depend on the initial data $u(x,0)$, that is there exist constants $C>0,\alpha>0$ such that for any $t>0$ $$\|u(x,t)\| \leq C e^{\alpha t} \|u(x,0)\|.\tag{2}$$

Solution of the IVPs

Fourier transform is a powerful tool to solve PDEs, see [1, 3] for an introduction. Assume $\hat{f}(\omega)= [\mathcal{F}f](\omega)$, then $$\hat{f}(\omega) = \frac{1}{(2\pi)^{d/2}} \int_{D} e^{-i \omega \cdot x}f(x)dx,\\ f(x) = \frac{1}{(2\pi)^{d/2}} \int_{D} e^{i \omega \cdot x}\hat{f}(\omega)d \omega.$$

Suppose initial value is some given mode, dented by $e^{i \omega \cdot x}\hat{f}(\omega)$, then the solution $u$ to (1) can be represented by $$u(x,t) = e^{i \omega \cdot x}e^{P(i \omega)t}\hat{f}(\omega),$$ where $e^{P(i \omega)t}$ is nothing but the Fourier transform of differential operator $P(\partial /\partial x)$. The principle of superposition allows us to obtain the exact solution to (1): $$u(x,t) = \frac{1}{(2\pi)^{d/2}} \int_{D} e^{i \omega \cdot x}e^{P(i \omega)t}\hat{f}(\omega) d \omega. \tag{3}$$

Well-posedness

If there exist constants $C>0,\alpha>0$ such that $$|e^{P(i \omega)t}| \leq C e^{\alpha t}, \quad \forall \omega \in D, t>0, \tag{4}$$ then we from Parseval’s identity obtain $$\|u(x,t)\|=\|e^{P(i \omega)t}\hat{f}(\omega)\| \leq C e^{\alpha t} \|\hat{f}(\omega)\| = C e^{\alpha t} \|f(x)\|.\tag{5}$$ It can be shown that the converse is also true. That is (4) is a necessary and sufficient condition for stability estimate (5). To see it, we need to show given $\|u(x,t)\|\leq C e^{\alpha t} \|f(x)\|$, there holds (4). We give full proof in the following.

Begin of the proof
To show (4) is to show the matrix operator $e^{P(i \omega)t}$ is bounded from above. General proof would start with the definition of an operator $T$: $\sup _{v} \frac{|Tv|}{|v|}=\gamma$. However, one could also show by choosing a special sequence $v_j$ such that $$(1-\epsilon_j) |T| |v_j| \leq \gamma |v_j|,$$ which also implies $|T|\leq \gamma$.

Another technique is that $\|\hat{f}(\omega)\|$ is continuous for any $\omega \in D$.we need to show $|e^{P(i \omega_0)t}| \leq C e^{\alpha t}$, where $C,\alpha$ are independent of $\omega_0$. Using (5) we hope $$\begin{split} C e^{\alpha t} \|f_j(x)\| &\geq \|u_j\| \quad (\text{from }(5))\\ &= \|e^{P(i \omega)t}\hat{f}_j\|\quad (\text{Parseval})\\ &= (1-\epsilon_j) e^{P(i \omega_0)t} \|\hat{f}_j\|\quad (\text{? }). \end{split}$$

To the end of the last step, we let $\omega_0$ be fixed, and from the continuity of $\|\cdot \|$, for any $\epsilon_j>0$ there exists $\delta_j$ s.t. $$\|e^{P(i \omega)t}\hat{f}_j\|\geq (1-\epsilon_j) \|e^{P(i \omega_0)t}\hat{f}_j(\omega)\|=(1-\epsilon_j) |e^{P(i \omega_0)t}|\|\hat{f}_j(\omega)\|,$$ where $\hat{f}_j(\omega)=v$ for $|\omega-\omega_0|<\delta_j$ and zero otherwise; and $v \in \mathbb{C}^n$ with $|v|=1$ satisfying $\|e^{P(i \omega_0)t}\|=\|e^{P(i \omega_0)t}v\|$.
end of the proof

Generalization to $L^2$ space

In the last section, we consider the initial data $f$ of interest to be especially smooth, however, this can be extended to $L^2$ space using density argument.

Let $f_j(x)\to f(x)$ in $L^2$-norm, and from the estimate (5) we know there are solutions $u_j(x,t)$ satisfy $$\|u_i-u_j\| \leq C e^{\alpha t} \|f_i-f_j\| \to 0,$$ which from weak convergence implies there is a solution $u \in L^2(D)$ such that $\|u_i-u\|\to 0$. Such solution $u(x, t)$ in $L^2$-space is called the generalized solution of IVPs (1).

We give an example in [3, Section 2.2.5] that illustrates the process.

Example 1. For one-dimensional differential equation $u_t+u_x=0$ with initial value $u(x,0)=f(x) = 1$ for $-1< x< 1$ and $f(x)=0$ otherwise. We construct a sequence $\left\{f_j \right\}$ that approximates $f$ by $$f_j(x) = \frac{1}{\sqrt{2}}\int_{-\infty}^{\infty}e^{i \omega x}\hat{f}_j(\omega)d \omega,$$ where $\hat{f}_j=\hat{f}$ for $-j<\omega< j$ and zero otherwise. Then the solution with initial data $f_j$ would be $$u_j(x,t) = \frac{1}{\sqrt{2}}\int_{-\infty }^{\infty} e^{i \omega x}e^{-i \omega t}\hat{f}_j(\omega)d \omega =f_j(x-t).$$ Take the limit and obtain $u=f(x-t)$, which is not smoot but still belongs to $L^2(\mathbb{R})$.

References

[1] Fourier 变换和 Laplace 变换.
[2] A short introduction to matrix exponentials. Numanal.com.
[3] H.-O. Kreiss and J. Lorenz, Initial-Boundary Value Problems and the Navier-Stokes Equations. Society for Industrial and Applied Mathematics, 2004. doi: 10.1137/1.9780898719130.