I'm reading Evans' book about PDEs by myself and I'm trying understand the motivation for the definition of weak solutions to parabolic equations of second order on pages $373$ and $374$.

First, we do preliminary assumptions:

$U \subset \mathbb{R}^n$ is an open bounded set and $U_T := U \times (0,T]$ for some fixed $T > 0$.

We will first study the initial/boundary-value problem

$$(1) \begin{cases} u_t + Lu = f \ \text{in} \ U_T\\ u = 0 \ \text{on} \ \partial U \times [0,T]\\ u = g \ \text{on} \ U \times \{ t = 0 \} \end{cases}.$$

The letter $L$ denotes for each time $t$ a second order partial differential operator , having the divergence form

$$(2) \ Lu = - \sum_{i,j=1}^n (a^{ij}(x,t)u_{x_i})_{x_j} + \sum_{i=1}^n b^i u_{x_i} + c(x,t)u.$$

$\textbf{b. Weak solutions.}$ Mimicking the developments in $\S 6.1.2$ for elliptic equations, we consider first the case that $L$ has divergence form $(2)$ and try to find an appropriate notion of weak solution for the initial/boundary-value problem $(1)$. We assume for now that

\begin{align*} (5) \ a^{ij}, b^i, c &\in L^{\infty} (U_T) \ (i,j = 1, \cdots, n)\\ (6) \ f &\in L^2(U_T),\\ (7) \ g &\in L^2(U) \end{align*}

We also suppose $a^{ij} = a^{ji} \ (i,j = 1, \cdots, n)$.

Let us now define, by analogy with the notation introduced in Chapter $6$, the time-dependent bilinear form

$$(8) \ B[u,v;t] := \int_U \sum_{i,j=1}^n a^{ij}(\cdot,t) u_{x_i}v_{x_j} + \sum_{i=1}^n b^i(\cdot, t) u_{x_i} v + c(\cdot,t)uv dx$$

for $u,v \in H^1_0(U)$ and a.e. $0 \leq t \leq T$.

Now, the motivation:

$\textbf{Motivation for definition of weak solution.}$ To make plausible the following definition of weak solution, let us first temporarily suppose that $u = u(x,t)$ is in fact a smooth solution in parabolic setting of our problem $(1)$. We know switch our pointview, by associating with $u$ a mapping

$$\textbf{u}: [0,T] \longrightarrow H^1_0(U)$$

defined by

$$[\textbf{u}(t)](x) := u(x,t) \ (x \in U, 0 \leq t \leq T).$$

In other words, we are going consider $u$ not as a function of $x$ and $t$ together, but rather as a mapping $\textbf{u}$ of $t$ into the space $H^1_0(U)$ of functions of $x$. This point of view will greatly clarify the following presentation.

Returning to the problem $(1)$, let us similary define

$$\textbf{f}: [0,T] \longrightarrow L^2(U)$$


$$[\textbf{f}(t)](x) := f(x,t) \ (x \in U, 0 \leq t \leq T).$$

Then if we fix a function $v \in H^1_0(U)$, we can multiply the PDE $\frac{\partial u}{\partial t} + Lu = f$ by $v$ and integrate by parts to find

$$(9) \ (\textbf{u}',v) + B[\textbf{u},v;t] = (\textbf{f},v) \ \left( ' = \frac{d}{dt} \right)$$

for each $0 \leq t \leq T$, the pairing $(,)$ denoting the inner product of $L^2(U)$. Next, observe that

$$(10) \ u_t = g^0 + \sum_{j=1}^n g^j_{x_j} \ \text{in} \ U_T$$

for $g^0 := f - \sum_{i=1}^n b^i u_{x_i} - cu$ and $g^j := \sum_{i=1}^n a^{ij}u_{x_i}$ $(j = 1, \cdots, n)$. Consequently and the definitions from $\S 5.9.1$ imply the right-hand side of $(10)$ lies in the Sobolev space $H^{-1}(U)$, with

$$||u_t||_{H^{-1}(U)} \leq \left( \sum_{j=1}^n ||g^j||^2_{L^2(U)} \right)^{\frac{1}{2}} \leq C \left( ||u||_{H^1_0(U)} + ||f||_{L^2(U)} \right).$$

This estimate suggests it may be reasonable to look for weak solution with $\textbf{u}' \in H^{-1}(U)$ for a.e $0 \leq t \leq T$ , in which case the first term in $(9)$ can be reexpressed as $\langle \textbf{u}',v \rangle$, $\langle , \rangle$ being the pairing of $H^{-1}(U)$ and $H^1_0(U)$.

My doubt is how the inequalities above are obtained? I can't see how derive them since we not suppose $\textbf{u}' \in H^{-1}(U)$ and, therefore, we can't use the result from $\S 5.9.1$, which is a caracterization of the space $H^{-1}$.

Thanks in advance!


1 Answer 1


The idea of the motivation is to conclude that $\mathbf{u}'\in H^{-1}(U)$ instead of take it as an assumption.

Equality (10) shows that $u_t\in L^2(U)\subset H^{-1}(U)$ and thus the first inequality follows from §5.9.1. Then $\mathbf{u}'(t)\in H^{-1}(U)$ (for each $t$) because $[\mathbf{u}'(t)](x)=u_t(t,x)$.


From the definition of $g^0$, $$\begin{align*} \|g^0\|_{L^2(U)} &= \left\|f - \sum_{i=1}^n b^i u_{x_i} - cu\right\|_{L^2(U)}\\ &\leq\|f\|_{L^2(U)}+\sum_{i=1}^n\|b^i\|_{L^\infty(U)}\|u_{x_i}\|_{L^2(U)}+\|c\|_{L^\infty(U)}\|u\|_{L^2(U)}\\ &\leq \|f\|_{L^2(U)}+c_0\sum_{i=1}^n\|u_{x_i}\|_{L^2(U)}+c_1\|u\|_{L^2(U)}\\ &\leq \|f\|_{L^2(U)}+c_2\|u\|_{H_0^1(U)}\\\end{align*}$$ and thus $$\|g^0\|_{L^2(U)}^2\leq c_3\left(\|f\|_{L^2(U)}+\|u\|_{H_0^1(U)}\right)^2.$$ Analogously, from the definition of $g^j$, $$\begin{align*} \|g^j\|_{L^2(U)}^2 &=\left\| \sum_{i=1}^n a^{ij}u_{x_i}\right\|_{L^2(U)}^2\\ &\leq\left(\sum_{i=1}^n\|a^{ij}\|_{L^\infty}\left\| u_{x_i}\right\|_{L^2(U)}\right)^2\\ &\leq \left(\sum_{i=1}^nc_4\left\| u_{x_i}\right\|_{L^2(U)}\right)^2\\ &\leq c_5\sum_{i=1}^n\left\| u_{x_i}\right\|_{L^2(U)}^2\\ &\leq c_5\| u\|_{H_0^1(U)}^2\\ &\leq c_5(\|f\|_{L^2(U)}+\|u\|_{H_0^1(U)})^2. \end{align*}$$ Therefore, $$\begin{align*} \|u_t\|_{H^{-1}(U)} &\leq \left( \sum_{j=0}^n \|g^j\|^2_{L^2(U)} \right)^{\frac{1}{2}} \\ &\leq \left( \sum_{j=1}^n c_6\left(\|f\|_{L^2(U)}+\|u\|_{H_0^1(U)}\right)^2\right)^{\frac{1}{2}} \\ &\leq C\left(\|f\|_{L^2(U)}+\|u\|_{H_0^1(U)}\right) \end{align*}$$

Edit 2

For the first inequality in your comment, note that $$\begin{align*} & \|u\|_{H_0^1}=\left(\|u\|_{L^2}^2+\|u_{x_1}\|_{L^2}^2+\cdots+\|u_{x_n}\|_{L^2}^2\right)^{1/2}\\ \Longrightarrow\qquad &\|u\|_{H_0^1}^2=\|u\|_{L^2}^2+\|u_{x_1}\|_{L^2}^2+\cdots+\|u_{x_n}\|_{L^2}^2\\ \Longrightarrow\qquad &\|u\|_{H_0^1}^2\geq\|u\|_{L^2}^2\quad\text{and}\quad \|u\|_{H_0^1}^2\geq\|u_{x_j}\|_{L^2}^2\;(j=1,...,n)\\ \Longrightarrow\qquad &\|u\|_{L^2}\leq \|u\|_{H_0^1}\quad\text{and}\quad \|u_{x_j}\|_{L^2}\leq \|u\|_{H_0^1}\;(j=1,...,n)\\ \Longrightarrow\qquad &c_0\sum_{i=1}^n\|u_{x_i}\|_{L^2}+c_1\|u\|_{L^2} \leq c_0 \sum_{i=1}^n\|u\|_{H_0^1}+c_1\|u\|_{H_0^1}=(nc_0+c_1)\|u\|_{H_0^1} \end{align*}$$ For the second, note that (as proved here) if $x_1,...,x_n\geq 0$ and $p>0$, then there exists a constant $k$ (which depends on $p$ and $n$) such that $$(x_1+\cdots+x_n)^p\leq k(x_1^p+\cdots +x_n^p).$$

  • $\begingroup$ Ok, I understand now how can the first inequality was obtained, but how the second inequality was obtained? The closest thing that I could think is use the theorem $3$ of $\S 5.9.2$, but this theorem states that if $\textbf{u} \in L^2(0,T;H^1_0(U))$ with $\textbf{u}' \in L^2(0,T;H^{-1}(U))$, then $$\max_{0 \leq t \leq T} ||\textbf{u}(t)||_{L^2(U)} \leq C \left( ||\textbf{u}||_{L^2(0,T;H^1_0(U))} + ||\textbf{u}'||_{L^2(0,T;H^{-1}(U))} \right),$$ the constant depending only on $T$, but I think this isn't help me since the estimative doesn't have $||f||_{L^2(U)}$. $\endgroup$
    – George
    Mar 5, 2019 at 15:08
  • $\begingroup$ @George For the second inequality, see my edit. $\endgroup$
    – Pedro
    Mar 5, 2019 at 16:28
  • $\begingroup$ I didn't understand the following inequalities in your edit: $$1. \|f\|_{L^2(U)}+c_0\sum_{i=1}^n\|u_{x_i}\|_{L^2(U)}+c_1\|u\|_{L^2(U)} \leq \|f\|_{L^2(U)}+c_2\|u\|_{H_0^1(U)}$$ $$2. \left(\sum_{i=1}^nc_4\left\| u_{x_i}\right\|_{L^2(U)}\right)^2 \leq c_5\sum_{i=1}^n\left\| u_{x_i}\right\|_{L^2(U)}^2$$ I didn't understand the first inequality, because $|| \cdot ||_{H^1_0(U)} = || \cdot ||_{W^{1,2}(U)}$ and this norm involves the square root of the sum of the square of $L^2$-norms of $u$ and its weak derivatives of order $1$, but you only have the sum of these $L^2$-norms. $\endgroup$
    – George
    Mar 5, 2019 at 18:26
  • $\begingroup$ While the first inequality I imagine that it's an inequality of real numbers like this: $$\left( \sum_{i=1}^n a_i \right)^2 \leq \sum_{i=1}^n a_i^2$$ with $a_i > 0$ real numbers for each $i = 1, \cdots, n$, but I can't see why this is true $\endgroup$
    – George
    Mar 5, 2019 at 18:32
  • $\begingroup$ @George See my Edit 2. $\endgroup$
    – Pedro
    Mar 5, 2019 at 18:56

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