# Motivation for the definition of weak solutions to parabolic equations of second order

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

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

by

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

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 $$H^{-1}$$.

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 $$[\mathbf{u}'(t)](x)=u_t(t,x)$$.

## Edit

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 $$(x_1+\cdots+x_n)^p\leq k(x_1^p+\cdots +x_n^p).$$

• 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 $||f||_{L^2(U)}$. – George Mar 5 '19 at 15:08
• @George For the second inequality, see my edit. – Pedro Mar 5 '19 at 16:28
• 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. – George Mar 5 '19 at 18:26
• 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 – George Mar 5 '19 at 18:32
• @George See my Edit 2. – Pedro Mar 5 '19 at 18:56