Week 13: Parabolic equations

We now turn our attention to parabolic equations. To this end, suppose $L$ and $f$ are as in Week 7 (in divergence or nondivergence form). A linear second-order parabolic equation is an equation of the form

where $u=u(x,t)$ is a function $u:\Omega\times I\rightarrow\mathbb{R}$, $I\subset\mathbb{R}$ an interval. Here we allow the coefficients $a_{ij}$, $b^{i}$ and $c$ of $L$ to depend on $t$ as well, i.e. they are also functions on $\Omega\times I$.

Motivation

Parabolic equations arise naturally in the study of heat conduction. For example, the work of Fourier implies that in an isotropic medium $\Omega\subset\mathbb{R}^{n}$, given an initial temperature distribution $u_{0}:\Omega\rightarrow\mathbb{R}$, the temperature $u(x,t)$ at $x\in\Omega$ and time $t>0$ should solve the initial-value problem

as we will see, the solution to this problem is uniquely determined by $u_{0}$ and appropriate boundary values of $u$, which corresponds to the temperature distribution outside $\Omega$. As $t\rightarrow\infty$, one would expect the temperature of $\Omega$ to reach an equilibrium so that formally, the limit $u_{\infty}:=\lim_{t\rightarrow\infty}u(\cdot,t)$ should solve the equation $-\Delta u_{\infty}=0$. This suggests that a study of parabolic equations could lead to insight into elliptic equations and be another route to proving existence of solutions.

Another motivation for parabolic equations is the variational aspect of elliptic problems (see Week 7). Before describing this connection, consider the classical problem of finding the minima of some $f\in C^{\infty}(\mathbb{R}^{n})$. As we know, if $x_{0}$ is a minimiser of $f$, then $\D f(x_{0})=0$. One way of finding $x_{0}$ is the method of steepest descent (or negative gradient method), which essentially boils down to considering the system of ODE

and analysing the limit $\lim_{t\rightarrow\infty}x(t)$, which ought to lead to a minimiser $x_{0}$, if one exists (and if $x(t)$ is defined for all $t$!). That this is promising is suggested by the fact that

so that $f\circ x$ is non-increasing. Note also that the solutions $x_{0}$ to $\D f(x_{0})=0$ are equilibrium points of the above ODE system.

Going back to the variational formulation of the equation $Lu=0$ in the case where $b^{i}\equiv 0$ and $u$ vanishes on $\partial\Omega$, recall that solutions to this equation arise as critical points of the energy

in the sense that the `derivative of $E$ at $u$ in direction $\varphi\in C_{0}^{\infty}(\Omega)$' vanishes for all $\varphi$, which we write formally as:

these inner products being the $L^{2}(\Omega)$ inner product. In analogy with the negative gradient method, we now look for appropriately regular $u:\Omega\times\left[0,\infty\right[\rightarrow \mathbb{R}$ vanishing on $\partial\Omega\times\left[0,\infty\right[$ solving the initial-value problem

As with the negative gradient method, we compute that for such a function $u$,

so that $u$ indeed tends to decrease the energy associated to the problem. Here we again expect that as $t\rightarrow\infty$, $u(\cdot,t)$ should appropriately tend to a solution to $Lu=0$.

Weak maximum principle, comparison principle and uniqueness

In this section we consider classical solutions $u\in C^{2,1}(\Omega\times\left]0,T\right[)\cap C^{0}(\overline\Omega\times [0,T])$ to the Dirichlet initial-boundary value problem

for given $u_{0}\in C^{0}(\overline\Omega)$ and $g\in C^{0}(\partial\Omega\times\left[0,T\right])$, where we shall assume that $L$ is an elliptic operator in nondivergence form:

with $a_{ij},b^{i},c\in C^{0}(\overline{\Omega}\times[0,T])$ for all $i,j\in\{1,\dots,n\}$. Let $Q_{T}:=\Omega\times\left]0,T\right[$ and define the parabolic boundary of $Q_{T}$ as

As before, we say that $u$ is a subsolution (or supersolution) to the problem (*) if the first equation is replaced with the inequality $\partial_{t}u+Lu\leq 0$ (resp. $\partial_{t}u + Lu\geq 0$).

Theorem 13.1 (weak maximum principle for $c\equiv 0$). Suppose $u\in C^{2,1}(\Omega\times\left]0,T\right[)\cap C^{0}(\overline{\Omega}\times\left[0,T\right])$ is a subsolution to the initial-boundary value problem (*) with $c\equiv 0$. Then

If instead $u$ is a supersolution to (*), then

Proof. We first consider the case where $u$ is a subsolution and $\partial_{t}u+Lu<0$.

• Fix $\eps>0$ small and let $(x_{0},t_{0})\in \overline{Q}_{T-\eps}$ be such that $\max_{\overline{Q}_{T-\eps}}u=u(x_{0},t_{0})$. We now consider cases.

  • If $(x_{0},t_{0})\in Q_{T-\eps}$, then we must have $\partial_{t}u(x_{0},t_{0})=0$, $\D u(x_{0},t_{0})=0$ and $\D^{2}u(x_{0},t_{0})$ non-positive definite, which altogether imply that

    $\partial_{t}u+Lu\geq 0,$

    contradicting the assumed inequality.

  • If $(x_{0},t_{0})\in\Omega\times\{T-\eps\}$, then we again have $\D u(x_{0},t_{0})=0$ and $\D^{2}u(x_{0},t_{0})$ non-positive definite, and since $u(x_{0},t_{0})\geq u(x_{0},t_{0}+h)$ for all small $h<0$, we obtain that $\partial_{t}u(x_{0},t_{0})\geq 0$, again leading to the inequality

    $\partial_{t}u+Lu\geq 0,$

    contradicting our assumption.

  We therefore deduce that $(x_{0},t_{0})\in \partial'Q_{T-\eps}$. Therefore, for any small $\eps>0$ and any $(x,t)\in Q_{T-\eps}$, we have that

  $u(x,t)\leq \max_{\partial'Q_{T-\eps}}u\leq \max_{\partial'Q_{T}}u.$

  Since for any $(x,t)\in Q_{T}$ we may find an $\eps_{0}>0$ such that $(x,t)\in Q_{T-\eps}$ for all $\eps<\eps_{0}$, we deduce that

  $\max_{\overline{Q}_{T}}u=\sup_{Q_{T}}u\leq \max_{\partial'Q_{T}}u.$

  Since $\partial'Q_{T}\subset \overline{Q}_{T}$, the opposite inequality is also true and we are done.

Now suppose $\partial_{t}u+Lu\leq 0$ and consider $v(x,t):=u(x,t) - \eps t$ for fixed $\eps>0$. We immediately see that

$\partial_{t}v + Lv = -\eps + \partial_{t}u + Lu \leq -\eps < 0$

so that we fall into the preceding case, whence for all $(x,t)\in Q_{T}$,

$u(x,t) - \eps t \leq \max_{\overline{Q}_{T}}v = \max_{\partial'Q_{T}}v = \max_{(x,t)\in\partial'Q_{T}}(u(x,t) - \eps t)\leq \max_{\partial'Q_{T}}u.$

Taking the limit $\eps\searrow 0$, we arrive at $u(x,t)\leq \max_{\partial'Q_{T}}u$, i.e. $\sup_{Q_{T}}\leq \max_{\partial'Q_{T}}u$, whence we are done. For the supersolution case, replace $u$ with $-u$.

Remark 13.2. It follows immediately using the same techniques that if $u$ solves (*) with $T=\infty$, then

and

These expressions are of course only finite if $g$ is appropriately bounded.

Just as with elliptic equations, we also have a maximum principle in the case where $c\geq 0$.

Theorem 13.3 (weak maximum principle). Suppose $u\in C^{2,1}(\Omega\times\left]0,T\right[)\cap C^{0}(\overline{\Omega}\times\left[0,T\right])$ is a subsolution to the initial-boundary value problem (*) with $c\geq 0$. Then

If instead $u$ is a supersolution to (*), then

In particular, if $u$ solves (*), then $\max_{Q_{T}}|u| = \max_{\partial'Q_{T}}|u|$.

Proof. Suppose $u$ is a subsolution. As in the case where $c\equiv 0$, we first assume that $\partial_{t}u + Lu<0$, and we may as well suppose $u\in C^{2,1}(\Omega\times[0,T])$. If $\max u\leq 0$, then we are clearly done, so suppose $\max u = u(x_{0},t_{0})>0$. If this point is in the interior of $Q_{T}$ or in $\Omega\times \{T\}$, we deduce as before that $\partial_{t}u(x_{0},t_{0})\geq 0$, $\D u(x_{0},t_{0})=0$ and $\D^{2}u(x_{0},t_{0})$ is nonpositive definite so that

$(\partial_{t} u + Lu)(x_{0},t_{0})=\partial_{t}u(x_{0},t_{0}) - \sum_{i,j=1}^{n}a_{ij}(x_{0},t_{0})\partial_{i}\partial_{j}u(x_{0},t_{0}) + \sum_{i=1}^{n}b^{i}(x_{0},t_{0})\partial_{i}u(x_{0},t_{0}) + c(x_{0},t_{0})u(x_{0},t_{0}) \geq 0,$

which yields a contradiction. In the more general case, we consider $v(x,t):= u(x,t) -\eps t$ and note that if $u$ has a positive maximum, then so does $v$ for sufficiently small $\eps$. This then establishes the claim after taking $\eps\searrow 0$ as in the case where $c\equiv 0$. The supersolution case follows from replacing $u$ with $-u$. Finally, if $u$ is a solution, then it is both a subsolution and supersolution so that the inequality $\min_{Q_{T}}|u|\leq \max_{\partial'Q_{T}}|u|$ (and thus the equality) immediately follows.

Remark 13.4. In contrast to the elliptic case, maximum principle-type theorems may still be established even if $c$ is not necessarily nonnegative. For instance, if $u$ is a supersolution to (*) with $u_{0},g\geq 0$, then the function $v(x,t)=u(x,t)e^{\sup c \cdot t}$ satisfies

which allows us to conclude using Theorem 13.3 that $v\geq 0$, which in turn implies that $u\geq 0$. If instead $u$ is a subsolution and $u_{0},g\geq 0$, we deduce similarly that $u\leq 0$. In this case boundedness of $c$ is crucial and follows as a result of our assumptions.

Corollary 13.5 (comparison principle). Suppose that $u^{i}\in C^{2,1}(\Omega\times\left]0,T\right[)\cap C^{0}(\overline{\Omega}\times\left[0,T\right])$ is a solution to the initial-boundary value problem

$f^{i}\in C^{2,1}(\Omega\times\left]0,T\right[)$, $u_{0}^{i}\in C^{0}(\overline{\Omega})$ and $g^{i}\in C^{0}(\partial\Omega\times\left[0,T\right])$ ($i\in\{1,2\}$). Suppose furthermore that the inequalities

hold. Under these conditions, $u^{1}\leq u_{2}$ on $\overline\Omega\times\left[0,T\right]$.

Proof. Set $u:=u^{1}-u^{2}$. It follows immediately that $u$ is a subsolution to (*) with initial-boundary data $u_{0}\leq 0$ and $g\leq 0$, whence the maximum principle (cf. Remark 13.4) implies that

$u^{1}-u^{2} = u \leq \max\{\max u_{0},\max g\}\leq 0$

on $Q_{T}$, which establishes the claim.

Remark 13.6. The comparison principle implies in particular that a function $u\in C^{2,1}(Q_{T})\cap C^{0}(\overline{Q_{T}})$ is uniquely determined by the values of $\partial_{t}u+Lu$ and $\left.u\right|_{\partial'Q_{T}}$.

Harnack inequality

Just as with elliptic equations, we have the following Harnack inequality.

Theorem 13.7 (parabolic Harnack inequality). Let $U\Subset\Omega$ be connected and $t_{1},t_{2}\in\left]0,T\right]$ with $t_{1}<t_{2}$. There exists a constant $C>0$ depending only on $U$, $t_{1}$, $t_{2}$ and $L$ such that whenever $u\in C^{2,1}(Q_{T})$ is a nonnegative solution to $\partial_{t}u + Lu=0$ on $Q_{T}$,

Strong maximum principle

Suppose throughout this section that $\Omega$ is connected.

Theorem 13.8 (strong maximum principle for $c\equiv 0$). Suppose $u\in C^{2,1}(\Omega\times\left]0,T\right[)\cap C^{0}(\overline{\Omega}\times\left[0,T\right])$ is a subsolution to the initial-boundary value problem (*) with $c\equiv 0$ and $\max_{\overline{Q}_{T}}u=u(x_{0},t_{0})$ for some $(x_{0},t_{0})\in Q_{T}$. Then $u$ is constant on $Q_{t_{0}}$.

If instead $u$ is a supersolution to (*), then if $\min_{\overline{Q}_{T}}u=u(x_{0},t_{0})$ for some $(x_{0},t_{0})\in Q_{T}$, we must have that $u$ is constant on $Q_{t_{0}}$.

Proof. As usual we suppose $u$ is a subsolution. Let

$\Omega'=\{x\in \Omega:\forall t\in \left]0,t_{0}\right[,\ u(x,t)=u(x_{0},t_{0}) \}$

This set is clearly closed in $\Omega$. We shall now show that it is open and nonempty. To this end, fix a open ball $U\Subset\Omega$ with $x_{0}\in U$ and let $v\in C^{2,1}(R_{T})\cap C^{0}(\overline{R}_{T})$ be a solution to the initial-boundary value problem

$\begin{aligned}\partial_{t}v + Lv&=0\ \textup{on }R_{T}\\\left.v\right|_{\partial'R_{T}}&=\left.u\right|_{\partial'R_{T}}, \end{aligned}$

where $R_{T}:=U\times \left]0,T\right[$. By the comparison principle, we have that $u\leq v\leq u(x_{0},t_{0})$ so that $v(x_{0},t_{0})=u(x_{0},t_{0})$ as well. Now, since $u(x_{0},t_{0})-v\geq 0$ also solves $\partial_{t}v+Lv=0$, we may employ the Harnack inequality on some $V\Subset U$ connected with $x_{0}\in V$ to deduce that for any $t<t_{0}$,

$\sup_{V}(u(x_{0},t) - v) \leq C(t)\inf_{V}(u(x_{0},t_{0}) - v) \leq C(t) (u(x_{0},t_{0})- v(x_{0},t_{0})) = 0$

so that $v\geq u(x_{0},t_{0})$, i.e. $v\equiv u(x_{0},t_{0})$ so that $\left.u\right|_{\partial'R_{t_{0}}}\equiv u(x_{0},t_{0})$. Since $V\ni x_{0}$ is otherwise arbitrary, it follows that $u\equiv u(x_{0},t_{0})$ on $R_{t_{0}}$, i.e. $U\subset \Omega'$. $\Omega'$ is therefore nonempty and, by the preceding argument, open in $\Omega$. By connectedness, we must have $\Omega'=\Omega$.

As in the elliptic case, we also obtain a strong maximum principle in the case where $c\geq 0$ under appropriate sign conditions.

Theorem 13.9 (string maximum principle for $c\geq 0$). Suppose $u\in C^{2,1}(\Omega\times\left]0,T\right[)\cap C^{0}(\overline{\Omega}\times\left[0,T\right])$ is a subsolution to the initial-boundary value problem (*) with $c\geq 0$ and $\max_{\overline{Q}_{T}}u=u(x_{0},t_{0})\geq 0$ for some $(x_{0},t_{0})\in Q_{T}$. Then $u$ is constant on $Q_{t_{0}}$.

If instead $u$ is a supersolution to (*), then if $\min_{\overline{Q}_{T}}u=u(x_{0},t_{0})\leq 0$ for some $(x_{0},t_{0})\in Q_{T}$, we must have that $u$ is constant on $Q_{t_{0}}$.

Proof. Let $\Omega'$ be as in the preceding proof and again suppose $u$ is a subsolution. Note that if $u(x_{0},t_{0})=0$, the above proof applies mutatis mutandis, so suppose that $u(x_{0},t_{0})>0$. Again let $U\Subset \Omega$ be an open ball with $x_{0}\in U$ and suppose $v\in C^{2,1}(R_{T})\cap C^{0}(\overline{R}_{T})$ solve the initial-boundary balue problem

$\begin{aligned}\partial_{t}v + (L-c)v&=0\ \textup{on }R_{T}\\ \left.v\right|_{\partial'R_{T}}&=\left.u^{+}\right|_{\partial'R_{T}}.\end{aligned}$

By the weak maximum principle, we see that $0\leq v\leq u(x_{0},t_{0})$. Moreover, on the set $\{u\geq 0\}$, we have that $\partial_{t}u + (L-c)u\leq -cu\leq 0$ so that by the comparison principle, $u\leq v$ on $\{u\geq 0\}\cap R_{T}$, thus also on all of $R_{T}$, whence $v(x_{0},t_{0})=u(x_{0},t_{0})$. Thus, since $u(x_{0},t_{0})-v\geq 0$ solves the equation $\partial_{t}v + (L-c)u=0$, we again employ the Harnack inequality on $V\Subset U$ connected with $x_{0}\in V$ to obtain that for any $t<t_{0}$,

$\sup_{V}(u(x_{0},t)- v) \leq C(t)\inf_{V}(u(x_{0},t_{0}) - v)\leq C(t)(u(x_{0},t_{0}) - v(x_{0},t_{0}))=0$

so that $v\geq u(x_{0},t_{0})$. Therefore, as before, we must have $v\equiv u(x_{0},t_{0})$ on $R_{t_{0}}$ so that $u\equiv u(x_{0},t_{0})>0$ on $\{u\geq 0\}\cap \partial'(V \times ]0,t_{0}[)$, whence we must have $u\equiv u(x_{0},t_{0})$ on all of $\partial'(V\times \left]0,t_{0}\right[$. We therefore deduce as before that $\left.u\right|_{R_{t_{0}}}\equiv u(x_{0},t_{0})$ i.e. $U\subset \Omega'$ so that arguing as before, $\Omega'=\Omega$.