Existence of Global-in-Time Solutions to a System of Fully Nonlinear Parabolic Equations

Abstract

We consider the Cauchy problem for a system of fully nonlinear parabolic equations. In this paper, we shall show the existence of global-in-time solutions to the problem. Our condition to ensure the global existence is specific to the fully nonlinear parabolic system.

Appendix A: Perron’s Method

In this appendix, we state Perron’s method and give its proof. Let $T>0$ and let $S$ be a nonempty subset of viscosity subsolutions of (1.1) in $\boldsymbol{R}^{N}\times (0,T)$.

Lemma A.1

Let $T>0$. Assume that $S\neq \emptyset$. For $i=1,2$, set

$$ u_{i} (x,t):= \sup \{v_{i}(x,t) | \ v=(v_{1},v_{2}) \in S\}, \quad x \in \boldsymbol{R}^{N}, t\in (0,T). $$

If

$$\sup _{K} |u_{i}| < \infty ,\quad i=1,2 $$

for any compact subset $K\subset \boldsymbol{R}^{N} \times (0,T)$, then $u=(u_{1},u_{2})$ is a viscosity subsolution of (1.1).

Proof

For $i=1,2$ and $\varphi \in C^{2,1}(\boldsymbol{R}^{N}\times (0,T))$, we assume that $u_{i}^{*}-\varphi$ attains its strict maximum at $(x_{0},t_{0})\in \boldsymbol{R}^{N} \times (0,T)$. Choose $r>0$ so that $[t_{0}-r,t_{0}+r]\subset (0,T)$. Then for any $\varepsilon >0$, we have

$$ u_{i}^{*}(x_{0},t_{0}) = \lim _{\varepsilon \to 0} \sup _{ \substack{x\in B(x_{0},\varepsilon ) \\ |t-t_{0}|< \varepsilon }} u_{i}(x,t) \leq \sup _{ \substack{x\in B(x_{0},\varepsilon ) \\ |t-t_{0}|< \varepsilon }} u_{i}(x,t), $$

where $u_{i}^{*}$ us defined in (2.1). For all $\tau >0$, there exist sequences $x_{\tau ,\varepsilon }\in B(x_{0},\varepsilon )$ and $t_{\tau ,\varepsilon }\in (t_{0}-r, t_{0}+r)$ such that

$$ \sup _{ \substack{x\in B(x_{0},\varepsilon ) \\ |t-t_{0}|< \varepsilon }} u_{i}(x,t) \leq u_{i}(x_{\tau ,\varepsilon },t_{\tau ,\varepsilon }) +\tau . $$

For any $\tau >0$ there exists $\delta _{\tau }>0$ such that for all $x\in \boldsymbol{R}^{N}$, $t\in (0,T)$, if $|x-x_{0}| + |t-t_{0}|<\delta _{\tau}$, then $|\varphi (x,t) - \varphi (x_{0},t_{0})|<\tau$. For each $k=1,2,\dots$ set

$$ \varepsilon =\min \left \{ \frac{\delta _{1/k}}{2}, \frac{1}{k},r \right \}. $$

There exist $x_{k} \in B_{r}(x_{0})$, $t_{k}\in (t_{0}-r,t+r)$ such that

$$ \begin{aligned} &x_{k} \to x_{0}, \quad t_{k}\to t_{0} \quad \mathrm{as} \ k\to \infty , \\ &u_{i}^{*}(x_{0},t_{0}) < u_{i}(x_{k},t_{k}) + \frac{1}{k}, \quad | \varphi (x_{k},t_{k})- \varphi (x_{0},t_{0})| < \frac{1}{k}. \end{aligned} $$

Moreover, by the definition of $u_{i}$, there exists $(u_{1}^{k},u_{2}^{k})\in S$ such that

$$ u_{i}(x_{k},t_{k}) < u_{i}^{k}(x_{k},t_{k}) +\frac{1}{k}. $$

Choose $(y_{k},s_{k})\in \overline{B}_{r}(x_{0})\times [t_{0}-r,t_{0}+r]$ so that $(u_{i}^{k})^{*} -\varphi$ attains its maximum at $(y_{k},s_{k})$. Taking a subsequence (still denoted by the same symbol), we see that as $k\to \infty$, $y_{k} \to \hat{y}$, $t_{k}\to \hat{t}$ for some $\hat{y}\in \overline{B}_{r}(x_{0})$, $t_{k}\to \hat{t}\in [t_{0}-r,t_{0}+r]$. We have

$$ \begin{aligned} (u_{i}^{*}-\varphi )(x_{0},t_{0}) & < (u_{i}^{k}-\varphi )(x_{k},t_{k}) +\frac{3}{k} \\ &\leq ((u_{i}^{k})^{*}-\varphi )(x_{k},t_{k}) +\frac{3}{k} \\ &\leq ((u_{i}^{k})^{*}-\varphi )(y_{k},s_{k}) +\frac{3}{k} \\ &\leq (u_{i}^{*}-\varphi )(y_{k},s_{k}) +\frac{3}{k}. \end{aligned} $$

(A.1)

Since $u_{i}^{*}$ is upper semicontinuous, we see that

$$ (u_{i}^{*} -\varphi )(x_{0},t_{0}) \leq (u_{i}^{*} - \varphi )( \hat{y},\hat{t}). $$

On the other hand, $u_{i}^{*} -\varphi$ has a strict maximum at $(x_{0},t_{0})$, hence $x_{0}=\hat{y}$ and $t_{0} =\hat{t}$. Therefore, $y_{k}\in B_{r}(x_{0})$, $t_{k}\in (t_{0}-r,t_{0}+r)$ for sufficiently large $k$. In addition, using (A.1) again, we have

$$ \lim _{k\to \infty}(u_{i}^{k})^{*}(y_{k},s_{k}) = u_{i}^{*}(x_{0},t_{0}) $$

and

$$ \begin{aligned} \limsup _{k\to \infty} (u_{i}^{k})^{*}(y_{k},s_{k}) \leq \limsup _{k \to \infty}u_{j}^{*}(y_{k},s_{k}) \leq u_{j}^{*}(x_{0},t_{0}), \end{aligned} $$

where $j\neq i$. Consequently, we obtain

$$ \partial _{t} \varphi (y_{k},s_{k}) + F_{i}(y_{k},D^{2}\varphi (y_{k},s_{k})) \leq |u_{j}^{*}(y_{k},s_{k})|^{p_{i} -1}u_{j}^{*}(y_{k},s_{k}), $$

hence

$$ \partial _{t} \varphi (x_{0},t_{0}) + F_{i}(x_{0},D^{2}\varphi (x_{0},t_{0})) \leq |u_{j}^{*}(x_{0},t_{0})|^{p_{i} -1}u_{j}^{*}(x_{0},t_{0}), $$

which completes the proof. □

Proposition A.1

Assume that $\xi =(\xi _{1},\xi _{2})\in (L^{\infty}_{\mathrm{loc}}(\boldsymbol{R}^{N} \times (0,T)))^{2}$ is a viscosity subsolution of (1.1) and $\eta =(\eta _{1},\eta _{2})\in (L^{\infty}_{\mathrm{loc}}(\boldsymbol{R}^{N} \times (0,T)))^{2}$ is a viscosity supersolution of (1.1) for some $T>0$. If $\xi$ and $\eta$ satisfy

$$\xi _{1} \leq \eta _{1}, \quad \xi _{2} \leq \eta _{2} \quad \mathrm{in} \ \boldsymbol{R}^{N} \times (0,T), $$

then

$$ u_{i}(x,t) := \sup \{v_{i}(x,t) \mid v=(v_{1},v_{2})\in S, \ \xi \leq v \leq \eta \} $$

is a viscosity solution of (1.1) in $\boldsymbol{R}^{N}\times (0,T)$. Here $\xi \leq v \leq \eta$ means that $\xi _{i} \leq v_{i} \leq \eta _{i}$ in $\boldsymbol{R}^{N}\times (0,T)$ for $i=1,2$.

Proof

By Lemma A.1, $u=(u_{1},u_{2})$ is a viscosity subsolution to (1.1). Suppose, contrary to our claim, that there exist $\varphi \in C^{2,1}$ and $(x_{0},t_{0})\in \boldsymbol{R}^{N}\times (0,T)$, $u_{i*} -\varphi$ attains a strict minimum at $(x_{0},t_{0})$, $(u_{i*}-\varphi )(x_{0},t_{0}) =0$ for some $i=1,2$ and exists $\theta >0$ such that

$$ \partial _{t} \varphi + F_{i}(x_{0},D^{2}\varphi ) - |u_{j*}|^{p_{i} -1}u_{j*} < -\theta $$

(A.2)

at $(x_{0},t_{0})$, where $i\neq j$ and $p_{1} =p$, $p_{2} =q$.

We firstly show that $\varphi (x_{0},t_{0}) <\eta _{i*}(x_{0},t_{0})$. In fact, we see that $\varphi \leq u_{i*} \leq \eta _{i*}$ and $u_{j*} \leq \eta _{j*}$ and $\eta _{i*} - \varphi$ attains a minimum at $(x_{0},t_{0})$ if $\varphi (x_{0},t_{0}) = \eta _{i*}(x_{0},t_{0})$, thus by the definition of the viscosity supersolution, we obtain

$$ \partial _{t} \varphi (x_{0},t_{0}) + F(x_{0},D^{2}\varphi (x_{0},t_{0})) \geq |\eta _{j*}|^{p_{j} - 1}\eta _{j*} (x_{0},t_{0}) \geq |u_{j*}|^{p_{j} -1}u_{j*}(x_{0},t_{0}). $$

This contradicts the assumption.

For any $\rho >0$, there exists $\varepsilon _{\rho}>0$ such that

$$ \begin{aligned} u_{j*}(x_{0},t_{0}) -\rho &= \sup _{\varepsilon >0}\inf _{ \substack{|x-x_{0}|< \varepsilon \\|t-t_{0}|< \varepsilon }} u_{j}(x,t) - \rho \\ &< \inf _{ \substack{|x-x_{0}|< \varepsilon _{\rho }\\|t-t_{0}|< \varepsilon _{\rho}}} u_{j}(x,t) \\ &\leq u_{j}(x,t) \end{aligned} $$

for $|x-x_{0}|<\varepsilon _{\rho}$ and $|t-t_{0}|<\varepsilon _{\rho}$. By using the mean value theorem, there exists $\hat{\theta}$ such that

$$ \begin{aligned} |u_{j*}|^{p_{i} -1}u_{j*} &\geq |u_{j*}(x_{0},t_{0}) -\rho |^{p_{i} -1}(u_{j*}(x_{0},t_{0})- \rho ) \\ &= |u_{j*}(x_{0},t_{0}) |^{p_{i} -1}u_{j*}(x_{0},t_{0}) \\ &- \rho p_{i} | \hat{\theta} u_{j*}(x_{0},t_{0})(x_{0},t_{0}) + (1- \hat{\theta})(u_{j*}(x_{0},t_{0})-\rho ) |^{p_{i} -1}. \end{aligned} $$

We can find $\rho >0$ and $s_{0}>0$ so small that

$$ \rho p_{i} | \hat{\theta} u_{j*}(x_{0},t_{0})(x_{0},t_{0}) + (1- \hat{\theta})(u_{j*}(x_{0},t_{0})-\rho ) |^{p_{i} -1} < \frac{\theta}{4} $$

and

$$ \vert \partial _{t}\varphi (x_{0},t_{0}) + F_{i}(x_{0},D^{2}\varphi (x_{0},t_{0})) - \partial _{t}\varphi (x,t) - F_{i}(x,D^{2}\varphi (x,t)) \vert < \frac{\theta}{4} $$

for $|x-x_{0}|< s_{0}$ and $|t-t_{0}|< s_{0}$. This together with (A.2) and the continuity of $\partial _{t}\varphi$ and $F_{i}(\cdot ,D^{2}\varphi )$ implies that

$$ \begin{aligned} -\theta &> \partial _{t} \varphi (x,t) +F_{i}(x,D^{2}\varphi (x,t)) - \frac{\theta}{4} -|u_{j*}|^{p_{i}-1}u_{j*}(x,t) -\frac{\theta}{4} \end{aligned} $$

for $|x-x_{0}| < s_{0}$ and $|t-t_{0}|< s_{0}$. Therefore,

$$ \partial _{t} \varphi (x,t) +F_{i}(x,D^{2}\varphi (x,t)) -|u_{j*}|^{p_{i}-1}u_{j*}(x,t) < -\frac{\theta}{2}. $$

(A.3)

It is already shown that $u_{i*}(x_{0},t_{0})=\varphi (x_{0},t_{0}) <\eta _{i*}(x_{0},t_{0})$. Set

$$ 3\hat{\tau} :=\eta _{i*}(x_{0},t_{0}) -u_{i*}(x_{0},t_{0})>0. $$

Since $\eta _{i*}$ is lower semicontinuous and $\varphi$ is continuous, we can find $s_{1}\in (0,s_{0})$ such that for all $|x-x_{0}| < s_{1}$ and $t\in (t_{0} -s_{1},t_{0}+s_{1})$,

$$ \eta _{i*}(x,t) -\varphi (x,t) > \eta _{i*}(x_{0},t_{0}) -\varphi (x_{0},t_{0}) - \hat{\tau} = 2 \hat{\tau}. $$

Therefore $\varphi (x,t) +2\hat{\tau} < \eta _{i*}(x,t)$ in $D$, where

$$ D:= B_{s_{1}}(x_{0}) \times (t_{0}-s_{1},t_{0}+s_{1}). $$

On the other hand, since $u_{i*}-\varphi$ attains a strict minimum at $(x_{0},t_{0})$ and $(u_{i*}-\varphi )(x_{0},t_{0})=0$, there exist $\varepsilon \in (0,s_{1} /2)$ and $\tau _{0} \in (0,\hat{\tau})$ such that

$$ (u_{i*}-\varphi )(x,t) \geq \min _{(x,t)\in A}\{ (u_{i*}-\varphi )(x,t) \} > \tau _{0}. $$

Here we have set

$$ A := \left ( \overline{B_{s_{1} /2 +\varepsilon }(x_{0})}\setminus B_{s_{1} /2 -\varepsilon }(x_{0}) \right ) \times \left \{ t \mid \frac{s_{1}}{2}-\varepsilon \leq |t-t_{0}| \leq \frac{s_{1}}{2} + \varepsilon \right \}. $$

We now define $(w_{1},w_{2})$ by

$$ \begin{aligned} w_{i}(x,t) &:= \textstyle\begin{cases} \max \{u_{i}(x,t),\varphi (x,t) + \tau _{0}\} \quad &\mathrm{in} \ D/2, \\ u_{i}(x,t) \quad \mathrm{in} \ (\boldsymbol{R}^{N}\times (0,T)) \setminus (D/2), \end{cases}\displaystyle \\ w_{j} (x,t) &:= u_{j}(x,t) \quad \mathrm{in} \ \boldsymbol{R}^{N}\times (0,T), \end{aligned} $$

where

$$ D/2 := B_{s_{1} /2}(x_{0}) \times \left (t_{0}- \frac{s_{1}}{2},t_{0}+ \frac{s_{1}}{2} \right ). $$

In what follows, we shall show that $(w_{1},w_{2})$ is a viscosity subsolution to (1.1) in $\boldsymbol{R}^{N}\times (0,T)$ satisfying $\xi _{k} \leq w_{k} \leq \eta _{k}$ for $k=1,2$. It follows from the definition of $w_{i}$ that we have $\xi _{i} \leq u_{i} \leq w_{i}$ in $\boldsymbol{R}^{N}\times (0,T)$. Since $\varphi (x,t) + \tau _{0} \leq \eta _{i*}$ in $D$, we see that $\varphi + \tau _{0} \leq \eta _{i}$ in $D$, hence $w_{k} \leq \eta _{k}$ in $\boldsymbol{R}^{N}\times (0,T)$ for $k=1,2$. Consequently, for $k=1,2$, we obtain

$$ \xi _{k} \leq w_{k} \leq \eta _{k} \quad \mathrm{in} \ \boldsymbol{R}^{N} \times (0,T). $$

We can find $n\in \{1,2,\dots \}$ sufficiently large so that

$$ \frac{1}{n} < \frac{s_{1}}{2} -\varepsilon \quad \mathrm{and} \quad \frac{1}{n} < \frac{\tau _{0}}{2} $$

and there exist $x_{n} \in B_{1/n}(x_{0})$ and $t_{n}\in \boldsymbol{R}$ with $|t_{0}-t_{n}|<1/n$ such that

$$ u_{i} (x_{n},t_{n}) < u_{i*}(x_{0},t_{0}) + \frac{1}{n}. $$

Moreover, it follows that

$$ u_{i*}(x_{0},t_{0}) + \frac{1}{n} < u_{i*}(x_{0},t_{0}) + \frac{\tau _{0}}{2} = \varphi (x_{0},t_{0}) + \frac{\tau _{0}}{2} < \varphi (x_{0},t_{0}) + \tau _{0}. $$

Note that $(x_{n},t_{n}) \in D/2$.

We next prove that $(w_{1},w_{2})$ is a viscosity subsolution to (1.1) in $\boldsymbol{R}^{n}\times (0,T)$. Let us take $\boldsymbol{R}^{N}\times (0,T)$ and $\psi \in C^{2,1}(\boldsymbol{R}^{N}\times (0,T))$ arbitrarily.

We firstly assume that $w_{i}^{*} - \psi$ attains a local maximum at $(\hat{x},\hat{t})$. Consider the first case $w_{i}^{*} (\hat{x},\hat{t}) = u_{i}^{*} (\hat{x},\hat{t})$. Then

$$ \begin{aligned} u_{i}^{*} (\hat{x},\hat{t}) -\psi (\hat{x},\hat{t}) &= w_{i}^{*}( \hat{x},\hat{t}) - \psi (\hat{x},\hat{t}) \\ &\geq w_{i}^{*} (x,t) -\psi (x,t) \\ &\geq u_{i}^{*} (x,t) - \psi (x,t) \end{aligned} $$

in $\boldsymbol{R}^{N}\times (0,T)$. Thus, $u_{i}^{*} -\psi$ attains its maximum at $(\hat{x},\hat{t})$. Moreover, since $(u_{1},u_{2})$ is a subsolution to (1.1) in $\boldsymbol{R}^{N}\times (0,T)$ and $u_{j} \equiv w_{j}$, we have

$$\partial _{t} \psi + F_{i}(\cdot ,D^{2}\psi ) \leq |u_{j}^{*}|^{p_{i} -1}u_{j}^{*} =|w_{j}^{*}|^{p_{i} -1}w_{j}^{*} $$

at $(\hat{x},\hat{t})$.

We now consider the second case $w_{i}^{*}(\hat{x},\hat{t}) =(\varphi +\tau _{0})^{*}(\hat{x},\hat{t}) =\varphi (\hat{x},\hat{t})+\tau _{0}$. Note that $(\hat{x},\hat{t})\in D/2$. The same argument above implies that $\varphi +\tau _{0}-\psi$ attains its maximum at $(\hat{x},\hat{t})$. Thus, we see that

$$ \partial _{t} \varphi (\hat{x},\hat{t}) = \partial _{t} \psi (\hat{x}, \hat{t}), \quad D\varphi (\hat{x},\hat{t}) = D\psi (\hat{x},\hat{t}), \quad D^{2} \varphi (\hat{x},\hat{t}) \leq D^{2} \psi (\hat{x}, \hat{t}). $$

It follows from (1.8) and (A.3) that

$$ \begin{aligned} \partial _{t} \psi + F_{i}(\cdot ,D^{2}\psi ) &\leq \partial _{t} \varphi + F_{i}(\cdot ,D^{2}\varphi ) \\ &\leq |{u_{j}}_{*}|^{p_{i} -1} {u_{j}}_{*} \\ &\leq |{u_{j}}^{*}|^{p_{i} -1} {u_{j}}^{*} \\ &= |{w_{j}}^{*}|^{p_{i} -1} {w_{j}}^{*} \end{aligned} $$

at $(\hat{x},\hat{t})$.

We secondly assume that $w_{j}^{*} - \psi$ attains a local maximum at $(\hat{x},\hat{t})$. Since $w_{j} =u_{j}$ in $\boldsymbol{R}^{N}\times (0,T)$, $u_{j}^{*}-\psi$ also attains its maximum at $(\hat{x},\hat{t})$. Therefore, we obtain

$$ \begin{aligned} \partial _{t} \psi + F_{j}(\cdot ,D^{2}\psi ) \leq |u_{i}^{*}|^{p_{j} -1}u_{i}^{*} \leq |w_{i}^{*}|^{p_{j} -1}w_{i}^{*} \end{aligned} $$

at $(\hat{x},\hat{t})$.

Consequently, $(w_{1},w_{2})$ is a viscosity subsolution to (1.1) in $\boldsymbol{R}^{N}\times (0,T)$. This contradicts the definition of $(u_{1},u_{2})$. □