Large-time behavior of unbounded solutions of viscous Hamilton-Jacobi Equations in $\mathbb{R}^{N}$.

The well-posedness of (9)-(11) is a consequence of the strong comparison result proven in [9], Theorem 3.1. Existence then follows by Perron’s method, provided there exist suitable sub- and supersolutions. The subsolution can be chosen as $\underline{u}(x,t)=\inf_{B_{R}}u_{0}+t\inf_{B_{R}}f$ while the construction of the supersolution is more involved because of the state-constraint boundary condition. ∎

The result follows from the fact the solution satisfying the state-constraints boundary condition (10) is the maximal subsolution of (9)-(11) (see e.g., [19], Section 7.C^′). We provide a constructive proof of the result, however, to illustrate this notion more clearly.

Observe that, for $R^{\prime}\geq R$, $u^{R^{\prime}}$ is in particular a subsolution of (9) in $B_{R}\times(0,T)$, satisfying also (11). Let $\epsilon>0$, and let $\rho_{\epsilon}$ denote the standard mollifier in $\mathbb{R}^{N+1}$. Since (9) is convex in $(u,Du)$, $u^{R^{\prime}}\ast\rho_{\epsilon}$ is a classical subsolution of (9) in $B_{R}$ (this is Lemma 2.7 in [10]). Define, for small ${\delta>0}$, $x\in\overline{B}_{R}$ and $t\geq 0$,

$$w(x,t)=(u^{R^{\prime}}\ast\rho_{\epsilon})(x,t)-\delta t.$$

(We have dropped the dependences of $w$ in $R^{\prime}$, $\epsilon$ and $\delta$ for the sake of notational simplicity.) We claim that $w$ is a smooth subsolution of (9) in $B_{R}\times(0,T)$. Indeed, for all $(x,t)\in B_{R}\times(0,T]$, we can compute

	$\displaystyle w_{t}(x,t)-\Delta w(x,t)+|Dw(x,t)|^{m}$
	$\displaystyle\quad=(u^{R^{\prime}}\ast\rho_{\epsilon})_{t}-\Delta(u^{R^{\prime}}\ast\rho_{\epsilon})(x,t)+|D(u^{R^{\prime}}\ast\rho_{\epsilon})(x,t)|^{m}-\delta$
	$\displaystyle\quad\leq f(x)-\delta<f(x).$		(12)

And this computation, which is only valid for smooth enough $u^{R^{\prime}}$ can be justified if $u^{R^{\prime}}$ is only a continuous viscosity subsolution of (9).

We set $\eta_{\epsilon}:=\min_{\overline{B_{R}}}(u_{0}-w(x,0))$. We remark that, by the continuity of $u^{R^{\prime}}$, $\eta_{\epsilon}\to 0$ as $\epsilon\to 0$. We now show that $u^{R}\geq w+\eta_{\epsilon}$ in ${\overline{B_{R}}}\times[0,T]$ for any $T>0$. We argue by contradiction assuming that

$$(u^{R}-w-\eta_{\epsilon})(x_{0},t_{0}):=\min_{\overline{B_{R}}\times[0,T]}(u^{R}-w-\eta_{\epsilon})<0.$$

By definition of $\eta_{\epsilon}$, we have $t_{0}>0$ and since $w+\eta_{\epsilon}$ is smooth, the definition of the state-constraints boundary condition (10) implies that

$$w_{t}(x_{0},t_{0})-\Delta w(x_{0},t_{0})+|Dw(x_{0},t_{0})|^{m}\geq f(x_{0}),$$

which contradicts (12). Therefore, $u^{R}\geq w+\eta_{\epsilon}$ in $\overline{B_{R}}\times[0,T]$. Taking the limit $\epsilon,\delta\rightarrow 0$, then $T\rightarrow\infty$, we conclude. ∎

We are going to obtain a solution of (7)-(8) as a locally uniform limit of the solutions $u^{R}$ of Lemma 1 as $R\to+\infty$.

To do so, we consider a fixed $\bar{R}$. For $R>2\bar{R}+1$, we have, assuming $u_{0},f\geq 0$

$$0\leq u^{R}\leq u^{2\bar{R}+1}\quad\hbox{in }\overline{B_{2\bar{R}}}\times[0,T]\;,$$

hence $\{u^{R}\}_{R>2\bar{R}+1}$ is uniformly bounded over $\overline{B_{2\bar{R}}}\times[0,T]$.

Furthermore, using Theorem 8 and Corollary 9 in the Appendix, the $C^{0,1/2}$-norm of $u^{R}$ on $\overline{B_{\bar{R}}}\times[0,T]$ remains also uniformly bounded for $R>2\bar{R}+1$. And we also recall that the $u^{R}$ are decreasing in $R$.

Thus, by using the Ascoli-Arzela Theorem together with the monotonicity of $(U^{R})_{R}$, we have the uniform convergence of $u^{R}$ on $\overline{B_{\bar{R}}}\times[0,T])$. Then, by a diagonal argument, we may extract a subsequence of $\{u^{R}\}_{R>0}$ that converges locally uniformly to some $u\in C(\mathbb{R}^{N}\times[0,T])$. By stability, it follows that $u$ is a viscosity solution of (7)-(8). ∎

To deal with the difficulty of $u$ and $v$ being unbounded, we define

$$z_{1}(x,t)=-e^{-u(x,t)},\qquad z_{2}(x,t)=-e^{-v(x,t)}.$$

Since $u$ and $v$ are bounded from below, $z_{1}$ and $z_{2}$ are bounded. Furthermore, we have that $z_{1}$ and $z_{2}$ are respectively a sub- and supersolution of

$$z_{t}-\Delta z+N(x,z,Dz)=0,$$

(13)

where $N:\mathbb{R}^{N}\times\mathbb{R}\times\mathbb{R}^{N}\to\mathbb{R}^{N}$ is given by

$$N(x,r,p)=r\left(f(x)+\left|\frac{p}{r}\right|^{2}-\left|\frac{p}{r}\right|^{m}\right).$$

(14)

Formally, this is a straightforward computation. Given the monotonicity of the transformation defining $z_{1},z_{2},$ there is no difficulty in passing the computation over to smooth test functions. Another consequence is that $z_{1}(x,t)\leq z_{2}(x,t)$ if and only if $u(x,t)\leq v(x,t)$. Hence Theorem 4 follows from proving the same comparison result for sub- and supersolutions of (13).

We proceed with the usual scheme of doubling variables, and for simplicity we first treat the case in which $Q$ is unbounded in $t$. Assume that the conclusion of the theorem is false and $M:=\sup_{Q}(z_{1}-z_{2})>0$. Define first, for $\delta>0$,

$$M_{\delta}:=\sup_{Q}\left(z_{1}(x,t)-z_{2}(x,t)-\delta(|x|^{2}+t)\right).$$

(15)

As the penalized function on the right-hand side of (15) is upper-semicontinuous and goes to $-\infty$ as either $|x|\to+\infty$ or $t\to+\infty$, the supremum in achieved at some $(\bar{x},\bar{t})\in\overline{Q}$. From standard arguments, we have that $M_{\delta}\to M$ as $\delta\to 0$ (see e.g., [19], Proposition 3.7), which implies that for small enough $\delta$, $M_{\delta}>0$. Moreover, since $\limsup_{(y,s)\to(x,t)}(z_{1}(y,s)-z_{2}(y,s))$ for all $(x,t)\in\partial_{p}Q$, we necessarily have $(\bar{x},\bar{t})\in Q$. Define now

$$M_{\delta,\alpha}:=\sup_{Q\times Q}\left\{z_{1}(x,t)-z_{2}(y,t)-\frac{\delta}{2}(|x|^{2}+|y|^{2}+t)-\frac{\alpha}{2}|x-y|^{2}\right\}.$$

Again the supremum above if achieved at a point $(\hat{x},\hat{y},\hat{t})$. Similarly, we know that as $\alpha\rightarrow\infty$ and $\delta$ remains fixed, $M_{\alpha,\delta}\rightarrow M_{\delta}$ and $\hat{x},\hat{y}\to\bar{x}$ and $\hat{t}\to\bar{t}$ for $\bar{x},\bar{t}$ as above.

Thus, an application of Ishii’s Lemma (see [19], Theorem 3.2) gives

$$N(\hat{x},z_{1}(\hat{x},\hat{t}),\alpha(\hat{x}-\hat{y})+\delta\hat{x})-N(\hat{y},z_{2}(\hat{y},\hat{t}),\alpha(\hat{x}-\hat{y})-\delta\hat{y})\leq\left(2n+\frac{1}{2}\right)\delta.$$

(16)

We aim to bound this difference from below by a positive constant independent of $\delta$. Let

$$h(s)=N(a(s),b(s),c(s)),$$

where

	$\displaystyle a(s)=s\hat{x}+(1-s)\hat{y},\qquad b(s)=sz_{1}(\hat{x},\hat{t})+(1-s)z_{2}(\hat{y},\hat{t}),$
	$\displaystyle c(s)=\alpha(\hat{x}-\hat{y})+\delta(s\hat{x}+(s-1)\hat{y}).$

We rewrite (16) as

		$\displaystyle N(\hat{x},z_{1}(\hat{x},\hat{t}),\alpha(\hat{x}-\hat{y})+\delta\hat{x})-N(\hat{y},z_{2}(\hat{y},\hat{t}),\alpha(\hat{x}-\hat{y})-\delta\hat{y})$
		$\displaystyle{}=N(a(1),b(1),c(1))-N(a(0),b(0),c(0))$
		$\displaystyle{}=h(1)-h(0)=\int_{0}^{1}h^{\prime}(s)\,ds.$		(17)

Computing

	$\displaystyle a^{\prime}(s)=\hat{x}-\hat{y},\qquad b^{\prime}(s)=z_{1}(\hat{x},\hat{t})-z_{2}(\hat{y},\hat{t}),$
	$\displaystyle c^{\prime}(s)=\delta(\hat{x}+\hat{y}),$

and, from (14),

	$\displaystyle\frac{\partial N}{\partial x}=r\,Df(x),\qquad\frac{\partial N}{\partial r}=f(x)-\left|\frac{p}{r}\right|^{2}+(m-1)\left|\frac{p}{r}\right|^{m},$
	$\displaystyle\frac{\partial N}{\partial p}=\left(\frac{2}{|r|^{2}}-\frac{m(m-2)|p|^{m-2}}{|r|^{m}}\right)p,$

we have

	$\displaystyle\int_{0}^{1}h^{\prime}(s)\,ds$	$\displaystyle{}=\int_{0}^{1}\frac{\partial N}{\partial x}(a(s),b(s),c(s))\cdot a^{\prime}(s)+\frac{\partial N}{\partial r}(a(s),b(s),c(s))b^{\prime}(s)$
		$\displaystyle\qquad{}+\frac{\partial N}{\partial p}(a(s),b(s),c(s))\cdot c^{\prime}(s)\,ds.$

We proceed to bound each term in the last expression

$\displaystyle\frac{\partial N}{\partial x}(a(s),b(s),c(s))\cdot a^{\prime}(s)\geq-\max\{\|z_{1}\|_{\infty},\|z_{2}\|_{\infty}\}\|Df\|_{\infty,B_{R(\delta)}}|\hat{x}-\hat{y}|,$

where $B_{R(\delta)}$ is a ball that contains $(\hat{x},\hat{y})$, the points at which the maximum of $\Phi=\Phi^{\delta,\alpha}(x,y)$ is achieved, considering $\delta>0$ is fixed; $B_{R(\delta)}$ is uniform in $\alpha$. Since $\hat{x}-\hat{y}\rightarrow 0$ as $\alpha\rightarrow\infty$, this implies

$$\frac{\partial N}{\partial x}(a(s),b(s),c(s))\cdot a^{\prime}(s)\rightarrow 0\quad\textrm{as }\alpha\rightarrow\infty.$$

We write $q:=\frac{p}{r}$. Young’s inequality gives

$$|q|^{2}\leq\frac{2}{m}|q|^{m}+\frac{m}{m-2},$$

hence

	$\displaystyle\frac{\partial N}{\partial r}$	$\displaystyle{}=f(x)-\left|\frac{p}{r}\right|^{2}+(m-1)\left|\frac{p}{r}\right|^{m}$
		$\displaystyle{}\geq f(x)+(m-1)|q|^{m}-\frac{2}{m}|q|^{m}-\frac{m}{m-2}$
		$\displaystyle{}\geq f(x)-\frac{m}{m-2}+\left(m-1-\frac{2}{m}\right)|q|^{m}$
		$\displaystyle{}\geq 1+C|q|^{m},$		(18)

taking $0<C<m-1-\frac{2}{m}$ in the last inequality. The bound

$$f(x)\geq 1+\frac{m}{m-2}\quad\textrm{for all }x\in\mathbb{R}^{N},$$

(19)

may be assumed without loss of generality by initially considering, instead of $u$ and $v$, the functions

$$u(x,t)+\left(1+\frac{m}{m-2}\right)t,\qquad v(x,t)+\left(1+\frac{m}{m-2}\right)t,$$

for all $(x,t)\in Q$ (see the comments at the beginning of the introduction). On the other hand, by direct computation,

	$\displaystyle\left|\frac{\partial N}{\partial p}\right|$	$\displaystyle{}=2\left|\frac{p}{r}\right|+m\left|\frac{p}{r}\right|^{m-1}=2|q|+m|q|^{m-1}$
		$\displaystyle{}\leq m(|q|+|q|^{m-1})\leq 2m(1+|q|^{m}).$

Here we’ve used that $m>2$ and that $|q|+|q|^{m-1}\leq 2(1+|q|^{m})$, which follows easily by considering the cases $|q|>1$ and $|q|\leq 1$ separately. We have therefore obtained

$$\left(1+\frac{2m}{C}\right)\frac{\partial N}{\partial r}\geq\left|\frac{\partial N}{\partial p}\right|.$$

(20)

Thus, using (2.2) and (20), for small $\delta>0$ we have

	$\displaystyle\liminf_{\alpha\rightarrow\infty}\int_{0}^{1}h^{\prime}(s)\,ds$	$\displaystyle{}\geq\liminf_{\alpha\rightarrow\infty}\int_{0}^{1}\frac{\partial N}{\partial r}(z_{1}(\hat{x},\hat{t})-z_{2}(\hat{y},\hat{t}))+\frac{\partial N}{\partial p}\cdot\delta(\hat{x}+\hat{y})\,ds$
		$\displaystyle{}\geq\liminf_{\alpha\rightarrow\infty}\int_{0}^{1}\frac{\partial N}{\partial r}\left(M_{\alpha,\delta}+\delta(|\hat{x}|^{2}+|\hat{y}|^{2}+\hat{t})+\frac{\alpha}{2}|\hat{x}-\hat{y}|^{2}\right)$
		$\displaystyle{}\quad+\frac{\partial N}{\partial p}\cdot\delta(\hat{x}+\hat{y})\,ds$
		$\displaystyle{}\geq\liminf_{\alpha\rightarrow\infty}\int_{0}^{1}\frac{\partial N}{\partial r}M_{\alpha,\delta}-\delta\left|\frac{\partial N}{\partial p}\right|(|\hat{x}|+|\hat{y}|)\,ds$
		$\displaystyle{}\geq\int_{0}^{1}\frac{\partial N}{\partial r}\left(M_{\delta}-2\delta\left(1+\frac{2m}{C}\right)|\bar{x}|\right)\,ds$
		$\displaystyle{}>\frac{M}{2}>0.$		(21)

Here we have also used the facts concerning the limits $\alpha\to\infty$, $\delta\to 0$ (taken in that order) mentioned at the outset of the argument. Together with (16) and (2.2), (21) gives the desired contradiction.

Finally, if $Q\subset\mathbb{R}^{N}\times[0,T]$ for some $T>0$, we note that the preceding argument equally applies, but the penalization term in $t$ is no longer necessary in (15). That is, we may define

$$M_{\delta}:=\sup_{Q}\left(z_{1}(x,t)-z_{2}(x,t)-\delta|x|^{2}\right),$$

instead of (15) and proceed in the same way. ∎

We note that, from Remark 3, any solution of (7)-(8) is bounded from below over $\mathbb{R}^{N}\times[0,T]$, since it is nonnegative. Therefore, combining the results of Proposition 3 and Theorem 4, there exists a unique solution of (7)-(8). Since $T>0$ is arbitrary (in particular, with no dependence on the data) the solution can be uniquely extended to a solution of (1)-(2), which is also nonnegative. Uniqueness then follows immediately from Theorem 4. And the case when $u_{0}\in W^{1,\infty}_{\textit{loc}}(\mathbb{R}^{N})$ is complete.

When $u_{0}\in C(\mathbb{R}^{N})$, we argue in the following way: by classical results, there exists a sequence $(u^{\epsilon}_{0})_{\epsilon}$ of functions of $W^{1,\infty}_{\textit{loc}}(\mathbb{R}^{N})$ such that, for any $\epsilon$, $|u^{\epsilon}_{0}(x)-u_{0}(x)|\leq\epsilon$ in $\mathbb{R}^{N}$. If $u^{\epsilon}$ denotes the unique solution of (1) associated to the initial data $u^{\epsilon}_{0}$, we have, by the comparison result

$$|u^{\epsilon}(x,t)-u^{\epsilon^{\prime}}(x,t)|\leq\epsilon^{\prime}+\epsilon\quad\textrm{in }\mathbb{R}^{N}\times[0,+\infty)\;,$$

since $|u^{\epsilon}_{0}(x)-u^{\epsilon^{\prime}}_{0}(x)|\leq\epsilon^{\prime}+\epsilon$ in $\mathbb{R}^{N}$. Therefore, the Cauchy sequence $(u^{\epsilon})_{\epsilon}$ converge uniformly in $\mathbb{R}^{N}\times[0,+\infty)$ to the unique viscosity solution of of (1)-(2) by Theorem 4. ∎

Let $\gamma=\frac{m-2}{m-1}$, and for $R>0$, define

$$v(x)=R^{-\gamma}\phi(Rx).$$

Since $\phi$ is, in particular, a solution of (6) in $B_{R}$, $v$ is a solution of

$$-\Delta v+|Dv|^{m}=R^{2-\gamma}(f(Rx)-\lambda^{*})\leq R^{2-\gamma}M_{R}\quad\textrm{in }B_{1},$$

Now set $K_{R}=R^{2-\gamma}M_{R}$ and

$$u(x)=K_{R}^{-\frac{1}{m}}v(x)$$

to obtain that $u$ is a solution of

$$-K_{R}^{\frac{1}{m}-1}\Delta u+|Du|^{m}\leq 1\quad\textrm{in }B_{1}.$$

(22)

Observe that $m>2$ implies that $\gamma<2$ and $\frac{1}{m}-1<0$. Since $f$ is (in particular) coercive, we have that $K_{R}\to 0$ as $R\to\infty$. Hence, for $R$ sufficiently large we obtain $0\leq K_{R}\leq 1$.

We are now in position to apply the estimate of Theorem (7) (in the appendix) to Equation (22): there exists a constant $C_{1}>0$, independent of $R>0$, such that

$$|u(x)-u(y)|\leq C_{1}|x-y|^{\gamma}\quad\textrm{for all }x,y\in\overline{B}_{1}.$$

In terms of $\phi$, for $|x|\leq 1,y=0$ this gives

$$K_{R}^{-\frac{1}{m}}R^{-\gamma}|\phi(Rx)-\phi(0)|\leq C_{1},$$

or, for $|x|\leq R$,

$$|\phi(x)-\phi(0)|\leq C_{1}K_{R}^{\frac{1}{m}}R^{\gamma}=C_{1}(R^{2-\gamma}M_{R})^{\frac{1}{m}}R^{-\gamma}=C_{1}M_{R}^{\frac{1}{m}}R\;.$$