On the viscosity solutions to a class of nonlinear degenerate parabolic differential equations

Abstract

In this work, we show existence and uniqueness of positive solutions of $H(Du, D^2u)+\chi (t)|Du|^\Gamma -f(u)u_t=0$ in $\Omega \times (0, T)$ and $u=h$ on its parabolic boundary. The operator H satisfies certain homogeneity conditions, $\Gamma >0$ and depends on the degree of homogeneity of $H, f>0$, increasing and meets a concavity condition. We also consider the case $f\equiv 1$ and prove existence of solutions without sign restrictions.

Appendix

We discuss a maximum principle that applies to the case where f is a positive continuous function. No sign conditions are imposed on the sub-solutions and super-solutions.

Recall Conditions A and B, (1.2)–(1.6). From (1.5), we have

$$\begin{aligned} m_{\min }(\lambda )=\inf _{|e|=1}H(e, I-\lambda e\otimes e),\;\;\mu _{\max }=\sup _{|e|=1}H(e, \lambda e\otimes e-I) \end{aligned}$$

and $m(\lambda )=\min (m_{\min }(\lambda ),\;-\mu _{\max }(\lambda )).$ In place of Condition C, we assume that

$$\begin{aligned} m(0)>0\;\;\text{ and }\;\;\lim _{\lambda \rightarrow -\infty }m(\lambda )=\infty . \end{aligned}$$

(8.1)

Recall the notation, $\hat{H}(p,X)=-H(p, -X),\;\forall (p,X)\in \mathbb {R}^n\times S^n$, see Remark 1.1.

Lemma 8.1

(Weak Maximum Principle) Let $\Omega \subset \mathbb {R}^n,\;n\ge 2$, be a bounded domain and $T>0$. Suppose that H satisfies Conditions A, B and (8.1). Suppose that $\chi :[0,T]\rightarrow \mathbb {R}$ and $f:\mathbb {R}\rightarrow [0,\infty )$ and $f\not \equiv 0,$ are continuous functions.

Let $\Gamma >0$ and $\phi \in usc(lsc)(\Omega _T\cup P_T)$ solve

$$\begin{aligned} H(D\phi , D^2\phi )+\chi (t)|D\phi |^\Gamma - f(\phi ) \phi _t\ge (\le ) 0,\;\;\text{ in } \Omega _T. \end{aligned}$$

• (a) If $\Gamma \ge k$ then $\sup _{\Omega _{T}}\phi \le \sup _{P_T} \phi =\sup _{\Omega _T\cup P_T}\phi \;(\inf _{\Omega _T} \phi \ge \inf _{P_T}\phi =\inf _{\Omega _T\cup P_T}\phi ).$

• (b) If $0<\Gamma <k$ and $\inf f>0$ then the conclusion in (a) holds.

• (c) If $\chi \equiv 0$ then the conclusion in (a) holds even if $\inf f=0$.

Proof

Let $0<{\hat{\tau }}<\tau <T, \Omega _{{\hat{\tau }},\tau }=\Omega \times [{\hat{\tau }}, \;\tau ]$ and P the parabolic boundary of $\Omega _{{\hat{\tau }},\tau }$. Our goal is to prove the weak maximum principle in $\Omega _{{\hat{\tau }},\tau }$ for any $0<{\hat{\tau }}<\tau <T$ and then extend it to $\Omega _T$. Note that u is bounded from above in ${{{\overline{\Omega }}}_{{\hat{\tau }},\tau }}$ since $u\in usc(\Omega _T\cup P_T)$.

Choose $z\in \mathbb {R}^n{\setminus }\Omega$ and $R>0$ such that $\Omega \subset B_{R}(z){{\setminus }} B_{R/2}(z)$. Call $r=|x-z|$; clearly, $R/2\le r\le R,\;\forall x\in \Omega$.

Set

$$\begin{aligned}&\vartheta =\sup _{\Omega _{{\hat{\tau }},\tau }} \phi ,\;\;\;\ell =\sup _{P}\phi ,\;\;\;\delta =\vartheta -\ell ,\;\;\;c=\sup _{{\overline{\Omega }}}\phi (x,\tau ),\;\;\eta =\max (\delta ,\;\;c-\ell )\;\nonumber \\&\quad \text{ and }\;\;\;\nu =\max (c,\; \vartheta ,\;\ell ). \end{aligned}$$

(8.2)

We recall from Remark 2.2(ii) and (6.4)(ii) that if $v=a-br^\beta$, where $b>0$ and $\beta >0$, then

$$\begin{aligned} -\frac{\left( b\beta \right) ^k}{ r^{\gamma -\beta k}}\mu (2-\beta )\le H(Dv, D^2v)\le -\frac{\left( b\beta \right) ^k}{ r^{\gamma -\beta k}} m(2-\beta ). \end{aligned}$$

(8.3)

We argue by contradiction and assume that $\delta >0.$ Since $\Omega _{{\hat{\tau }},\tau }$ is an open set there is a point $(\xi ,\theta )\in {\Omega _{{\hat{\tau }},\tau } }$ such that $\phi (\xi ,\theta )>\ell +3\delta /4$ and $0<{\hat{\tau }}<\theta <\tau$. Define

$$\begin{aligned} g(t)=0,\;\;\forall t\in [{\hat{\tau }},\;\theta ]\;\;\;\text{ and }\;\;\;g(t)=(t-\theta )^4/(\tau -\theta )^4,\;\;\forall t\in [\theta ,\;\tau ]. \end{aligned}$$

Select $0<\varepsilon \le \min (0.5,\;\delta /4).$ For $\beta >0$, set

$$\begin{aligned} \psi (x,t)=\psi (r,t)=\ell +\frac{\varepsilon }{4} +\eta g(t)-\frac{\varepsilon r^\beta }{32R^\beta },\;\;\forall (x,t)\in {\overline{\Omega }}_{{\hat{\tau }},\tau }. \end{aligned}$$

Thus, $\psi (x,t)\ge \ell +\varepsilon /8,\;\forall (x,t)\in {\overline{\Omega }}_{{\hat{\tau }},\tau },$ and $\psi (x,\tau )\ge \ell +\eta +\varepsilon /8\ge c+\varepsilon /8,\;\forall x\in {\overline{\Omega }}$. Moreover,

$$\begin{aligned} \phi (\xi ,\theta )-\psi (\xi , \theta )\ge \ell +\frac{3\delta }{4}-\ell -\frac{\varepsilon }{4}=\frac{3\delta }{4}-\frac{\varepsilon }{4} \ge \frac{\delta }{4}>0. \end{aligned}$$

(8.4)

Since $\phi -\psi \le 0$ on $\partial {\overline{\Omega }}_{{\hat{\tau }},\tau }$ and $(\phi -\psi )(\xi , \theta )>0$, the function $\phi -\psi$ has a positive maximum at some point $(y,s)\in \Omega _{{\hat{\tau }},\tau }$.

Set $B_0=\sup _{[0,T]}|\chi (t)|$, call $\rho =|y-z|$ and use (8.3) to get

$$\begin{aligned}&H\left( D\psi (y,s), D^2\psi (y,s)\right) +\chi (s)|D\psi (y,s)|^\Gamma \nonumber \\&\quad \le -\left( \frac{\varepsilon \beta }{32 R^\beta }\right) ^k \frac{m(2-\beta )}{\rho ^{\gamma -\beta k}}+{B_0} \left( \frac{\varepsilon \beta }{32R^\beta }\right) ^\Gamma \rho ^{(\beta -1)\Gamma }\nonumber \\&\quad =\rho ^{\beta k-\gamma }\left( \frac{\varepsilon \beta }{32 R^\beta }\right) ^k\left[ {B_0} \left( \frac{\varepsilon \beta }{32R^\beta }\right) ^{\Gamma -k} \rho ^{\gamma -\Gamma +\beta (\Gamma -k)}-m(2-\beta )\right] \nonumber \\&\quad =\rho ^{\beta k-\gamma }\left( \frac{\varepsilon \beta }{32 R^\beta }\right) ^k\left[ {B_0} \left( \frac{\varepsilon \beta }{32} \left( \frac{\rho }{R}\right) ^\beta \right) ^{\Gamma -k} \rho ^{\gamma -\Gamma }-m(2-\beta )\right] \end{aligned}$$

(8.5)

Call I the right hand side of the third line in (8.5) and note that $1/2\le \rho /R\le 1$. We now show part (a) of the lemma. Note that $\psi _t(y,s)\ge 0$.

• (i) If $\Gamma >k$ then taking $\beta =2$ (see (8.1)) and $\varepsilon$ small enough we can make $I<0$. We conclude from (8.5) that $I<0\le f(\phi (y,s))\psi _t(y,s)$ implying that the lemma holds for $0<{\hat{\tau }}<\tau <T.$

• (ii) If $\Gamma =k$ then $\gamma -k=k_2$ (see (1.3) and (1.4)). Taking $\beta$ large and using (8.1) we can make $I<0$. We conclude from (8.5) that $I<0\le f(\phi (y,s))\psi _t(y,s)$ implying that the lemma holds for $0<{\hat{\tau }}<\tau <T.$

Taking $B_0=0$ and arguing as above we get part (c) of the lemma.

To see part (b), set $\omega =\inf f$ and modify $g(t)=(t/\tau )^\alpha$, where $\alpha$ is large so that $\eta g(\theta )\le \varepsilon /8$. Since $\varepsilon \le \delta /4$, this ensures that in (8.4)

$$\begin{aligned} \phi (\xi ,\theta )-\psi (\xi , \theta )\ge \ell +\frac{3\delta }{4}-\left( \ell +\frac{\varepsilon }{4}+\eta g(\theta )\right) \ge \frac{3\delta }{4}-\frac{\varepsilon }{4} -\frac{\varepsilon }{8}\ge \frac{\delta }{4}>0. \end{aligned}$$

Using (8.5) estimate I (disregard the second term in the parenthesis) as

$$\begin{aligned} I\le A \left( \frac{\varepsilon \beta }{32R^\beta }\right) ^\Gamma \rho ^{(\beta -1)\Gamma }=\frac{A}{\rho ^\Gamma }\left( \frac{\varepsilon \beta \rho ^\beta }{32R^\beta }\right) ^\Gamma \le 2^\Gamma A \left( \frac{\beta }{32}\right) ^\Gamma \frac{\varepsilon ^\Gamma }{R^\Gamma }. \end{aligned}$$

Next, $\psi _t(y,s)=\alpha \eta s^{\alpha -1}/\tau ^\alpha \ge \alpha \eta {\hat{\tau }}^{\alpha -1}/\tau ^\alpha$ implying that

$$\begin{aligned} I-f(\phi (y,s))\psi _t(y,s)\le 2^\Gamma A\left( \frac{\beta }{32}\right) ^\Gamma \frac{\varepsilon ^\Gamma }{R^\Gamma }-\alpha {\omega } \eta \left( {\hat{\tau }}^{\alpha -1}/\tau ^\alpha \right) <0, \end{aligned}$$

if R is chosen large enough. Using (8.5), we get a contradiction and $\phi \le \ell$ in $\Omega _{{\hat{\tau }}, \tau }.$

If $\sup _{\Omega _T}\phi >\sup _{P_T} \phi$ then there is a point $(y,s)\in \Omega _T$ (with $0<s<T$) such that $\phi (y,s)>\sup _{P_T} \phi$. Select $0<\hat{s}<s<{\bar{s}}<T$ and call P the parabolic boundary of $\Omega _{\hat{s},{\bar{s}}}$. Then, $\sup _{P_T}\phi <\phi (y,s)\le \sup _{\Omega _{\hat{s},{\bar{s}}}}\phi \le \sup _{P}\phi \le \sup _{P_T}\phi .$ This is a contradiction and the lemma holds.

To show the weak minimum principle, take $v=-\phi$ and conclude that $H(-Dv, -D^2v)\le f(-v)(-v_t)$. If $\hat{f}(v)=f(-v)$ then (1.3) shows that $\hat{H}(Dv, D^2v)\ge {\tilde{f}}(v)v_t$. As noted in Remark 1.1, $\hat{H}$ satisfies Conditions A, B and (8.1) and the minimum principle follows. $\square$

Remark 8.2

Suppose that u solves $H(Du, D^2u)+\chi (t)|Du|^\Gamma -f(u)u_t=g(x,t),$ where $L=\sup _{\Omega _T}|g|<\infty$ and $\omega =\inf _{\mathbb {R}} f>0$. Using $u\pm \ell t, \ell \ge L/\omega$ large, one gets $\inf _{P_T}(u+\ell t)-\ell t\le u\le \sup _{P_T}(u-\ell t)+\ell t.$