Neumann Boundary Problems for Parabolic Partial Differential Equations with Divergence Terms

Abstract

In this paper, we give a probabilistic interpretation for solutions to the Neumann boundary problems for a class of semi-linear parabolic partial differential equations (PDEs for short) with singular non-linear divergence terms. This probabilistic approach leads to the study on a new class of backward stochastic differential equations (BSDEs for short). A connection between this class of BSDEs and semi-linear PDEs is established.

Appendix

In this section we give the proof of Theorem 1, for which the following estimate is needed.

Lemma 5

Assume that (H1), (H2) and (H4) are satisfied and the coefficient g only depends on (t, x). If u is the solution of PDE (1), then it holds that

$$ \begin{array}{llll} &\displaystyle \sup\limits_{t\in[0,T]}\|u_{t}\|^{2}+{{\int}_{0}^{T}} \|u_{t}\|^{2}_{1} dt\\ \leq &\displaystyle C\Big(\|{\Phi}\|^{2}+{{\int}_{0}^{T}}{\int}_{\partial D}|h_{r}(x,0)|^{2} d\sigma(x)dr+{{\int}_{0}^{T}}{\int}_{D}|g_{r}(x)|^{2}+|f_{r}(x,0,0)|^{2} dxdr\Big), \end{array} $$

(24)

where the constant C > 0 depends on terminal time T, domain D, Lipschitz constants of f, g, h and boundedness of b.

Proof

By the Lipschitz and integrability conditions, we have

$$ \begin{array}{@{}rcl@{}} \displaystyle \|u_{t}\|^{2}+{{\int}_{t}^{T}} \|\nabla u_{r}\|^{2} dr &=&\displaystyle \|{\Phi}\|^{2}+2{{\int}_{t}^{T}}{\int}_{D} f_{r}(x,u_{r},\nabla u_{r})u_{r}(x)dxdr\\ &\displaystyle &+2{{\int}_{t}^{T}}{\int}_{D} \langle g_{r}(x), \nabla u_{r}(x)\rangle dxdr\\&\displaystyle &+{{\int}_{t}^{T}}{\int}_{\partial D} h_{r}(x,u_{r})u_{r}(x) d\sigma(x)dr\\&&\displaystyle +2{{\int}_{t}^{T}}{\int}_{D} \langle b,\nabla u_{r}\rangle(x) u_{r}(x)dxdr\\ &\leq&\displaystyle \|{\Phi}\|^{2}+(2C+\frac{1}{\epsilon_{2}}+1) {{\int}_{t}^{T}}\|u_{r}\|^{2} dr+C^{2}\epsilon_{2}{{\int}_{t}^{T}} \|\nabla u_{r}\|^{2} dr\\ &&\displaystyle +{{\int}_{t}^{T}}\|f_{r}(0,0)\|^{2} dr+\epsilon_{1}{{\int}_{t}^{T}} \|\nabla u_{r}\|^{2} dr+\frac{1}{\epsilon_{1}}{{\int}_{t}^{T}}\|g_{r}\|^{2} dr\\ &&\displaystyle +\|Tr\|^{2}\epsilon_{3}{{\int}_{t}^{T}} \|u_{r}\|_{1}^{2} dr +{\alpha\|Tr\|^{2}}{{\int}_{t}^{T}} \|u_{r}\|_{1}^{2} dr\\&&\displaystyle +\frac{1}{4\epsilon_{3}}{{\int}_{t}^{T}}{\int}_{\partial D}|h_{r}(x,0)|^{2}dxdr+\frac{1}{\epsilon_{4}}{{\int}_{t}^{T}}\|u_{r}\|^{2} dr\\&&\displaystyle +M^{2}\epsilon_{4} {{\int}_{t}^{T}}\|\nabla u_{r}\|^{2} dr, \end{array} $$

where ∥Tr∥ is the norm of the trace operator Tr and constant $M=\sup _{x\in D}|b(x)|$.

By further calculation, we obtain

$$ \begin{array}{llll} &\displaystyle \|u_{t}\|^{2}+(1-\epsilon_{1}-C^{2}\epsilon_{2}-\|Tr\|^{2}\epsilon_{3}-\alpha\|Tr\|^{2}-M^{2}\epsilon_{4}){{\int}_{t}^{T}} \|\nabla u_{r}\|^{2} dr\\ \leq&\displaystyle \|{\Phi}\|^{2}+(2C+\frac{1}{\epsilon_{2}}+1+\|Tr\|^{2}\epsilon_{3}+\alpha\|Tr\|^{2}+\frac{1}{\epsilon_{4}}){{\int}_{t}^{T}} \|u_{r}\|^{2} dr\\ &\displaystyle +{{\int}_{t}^{T}}\|f_{r}(0,0)\|^{2} dr+\frac{1}{\epsilon_{1}}{{\int}_{t}^{T}}\|g_{r}\|^{2} dr+\frac{1}{4\epsilon_{3}}{{\int}_{t}^{T}}{\int}_{\partial D}|h_{r}(x,0)|^{2}d\sigma(x) dr. \end{array} $$

We can choose 𝜖₁,𝜖₂,𝜖₃,𝜖₄ small enough such that 𝜖₁ + C²𝜖₂ + ∥Tr∥²𝜖₃ + α∥Tr∥² + M²𝜖₄ < 1, since α∥Tr∥² < 1. Thus by Gronwall’s inequality and a further calculation, we obtain the desired estimate. □

Set the space ${\mathscr{H}}_{T}:=C([0,T];L^{2}(D))\cap L^{2}([0,T];H^{1}(D))$. We first consider the situation that the coefficients f, g, h only depends on the variables t and x, that is, we suppose that f, g and h are in $L^{2}([0,T]\times D;\mathbb {R})$, $L^{2}([0,T]\times D;\mathbb {R}^{N})$ and $L^{2}([0,T]\times \partial D;\mathbb {R})$, respectively. The following existence and uniqueness result holds.

Proposition 6

There exists a unique weak solution $u\in {\mathscr{H}}_{T}$ to the linear PDE (23).

Proof

Assume that $\{g^{n}\}_{n\in \mathbb {N}^{*}}$ is a sequence of smooth functions with $g^{n}\in C^{\infty }[0,T]\otimes C^{1}({\bar {D}})$, which approaches to g in L²([0,T] × D). It is known that the following Neumann boundary problem without singular coefficient

$$ \left\{ \begin{array}{llll} &\displaystyle \partial_{t} {u^{n}_{t}}+ \frac{1}{2}{\Delta} {u^{n}_{t}}+\langle b,\nabla {u^{n}_{t}}\rangle -\text{div}({g^{n}_{t}})+f_{t}=0 \ \ \quad \text{on}\ \ [0,T]\times D\\ &\displaystyle {u^{n}_{T}}={\Phi} \ \ \quad \text{on}\ \ D\\ &\displaystyle \langle\nabla {u^{n}_{t}}, \vec{n} \rangle=2\langle {g^{n}_{t}}, \vec{n} \rangle-h_{t} \ \ \quad \text{on}\ \ [0,T]\times\partial D, \end{array} \right. $$

has a unique weak solution $u^{n}\in {\mathscr{H}}_{T}$ [11, Theorem 6.39] satisfying that

$$ \begin{array}{llll} &\displaystyle ({u^{n}_{T}},\phi_{T})-({u^{n}_{0}},\phi_{0})-{{\int}_{0}^{T}}({u^{n}_{t}},\partial_{t}\phi_{t}) dt={{\int}_{0}^{T}}(\nabla {u^{n}_{t}}, \nabla \phi_{t}) dt-{{\int}_{0}^{T}} ({g^{n}_{t}},\nabla\phi_{t}) dt\\&\displaystyle -{{\int}_{0}^{T}}{\int}_{D}\langle b,\nabla {u^{n}_{t}}\rangle(x)\phi_{t}(x) dxdt -{{\int}_{0}^{T}}(f_{t},\phi_{t}) dt-\frac{1}{2}{{\int}_{0}^{T}}{\int}_{\partial D} h_{t}(x)\phi_{t}(x) d\sigma(x) dt. \end{array} $$

(25)

For any positive integers n < m, we have

$$ \left\{ \begin{array}{llll} &\displaystyle \partial_{t} ({u^{n}_{t}}-{u^{m}_{t}})+ \frac{1}{2}{\Delta} ({u^{n}_{t}}-{u^{m}_{t}})+\langle b,\nabla ({u^{n}_{t}}-{u^{m}_{t}})\rangle-\text{div}({g^{n}_{t}}-{g^{m}_{t}})=0 \ \ \text{on}\ \ [0,T]\times D,\\ &\displaystyle {u^{n}_{T}}-{u^{m}_{T}}=0 \ \ \quad \text{on}\ \ D\\ &\displaystyle \langle \nabla ({u^{n}_{t}}-{u^{m}_{t}}), \vec{n}\rangle=2\langle {g^{n}_{t}}-{g^{m}_{t}}, \vec{n}\rangle \ \ \quad \text{on}\ \ [0,T]\times\partial D. \end{array} \right. $$

By the similar proof of Lemma 5, we find that $\{u^{n}\}_{n\in \mathbb {N}^{*}}$ is a Cauchy sequence in ${\mathscr{H}}_{T}$, of which the limit is denoted by $u\in {\mathscr{H}}_{T}$. By passing limits on both sides of Eq. 25, we find u is a weak solution of Eq. 23. The uniqueness can be obtained by linearity of the equation and the similar estimate in Lemma 5. □

Proof of Theorem 1

Existence: For simplicity, we set ${g^{n}_{t}}(x)=g_{t}(x,{u^{n}_{t}}(x),\nabla {u^{n}_{t}}(x))$, ${h^{n}_{t}}(x)=h_{t}(x,{u^{n}_{t}}(x))$ and ${f^{n}_{t}}(x)=f_{t}(x,{u^{n}_{t}}(x),\nabla {u^{n}_{t}}(x))$. Choosing 𝜃 > 0, we have

$$\begin{array}{llll} &{}\displaystyle \|u^{n+1}_{0}-{u^{n}_{0}}\|^{2}+{{\int}_{0}^{T}} e^{\theta s}\|\nabla(u^{n+1}_{s}-{u^{n}_{s}})\|^{2}ds=-\theta {{\int}_{0}^{T}} e^{\theta s}\|u^{n+1}_{s}-{u^{n}_{s}}\|^{2}ds\\& +\displaystyle 2{{\int}_{0}^{T}}{\int}_{D} e^{\theta s}\langle b,\nabla (u^{n+1}_{s}-{u^{n}_{s}})\rangle(x)(u^{n+1}_{s}(x)-{u^{n}_{s}}(x))dxds\\& +\displaystyle 2{{\int}_{0}^{T}}e^{\theta s}({f^{n}_{s}}-f^{n-1}_{s}, u^{n+1}_{s}-{u^{n}_{s}})ds \\&\displaystyle +2{{\int}_{0}^{T}} {\int}_{D} e^{\theta s}\langle {g_{s}^{n}}-g_{s}^{n-1},\nabla(u^{n+1}_{s}-{u^{n}_{s}})\rangle dxds\\ &\displaystyle +{{\int}_{0}^{T}} {\int}_{\partial D}e^{\theta s}({h_{s}^{n}}-h_{s}^{n-1})(u^{n+1}_{s}-{u^{n}_{s}})d\sigma(x)ds. \end{array} $$

By using Lipschitz conditions on the coefficients and Cauchy-Schwarz inequality, we have the following estimates:

$$\begin{array}{llll} &\displaystyle 2{{\int}_{0}^{T}} {\int}_{D} e^{\theta s}\langle {g_{s}^{n}}-g_{s}^{n-1},\nabla(u^{n+1}_{s}-{u^{n}_{s}})\rangle dxds\\ \leq&\displaystyle 2{{\int}_{0}^{T}}e^{\theta s}\left( \beta\|{u^{n}_{s}}-u^{n-1}_{s}\|+\gamma\|\nabla ({u^{n}_{s}}-u^{n-1}_{s})\|\right) \|\nabla(u^{n+1}_{s}-{u^{n}_{s}})\| ds\\ \leq&\displaystyle \beta \epsilon {{\int}_{0}^{T}} e^{\theta s}\|\nabla(u^{n+1}_{s}-{u^{n}_{s}})\|^{2} ds +\frac{\beta}{\epsilon}{{\int}_{0}^{T}} e^{\theta s}\|{u^{n}_{s}}-u^{n-1}_{s}\|^{2} ds\\&\displaystyle +{\gamma{\int}_{0}^{T}} e^{\theta s}\|\nabla({u_{s}^{n}}-u_{s}^{n-1})\|^{2}ds +{\gamma{\int}_{0}^{T}} e^{\theta s}\|\nabla(u_{s}^{n+1}-{u_{s}^{n}})\|^{2}ds \end{array}$$

and

$$\begin{array}{llll} &\displaystyle 2{{\int}_{0}^{T}}{\int}_{D} e^{\theta s}\langle b,\nabla (u^{n+1}_{s}-{u^{n}_{s}})\rangle(x)(u^{n+1}_{s}(x)-{u^{n}_{s}}(x))dxds\\ \leq&\displaystyle \frac{1}{\epsilon}{{\int}_{0}^{T}} e^{\theta s}\|u^{n+1}_{s}-{u^{n}_{s}}\|^{2}ds+M^{2}{\epsilon{\int}_{0}^{T}}e^{\theta s}\|\nabla (u^{n+1}_{s}-{u^{n}_{s}})\|^{2}ds \end{array}$$

and

$$\begin{array}{llll} &\displaystyle 2{{\int}_{0}^{T}}e^{\theta s}({f^{n}_{s}}-f^{n-1}_{s}, u^{n+1}_{s}-{u^{n}_{s}})ds\\ \leq&\displaystyle \frac{1}{\epsilon}{{\int}_{0}^{T}} e^{\theta s}\|u^{n+1}_{s}-{u^{n}_{s}}\|^{2}ds+C^{2}{\epsilon{\int}_{0}^{T}} e^{\theta s}\|{u^{n}_{s}}-u^{n-1}_{s}\|^{2}_{1} ds \end{array}$$

and

$$\begin{array}{llll} &\displaystyle {{\int}_{0}^{T}} {\int}_{\partial D} e^{\theta s}({h_{s}^{n}}-h_{s}^{n-1})(u^{n+1}_{s}-{u^{n}_{s}})d\sigma(x)ds\\ \leq&\displaystyle \frac{1}{2}\alpha\|Tr\|^{2} {{\int}_{0}^{T}} e^{\theta s}\|u^{n+1}_{s}-{u^{n}_{s}}\|_{1}^{2} ds+\frac{1}{2}\alpha\|Tr\|^{2} {{\int}_{0}^{T}} e^{\theta s}\|{u^{n}_{s}}-u^{n-1}_{s}\|_{1}^{2} ds. \end{array}$$

Therefore, it follows that

$$\begin{array}{llll} &\displaystyle (\theta-\frac{2}{\epsilon}-\frac{1}{2}\alpha\|Tr\|^{2}){{\int}_{0}^{T}} e^{\theta s}\|u^{n+1}_{s}-{u^{n}_{s}}\|^{2}ds \\&\displaystyle \qquad\quad+(1-\beta\epsilon-\gamma-M^{2}\epsilon-\frac{1}{2}\alpha\|Tr\|^{2}){{\int}_{0}^{T}} e^{\theta s}\|\nabla (u^{n+1}_{s}-{u^{n}_{s}})\|^{2}ds\\ \leq&\displaystyle (C^{2}\epsilon+\frac{\beta}{\epsilon}+\frac{1}{2}\alpha\|Tr\|^{2}){{\int}_{0}^{T}} e^{\theta s}\|{u^{n}_{s}}-u^{n-1}_{s}\|^{2}ds \\&\displaystyle \qquad\quad+(C^{2}\epsilon+\gamma+\frac{1}{2}\alpha\|Tr\|^{2}){{\int}_{0}^{T}} e^{\theta s}\|\nabla({u^{n}_{s}}-u^{n-1}_{s})\|^{2}ds. \end{array}$$

We choose 𝜖 small enough and then 𝜃 such that

$$C^{2}\epsilon+\gamma+\frac{1}{2}\alpha\|Tr\|^{2}< 1-\beta\epsilon-\gamma-M^{2}\epsilon-\frac{1}{2}\alpha\|Tr\|^{2}$$

and

$$\frac{\theta-\frac{2}{\epsilon}-\frac{1}{2}\alpha\|Tr\|^{2}}{1-\beta\epsilon-\gamma-M^{2}\epsilon-\frac{1}{2}\alpha\|Tr\|^{2}}=\frac{C^{2}\epsilon+\frac{\beta}{\epsilon}+\frac{1}{2}\alpha\|Tr\|^{2}}{C^{2}\epsilon+\gamma+\frac{1}{2}\alpha\|Tr\|^{2}}.$$

Setting $\rho =\frac {\theta -\frac {2}{\epsilon }-\frac {1}{2}\alpha \|Tr\|^{2}}{1-\beta \epsilon -\gamma -M^{2}\epsilon -\frac {1}{2}\alpha \|Tr\|^{2}}$, we find that

$$\begin{array}{llll} \|u^{n+1}-u^{n}\|_{\theta,\rho}&\leq\frac{C^{2}\epsilon+\gamma+\frac{1}{2}\alpha\|Tr\|^{2}}{1-\beta\epsilon-\gamma-M^{2}\epsilon-\frac{1}{2}\alpha\|Tr\|^{2}}\|u^{n}-u^{n-1}\|_{\theta,\rho}\\&\leq\cdots\leq\left( \frac{C^{2}\epsilon+\gamma+\frac{1}{2}\alpha\|Tr\|^{2}}{1-\beta\epsilon-\gamma-M^{2}\epsilon-\frac{1}{2}\alpha\|Tr\|^{2}}\right)^{n}\|u^{1}\|_{\theta,\rho} \end{array}$$

with the norm $\|v\|_{\theta ,\rho }:={{\int \limits }_{0}^{T}}e^{\theta s}(\rho \|v_{s}\|^{2}+\|\nabla v_{s}\|^{2})ds$ for v ∈ L²([0,T]; H¹(D)).

We have $\left (\frac {C^{2}\epsilon +\gamma +\frac {1}{2}\alpha \|Tr\|^{2}}{1-\beta \epsilon -\gamma -M^{2}\epsilon -\frac {1}{2}\alpha \|Tr\|^{2}}\right )^{n}\rightarrow 0$, as $n\rightarrow \infty$. It means that $\{u^{n}\}_{n}$ is a Cauchy sequence in L²([0,T]; H¹(D)), and its limit is denoted by u.

For any test function ϕ, we have

$$ \begin{array}{@{}rcl@{}} ({u^{n}_{T}},\phi_{T})\!-({u^{n}_{0}},\phi_{0})& -&{{\int}_{0}^{T}}({u^{n}_{t}},\partial_{t}\phi_{t}) dt={{\int}_{0}^{T}}(\nabla {u^{n}_{t}},\nabla\phi_{t}) dt-{{\int}_{0}^{T}} (\langle b,\nabla {u^{n}_{t}}\rangle, \phi_{t} ) dt\\ & -&{{\int}_{0}^{T}}\! ({g^{n}_{t}},\nabla\phi_{t}) dt - {{\int}_{0}^{T}}\!({f^{n}_{t}},\phi_{t}) dt - \frac{1}{2}{{\int}_{0}^{T}}\!\!{\int}_{\partial D}\! {h^{n}_{t}}(x)\phi_{t}(x) \sigma(dx)dt. \end{array} $$

Taking limits on both sides of the above equation, we obtain

$$ \begin{array}{@{}rcl@{}} (u_{T},\phi_{T})-(u_{0},\phi_{0})& -&{{\int}_{0}^{T}}(u_{t},\partial_{t}\phi_{t}) dt={{\int}_{0}^{T}}(\nabla u_{t},\nabla\phi_{t}) dt-{{\int}_{0}^{T}} (\langle b,\nabla u_{t}\rangle, \phi_{t} ) dt\\ & -&\!\!{{\int}_{0}^{T}} (g_{t},\nabla\phi_{t}) dt - {{\int}_{0}^{T}}(f_{t},\phi_{t}) dt - \frac{1}{2}{{\int}_{0}^{T}}{\int}_{\partial D} h_{t}(x)\phi_{t}(x) \sigma(dx)dt. \end{array} $$

Hence, we prove u is the weak solution of PDE (1).

Uniqueness: Suppose $u,\bar u\in L^{2}([0,T];H^{1}(D))$ are two solutions, we have

$$\begin{array}{llll} &\displaystyle \|u_{0}-\bar u_{0}\|^{2}+{{\int}_{0}^{T}} e^{\theta s}\|\nabla(u_{s}-\bar u_{s})\|^{2} ds\\ =&\displaystyle -\theta {{\int}_{0}^{T}} e^{\theta s}\|u_{s}-\bar u_{s}\|^{2} ds +2{{\int}_{0}^{T}}e^{\theta s}(f_{s}(u_{s},\nabla u_{s})-f_{s}(\bar u_{s},\nabla \bar u_{s}), u_{s}-\bar u_{s}) ds\\ &\displaystyle +2{{\int}_{0}^{T}} {\int}_{D} e^{\theta s}\langle g_{s}(x,u_{s},\nabla u_{s})-g_{s}(x,\bar u_{s},\nabla \bar u_{s}),\nabla(u_{s}(x)-\bar u_{s}(x))\rangle dxds\\ &\displaystyle +{{\int}_{0}^{T}} {\int}_{\partial D}e^{\theta s}(h_{s}(x,u)-h_{s}(x,\bar u))(u_{s}(x)-\bar u_{s}(x)) d\sigma(x) ds\\ &\displaystyle +2{{\int}_{0}^{T}}{\int}_{D} e^{\theta s}\langle b,\nabla (u_{s}-\bar{u}_{s})\rangle(x)(u_{s}(x)-\bar{u}_{s}(x)) dxds. \end{array} $$

By the same calculation in the proof of existence, there is a positive constant k < 1, such that

$$ \|u-\bar u\|^{2}_{\theta, \rho}\leq k \|u-\bar u\|^{2}_{\theta, \rho}. $$

Hence, it follows that $\|u-\bar u\|_{\theta , \rho }=0$, which implies $u=\bar u$, a.e.. □