A nonlocal diffusion problem that approximates the heat equation with Neumann boundary conditions

Remark 1.1

The main advantage of the model given by (1.4) compared with (1.3) is that when one deals with a nonlinear datum of the form $g(y\text{,}t)=f(u(y\text{,}t))$ it is necessary to use an extension of the solution u from $\mathrm{\Omega }$ to ${\mathbb{R}}^N⧹\mathrm{\Omega }$ in the case of (1.3) (notice that such an extension is not trivial since g depends on the solution itself). However, it is not necessary to perform such extension when dealing with (1.6).

Theorem 1.1

For every $u_0\in L^1(\mathrm{\Omega })$ and every $g\in L_{\mathit{loc}}^{\infty }[(0\text{,}\infty )\text{;}{L}^1(\partial \mathrm{\Omega })]$ there exists a unique solution $w^{∊}\in C[[0\text{,}\infty )\text{;}{L}^1(\mathrm{\Omega })]$ to problem (1.6).

Theorem 1.2

Let $\mathrm{\Omega }$ be a bounded $C^{2+\alpha }$ domain, $g\in C^{1+\alpha \text{,}\frac{1+\alpha }{2}}(\overline{({\mathbb{R}}^N⧹\mathrm{\Omega })}\times [0\text{,}T])$ , let v the solution to (1.2) and assume that $v\in C^{2+\alpha \text{,}\frac{\alpha }{2}}(\bar{\mathrm{\Omega }}\times [0\text{,}T])$ , for some $0<\alpha <1$ . Let $w^{∊}$ be the solution to (1.6). Then, for each $t\in [0\text{,}T]$ $$w^{∊}(x\text{,}t)\rightharpoonup v(x\text{,}t)\mathrm{as}∊\to 0\ast -\text{weakly in}{L}^{\infty }(\mathrm{\Omega })\text{.}$$

Theorem 2.1

Let $\mathrm{\Omega }$ be a bounded $C^{2+\alpha }$ domain, $g\in C^{1+\alpha \text{,}\frac{1+\alpha }{2}}(\overline{({\mathbb{R}}^N⧹\mathrm{\Omega })}\times [0\text{,}T])\text{,}v\in C^{2+\alpha \text{,}\frac{\alpha }{2}}(\bar{\mathrm{\Omega }}\times [0\text{,}T])$ the solution to (1.2), for some $0<\alpha <1$ . Let $u^{∊}$ be a solution to (1.3). Then, there is an adequate constant K such that, for each $t\in [0\text{,}T]$ , $$u^{∊}(x\text{,}t)\rightharpoonup v(x\text{,}t)\ast -\text{weakly in}{L}^{\infty }(\mathrm{\Omega })\text{.}$$

Proof

Proof Proof of Theorem 1.2

With the aim to prove Theorem 1.2, we just want to obtain that $w^{∊}(x\text{,}t)$ the solution to (1.6) is close to $u^{∊}(x\text{,}t)$ the solution to (1.3). To this end we consider equation verified by the difference $$u^{∊}(x\text{,}t)-w^{∊}(x\text{,}t)$$ that is,

$$\begin{array}{cc}\hfill & {(u^{∊}(x\text{,}t)-w^{∊}(x\text{,}t))}_t=\frac{1}{{∊}^{N+2}}{\int }_{\mathrm{\Omega }}J\left(\frac{x-y}{∊}\right)((u^{∊}(y\text{,}t)-w^{∊}(y\text{,}t))-(u^{∊}(x\text{,}t)-w^{∊}(x\text{,}t)))\mathit{dy}+A_{∊}(x\text{,}t)\hfill \\ \hfill & (u^{∊}(x\text{,}0)-w^{∊}(x\text{,}0))=0\text{,}\hfill \end{array}$$ where, from the fact that $w^{∊}\text{,}{u}^{∊}$ satisfies (1.6) and (1.3) respectively, we get that $$A_{∊}(x\text{,}t)=\frac{K}{{∊}^{N+1}}{\int }_{{\mathbb{R}}^N⧹\mathrm{\Omega }}J\left(\frac{x-y}{∊}\right)g(y)\mathit{dy}-\frac{C_1}{{∊}^N}{\int }_{\partial \mathrm{\Omega }}J\left(\frac{x-y}{∊}\right)g(y){\mathit{dS}}_y\text{.}$$

After integration in $\mathrm{\Omega }$ of (2.1), we consider $A_{∊}(x\text{,}t)$ and we decompose it as follows:

$${\int }_{\mathrm{\Omega }}{A}_{∊}(x\text{,}t)=I_1-I_2\text{,}$$ with $$I_1=\frac{K}{{∊}^{N+1}}{\int }_{\mathrm{\Omega }}{\int }_{{\mathbb{R}}^N-⧹\mathrm{\Omega }}J\left(\frac{x-y}{∊}\right)g(y)\mathit{dy}\mathit{dx}\text{,}$$ and $$I_2=\frac{C_1}{{∊}^N}{\int }_{\mathrm{\Omega }}{\int }_{\partial \mathrm{\Omega }}J\left(\frac{x-y}{∊}\right)g(y){\mathit{dS}}_y\mathit{dx}\text{.}$$

Let $\delta$ such that $2∊<\delta$ . We consider the following sets: $${\mathrm{\Gamma }}_{\delta }^{+}=\{\tilde{x}\in \mathrm{\Omega }/d(\tilde{x}\text{,}\partial \mathrm{\Omega })⩽\delta \}\text{,}$$ $${\mathrm{\Gamma }}_{\delta }^{-}=\{\tilde{x}\in ({\mathbb{R}}^N⧹\mathrm{\Omega })/d(\tilde{x}\text{,}\partial \mathrm{\Omega })⩽\delta \}\text{,}$$ and $${\mathrm{\Gamma }}_{[-\delta \text{,}\delta ]}=\{\tilde{x}\in {\mathbb{R}}^N/d(\tilde{x}\text{,}\partial \mathrm{\Omega })⩽\delta \}={\mathrm{\Gamma }}_{\delta }^{+}\cup {\mathrm{\Gamma }}_{\delta }^{-}\text{.}$$

Note that $$I_1=\frac{K}{{∊}^{N+1}}{\int }_{{\mathrm{\Gamma }}_{∊}^{+}}{\int }_{{\mathrm{\Gamma }}_{∊}^{-}}J\left(\frac{x-y}{∊}\right)g(y)\mathit{dy}\mathit{dx}\text{.}$$

When $\tilde{y}\in \partial \mathrm{\Omega }$ and $s\in [-\delta \text{,}\delta ]$ , we use the change of variables $$y=\tilde{y}+n(\tilde{y})s$$ being $n(\tilde{y})$ the normal vector at $\tilde{y}$ to obtain

$$I_1=\frac{1}{{∊}^N}{\int }_{{\mathrm{\Gamma }}_{∊}^{+}}\left[\frac{1}{∊}{\int }_0^{∊}{\int }_{\partial \mathrm{\Omega }}\mathit{KJ}\left(\frac{x-(\tilde{y}+n(\tilde{y})s)}{∊}\right)g(\tilde{y}+n(\tilde{y})s)|\bar{J}(\tilde{y}\text{,}s)|{\mathit{dS}}_{\tilde{y}}\mathit{ds}\right]\mathit{dx}\text{,}$$ where $|\bar{J}(\tilde{y}\text{,}s)|$ is the Jacobian of the change of variable. In the same way we can transform $I_2$ and obtain

(2.4)

$$I_2=\frac{C_1}{{∊}^N}{\int }_{{\mathrm{\Gamma }}_{∊}^{+}}{\int }_{\partial \mathrm{\Omega }}J\left(\frac{x-\tilde{y}}{∊}\right)g(\tilde{y}){\mathit{dS}}_{\tilde{y}}\mathit{dx}\text{.}$$

Now we consider the extension of g given by $$g(\tilde{y}+n(\tilde{y})s)=g(\tilde{y})$$ and from (2.3) and (2.4) we get

$$I_1-I_2=\frac{1}{{∊}^N}{\int }_{{\mathrm{\Gamma }}_{∊}^{+}}\left[{\int }_{\partial \mathrm{\Omega }}\left(I_3\right)d\tilde{y}\right]\mathit{dx}\text{,}$$ being $$I_3=\frac{1}{∊}{\int }_0^{∊}\mathit{KJ}\left(\frac{x-\tilde{y}}{∊}-n(\tilde{y})\frac{s}{∊}\right)g(\tilde{y}+n(\tilde{y})s)|\bar{J}(\tilde{y}\text{,}s)|\mathit{ds}-C_{1}J\left(\frac{x-\tilde{y}}{∊}\right)g(\tilde{y}+n(\tilde{y})s)\text{.}$$

Using the change of variables $$\tilde{s}=\frac{s}{∊}$$ in the first integral we obtain

$$I_3={\int }_0^1\mathit{KJ}\left(\frac{x-\tilde{y}}{∊}-n(\tilde{y})\tilde{s}\right)g(\tilde{y}+n(\tilde{y})s)|\bar{J}(\tilde{y}\text{,}∊\tilde{s})|d\tilde{s}-C_{1}J\left(\frac{x-\tilde{y}}{∊}\right)g(\tilde{y}+n(\tilde{y})s)\text{.}$$

Taking into account (2.2), (2.5) and (2.6), making the change of variables $$z=\frac{x-\tilde{y}}{∊}$$ and considering that $$x=\tilde{y}+∊z$$ we obtain

$${\int }_{\mathrm{\Omega }}{A}_{∊}(x\text{,}t)={\int }_{\partial \mathrm{\Omega }}\left[{\int }_{\{z/(\tilde{y}+∊z)\in \mathrm{\Omega }\}}\left({\int }_0^1\mathit{KJ}(z-n(\tilde{y})\tilde{s})|\bar{J}(\tilde{y}\text{,}∊\tilde{s})|d\tilde{s}-C_{1}J(z)\right)\mathit{dz}\right]g(\tilde{y}){\mathit{dS}}_{\tilde{y}}\text{.}$$

Choosing $C_1$ appropriately in such a way that $${\displaystyle \underset{∊\to 0}{\lim }}{\int }_{\{z/(\tilde{y}+∊z)\in \mathrm{\Omega }\}}\left({\int }_0^1\mathit{KJ}(z-n(\tilde{y})\tilde{s})|\bar{J}(\tilde{y}\text{,}∊\tilde{s})|d\tilde{s}-C_{1}J(z)\right)\mathit{dz}=0\text{,}$$ we conclude

$${\displaystyle \underset{∊\to 0}{\lim }}{\int }_{\mathrm{\Omega }}{A}_{∊}(x\text{,}t)=0\text{.}$$

With this estimate we can control $$u^{∊}(x\text{,}t)-w^{∊}(x\text{,}t)\text{.}$$

In fact, let $z_{∊}$ be a nonnegative solution of the problem

$$\begin{array}{cc}{z}_t^{∊}(x\text{,}t)=\frac{1}{{∊}^{N+2}}{\int }_{\mathrm{\Omega }}J\left(\frac{x-y}{∊}\right)(z^{∊}(y\text{,}t)-z^{∊}(x\text{,}t))\mathit{dy}+|A_{∊}(x\text{,}t)|\hfill & (x\text{,}t)\in \bar{\mathrm{\Omega }}\times (0\text{,}T)\text{,}\hfill \\ z(x\text{,}0)=0\text{,}\hfill & x\in \bar{\mathrm{\Omega }}\text{.}\hfill \end{array}$$

Integrating (2.9) in $\mathrm{\Omega }$ and using the symmetry of J we have $$\frac{\partial }{\partial t}{\int }_{\mathrm{\Omega }}{z}^{∊}(x\text{,}t)\mathit{dx}={\int }_{\mathrm{\Omega }}|A_{∊}(x\text{,}t)|\mathit{dx}$$ and therefore $${\int }_{\mathrm{\Omega }}{z}^{∊}(x\text{,}t)\mathit{dx}={\int }_0^T{\int }_{\mathrm{\Omega }}|A_{∊}(x\text{,}t)|\mathit{dxdt}\text{.}$$

From (2.8) we obtain that

$${\int }_{\mathrm{\Omega }}{z}^{∊}(x\text{,}t)\mathit{dx}\to 0\mathrm{when}∊\to 0\text{.}$$

Now, since $$A_{∊}(x\text{,}t)⩽|A_{∊}(x\text{,}t)|\text{,}$$ we get that $$|u^{∊}(x\text{,}t)-w^{∊}(x\text{,}t)|⩽z_{∊}(x\text{,}t)\text{.}$$

Hence, from (2.10), it holds that

$${\int }_{\mathrm{\Omega }}|u^{∊}(x\text{,}t)-w^{∊}(x\text{,}t)|\mathit{dx}\to 0\text{,}\mathrm{when}∊\to 0\text{.}$$

To finish the proof of the theorem, let $\theta \in L^{\infty }(\mathrm{\Omega })$ , then $${\int }_{\mathrm{\Omega }}{w}^{∊}\theta \mathit{dx}⩽{\int }_{\mathrm{\Omega }}|w^{∊}(x\text{,}t)-u^{∊}(x\text{,}t)||\theta (x)|\mathit{dx}+{\int }_{\mathrm{\Omega }}{u}^{∊}(x\text{,}t)\theta (x)\mathit{dx}\text{.}$$

Taking into account (2.11) and Theorem 2.1 we conclude that $${\int }_{\mathrm{\Omega }}{w}^{∊}\theta \mathit{dx}\to {\int }_{\mathrm{\Omega }}v(x\text{,}t)\theta \mathit{dx}\text{,}\mathrm{when}∊\to 0\text{,}$$ as we wanted to show. □