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Let $$ \mathcal{L}\equiv\sum_{i,j}^{n}a_{i,j}\frac{\partial^{2}}{\partial x_{i}\,\partial x_{j}}+\sum_{i}^{n}b_{i}\frac{\partial}{\partial x_{i}}+c $$ be uniformly elliptic on $\Omega\equiv\mathbb{R}^{n}\times\left(0,T\right)$. Let $$ \mathcal{M\equiv L}-\frac{\partial}{\partial t} $$ Consider the parabolic equality $$ \mathcal{M}u=f\text{ on }\Omega. $$ Lastly, suppose $u\left(\mathbf{x},0\right)$ is monotone increasing in $\mathbf{x}$ and that $f\left(\mathbf{x},t\right)$ is monotone decreasing in $\mathbf{x}$ for all times $t$. A function $g:\mathbb{R}^n \rightarrow \mathbb{R}$ is monotone increasing if for all $\mathbf{x},\mathbf{y}\in\mathbb{R}^n$, $g\left(\mathbf{x}\right)\leq g\left(\mathbf{y}\right)$ whenever $\mathbf{x}\leq \mathbf{y}$. Monotone decreasing is defined similarly.

First, consider the case of $a_{i,j}$, $b_{i}$, $c$ being functions of $t$ only. Define $v\left(\mathbf{x},t\right)=u\left(\mathbf{x}-\epsilon,t\right)$ (where $\epsilon \in \mathbb{R}^n$ with $\epsilon > 0$) and note that $$ \mathcal{M}v=f\left(\mathbf{x}-\epsilon,t\right)\text{ on }\Omega. $$ Let $w=u-v$ so that $$ \mathcal{M}w=f\left(\mathbf{x},t\right)-f\left(\mathbf{x}-\epsilon,t\right)\leq0\text{ on }\Omega. $$ Since $u$ is monotone at time zero, we have $$w\left(\mathbf{x},0\right)=u\left(\mathbf{x},0\right)-v\left(\mathbf{x},0\right)=u\left(\mathbf{x},0\right)-u\left(\mathbf{x}-\epsilon,0\right)\geq0 \text{ on } \mathbb{R}^n.$$ These last two inequalities yield $w\geq0$ (and hence $u$ is monotone in $\mathbf{x}$) everywhere in $\Omega$.

However, this argument is rather restrictive, as it requires the assumption that the coefficients $a_{i,j}$, $b_{i}$, and $c$ are constant in space (i.e. w.r.t. $\mathbf{x}$). Does anyone know of any relaxations to the above? (see edit below)

EDIT: Some things were forgotten that are important: We need to assume the boundedness of the coefficients in the operator, along with $$\liminf_{\left|\mathbf{x}\right|\rightarrow\infty} \mathbf{u} \geq 0$$ for all times.

EDIT: I have found a potentially interesting relaxation. Suppose $\mathbf{u}$ is convex in $\mathbf{x}$ on $\Omega$. Let $A\in\mathbb{R}^{n\times n}$ be the matrix with $a_{j,k}$ in the $j^{\text{th}}$ row, $k^{\text{th}}$ column. Since $\mathcal{L}$ is assumed to be uniformly elliptic, $A$ is positive semidefinite. Denote this by $A\succeq0$. Now, relax the requirement on the coefficients $a_{j,k}$ as follows: $A\left(\mathbf{x}+\epsilon,t\right)-A\left(\mathbf{x},t\right)\succeq0$ on $\Omega$ for all $\epsilon>0$. Note that in the one-dimensional case (i.e. $n=1$), this requirement becomes $a\equiv a_{1,1}$ is monotone increasing in space. Noting that $$ \left(A\left(\mathbf{x}+\epsilon,t\right)-A\left(\mathbf{x},t\right)\right)\circ\left[\nabla_{\mathbf{x}}u_{i}\right]\left(\mathbf{x},t\right)\succeq0 $$ where $\nabla$ denotes the Hessian operator and $\circ$ the Hadamard (entrywise) product, we get $$ \sum_{i,j}^{n}a_{i,j}\left(\mathbf{x},t\right)\frac{\partial^{2}u}{\partial x_{i}\,\partial x_{j}}\leq\sum_{i,j}^{n}a_{i,j}\left(\mathbf{x}+\epsilon,t\right)\frac{\partial^{2}u}{\partial x_{i}\,\partial x_{j}}. $$ Using the above, we arrive once again at $\mathcal{M}w\leq 0$. This brings me to a more interesting question...

Are there better relaxations than the above when $u\left(\mathbf{x},t\right)$ is convex in $\mathbf{x}$?

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What is $\epsilon$ for the case $n>1$ and how monotonicity in this case is defined? –  Andrew Jun 27 '13 at 20:23
    
Good question, sorry for the ambiguity. $\epsilon$ is in $\mathbb{R}^n$ with $\epsilon > 0$. A function $f:A \rightarrow B$ defined on partially ordered sets $A$ and $B$ is monotone (increasing) if for all $x,y\in A$ $f\left(x\right)\leq f\left(y\right)$ whenever $x\leq y$. Monotone decreasing is defined similarly. (I have now added these edits to the original question) –  par Jun 27 '13 at 20:27
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There seem to be some assumptions on your partial order which are not explicit. Is it compatible with sums for instance ? With scaling by positive constants ? If you said which order you have in mind, your question would be clearer I think. –  Thomas Richard Jul 10 '13 at 18:07
    
@ThomasRichard: Sorry for the late response. I did not see your comment for some reason. I realize now that my wording was not very clear; I was really referring to the natural order on $\mathbb{R}^n$: i.e. if for each $i$, $x_i \geq y_i$, then $\mathbf{x} \geq \mathbf{y}$. So yes, it is compatible with summation and scaling by positive constants. –  par Jul 18 '13 at 9:04

1 Answer 1

I am not convinced the argument works under the current specifications you have given (it can, but you need to make them). You claim that the last two inequalities lead to the desired result (I assume you are using a maximum principle to achieve this). However, not only is the domain unbounded in this problem, but potentially, the coefficient functions (for example $c(t) = 1/t$ for all $t>0$) and the solutions. I am not aware of any maximum principles for parabolic operators that can handle arbitrarily large solutions (or the function $c$ above). Of course, if you are using some other means to obtain these inequalities then I admit that this answer is probably unhelpful.

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You're right. I do have to assume that the coefficients appearing in the operator are at least continuous and bounded. I don't think the (un)boundedness of the domain is an issue. –  par Jul 12 '13 at 10:01
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You also need to specify the behaviour of the solution to the differential inequality as $|x|\to\infty$ (namely, how the solution can blow up ... or not blow up). See the link below for a good reason why this is required. Additionally, with regard to max principles you are intending to use, you can probably include spatial inhomogeneity in the coefficients without messing things up too much. mathoverflow.net/questions/82408/… –  JCM Jul 12 '13 at 14:04
    
Oh right. I forgot to mention that. You are absolutely correct. I think something weaker than a growth condition is sufficient here, such as $\liminf_{\left|x\right|\rightarrow\infty} \mathbf{u} \geq 0$. Adding to the edits. –  par Jul 13 '13 at 2:27
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Now your results seem reasonable to me, good. –  JCM Jul 16 '13 at 16:25
    
Thank you for your help :) –  par Jul 18 '13 at 9:00

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