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I have a question regarding based on text regarding 2nd-order partial differential equations.

Consider the quasilinear 2nd-order partial differential equation: $$-\text{div}(a(x,y,\nabla u)) + c(x,u,\nabla u) = g.$$ I have some questions regarding a paragraph describing $a(x,u,\frac{}{})$. The paragraph is as follows:

$p'' \in (1, + \infty)$ will denote the growth of the leading nonlinearity $a(x,u,\frac{}{})$ which essentially determines the setting and the other data qualification. Also, $a(x,u,\frac{}{})$ will be assumed to behave monotonically which is related to the adjective 'elliptic'. For the linear case $a(x,r,s) = \mathcal{A}s $, the montonicity and coercivity below implies the matrix $\mathcal{A}$ is positive definite, which is conventionally called "elliptic", contrary to $\mathcal{A}$ indefinte which is addressed as hyperbolic(resp parabolic). Questions:

1.I first want to confirm 'nonlinearity' here simply means that the mapping $a: \Omega \times \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is a nonlinear mapping?

2.What is $\mathcal{A}s$ explicitly?

3.Why is $a(x,r,s) = \mathcal{A}s$ only if $a(x,r,s)$ is linear?

4.How would elliplicity be defined for the nonlinear case?

Thanks a lot for assistance.

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  • $\begingroup$ What book is it from? Need a bit more context $\endgroup$
    – Erik
    Jan 6, 2014 at 20:47

2 Answers 2

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'nonlinearity' here simply means that the mapping $a:\Omega \times \mathbb R\times \mathbb R^n \to \mathbb R^n$ is a nonlinear mapping?

Close, but not quite. For example, the equation $\operatorname{div}((1+x^2)\nabla u) =0$ is linear. What matters is how the PDE involves the unknown $u$, the dependence on $x$ need not be linear. Precisely, we need $$ a(x,\alpha u+\beta v,\nabla (\alpha u+\beta v)) = \alpha a(x,u,\nabla u) +\beta a(x,v,\nabla v)\tag{1} $$ to hold. And similarly for $c$. That is, the PDE is linear if and only if

  • $(r,s)\mapsto a(x,r,s)$ is a linear map for every fixed $x$.
  • $(r,s)\mapsto c(x,r,s)$ is a linear map for every fixed $x$.

How $a$ and $c$ depend on $x$ is not important for linearity of the PDE. However, if they are independent of $x$ we get a nice linear equation with constant coefficients.

What is $\mathbb As$ explicitly?

$\mathbb A$ is a constant matrix; $s$ is a vector. Thus, $\mathbb As$ is the result of applying the matrix to a vector.

Why is $a(x,r,s)=\mathbb As$ only if $a(x,r,s)$ is linear?

Every linear map is represented by a matrix. A nonlinear map cannot be represented in this way.

How would ellipticity be defined for the nonlinear case?

However the author of your book chooses to. Definitions vary a lot for nonlinear, possibly degenerate, PDE. An important property that $a$ should have is monotonicity: $\langle( a(x, s)-a(x,s') ,s-s'\rangle$ should be positive. This allows one to at least hope for the uniqueness of the Dirichlet boundary problem, if the lower order terms cooperate. But more assumptions are needed to say something about the PDE, and what they will be depends on what you want to do.

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  • $\begingroup$ Thanks for your response. I know that $\mathbb{A}$ is a matrix and $s$ is a vector since $s: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$, what I wanted to know is how exactly does this matrix $\mathbb{A}$ look? Also in your answer to question three you said "every linear map is represented by a matrix" but above you said that $a(x,r,s)$ is not assumed linear in the normal sense of a linear mapping but rather in the PDE sense which you described above? Is the author referring to two different ideas of linearity in this text? $\endgroup$
    – user103184
    Jan 7, 2014 at 12:12
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    $\begingroup$ @Moses How $\mathbb A$ looks: it should be positive definite. The closer it is to the identity matrix, the closer the equation is to the Laplace equation, which the the ideal of all elliptic equations. Yes, there are multiple notions of linearity here. No surprise: when we talk about a function with three arguments, like $a$, there are already several ways for it to be linear: it could be in one argument, in every argument, or in some of them. I think the author mostly means the PDE being linear, which is translated, without explicit mention, into $a$ being linear in the unknown function. $\endgroup$ Jan 7, 2014 at 16:14
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    $\begingroup$ Okay, I still have a couple of questions, sorry to bother so much(I could post as a new question if you prefer). So you are saying that given $a(x,r,s)$ is linear in the PDE sense it follow that $a(x,r,s)$ is linear in $s$ in the usual mapping sense? Secondly what is the general connection between linearity in PDE and usual mapping linearity, does one imply the other? Lastly, I read somewhere that every PDE operator is a linear transformation, can you confirm if this is true? Thanks. $\endgroup$
    – user103184
    Jan 7, 2014 at 16:28
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    $\begingroup$ @Moses I added details to the post. Concerning "every PDE operator is a linear transformation" -- please understand that people make definitions for their own convenience. If someone writes a book and finds it convenient to define "PDE operator is a linear transformation of some kind", then that statement is correct in their book. In another book you will find a different definition of a PDE operator, which will include nonlinear operators as well. $\endgroup$ Jan 7, 2014 at 17:36
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I agree with Moses in that the term 'linear' doesn't seem standard even when used for PDEs. Let me give my definition based on Post No bulls answer and the book by Evans 'Partial Differential Equations'. Any comments and corrections would be welcome.

This is the definition of a PDE given in Evans book, it is a function of the form:

$F(D^{k}u(x),D^{k-1}u(x),...,Du(x),u(x),x) = 0$

where $F: \mathbb{R}^{n^{k}} \times \mathbb{R}^{n^{k-1}} \times ...\times \mathbb{R}^{n} \times U \rightarrow \mathbb{R}$ is given and $u: U \rightarrow \mathbb{R}$ is the unknown.

Then in Evans book a PDE is linear if it has the form $\sum_{|\alpha| \leq k}a_{\alpha}(x)D^{\alpha}u = f(x)$ for given functions $a_{\alpha},f$.

So by Post Bulls explanation and Evans book's definition I would say in conclusion that a linear PDE is one that it a linear transformation(satisfies additivity and homogeneity of degree 1) for all independent variables of $F$ other than $x$.

That seems to be the most general notion of linearity for PDEs, there are obviously weaker forms of linearity such as quasilinearity and semi-linearity as well.

I welcome any comments or criticisms regarding this.

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