The motivation behind weak solutions to partial differential equations (PDEs) in divergence form, based on the formula for integration by parts, is well known. In this way, one can develop the notion of weak derivative for suitable classes of Lebesgue integrable functions. This section discusses some fine properties of functions found within one such class, the so-called Sobolev space with given , , and domain .

#### Weak Maximum Principles

In studying the existence of weak solutions to quasilinear PDE, one can consider the method of sub and super solutions. It is a constructive approach in which a weak maximum principle plays a key role. An example of this principle reads as follows.

**Theorem 1. **(Weak Maximum Principle) *Let be a bounded domain with Lipschitz boundary, and let be such that on in the sense that . Furthermore, suppose that in weakly, for some , which is to say that *

*for all with a.e in . Then a.e in .*

I met this result when I attended The University of Oxford’s course on Fixed Point Methods for Nonlinear PDEs (HT 2020). The conclusion follows by taking in the weak definition of as given in the statement of the theorem, to then conclude , which is obviously non-negative a.e. This idea can be taken further in the setting of slightly more elaborate PDEs formulated over bounded open sets where the constant may be negative, which the following result of mine shows.

**Theorem 2. **(A Generalisation) *Let be a bounded open set. Consider a pair of real constants such that and . Further, let be a vector field for which* *Suppose is such that on in the sense that , and holds in weakly, which is to say*

*for all with a.e in . Then a.e in .*

**Note:** Here we define , while denotes the Poincare constant associated with Poincare’s inequality on .

*Proof.* Write where we let and . Then, by assumption we know that and is non-negative a.e in . Consequently, the weak interpretation of implies

which simplifies to

Since we find

giving

By Poincare’s inequality on , which reads for all , we arrive at

or

Now, if , then we see immediately that a.e in and so a.e in , by virtue of Poincare’s inequality. And so, which is non-negative a.e. On the other hand, if we find by Poincare’s inequality,

Because , it follows that we must have a.e in once more. This completes the proof of the theorem.

**Remark 1.** *Of course, in the above result we may replace with without changing the conclusion. This result, together with the Lax-Milgram Theorem which one could use to prove a solution operator is well-defined, we can develop examples of quasilinear PDE such as *

*formulated over a bounded open set , with*

*homogeneous Dirichlet boundary condition on ,**essentially bounded vector field as in Theorem 2,**suitably negative as given by Theorem 2, and**a nonlinear function,*

*for which the method of sub and super solutions applies to prove the existence of a weak solution.*

Let be a matrix of measurable functions defined almost-everywhere over an open set . If there exists a constant such that

for all and almost every , we’ll say is *-coercive over *. Assume is an essentially bounded function over with for some . With this assumption, let’s proceed to generalise Theorem 2 to operators in divergence form given by

**Theorem 3.** *Let be a bounded open set. Consider a triple of real constants such that the matrix is -coercive over and . Further, assume is a vector field for which , and let be an essentially bounded function over with . Suppose is such that on in the sense that , and that holds in weakly, which is to say*

*for all with a.e in . Then a.e in .**Proof.* Write where we let and . Then, by assumption we know that and is non-negative a.e in . Consequently, the weak interpretation of implies

which simplifies to

Since we find

By coercivity of we have

Therefore,

By Poincare’s inequality on , we arrive at

or

Since almost-everywhere in , we find

and so Poincare’s inequality implies

But, it is assumed that the triple satisfies . Thus, we see immediately a.e in and so a.e in , by virtue of Poincare’s inequality. Consequently, which is non-negative a.e in , as required.

**Remark 2.** *In both results the conditions on the constants can be interpreted geometrically. In the first instance we have inequality constraints on that yield an unbounded region delineated by three straight lines in : , and . On the other hand, in the second result the inequality constraints on define an unbounded region within the first quadrant of with part of its boundary found in a plane passing through the origin with normal . Notably, projecting this plane in the horizontal plane where , we observe an unbounded region which contains the one given by the constraints on in Theorem 2.*

Copyright © 2021 Yohance Osborne