Partial Differential Equations/Test functions

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Motivation

Before we dive deeply into the chapter, let's first motivate the notion of a test function. Let's consider two functions which are piecewise constant on the intervals [0,1),[1,2),[2,3),[3,4),[4,5) and zero elsewhere; like, for example, these two:

Let's call the left function f1, and the right function f2.

Of course we can easily see that the two functions are different; they differ on the interval [4,5); however, let's pretend that we are blind and our only way of finding out something about either function is evaluating the integrals

φ(x)f1(x)dx and φ(x)f2(x)dx

for functions φ in a given set of functions 𝒳.

We proceed with choosing 𝒳 sufficiently clever such that five evaluations of both integrals suffice to show that f1f2. To do so, we first introduce the characteristic function. Let A be any set. The characteristic function of A is defined as

χA(x):={1xA0xA

With this definition, we choose the set of functions 𝒳 as

𝒳:={χ[0,1),χ[1,2),χ[2,3),χ[3,4),χ[4,5)}

It is easy to see (see exercise 1), that for n{1,2,3,4,5}, the expression

χ[n1,n)(x)f1(x)dx

equals the value of f1 on the interval [n1,n), and the same is true for f2. But as both functions are uniquely determined by their values on the intervals [n1,n),n{1,2,3,4,5} (since they are zero everywhere else), we can implement the following equality test:

f1=f2φ𝒳:φ(x)f1(x)dx=φ(x)f2(x)dx

This obviously needs five evaluations of each integral, as #𝒳=5.

Since we used the functions in 𝒳 to test f1 and f2, we call them test functions. What we ask ourselves now is if this notion generalises from functions like f1 and f2, which are piecewise constant on certain intervals and zero everywhere else, to continuous functions. The following chapter shows that this is true.

Bump functions

In order to write down the definition of a bump function more shortly, we need the following two definitions: Template:TextBox

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Now we are ready to define a bump function in a brief way:

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These two properties make the function really look like a bump, as the following example shows:

The standard mollifier η in dimension d=1

Example 3.4: The standard mollifier η, given by

η:d,η(x)=1c{e11x2 if x2<10 if x21

, where c:=B1(0)e11x2dx, is a bump function (see exercise 2).

Schwartz functions

As for the bump functions, in order to write down the definition of Schwartz functions shortly, we first need two helpful definitions. Template:TextBox

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Now we are ready to define a Schwartz function.

Template:TextBox By xαβϕ we mean the function xxαβϕ(x).

f(x,y)=ex2y2

Example 3.8: The function

f:2,f(x,y)=ex2y2

is a Schwartz function.

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This means for example that the standard mollifier is a Schwartz function.

Proof:

Let φ be a bump function. Then, by definition of a bump function, φ𝒞(d). By the definition of bump functions, we choose R>0 such that

supp φBR(0)

, as in d, a set is compact iff it is closed & bounded. Further, for α,β0d arbitrary,

xαβφ(x):=supxd|xαβφ(x)|=supxBR(0)|xαβφ(x)|supp φBR(0)=supxBR(0)(|xα||βφ(x)|)rules for absolute valuesupxBR(0)(R|α||βφ(x)|)i{1,,d},(x1,,xd)BR(0):|xi|R<Extreme value theorem

Convergence of bump and Schwartz functions

Now we define what convergence of a sequence of bump (Schwartz) functions to a bump (Schwartz) function means.

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Proof:

Let Od be open, and let (φl)l be a sequence in 𝒟(O) such that φlφ𝒟(O) with respect to the notion of convergence of 𝒟(O). Let thus Kd be the compact set in which all the supp φl are contained. From this also follows that supp φK, since otherwise φlφ|c|, where c is any nonzero value φ takes outside K; this would contradict φlφ with respect to our notion of convergence.

In d, ‘compact’ is equivalent to ‘bounded and closed’. Therefore, KBR(0) for an R>0. Therefore, we have for all multiindices α,β0d:

xαβφlxαβφ=supxd|xαβφl(x)xαβφ(x)| definition of the supremum norm=supxBR(0)|xαβφl(x)xαβφ(x)| as supp φl,supp φKBR(0)R|α|supxBR(0)|βφl(x)βφ(x)|i{1,,d},(x1,,xd)BR(0):|xi|R=R|α|supxd|βφl(x)βφ(x)| as supp φl,supp φKBR(0)=R|α|βφl(x)βφ(x) definition of the supremum norm0,l since φlφ in 𝒟(O)

Therefore the sequence converges with respect to the notion of convergence for Schwartz functions.

The ‘testing’ property of test functions

In this section, we want to show that we can test equality of continuous functions f,g by evaluating the integrals

df(x)φ(x)dx and dg(x)φ(x)dx

for all φ𝒟(O) (thus, evaluating the integrals for all φ𝒮(d) will also suffice as 𝒟(O)𝒮(d) due to theorem 3.9).

But before we are able to show that, we need a modified mollifier, where the modification is dependent of a parameter, and two lemmas about that modified mollifier.

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Proof:

From the definition of η follows

supp η=B1(0).

Further, for R>0

xRB1(0)xR1xRxBR(0)

Therefore, and since

xsupp ηRxRsupp η

, we have:

xsupp ηRxBR(0)

In order to prove the next lemma, we need the following theorem from integration theory:

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We will omit the proof, as understanding it is not very important for understanding this wikibook.

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Proof:

dηR(x)dx=dη(xR)/RddxDef. of ηR=dη(x)dxintegration by substitution using xRx=B1(0)η(x)dxDef. of η=B1(0)e11xdxB1(0)e11xdxDef. of η=1

Now we are ready to prove the ‘testing’ property of test functions:

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Proof:

Let xd be arbitrary, and let ϵ>0. Since f is continuous, there exists a δ>0 such that

yBδ(x):|f(x)f(y)|<ϵ

Then we have

|f(x)df(y)ηδ(xy)dy|=|d(f(x)f(y))ηδ(xy)dy|lemma 3.16d|f(x)f(y)|ηδ(xy)dytriangle ineq. for the  and ηδ0=Bδ(0)|f(x)f(y)|ηδ(xy)dylemma 3.14Bδ(0)ϵηδ(xy)dymonotony of the ϵlemma 3.16 and ηδ0

Therefore, df(y)ηδ(xy)dyf(x),δ0. An analogous reasoning also shows that dg(y)ηδ(xy)dyg(x),δ0. But due to the assumption, we have

δ>0:dg(y)ηδ(xy)dy=df(y)ηδ(xy)dy

As limits in the reals are unique, it follows that f(x)=g(x), and since xd was arbitrary, we obtain f=g.

Remark 3.18: Let f,g:d be continuous. If

φ𝒮(d):dφ(x)f(x)dx=dφ(x)g(x)dx,

then f=g.

Proof:

This follows from all bump functions being Schwartz functions, which is why the requirements for theorem 3.17 are met.

Exercises

  1. Let b and f: be constant on the interval [b1,b). Show that

    y[b1,b):χ[b1,b)(x)f(x)dx=f(y)
  2. Prove that the standard mollifier as defined in example 3.4 is a bump function by proceeding as follows:
    1. Prove that the function

      x{e1xx>00x0

      is contained in 𝒞().

    2. Prove that the function

      x1x

      is contained in 𝒞(d).

    3. Conclude that η𝒞(d).
    4. Prove that supp η is compact by calculating supp η explicitly.
  3. Let Od be open, let φ𝒟(O) and let ϕ𝒮(d). Prove that if α,β0d, then αφ𝒟(O) and xαβϕ𝒮(d).
  4. Let Od be open, let φ1,,φn𝒟(O) be bump functions and let c1,,cn. Prove that j=1ncjφj𝒟(O).
  5. Let ϕ1,,ϕn be Schwartz functions functions and let c1,,cn. Prove that j=1ncjϕj is a Schwartz function.
  6. Let α0d, let p(x):=ςαcςxς be a polynomial, and let ϕlϕ in the sense of Schwartz functions. Prove that pϕlpϕ in the sense of Schwartz functions.

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