Find the set of all functions that satisfy the inequality

In summary, the problem is to find all harmonic functions u(x,y,z) that satisfy the inequality |u(x,y,z)| \leq A(1 + x^2 + y^2 + z^2), where A is a nonzero constant. The work involved rewriting the expression and considering solutions of the form u = e^{\mathbf{k} \cdot \mathbf{x}}, which is harmonic if and only if \sum k_i^2 = 0. The solution is given by the polynomial solutions of the form u = a + \mathbf{b} \cdot \mathbf{x} + \mathbf{x}^T C \mathbf{x} for some scalar a, vector \mathbf{b
  • #1
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Homework Statement
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Relevant Equations
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Problem:

Find the set of all harmonic functions ##u(x,y,z)## that satisfy the following inequality in all of ##R^3##

$$|u(x,y,z)|\leq A+A(x^2+y^2+z^2)$$
where ##A## is a nonzero constant.

Work:

I removed the absolute value bars by re-writing the expression
$$-C-C(x^2+y^2+z^2)\leq u\leq C+C(x^2+y^2+z^2)$$

I am looking for a function ##u## that satisfies ##\Delta u = 0## in ##R^3##, because we can take ##A## as large as needed to satisfy the inequality for any ##u##

List of solutions: all harmonic functions, too many to list.

I suppose one obvious solution is obtained by setting ##\leq \Rightarrow =##

$$\pm u(x,y,z)=C+C(x^2+y^2+z^2)$$

I think the solution is supposed to be in abstract form to describe many types of functions. I don't understand why the term on the right side of the inequality is ##C+C(x^2+y^2+z^2)##. There must be a clue that I am oblivious to. Thank you in advance for any advice.
 
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  • #2
Hi. I find it an interesting question.
u(x,y,z) has no singularity except infinite far ends.
As an obvious example say
[tex]u(x,y,z)=A e^{x+iy}[/tex], u is harmonic function,
[tex] |u|=A[/tex]
which satisfies the condition. Similar for ##Ae^{y+iz}## and ##Ae^{z+ix}##.
 
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  • #3
mitochan said:
Hi. I find it an interesting question.
u(x,y,z) has no singularity except infinite far ends.
As an obvious example say
[tex]u(x,y,z)=A e^{x+iy}[/tex], u is harmonic function,
[tex] |u|=A[/tex]
which satisfies the condition. Similar for ##Ae^{y+iz}## and ##Ae^{z+ix}##.

[itex]|Ae^{x + iy}| = Ae^x[/itex]. Does that satisfy the bound as [itex]x \to \infty[/itex]?

Consider solutions of the form [itex]u = e^{\mathbf{k} \cdot \mathbf{x}}[/itex] where [itex]\mathbf{k} = (k_1,k_2,k_3) \in \mathbb{C}^3[/itex]. This is harmonic if and only if [itex]k_1^2 + k_2^2 + k_3^2 = 0[/itex]. So either [itex]\mathbf{k} = 0[/itex] and [itex]u[/itex] is constant, or you can find a [itex]\mathbf{v} \in \mathbb{R}^3[/itex] such that [itex]\mathbf{k} \cdot \mathbf{v} > 0[/itex]. Consider [itex]|u(t\mathbf{v})|[/itex] as [itex]t \to \infty[/itex]. Does this satisfy the bound?

(EDIT: It is only necessary that [itex]\operatorname{Re}(\mathbf{k})\cdot \mathbf{v} > 0[/itex].)

The other solutions are polynonial solutions of the form [tex]u = a + \mathbf{b} \cdot \mathbf{x} + \mathbf{x}^T C \mathbf{x}[/tex] for some scalar [itex]a[/itex], vector [itex]\mathbf{b}[/itex] and symmetric matrix [itex]C[/itex].
 
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  • #4
@pasmith

Thank you for your comment, may I ask a few questions about your comment?

pasmith said:
|Aex+iy|=Aex. Does that satisfy the bound as x→∞?

I think [itex]|Ae^{x + iy}|[/itex] does not satisfy the bound because exponential functions increase faster than quadratic functions as [itex]x \to \infty[/itex]. is this true?

pasmith said:
Consider solutions of the form ##u=e^{k⋅x}## where ##k=(k_1,k_2,k_3)∈C^3##. This is harmonic if and only if ##k_1^2+k_2^2+k_3^2=0##. So either ##k=0## and ##u## is constant, or you can find a ##v∈R^3## such that ##k⋅v>0##. Consider ##|u(tv)|## as ##t→∞##. Does this satisfy the bound?

Could you please briefly explain how ##Re(k)⋅v>0## makes a harmonic ##u##?

When ##u=e^{k⋅v}## where k and v are both vectors, isn't ##u## a constant which is included in the polynomial solution?

For ##|u(tv)|##, doesn't this restrict our domain from ##R^3## to lines spanned by ##v##? I'm pretty confused by this and I will be happy if you can give me any explanation at all. Thank you.
 
  • #5
docnet said:
is this true?
Yea, it's true.[tex]k_1^2+k_2^2+k_3^2=0[/tex]
means at least one of them has real parts, say ##k_1##, which makes ##e^{k_1x_1} > A+A(x^2+y^2+z^2)## for large x_1 or -x_1.
 
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  • #6
docnet said:
@pasmith

Thank you for your comment, may I ask a few questions about your comment?

I think [itex]|Ae^{x + iy}|[/itex] does not satisfy the bound because exponential functions increase faster than quadratic functions as [itex]x \to \infty[/itex]. is this true?

Yes.

Could you please briefly explain how ##Re(k)⋅v>0## makes a harmonic ##u##?

When ##u=e^{k⋅v}## where k and v are both vectors, isn't ##u## a constant which is included in the polynomial solution?

For ##|u(tv)|##, doesn't this restrict our domain from ##R^3## to lines spanned by ##v##? I'm pretty confused by this and I will be happy if you can give me any explanation at all. Thank you.

The condition for [itex]e^{\mathbf{k} \cdot \mathbf{x}}[/itex] to be harmonic is that [itex]\sum k_i^2 = 0[/itex].

You are asked to find functions which satisfy [itex]|u(x,y,z)| \leq A(1 + x^2 + y^2 + z^2)[/itex] for all [itex](x,y,z) \in \mathbb{R}^3[/itex]. If this bound fails to hold at even a single point, [itex]u[/itex] is not one of the functions you are looking for.

In this case, if [itex]\mathbf{k}[/itex] is not zero then you can find a (non-zero) [itex]\mathbf{v} \in \mathbb{R}^3[/itex] such that [itex]\operatorname{Re}(\mathbf{k}) \cdot \mathbf{v} > 0[/itex]. This guarantees that [tex]
|u(\mathbf{v}t)| = |\exp(\operatorname{Re}(\mathbf{k}) \cdot \mathbf{v} t)| >
A(1 + \|\mathbf{v}\|^2t^2)[/tex] as [itex]t \to \infty[/itex]. This means that [itex]u(x,y,z) \leq A(1 + x^2 + y^2 + z^2)[/itex] doesn't hold on all of [itex]\mathbb{R}^3[/itex] for this choice of [itex]u[/itex].
 
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  • #7
pasmith said:
The other solutions are polynonial solutions of the form for some scalar , vector and symmetric matrix .

Following this way for ##ax^2,by^2,cz^2## terms to be harmonic
[tex]a+b+c=0[/tex]
for general u(x,y,z) of
[tex]u(x,y,z)=ax^2+by^2+cz^2+dxy+eyz+fzx+g[/tex]
In order to satisfy the magnitude condition roughly
[tex]|h|\leq A, h=\{a,b,c,d,e,f,g\}[/tex]
I have not gone into detailed investigation.
 
  • #8
[EDIT]
I forgot x,y,z terms
[tex]u(x,y,z)=A(ax^2+by^2+cz^2+dxy+eyz+fzx+gx+hy+iz+j)[/tex]
where
[tex]a+b+c=0[/tex]
Rough estimation is
[tex]|k| \leq 1, k=\{a,b,c,d,e,f,g,h,i,j\}[/tex]
 
  • #9
For simple 1 dimension case ##u(x)=A(ax^2+bx+c)##, a=0,
[tex]|bx+c|\leq 1+x^2[/tex]
[tex]b^2+c^2 \leq 1[/tex]

Two dimension case ##u(x,y)=A(ax^2-ay^2+bxy+cx+dy+e)##,
[tex]|ax^2-ay^2+bxy+cx+dy+e|<1+x^2+y^2[/tex]
Is is not easy.
 
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  • #10
pasmith said:
The other solutions are polynonial solutions of the form u=a+b⋅x+xTCx for some scalar a, vector b and symmetric matrix C.

May I ask how you need C to be symmetric?

I believe the requirement on C can be reduced to the following:

##u(x_1,x_2,x_3)=a+\vec{b}⋅x+x^tCx##

where ##x^t=(x_1, x_2, x_3)## and ##C∈Mat(3;R)## with ##Tr(C)=0##

mitochan said:
For simple 1 dimension case ##u(x)=A(ax^2+bx+c)##, a=0,
[tex]|bx+c|\leq 1+x^2[/tex]
[tex]b^2+c^2 \leq 1[/tex]

Two dimension case ##u(x,y)=A(ax^2-ay^2+bxy+cx+dy+e)##,
[tex]|ax^2-ay^2+bxy+cx+dy+e|<1+x^2+y^2[/tex]
Is is not easy.

Thank you, your posts are amazing. I believe you can avoid the work of setting the limits on coefficients, because you can make the assumption that A can be taken arbitrarily large, larger than any constant. so the solution is simpler. The cross-terms can be expressed by the matrix C, so we avoid doing even more writing.
 

Related to Find the set of all functions that satisfy the inequality

1. What does it mean to "find the set of all functions"?

When asked to find the set of all functions, it means to determine all possible mathematical expressions or rules that satisfy a given condition or set of conditions. In this case, the condition is an inequality.

2. How do you determine if a function satisfies an inequality?

To determine if a function satisfies an inequality, you must plug in different values for the independent variable and see if the resulting output values satisfy the given inequality. If the output values satisfy the inequality for all possible input values, then the function satisfies the inequality.

3. Can there be more than one function that satisfies an inequality?

Yes, there can be multiple functions that satisfy an inequality. This is because there are infinite possible mathematical expressions or rules that can satisfy a given condition or set of conditions. However, some inequalities may only have one solution.

4. What is the importance of finding the set of all functions that satisfy an inequality?

Finding the set of all functions that satisfy an inequality allows us to determine all possible solutions to a given problem. This can help us understand the behavior of a system or make predictions about the outcomes of certain situations.

5. Are there any limitations to finding the set of all functions that satisfy an inequality?

Yes, there may be limitations depending on the complexity of the given inequality. Some inequalities may have an infinite number of solutions, making it impossible to list out all possible functions. In these cases, we may need to use mathematical tools or techniques to represent the set of solutions in a more concise way.

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