Find the set of all functions that satisfy the inequality

[docnet]

Homework Statement

please see below

Relevant Equations

please see below

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.

 

[mitochan]

Hi. I find it an interesting question.
u(x,y,z) has no singularity except infinite far ends.
As an obvious example say
u(x,y,z)=A e^{x+iy}, u is harmonic function,
|u|=A
which satisfies the condition. Similar for $Ae^{y+iz}$ and $Ae^{z+ix}$.

 

[pasmith]

Hi. I find it an interesting question.
u(x,y,z) has no singularity except infinite far ends.
As an obvious example say
u(x,y,z)=A e^{x+iy}, u is harmonic function,
|u|=A
which satisfies the condition. Similar for $Ae^{y+iz}$ and $Ae^{z+ix}$.

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

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

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

The other solutions are polynonial solutions of the form u = a + \mathbf{b} \cdot \mathbf{x} + \mathbf{x}^T C \mathbf{x} for some scalar a, vector \mathbf{b} and symmetric matrix C.

 

[docnet]

@pasmith

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

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

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

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.

 

[mitochan]

is this true?

Yea, it's true.k_1^2+k_2^2+k_3^2=0
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.

 

[pasmith]

@pasmith

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

I think |Ae^{x + iy}| does not satisfy the bound because exponential functions increase faster than quadratic functions as x \to \infty. 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 e^{\mathbf{k} \cdot \mathbf{x}} to be harmonic is that \sum k_i^2 = 0.

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

In this case, if \mathbf{k} is not zero then you can find a (non-zero) \mathbf{v} \in \mathbb{R}^3 such that \operatorname{Re}(\mathbf{k}) \cdot \mathbf{v} &gt; 0. This guarantees that <br /> |u(\mathbf{v}t)| = |\exp(\operatorname{Re}(\mathbf{k}) \cdot \mathbf{v} t)| &gt; <br /> A(1 + \|\mathbf{v}\|^2t^2) as t \to \infty. This means that u(x,y,z) \leq A(1 + x^2 + y^2 + z^2) doesn't hold on all of \mathbb{R}^3 for this choice of u.

 

[mitochan]

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
a+b+c=0
for general u(x,y,z) of
u(x,y,z)=ax^2+by^2+cz^2+dxy+eyz+fzx+g
In order to satisfy the magnitude condition roughly
|h|\leq A, h=\{a,b,c,d,e,f,g\}
I have not gone into detailed investigation.

 

[mitochan]

[EDIT]
I forgot x,y,z terms
u(x,y,z)=A(ax^2+by^2+cz^2+dxy+eyz+fzx+gx+hy+iz+j)
where
a+b+c=0
Rough estimation is
|k| \leq 1, k=\{a,b,c,d,e,f,g,h,i,j\}

 

[mitochan]

For simple 1 dimension case $u(x)=A(ax^2+bx+c)$, a=0,
|bx+c|\leq 1+x^2
b^2+c^2 \leq 1

Two dimension case $u(x,y)=A(ax^2-ay^2+bxy+cx+dy+e)$,
|ax^2-ay^2+bxy+cx+dy+e|&lt;1+x^2+y^2
Is is not easy.

 

[docnet]

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 $Tr(C)=0$

For simple 1 dimension case $u(x)=A(ax^2+bx+c)$, a=0,
|bx+c|\leq 1+x^2
b^2+c^2 \leq 1

Two dimension case $u(x,y)=A(ax^2-ay^2+bxy+cx+dy+e)$,
|ax^2-ay^2+bxy+cx+dy+e|&lt;1+x^2+y^2
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.