If $u(x)$ is harmonic and equal to $phi(|x|)$, is $phi$ continuously differentiable?
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I was trying to show that radial harmonic functions on the unit ball (in $mathbb{R}^n$) are constant. To this end, I suppose that $u$ is a radial harmonic function on the unit ball and write
$$
u(x) = phi(lvert xrvert)
$$
for some function $phi$ defined on $[0, 1)$. I got stuck when trying to rigorously prove that $phi$ has to be differentiable.
real-analysis calculus pde
asked Dec 11 '18 at 20:49
QuokaQuoka
1,570316
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add a comment |
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I was trying to show that radial harmonic functions on the unit ball (in $mathbb{R}^n$) are constant. To this end, I suppose that $u$ is a radial harmonic function on the unit ball and write
$$
u(x) = phi(lvert xrvert)
$$
for some function $phi$ defined on $[0, 1)$. I got stuck when trying to rigorously prove that $phi$ has to be differentiable.
real-analysis calculus pde
asked Dec 11 '18 at 20:49
QuokaQuoka
1,570316
$endgroup$
add a comment |
$begingroup$
I was trying to show that radial harmonic functions on the unit ball (in $mathbb{R}^n$) are constant. To this end, I suppose that $u$ is a radial harmonic function on the unit ball and write
$$
u(x) = phi(lvert xrvert)
$$
for some function $phi$ defined on $[0, 1)$. I got stuck when trying to rigorously prove that $phi$ has to be differentiable.
real-analysis calculus pde
asked Dec 11 '18 at 20:49
QuokaQuoka
1,570316
$endgroup$
I was trying to show that radial harmonic functions on the unit ball (in $mathbb{R}^n$) are constant. To this end, I suppose that $u$ is a radial harmonic function on the unit ball and write
$$
u(x) = phi(lvert xrvert)
$$
for some function $phi$ defined on $[0, 1)$. I got stuck when trying to rigorously prove that $phi$ has to be differentiable.
real-analysis calculus pde
real-analysis calculus pde
asked Dec 11 '18 at 20:49
QuokaQuoka
1,570316
asked Dec 11 '18 at 20:49
QuokaQuoka
1,570316
asked Dec 11 '18 at 20:49
QuokaQuoka
1,570316
asked Dec 11 '18 at 20:49
QuokaQuoka
1,570316
asked Dec 11 '18 at 20:49
QuokaQuoka
1,570316
1,570316
add a comment |
add a comment |
2 Answers
2
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Let $s$ be any vector with norm $1$. Then $phi(t)=u(ts)$, thus $phi$ is continuously differentiable, since $u$ is.
answered Dec 11 '18 at 23:08
MindlackMindlack
4,910211
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add a comment |
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Hint: Use the MVP to see $u(0) = phi(r)$ for $0le r <1.$
answered Dec 11 '18 at 23:08
zhw.zhw.
74.8k43275
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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$begingroup$
Let $s$ be any vector with norm $1$. Then $phi(t)=u(ts)$, thus $phi$ is continuously differentiable, since $u$ is.
answered Dec 11 '18 at 23:08
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
Let $s$ be any vector with norm $1$. Then $phi(t)=u(ts)$, thus $phi$ is continuously differentiable, since $u$ is.
answered Dec 11 '18 at 23:08
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
Let $s$ be any vector with norm $1$. Then $phi(t)=u(ts)$, thus $phi$ is continuously differentiable, since $u$ is.
answered Dec 11 '18 at 23:08
MindlackMindlack
4,910211
$endgroup$
Let $s$ be any vector with norm $1$. Then $phi(t)=u(ts)$, thus $phi$ is continuously differentiable, since $u$ is.
answered Dec 11 '18 at 23:08
MindlackMindlack
4,910211
answered Dec 11 '18 at 23:08
MindlackMindlack
4,910211
answered Dec 11 '18 at 23:08
MindlackMindlack
4,910211
answered Dec 11 '18 at 23:08
MindlackMindlack
4,910211
4,910211
add a comment |
add a comment |
$begingroup$
Hint: Use the MVP to see $u(0) = phi(r)$ for $0le r <1.$
answered Dec 11 '18 at 23:08
zhw.zhw.
74.8k43275
$endgroup$
add a comment |
$begingroup$
Hint: Use the MVP to see $u(0) = phi(r)$ for $0le r <1.$
answered Dec 11 '18 at 23:08
zhw.zhw.
74.8k43275
$endgroup$
add a comment |
$begingroup$
Hint: Use the MVP to see $u(0) = phi(r)$ for $0le r <1.$
answered Dec 11 '18 at 23:08
zhw.zhw.
74.8k43275
$endgroup$
Hint: Use the MVP to see $u(0) = phi(r)$ for $0le r <1.$
answered Dec 11 '18 at 23:08
zhw.zhw.
74.8k43275
answered Dec 11 '18 at 23:08
zhw.zhw.
74.8k43275
answered Dec 11 '18 at 23:08
zhw.zhw.
74.8k43275
answered Dec 11 '18 at 23:08
zhw.zhw.
74.8k43275
74.8k43275
add a comment |
add a comment |
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