If partial derivatives of a harmonic function are constant, is the function linear?Harmonic function with...
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If partial derivatives of a harmonic function are constant, is the function linear?
Harmonic function with bounded preimageHarmonic function, existence of a constantHarmonic functionOn the partial derivatives of a harmonic functionAll second partial derivatives of harmonic function are $0$Show : A holomorphic function is harmonic if $frac{partial f}{partial overline{z}}=0$Prove that partial derivative of the harmonic function is greater than zero if the function takes the maximum at a specific point in the boundaryMean value integral of a harmonic function in the exterior of a ballHarmonic function radial on boundaryLinear harmonic functions
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Let $u: mathbb{R}^2 to mathbb{R}$ be a harmonic function.
If $frac{partial u(x,y)}{partial x} = k_1$ and $frac{partial u(x,y)}{partial y} = k_2,forall (x,y) in mathbb{R}^2, k_1, k_2 in mathbb{R}$, may I say that $u$ is a linear function?
real-analysis pde harmonic-functions
asked 5 hours ago
user71487user71487
1178
$endgroup$
add a comment |
$begingroup$
Let $u: mathbb{R}^2 to mathbb{R}$ be a harmonic function.
If $frac{partial u(x,y)}{partial x} = k_1$ and $frac{partial u(x,y)}{partial y} = k_2,forall (x,y) in mathbb{R}^2, k_1, k_2 in mathbb{R}$, may I say that $u$ is a linear function?
real-analysis pde harmonic-functions
asked 5 hours ago
user71487user71487
1178
$endgroup$
1
$begingroup$
Shouldn't it be $frac{partial u(x,y)}{partial y} = k_2$?
$endgroup$
– Charles Hudgins
5 hours ago
$begingroup$
It seems like Liouville's theorem or a modification of it should do the trick.
$endgroup$
– Charles Hudgins
5 hours ago
add a comment |
$begingroup$
Let $u: mathbb{R}^2 to mathbb{R}$ be a harmonic function.
If $frac{partial u(x,y)}{partial x} = k_1$ and $frac{partial u(x,y)}{partial y} = k_2,forall (x,y) in mathbb{R}^2, k_1, k_2 in mathbb{R}$, may I say that $u$ is a linear function?
real-analysis pde harmonic-functions
asked 5 hours ago
user71487user71487
1178
$endgroup$
Let $u: mathbb{R}^2 to mathbb{R}$ be a harmonic function.
If $frac{partial u(x,y)}{partial x} = k_1$ and $frac{partial u(x,y)}{partial y} = k_2,forall (x,y) in mathbb{R}^2, k_1, k_2 in mathbb{R}$, may I say that $u$ is a linear function?
real-analysis pde harmonic-functions
real-analysis pde harmonic-functions
asked 5 hours ago
user71487user71487
1178
asked 5 hours ago
user71487user71487
1178
edited 5 hours ago
user71487
asked 5 hours ago
user71487user71487
1178
asked 5 hours ago
user71487user71487
1178
asked 5 hours ago
user71487user71487
1178
1178
1
$begingroup$
Shouldn't it be $frac{partial u(x,y)}{partial y} = k_2$?
$endgroup$
– Charles Hudgins
5 hours ago
$begingroup$
It seems like Liouville's theorem or a modification of it should do the trick.
$endgroup$
– Charles Hudgins
5 hours ago
add a comment |
1
$begingroup$
Shouldn't it be $frac{partial u(x,y)}{partial y} = k_2$?
$endgroup$
– Charles Hudgins
5 hours ago
$begingroup$
It seems like Liouville's theorem or a modification of it should do the trick.
$endgroup$
– Charles Hudgins
5 hours ago
1
1
$begingroup$
Shouldn't it be $frac{partial u(x,y)}{partial y} = k_2$?
$endgroup$
– Charles Hudgins
5 hours ago
$begingroup$
Shouldn't it be $frac{partial u(x,y)}{partial y} = k_2$?
$endgroup$
– Charles Hudgins
5 hours ago
$begingroup$
It seems like Liouville's theorem or a modification of it should do the trick.
$endgroup$
– Charles Hudgins
5 hours ago
$begingroup$
It seems like Liouville's theorem or a modification of it should do the trick.
$endgroup$
– Charles Hudgins
5 hours ago
add a comment |
2 Answers
2
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oldest
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$begingroup$
No.
If $u(x,y) = k_{1}x+k_{2}y+C$, then:
$$dfrac{partial u(x,y)}{partial x} = k_{1}$$
$$dfrac{partial u(x,y)}{partial y} = k_{2}$$
But $u(x,y)$ is not linear - this kind of function is an affine function. $u$ is linear iff $C = 0$.
answered 5 hours ago
user458276user458276
7581315
$endgroup$
1
$begingroup$
Might be good to say $u$ is linear iff $C=0.$
$endgroup$
– zhw.
4 hours ago
add a comment |
$begingroup$
A simple counterexample is $uequiv 1.$
The general result here, which has nothing to do with harmonicity, is this: If $a,bin mathbb R,$ and $partial u/partial x equiv a,$ $partial u/partial y equiv b,$ then $u(x,y) = ax + by + u(0,0)$ for all $(x,y).$
Proof: $u(x,y)- u(0,0) = u(x,y)- u(x,0) +u(x,0)-u(0,0).$ Apply the mean value theorem to each difference.
answered 4 hours ago
zhw.zhw.
75.8k43376
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
No.
If $u(x,y) = k_{1}x+k_{2}y+C$, then:
$$dfrac{partial u(x,y)}{partial x} = k_{1}$$
$$dfrac{partial u(x,y)}{partial y} = k_{2}$$
But $u(x,y)$ is not linear - this kind of function is an affine function. $u$ is linear iff $C = 0$.
answered 5 hours ago
user458276user458276
7581315
$endgroup$
1
$begingroup$
Might be good to say $u$ is linear iff $C=0.$
$endgroup$
– zhw.
4 hours ago
add a comment |
$begingroup$
No.
If $u(x,y) = k_{1}x+k_{2}y+C$, then:
$$dfrac{partial u(x,y)}{partial x} = k_{1}$$
$$dfrac{partial u(x,y)}{partial y} = k_{2}$$
But $u(x,y)$ is not linear - this kind of function is an affine function. $u$ is linear iff $C = 0$.
answered 5 hours ago
user458276user458276
7581315
$endgroup$
1
$begingroup$
Might be good to say $u$ is linear iff $C=0.$
$endgroup$
– zhw.
4 hours ago
add a comment |
$begingroup$
No.
If $u(x,y) = k_{1}x+k_{2}y+C$, then:
$$dfrac{partial u(x,y)}{partial x} = k_{1}$$
$$dfrac{partial u(x,y)}{partial y} = k_{2}$$
But $u(x,y)$ is not linear - this kind of function is an affine function. $u$ is linear iff $C = 0$.
answered 5 hours ago
user458276user458276
7581315
$endgroup$
No.
If $u(x,y) = k_{1}x+k_{2}y+C$, then:
$$dfrac{partial u(x,y)}{partial x} = k_{1}$$
$$dfrac{partial u(x,y)}{partial y} = k_{2}$$
But $u(x,y)$ is not linear - this kind of function is an affine function. $u$ is linear iff $C = 0$.
answered 5 hours ago
user458276user458276
7581315
edited 4 hours ago
answered 5 hours ago
user458276user458276
7581315
answered 5 hours ago
user458276user458276
7581315
answered 5 hours ago
user458276user458276
7581315
7581315
1
$begingroup$
Might be good to say $u$ is linear iff $C=0.$
$endgroup$
– zhw.
4 hours ago
add a comment |
1
$begingroup$
Might be good to say $u$ is linear iff $C=0.$
$endgroup$
– zhw.
4 hours ago
1
1
$begingroup$
Might be good to say $u$ is linear iff $C=0.$
$endgroup$
– zhw.
4 hours ago
$begingroup$
Might be good to say $u$ is linear iff $C=0.$
$endgroup$
– zhw.
4 hours ago
add a comment |
$begingroup$
A simple counterexample is $uequiv 1.$
The general result here, which has nothing to do with harmonicity, is this: If $a,bin mathbb R,$ and $partial u/partial x equiv a,$ $partial u/partial y equiv b,$ then $u(x,y) = ax + by + u(0,0)$ for all $(x,y).$
Proof: $u(x,y)- u(0,0) = u(x,y)- u(x,0) +u(x,0)-u(0,0).$ Apply the mean value theorem to each difference.
answered 4 hours ago
zhw.zhw.
75.8k43376
$endgroup$
add a comment |
$begingroup$
A simple counterexample is $uequiv 1.$
The general result here, which has nothing to do with harmonicity, is this: If $a,bin mathbb R,$ and $partial u/partial x equiv a,$ $partial u/partial y equiv b,$ then $u(x,y) = ax + by + u(0,0)$ for all $(x,y).$
Proof: $u(x,y)- u(0,0) = u(x,y)- u(x,0) +u(x,0)-u(0,0).$ Apply the mean value theorem to each difference.
answered 4 hours ago
zhw.zhw.
75.8k43376
$endgroup$
add a comment |
$begingroup$
A simple counterexample is $uequiv 1.$
The general result here, which has nothing to do with harmonicity, is this: If $a,bin mathbb R,$ and $partial u/partial x equiv a,$ $partial u/partial y equiv b,$ then $u(x,y) = ax + by + u(0,0)$ for all $(x,y).$
Proof: $u(x,y)- u(0,0) = u(x,y)- u(x,0) +u(x,0)-u(0,0).$ Apply the mean value theorem to each difference.
answered 4 hours ago
zhw.zhw.
75.8k43376
$endgroup$
A simple counterexample is $uequiv 1.$
The general result here, which has nothing to do with harmonicity, is this: If $a,bin mathbb R,$ and $partial u/partial x equiv a,$ $partial u/partial y equiv b,$ then $u(x,y) = ax + by + u(0,0)$ for all $(x,y).$
Proof: $u(x,y)- u(0,0) = u(x,y)- u(x,0) +u(x,0)-u(0,0).$ Apply the mean value theorem to each difference.
answered 4 hours ago
zhw.zhw.
75.8k43376
edited 1 hour ago
answered 4 hours ago
zhw.zhw.
75.8k43376
answered 4 hours ago
zhw.zhw.
75.8k43376
answered 4 hours ago
zhw.zhw.
75.8k43376
75.8k43376
add a comment |
add a comment |
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1
$begingroup$
Shouldn't it be $frac{partial u(x,y)}{partial y} = k_2$?
$endgroup$
– Charles Hudgins
5 hours ago
$begingroup$
It seems like Liouville's theorem or a modification of it should do the trick.
$endgroup$
– Charles Hudgins
5 hours ago