Uniqueness of solutions of boundary value problem
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I know that $u_1 = ln|x^2+y^2|$ is harmonic. Knowing that $u_2 = 0$ is harmonic, I can see that boundary values for $u_1$ on unit circle coincide with boundary values of $u_2$.
My question is does this contradict the uniqueness of solutions to boundary value problems? I am not quite sure if 0 could be counted as a solution nor am I sure about whether absolute values have any role. If you could let me know, I would greatly appreciate it. Thanks.
ordinary-differential-equations harmonic-functions
asked Dec 5 '18 at 6:40
dmsj djsldmsj djsl
35517
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add a comment |
$begingroup$
I know that $u_1 = ln|x^2+y^2|$ is harmonic. Knowing that $u_2 = 0$ is harmonic, I can see that boundary values for $u_1$ on unit circle coincide with boundary values of $u_2$.
My question is does this contradict the uniqueness of solutions to boundary value problems? I am not quite sure if 0 could be counted as a solution nor am I sure about whether absolute values have any role. If you could let me know, I would greatly appreciate it. Thanks.
ordinary-differential-equations harmonic-functions
asked Dec 5 '18 at 6:40
dmsj djsldmsj djsl
35517
$endgroup$
add a comment |
$begingroup$
I know that $u_1 = ln|x^2+y^2|$ is harmonic. Knowing that $u_2 = 0$ is harmonic, I can see that boundary values for $u_1$ on unit circle coincide with boundary values of $u_2$.
My question is does this contradict the uniqueness of solutions to boundary value problems? I am not quite sure if 0 could be counted as a solution nor am I sure about whether absolute values have any role. If you could let me know, I would greatly appreciate it. Thanks.
ordinary-differential-equations harmonic-functions
asked Dec 5 '18 at 6:40
dmsj djsldmsj djsl
35517
$endgroup$
I know that $u_1 = ln|x^2+y^2|$ is harmonic. Knowing that $u_2 = 0$ is harmonic, I can see that boundary values for $u_1$ on unit circle coincide with boundary values of $u_2$.
My question is does this contradict the uniqueness of solutions to boundary value problems? I am not quite sure if 0 could be counted as a solution nor am I sure about whether absolute values have any role. If you could let me know, I would greatly appreciate it. Thanks.
ordinary-differential-equations harmonic-functions
ordinary-differential-equations harmonic-functions
asked Dec 5 '18 at 6:40
dmsj djsldmsj djsl
35517
asked Dec 5 '18 at 6:40
dmsj djsldmsj djsl
35517
asked Dec 5 '18 at 6:40
dmsj djsldmsj djsl
35517
asked Dec 5 '18 at 6:40
dmsj djsldmsj djsl
35517
asked Dec 5 '18 at 6:40
dmsj djsldmsj djsl
35517
35517
add a comment |
add a comment |
1 Answer
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I think that by "unit circle" you mean $D={(x,y): x^2+y^2 le 1}$. But your function $u_1$ is only defined for $(x,y) ne (0,0)$.
Let $D_0:= D setminus {(0,0)}$. Then $u_1$ and $u_2$ are harmonic on $D_0$ but $u_1 ne u_2$ on $ partial D_0$, since $(0,0) in partial D_0$, and $u_1$ is not defined in this point.
answered Dec 5 '18 at 6:54
FredFred
44.4k1845
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1 Answer
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1 Answer
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$begingroup$
I think that by "unit circle" you mean $D={(x,y): x^2+y^2 le 1}$. But your function $u_1$ is only defined for $(x,y) ne (0,0)$.
Let $D_0:= D setminus {(0,0)}$. Then $u_1$ and $u_2$ are harmonic on $D_0$ but $u_1 ne u_2$ on $ partial D_0$, since $(0,0) in partial D_0$, and $u_1$ is not defined in this point.
answered Dec 5 '18 at 6:54
FredFred
44.4k1845
$endgroup$
add a comment |
$begingroup$
I think that by "unit circle" you mean $D={(x,y): x^2+y^2 le 1}$. But your function $u_1$ is only defined for $(x,y) ne (0,0)$.
Let $D_0:= D setminus {(0,0)}$. Then $u_1$ and $u_2$ are harmonic on $D_0$ but $u_1 ne u_2$ on $ partial D_0$, since $(0,0) in partial D_0$, and $u_1$ is not defined in this point.
answered Dec 5 '18 at 6:54
FredFred
44.4k1845
$endgroup$
add a comment |
$begingroup$
I think that by "unit circle" you mean $D={(x,y): x^2+y^2 le 1}$. But your function $u_1$ is only defined for $(x,y) ne (0,0)$.
Let $D_0:= D setminus {(0,0)}$. Then $u_1$ and $u_2$ are harmonic on $D_0$ but $u_1 ne u_2$ on $ partial D_0$, since $(0,0) in partial D_0$, and $u_1$ is not defined in this point.
answered Dec 5 '18 at 6:54
FredFred
44.4k1845
$endgroup$
I think that by "unit circle" you mean $D={(x,y): x^2+y^2 le 1}$. But your function $u_1$ is only defined for $(x,y) ne (0,0)$.
Let $D_0:= D setminus {(0,0)}$. Then $u_1$ and $u_2$ are harmonic on $D_0$ but $u_1 ne u_2$ on $ partial D_0$, since $(0,0) in partial D_0$, and $u_1$ is not defined in this point.
answered Dec 5 '18 at 6:54
FredFred
44.4k1845
answered Dec 5 '18 at 6:54
FredFred
44.4k1845
answered Dec 5 '18 at 6:54
FredFred
44.4k1845
answered Dec 5 '18 at 6:54
FredFred
44.4k1845
44.4k1845
add a comment |
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