Mixing
Boundary problem with elliptic 2nd order pde
$begingroup$
I am reading some notes about harmonic functions and associated boundary problems and I came through a proposition stating that the problem
$u_{xx}+u_{yy}=0$
$u(0,y)=u(1,y)=u(x,1)=u(x,0)=0$ , where $x,yin[0,1]$
has only the 0 solution.
I would like kindly to ask you if the following problem
$u_{xx}+u_{yy}+au_{y}+bu_{x}=0$
$u(0,y)=u(1,y)=u(x,1)=u(x,0)=0$ , where $x,yin[0,1]$
has also only the $0$ solution. Any hints or refferences are welcome. Thanks in advance.
pde boundary-value-problem elliptic-equations
asked Jan 24 at 19:18
dmtridmtri
1,6612521
$endgroup$
add a comment |
$begingroup$
I am reading some notes about harmonic functions and associated boundary problems and I came through a proposition stating that the problem
$u_{xx}+u_{yy}=0$
$u(0,y)=u(1,y)=u(x,1)=u(x,0)=0$ , where $x,yin[0,1]$
has only the 0 solution.
I would like kindly to ask you if the following problem
$u_{xx}+u_{yy}+au_{y}+bu_{x}=0$
$u(0,y)=u(1,y)=u(x,1)=u(x,0)=0$ , where $x,yin[0,1]$
has also only the $0$ solution. Any hints or refferences are welcome. Thanks in advance.
pde boundary-value-problem elliptic-equations
asked Jan 24 at 19:18
dmtridmtri
1,6612521
$endgroup$
2
$begingroup$
My gut reaction: the key feature of the first problem that makes the trivial solution unique is its linearity. As the same is true of the second problem, I'm guessing it will also have to be unique.
$endgroup$
– Adrian Keister
Jan 24 at 19:37
add a comment |
$begingroup$
I am reading some notes about harmonic functions and associated boundary problems and I came through a proposition stating that the problem
$u_{xx}+u_{yy}=0$
$u(0,y)=u(1,y)=u(x,1)=u(x,0)=0$ , where $x,yin[0,1]$
has only the 0 solution.
I would like kindly to ask you if the following problem
$u_{xx}+u_{yy}+au_{y}+bu_{x}=0$
$u(0,y)=u(1,y)=u(x,1)=u(x,0)=0$ , where $x,yin[0,1]$
has also only the $0$ solution. Any hints or refferences are welcome. Thanks in advance.
pde boundary-value-problem elliptic-equations
asked Jan 24 at 19:18
dmtridmtri
1,6612521
$endgroup$
I am reading some notes about harmonic functions and associated boundary problems and I came through a proposition stating that the problem
$u_{xx}+u_{yy}=0$
$u(0,y)=u(1,y)=u(x,1)=u(x,0)=0$ , where $x,yin[0,1]$
has only the 0 solution.
I would like kindly to ask you if the following problem
$u_{xx}+u_{yy}+au_{y}+bu_{x}=0$
$u(0,y)=u(1,y)=u(x,1)=u(x,0)=0$ , where $x,yin[0,1]$
has also only the $0$ solution. Any hints or refferences are welcome. Thanks in advance.
pde boundary-value-problem elliptic-equations
pde boundary-value-problem elliptic-equations
asked Jan 24 at 19:18
dmtridmtri
1,6612521
asked Jan 24 at 19:18
dmtridmtri
1,6612521
edited Jan 25 at 4:16
dmtri
asked Jan 24 at 19:18
dmtridmtri
1,6612521
asked Jan 24 at 19:18
dmtridmtri
1,6612521
asked Jan 24 at 19:18
dmtridmtri
1,6612521
1,6612521
2
$begingroup$
My gut reaction: the key feature of the first problem that makes the trivial solution unique is its linearity. As the same is true of the second problem, I'm guessing it will also have to be unique.
$endgroup$
– Adrian Keister
Jan 24 at 19:37
add a comment |
2
$begingroup$
My gut reaction: the key feature of the first problem that makes the trivial solution unique is its linearity. As the same is true of the second problem, I'm guessing it will also have to be unique.
$endgroup$
– Adrian Keister
Jan 24 at 19:37
2
2
$begingroup$
My gut reaction: the key feature of the first problem that makes the trivial solution unique is its linearity. As the same is true of the second problem, I'm guessing it will also have to be unique.
$endgroup$
– Adrian Keister
Jan 24 at 19:37
$begingroup$
My gut reaction: the key feature of the first problem that makes the trivial solution unique is its linearity. As the same is true of the second problem, I'm guessing it will also have to be unique.
$endgroup$
– Adrian Keister
Jan 24 at 19:37
add a comment |
1 Answer
1
active
oldest
votes
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@dmitri Yes, the solution to your problem is unique. This follows from the (strong) maximum principle. The basic idea is that if you had a non-zero solution you would have a point in the square where it attains a positive maximum or a negative minimum. At that point your first derivatives would be zero and the second ones would have a definite sign. The degenerate case can be dealt with using a barrier type of argument. You can find more info here:
http://www.mi.uni-koeln.de/~gsweers/pdf/maxprinc.pdf
Check Corollary 7.
Hope this helps.
answered Jan 24 at 19:51
GReyesGReyes
2,09815
$endgroup$
$begingroup$
Nice refferenece! Thanks a lot. I thought that the strong maximum principle is valid only for the parabolic equations, like the heat equation.
$endgroup$
– dmtri
Jan 24 at 20:03
1
$begingroup$
Glad to have helped. There is an elliptic and a parabolic version of it.
$endgroup$
– GReyes
Jan 24 at 20:08
add a comment |
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1 Answer
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$begingroup$
@dmitri Yes, the solution to your problem is unique. This follows from the (strong) maximum principle. The basic idea is that if you had a non-zero solution you would have a point in the square where it attains a positive maximum or a negative minimum. At that point your first derivatives would be zero and the second ones would have a definite sign. The degenerate case can be dealt with using a barrier type of argument. You can find more info here:
http://www.mi.uni-koeln.de/~gsweers/pdf/maxprinc.pdf
Check Corollary 7.
Hope this helps.
answered Jan 24 at 19:51
GReyesGReyes
2,09815
$endgroup$
$begingroup$
Nice refferenece! Thanks a lot. I thought that the strong maximum principle is valid only for the parabolic equations, like the heat equation.
$endgroup$
– dmtri
Jan 24 at 20:03
1
$begingroup$
Glad to have helped. There is an elliptic and a parabolic version of it.
$endgroup$
– GReyes
Jan 24 at 20:08
add a comment |
$begingroup$
@dmitri Yes, the solution to your problem is unique. This follows from the (strong) maximum principle. The basic idea is that if you had a non-zero solution you would have a point in the square where it attains a positive maximum or a negative minimum. At that point your first derivatives would be zero and the second ones would have a definite sign. The degenerate case can be dealt with using a barrier type of argument. You can find more info here:
http://www.mi.uni-koeln.de/~gsweers/pdf/maxprinc.pdf
Check Corollary 7.
Hope this helps.
answered Jan 24 at 19:51
GReyesGReyes
2,09815
$endgroup$
$begingroup$
Nice refferenece! Thanks a lot. I thought that the strong maximum principle is valid only for the parabolic equations, like the heat equation.
$endgroup$
– dmtri
Jan 24 at 20:03
1
$begingroup$
Glad to have helped. There is an elliptic and a parabolic version of it.
$endgroup$
– GReyes
Jan 24 at 20:08
add a comment |
$begingroup$
@dmitri Yes, the solution to your problem is unique. This follows from the (strong) maximum principle. The basic idea is that if you had a non-zero solution you would have a point in the square where it attains a positive maximum or a negative minimum. At that point your first derivatives would be zero and the second ones would have a definite sign. The degenerate case can be dealt with using a barrier type of argument. You can find more info here:
http://www.mi.uni-koeln.de/~gsweers/pdf/maxprinc.pdf
Check Corollary 7.
Hope this helps.
answered Jan 24 at 19:51
GReyesGReyes
2,09815
$endgroup$
@dmitri Yes, the solution to your problem is unique. This follows from the (strong) maximum principle. The basic idea is that if you had a non-zero solution you would have a point in the square where it attains a positive maximum or a negative minimum. At that point your first derivatives would be zero and the second ones would have a definite sign. The degenerate case can be dealt with using a barrier type of argument. You can find more info here:
http://www.mi.uni-koeln.de/~gsweers/pdf/maxprinc.pdf
Check Corollary 7.
Hope this helps.
answered Jan 24 at 19:51
GReyesGReyes
2,09815
answered Jan 24 at 19:51
GReyesGReyes
2,09815
answered Jan 24 at 19:51
GReyesGReyes
2,09815
answered Jan 24 at 19:51
GReyesGReyes
2,09815
2,09815
$begingroup$
Nice refferenece! Thanks a lot. I thought that the strong maximum principle is valid only for the parabolic equations, like the heat equation.
$endgroup$
– dmtri
Jan 24 at 20:03
1
$begingroup$
Glad to have helped. There is an elliptic and a parabolic version of it.
$endgroup$
– GReyes
Jan 24 at 20:08
add a comment |
$begingroup$
Nice refferenece! Thanks a lot. I thought that the strong maximum principle is valid only for the parabolic equations, like the heat equation.
$endgroup$
– dmtri
Jan 24 at 20:03
1
$begingroup$
Glad to have helped. There is an elliptic and a parabolic version of it.
$endgroup$
– GReyes
Jan 24 at 20:08
$begingroup$
Nice refferenece! Thanks a lot. I thought that the strong maximum principle is valid only for the parabolic equations, like the heat equation.
$endgroup$
– dmtri
Jan 24 at 20:03
$begingroup$
Nice refferenece! Thanks a lot. I thought that the strong maximum principle is valid only for the parabolic equations, like the heat equation.
$endgroup$
– dmtri
Jan 24 at 20:03
1
1
$begingroup$
Glad to have helped. There is an elliptic and a parabolic version of it.
$endgroup$
– GReyes
Jan 24 at 20:08
$begingroup$
Glad to have helped. There is an elliptic and a parabolic version of it.
$endgroup$
– GReyes
Jan 24 at 20:08
add a comment |
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My gut reaction: the key feature of the first problem that makes the trivial solution unique is its linearity. As the same is true of the second problem, I'm guessing it will also have to be unique.
$endgroup$
– Adrian Keister
Jan 24 at 19:37