Mixing
Laplace's Equation in the semi infinite strip
$begingroup$
The problem is $$Delta u = 0$$ in the semi infinite strip $S = {(x,y)|0<x<1,y>0}$ subject to the boundary conditions $$
begin{cases}
u(0,y) = 0, yge0\
u(1,y) = 0, yge0\
u(x,0) = f(x), 0le x le 1
end{cases}
$$
where $f$ is a given function with $f(0) = f(1) = 0$. Write $$f(x) = sum_{n=1}^{infty} a_nsin(npi x)$$ and expand the general solution in terms of the special solutions given by $u_n(x,y) = e^{-n pi y}sin(n pi x)$.
The above is the problem.
So by using separation of variables and analyzing the first two boundary conditions, I got $u(x,y) = sum_{n=1}^{infty} sin(npi x)(C_n e^{-n pi y}+D_n e^{n pi y})$, where $C_n$ and $D_n$ are constants. My question is if I could get rid of the $e^{n pi y}$ part from the general solution. I've seen someone did so but usually the boundary conditions would include for example $lim_{ytoinfty} u(x,y)$ is finite. But here the boundary conditions do not say something like this. Can I still get rid of that part based on the give information? I'm a bit confused since the later instructions of the problem seem to suggest only $e^{-n pi y}$ part remains.
Thank you so much for any insights.
pde boundary-value-problem
asked Feb 3 at 6:09
xf16xf16
937
$endgroup$
add a comment |
$begingroup$
The problem is $$Delta u = 0$$ in the semi infinite strip $S = {(x,y)|0<x<1,y>0}$ subject to the boundary conditions $$
begin{cases}
u(0,y) = 0, yge0\
u(1,y) = 0, yge0\
u(x,0) = f(x), 0le x le 1
end{cases}
$$
where $f$ is a given function with $f(0) = f(1) = 0$. Write $$f(x) = sum_{n=1}^{infty} a_nsin(npi x)$$ and expand the general solution in terms of the special solutions given by $u_n(x,y) = e^{-n pi y}sin(n pi x)$.
The above is the problem.
So by using separation of variables and analyzing the first two boundary conditions, I got $u(x,y) = sum_{n=1}^{infty} sin(npi x)(C_n e^{-n pi y}+D_n e^{n pi y})$, where $C_n$ and $D_n$ are constants. My question is if I could get rid of the $e^{n pi y}$ part from the general solution. I've seen someone did so but usually the boundary conditions would include for example $lim_{ytoinfty} u(x,y)$ is finite. But here the boundary conditions do not say something like this. Can I still get rid of that part based on the give information? I'm a bit confused since the later instructions of the problem seem to suggest only $e^{-n pi y}$ part remains.
Thank you so much for any insights.
pde boundary-value-problem
asked Feb 3 at 6:09
xf16xf16
937
$endgroup$
1
$begingroup$
Unfortunately you need two conditions to determine two constants. The solution may be implied to be bounded at $y to infty$ which would give $D_n=0$
$endgroup$
– Dylan
Feb 3 at 9:00
add a comment |
$begingroup$
The problem is $$Delta u = 0$$ in the semi infinite strip $S = {(x,y)|0<x<1,y>0}$ subject to the boundary conditions $$
begin{cases}
u(0,y) = 0, yge0\
u(1,y) = 0, yge0\
u(x,0) = f(x), 0le x le 1
end{cases}
$$
where $f$ is a given function with $f(0) = f(1) = 0$. Write $$f(x) = sum_{n=1}^{infty} a_nsin(npi x)$$ and expand the general solution in terms of the special solutions given by $u_n(x,y) = e^{-n pi y}sin(n pi x)$.
The above is the problem.
So by using separation of variables and analyzing the first two boundary conditions, I got $u(x,y) = sum_{n=1}^{infty} sin(npi x)(C_n e^{-n pi y}+D_n e^{n pi y})$, where $C_n$ and $D_n$ are constants. My question is if I could get rid of the $e^{n pi y}$ part from the general solution. I've seen someone did so but usually the boundary conditions would include for example $lim_{ytoinfty} u(x,y)$ is finite. But here the boundary conditions do not say something like this. Can I still get rid of that part based on the give information? I'm a bit confused since the later instructions of the problem seem to suggest only $e^{-n pi y}$ part remains.
Thank you so much for any insights.
pde boundary-value-problem
asked Feb 3 at 6:09
xf16xf16
937
$endgroup$
The problem is $$Delta u = 0$$ in the semi infinite strip $S = {(x,y)|0<x<1,y>0}$ subject to the boundary conditions $$
begin{cases}
u(0,y) = 0, yge0\
u(1,y) = 0, yge0\
u(x,0) = f(x), 0le x le 1
end{cases}
$$
where $f$ is a given function with $f(0) = f(1) = 0$. Write $$f(x) = sum_{n=1}^{infty} a_nsin(npi x)$$ and expand the general solution in terms of the special solutions given by $u_n(x,y) = e^{-n pi y}sin(n pi x)$.
The above is the problem.
So by using separation of variables and analyzing the first two boundary conditions, I got $u(x,y) = sum_{n=1}^{infty} sin(npi x)(C_n e^{-n pi y}+D_n e^{n pi y})$, where $C_n$ and $D_n$ are constants. My question is if I could get rid of the $e^{n pi y}$ part from the general solution. I've seen someone did so but usually the boundary conditions would include for example $lim_{ytoinfty} u(x,y)$ is finite. But here the boundary conditions do not say something like this. Can I still get rid of that part based on the give information? I'm a bit confused since the later instructions of the problem seem to suggest only $e^{-n pi y}$ part remains.
Thank you so much for any insights.
pde boundary-value-problem
pde boundary-value-problem
asked Feb 3 at 6:09
xf16xf16
937
asked Feb 3 at 6:09
xf16xf16
937
asked Feb 3 at 6:09
xf16xf16
937
asked Feb 3 at 6:09
xf16xf16
937
asked Feb 3 at 6:09
xf16xf16
937
937
1
$begingroup$
Unfortunately you need two conditions to determine two constants. The solution may be implied to be bounded at $y to infty$ which would give $D_n=0$
$endgroup$
– Dylan
Feb 3 at 9:00
add a comment |
1
$begingroup$
Unfortunately you need two conditions to determine two constants. The solution may be implied to be bounded at $y to infty$ which would give $D_n=0$
$endgroup$
– Dylan
Feb 3 at 9:00
1
1
$begingroup$
Unfortunately you need two conditions to determine two constants. The solution may be implied to be bounded at $y to infty$ which would give $D_n=0$
$endgroup$
– Dylan
Feb 3 at 9:00
$begingroup$
Unfortunately you need two conditions to determine two constants. The solution may be implied to be bounded at $y to infty$ which would give $D_n=0$
$endgroup$
– Dylan
Feb 3 at 9:00
add a comment |
1 Answer
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oldest
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$begingroup$
Yes, there is one condition missing, as your semi-infinite strip has, very simply put, four boundary segments, at $x in {0,1}$ and at $y in {0,infty}$. But you have prescribed conditions on only three of the boundary segments so far, which leaves you with arbitrary constants in the solution (you obtain $C_n + D_n = a_n$ where only $a_n$ is known).
The boundedness of the solution as $y rightarrow infty$ usually follows from some physical considerations.
answered Feb 3 at 7:36
ChristophChristoph
59616
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add a comment |
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$begingroup$
Yes, there is one condition missing, as your semi-infinite strip has, very simply put, four boundary segments, at $x in {0,1}$ and at $y in {0,infty}$. But you have prescribed conditions on only three of the boundary segments so far, which leaves you with arbitrary constants in the solution (you obtain $C_n + D_n = a_n$ where only $a_n$ is known).
The boundedness of the solution as $y rightarrow infty$ usually follows from some physical considerations.
answered Feb 3 at 7:36
ChristophChristoph
59616
$endgroup$
add a comment |
$begingroup$
Yes, there is one condition missing, as your semi-infinite strip has, very simply put, four boundary segments, at $x in {0,1}$ and at $y in {0,infty}$. But you have prescribed conditions on only three of the boundary segments so far, which leaves you with arbitrary constants in the solution (you obtain $C_n + D_n = a_n$ where only $a_n$ is known).
The boundedness of the solution as $y rightarrow infty$ usually follows from some physical considerations.
answered Feb 3 at 7:36
ChristophChristoph
59616
$endgroup$
add a comment |
$begingroup$
Yes, there is one condition missing, as your semi-infinite strip has, very simply put, four boundary segments, at $x in {0,1}$ and at $y in {0,infty}$. But you have prescribed conditions on only three of the boundary segments so far, which leaves you with arbitrary constants in the solution (you obtain $C_n + D_n = a_n$ where only $a_n$ is known).
The boundedness of the solution as $y rightarrow infty$ usually follows from some physical considerations.
answered Feb 3 at 7:36
ChristophChristoph
59616
$endgroup$
Yes, there is one condition missing, as your semi-infinite strip has, very simply put, four boundary segments, at $x in {0,1}$ and at $y in {0,infty}$. But you have prescribed conditions on only three of the boundary segments so far, which leaves you with arbitrary constants in the solution (you obtain $C_n + D_n = a_n$ where only $a_n$ is known).
The boundedness of the solution as $y rightarrow infty$ usually follows from some physical considerations.
answered Feb 3 at 7:36
ChristophChristoph
59616
edited Feb 3 at 7:54
answered Feb 3 at 7:36
ChristophChristoph
59616
answered Feb 3 at 7:36
ChristophChristoph
59616
answered Feb 3 at 7:36
ChristophChristoph
59616
59616
add a comment |
add a comment |
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Unfortunately you need two conditions to determine two constants. The solution may be implied to be bounded at $y to infty$ which would give $D_n=0$
$endgroup$
– Dylan
Feb 3 at 9:00