Mixing
PDE: BVP of Laplace equation on $mathbbRtimes [0,+infty)$
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Solve the BVP:
$$u_xx+u_yy=0,~~xin mathbbR,~y>0,~~~u(x,0)=0,~~~u_y(x,0)=asin(x/a),~xin mathbbR,$$
where $a$ is a positive constant.
Attempt. By separation of variables we get $u=XY$,
where $X''Y+XY''=0$, i.e. $$X''-cX=0, ~Y''+cY=0.$$
The boundary condition $u(x,0)=0$ gives $Y(0)=0$ and the only bounded solutions for $y>0$ are $Y(y)=sin(ky),$ for $c=k^2>0,~k>0$. Then $X=Ae^-k+Be^x$ and in order to have bounded solutions for $x$ we get $X=Ae^-k$. Thus, our solutions are $u_k=X_kY_k$, for $k>0$.
Since $k>0$, $u_k's$ are uncountable. How can we get a sum of the form
$sum_kE_ke^-ksin(ky)$ in order to use the boundary condition $u_y(x,0)=asin(x/a)$?
Thanks in advance.
pde laplacian
asked Jul 24 at 13:43
Nikolaos Skout
1,810416
add a comment |Â
up vote
1
down vote
favorite
Solve the BVP:
$$u_xx+u_yy=0,~~xin mathbbR,~y>0,~~~u(x,0)=0,~~~u_y(x,0)=asin(x/a),~xin mathbbR,$$
where $a$ is a positive constant.
Attempt. By separation of variables we get $u=XY$,
where $X''Y+XY''=0$, i.e. $$X''-cX=0, ~Y''+cY=0.$$
The boundary condition $u(x,0)=0$ gives $Y(0)=0$ and the only bounded solutions for $y>0$ are $Y(y)=sin(ky),$ for $c=k^2>0,~k>0$. Then $X=Ae^-k+Be^x$ and in order to have bounded solutions for $x$ we get $X=Ae^-k$. Thus, our solutions are $u_k=X_kY_k$, for $k>0$.
Since $k>0$, $u_k's$ are uncountable. How can we get a sum of the form
$sum_kE_ke^-ksin(ky)$ in order to use the boundary condition $u_y(x,0)=asin(x/a)$?
Thanks in advance.
pde laplacian
asked Jul 24 at 13:43
Nikolaos Skout
1,810416
Perhaps the easiest way to solve your (Neumann) BVP is to use the Green function for the half plane. The separation of variables and solution by using the classical 2nd order ODE works well if you have a bounded domain, but if you are dealing with unbounded ones, the periodicity of the eigenfunctions is of no help at all.
– Daniele Tampieri
Jul 24 at 15:04
The problem I am working on insists on separating the variables.
– Nikolaos Skout
Jul 24 at 15:08
1
If you allow for unbounded solutions then $u = a^2sin(x/a)sinh(y/a)$ is a solution
– Dylan
Jul 24 at 15:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Solve the BVP:
$$u_xx+u_yy=0,~~xin mathbbR,~y>0,~~~u(x,0)=0,~~~u_y(x,0)=asin(x/a),~xin mathbbR,$$
where $a$ is a positive constant.
Attempt. By separation of variables we get $u=XY$,
where $X''Y+XY''=0$, i.e. $$X''-cX=0, ~Y''+cY=0.$$
The boundary condition $u(x,0)=0$ gives $Y(0)=0$ and the only bounded solutions for $y>0$ are $Y(y)=sin(ky),$ for $c=k^2>0,~k>0$. Then $X=Ae^-k+Be^x$ and in order to have bounded solutions for $x$ we get $X=Ae^-k$. Thus, our solutions are $u_k=X_kY_k$, for $k>0$.
Since $k>0$, $u_k's$ are uncountable. How can we get a sum of the form
$sum_kE_ke^-ksin(ky)$ in order to use the boundary condition $u_y(x,0)=asin(x/a)$?
Thanks in advance.
pde laplacian
asked Jul 24 at 13:43
Nikolaos Skout
1,810416
Solve the BVP:
$$u_xx+u_yy=0,~~xin mathbbR,~y>0,~~~u(x,0)=0,~~~u_y(x,0)=asin(x/a),~xin mathbbR,$$
where $a$ is a positive constant.
Attempt. By separation of variables we get $u=XY$,
where $X''Y+XY''=0$, i.e. $$X''-cX=0, ~Y''+cY=0.$$
The boundary condition $u(x,0)=0$ gives $Y(0)=0$ and the only bounded solutions for $y>0$ are $Y(y)=sin(ky),$ for $c=k^2>0,~k>0$. Then $X=Ae^-k+Be^x$ and in order to have bounded solutions for $x$ we get $X=Ae^-k$. Thus, our solutions are $u_k=X_kY_k$, for $k>0$.
Since $k>0$, $u_k's$ are uncountable. How can we get a sum of the form
$sum_kE_ke^-ksin(ky)$ in order to use the boundary condition $u_y(x,0)=asin(x/a)$?
Thanks in advance.
pde laplacian
asked Jul 24 at 13:43
Nikolaos Skout
1,810416
asked Jul 24 at 13:43
Nikolaos Skout
1,810416
asked Jul 24 at 13:43
Nikolaos Skout
1,810416
asked Jul 24 at 13:43
Nikolaos Skout
1,810416
1,810416
Perhaps the easiest way to solve your (Neumann) BVP is to use the Green function for the half plane. The separation of variables and solution by using the classical 2nd order ODE works well if you have a bounded domain, but if you are dealing with unbounded ones, the periodicity of the eigenfunctions is of no help at all.
– Daniele Tampieri
Jul 24 at 15:04
The problem I am working on insists on separating the variables.
– Nikolaos Skout
Jul 24 at 15:08
1
If you allow for unbounded solutions then $u = a^2sin(x/a)sinh(y/a)$ is a solution
– Dylan
Jul 24 at 15:59
add a comment |Â
Perhaps the easiest way to solve your (Neumann) BVP is to use the Green function for the half plane. The separation of variables and solution by using the classical 2nd order ODE works well if you have a bounded domain, but if you are dealing with unbounded ones, the periodicity of the eigenfunctions is of no help at all.
– Daniele Tampieri
Jul 24 at 15:04
The problem I am working on insists on separating the variables.
– Nikolaos Skout
Jul 24 at 15:08
1
If you allow for unbounded solutions then $u = a^2sin(x/a)sinh(y/a)$ is a solution
– Dylan
Jul 24 at 15:59
Perhaps the easiest way to solve your (Neumann) BVP is to use the Green function for the half plane. The separation of variables and solution by using the classical 2nd order ODE works well if you have a bounded domain, but if you are dealing with unbounded ones, the periodicity of the eigenfunctions is of no help at all.
– Daniele Tampieri
Jul 24 at 15:04
Perhaps the easiest way to solve your (Neumann) BVP is to use the Green function for the half plane. The separation of variables and solution by using the classical 2nd order ODE works well if you have a bounded domain, but if you are dealing with unbounded ones, the periodicity of the eigenfunctions is of no help at all.
– Daniele Tampieri
Jul 24 at 15:04
The problem I am working on insists on separating the variables.
– Nikolaos Skout
Jul 24 at 15:08
The problem I am working on insists on separating the variables.
– Nikolaos Skout
Jul 24 at 15:08
1
1
If you allow for unbounded solutions then $u = a^2sin(x/a)sinh(y/a)$ is a solution
– Dylan
Jul 24 at 15:59
If you allow for unbounded solutions then $u = a^2sin(x/a)sinh(y/a)$ is a solution
– Dylan
Jul 24 at 15:59
add a comment |Â
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Perhaps the easiest way to solve your (Neumann) BVP is to use the Green function for the half plane. The separation of variables and solution by using the classical 2nd order ODE works well if you have a bounded domain, but if you are dealing with unbounded ones, the periodicity of the eigenfunctions is of no help at all.
– Daniele Tampieri
Jul 24 at 15:04
The problem I am working on insists on separating the variables.
– Nikolaos Skout
Jul 24 at 15:08
1
If you allow for unbounded solutions then $u = a^2sin(x/a)sinh(y/a)$ is a solution
– Dylan
Jul 24 at 15:59