Poisson equation on semi-infinite strip
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Function $u(x,y)$ satisfy the equation:
$$Delta u = e^{-2y}sin x$$
in the semi-infinite strip:
$$0<x<pi, y>0$$
and the boundary condition:
$$u(0,y) = u(pi,y),text{ }u(x,0)=sin(3x),text{ } lim_{ytoinfty} u =0$$
Find $u_y(x,0)$
I've seen someone using Fourier tranform to solve similar poisson equation in the infinite strip, and separation of variables to solve Laplace in semi-infinite strip. So I'm wondering how to solve this one? (How do we usually decide when to use separation of variables and Fourier) Or the questions only asks for $u_y(x,0)$, so is it possible to find $u_y$ without solving the equation?
pde poissons-equation
asked Nov 19 at 23:02
QD666
1276
add a comment |
up vote
2
down vote
favorite
Function $u(x,y)$ satisfy the equation:
$$Delta u = e^{-2y}sin x$$
in the semi-infinite strip:
$$0<x<pi, y>0$$
and the boundary condition:
$$u(0,y) = u(pi,y),text{ }u(x,0)=sin(3x),text{ } lim_{ytoinfty} u =0$$
Find $u_y(x,0)$
I've seen someone using Fourier tranform to solve similar poisson equation in the infinite strip, and separation of variables to solve Laplace in semi-infinite strip. So I'm wondering how to solve this one? (How do we usually decide when to use separation of variables and Fourier) Or the questions only asks for $u_y(x,0)$, so is it possible to find $u_y$ without solving the equation?
pde poissons-equation
asked Nov 19 at 23:02
QD666
1276
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Function $u(x,y)$ satisfy the equation:
$$Delta u = e^{-2y}sin x$$
in the semi-infinite strip:
$$0<x<pi, y>0$$
and the boundary condition:
$$u(0,y) = u(pi,y),text{ }u(x,0)=sin(3x),text{ } lim_{ytoinfty} u =0$$
Find $u_y(x,0)$
I've seen someone using Fourier tranform to solve similar poisson equation in the infinite strip, and separation of variables to solve Laplace in semi-infinite strip. So I'm wondering how to solve this one? (How do we usually decide when to use separation of variables and Fourier) Or the questions only asks for $u_y(x,0)$, so is it possible to find $u_y$ without solving the equation?
pde poissons-equation
asked Nov 19 at 23:02
QD666
1276
Function $u(x,y)$ satisfy the equation:
$$Delta u = e^{-2y}sin x$$
in the semi-infinite strip:
$$0<x<pi, y>0$$
and the boundary condition:
$$u(0,y) = u(pi,y),text{ }u(x,0)=sin(3x),text{ } lim_{ytoinfty} u =0$$
Find $u_y(x,0)$
I've seen someone using Fourier tranform to solve similar poisson equation in the infinite strip, and separation of variables to solve Laplace in semi-infinite strip. So I'm wondering how to solve this one? (How do we usually decide when to use separation of variables and Fourier) Or the questions only asks for $u_y(x,0)$, so is it possible to find $u_y$ without solving the equation?
pde poissons-equation
pde poissons-equation
asked Nov 19 at 23:02
QD666
1276
asked Nov 19 at 23:02
QD666
1276
asked Nov 19 at 23:02
QD666
1276
asked Nov 19 at 23:02
QD666
1276
asked Nov 19 at 23:02
QD666
1276
1276
add a comment |
add a comment |
1 Answer
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1
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Let $v=u-frac{1}{3}e^{-2y}sin(x)$, where $u$ is a solution of the stated problem. Then $v$ is a solution of
$$
Delta v = Delta u - e^{-2y}sin(x)= 0,
$$
with conditions
$$
v(0,y)=v(pi,y) \
v(x,0)=u(x,0)-frac{1}{3}sin(x)=sin(3x)-frac{1}{3}sin(x)
$$
The solutions $v$ is
$$
v(x,y)=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}
$$
So,
begin{align}
u(x,y)&=v(x,y)+frac{1}{3}e^{-2y}sin(x) \ &=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}+frac{1}{3}sin(x)e^{-2y}.
end{align}
And,
$$
u_y(x,0)=-3sin(3x)+frac{1}{3}sin(x)-frac{2}{3}sin(x) \
=-3sin(3x)-frac{1}{3}sin(x).
$$
answered Nov 20 at 2:58
DisintegratingByParts
58.3k42478
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $v=u-frac{1}{3}e^{-2y}sin(x)$, where $u$ is a solution of the stated problem. Then $v$ is a solution of
$$
Delta v = Delta u - e^{-2y}sin(x)= 0,
$$
with conditions
$$
v(0,y)=v(pi,y) \
v(x,0)=u(x,0)-frac{1}{3}sin(x)=sin(3x)-frac{1}{3}sin(x)
$$
The solutions $v$ is
$$
v(x,y)=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}
$$
So,
begin{align}
u(x,y)&=v(x,y)+frac{1}{3}e^{-2y}sin(x) \ &=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}+frac{1}{3}sin(x)e^{-2y}.
end{align}
And,
$$
u_y(x,0)=-3sin(3x)+frac{1}{3}sin(x)-frac{2}{3}sin(x) \
=-3sin(3x)-frac{1}{3}sin(x).
$$
answered Nov 20 at 2:58
DisintegratingByParts
58.3k42478
add a comment |
up vote
1
down vote
accepted
Let $v=u-frac{1}{3}e^{-2y}sin(x)$, where $u$ is a solution of the stated problem. Then $v$ is a solution of
$$
Delta v = Delta u - e^{-2y}sin(x)= 0,
$$
with conditions
$$
v(0,y)=v(pi,y) \
v(x,0)=u(x,0)-frac{1}{3}sin(x)=sin(3x)-frac{1}{3}sin(x)
$$
The solutions $v$ is
$$
v(x,y)=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}
$$
So,
begin{align}
u(x,y)&=v(x,y)+frac{1}{3}e^{-2y}sin(x) \ &=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}+frac{1}{3}sin(x)e^{-2y}.
end{align}
And,
$$
u_y(x,0)=-3sin(3x)+frac{1}{3}sin(x)-frac{2}{3}sin(x) \
=-3sin(3x)-frac{1}{3}sin(x).
$$
answered Nov 20 at 2:58
DisintegratingByParts
58.3k42478
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $v=u-frac{1}{3}e^{-2y}sin(x)$, where $u$ is a solution of the stated problem. Then $v$ is a solution of
$$
Delta v = Delta u - e^{-2y}sin(x)= 0,
$$
with conditions
$$
v(0,y)=v(pi,y) \
v(x,0)=u(x,0)-frac{1}{3}sin(x)=sin(3x)-frac{1}{3}sin(x)
$$
The solutions $v$ is
$$
v(x,y)=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}
$$
So,
begin{align}
u(x,y)&=v(x,y)+frac{1}{3}e^{-2y}sin(x) \ &=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}+frac{1}{3}sin(x)e^{-2y}.
end{align}
And,
$$
u_y(x,0)=-3sin(3x)+frac{1}{3}sin(x)-frac{2}{3}sin(x) \
=-3sin(3x)-frac{1}{3}sin(x).
$$
answered Nov 20 at 2:58
DisintegratingByParts
58.3k42478
Let $v=u-frac{1}{3}e^{-2y}sin(x)$, where $u$ is a solution of the stated problem. Then $v$ is a solution of
$$
Delta v = Delta u - e^{-2y}sin(x)= 0,
$$
with conditions
$$
v(0,y)=v(pi,y) \
v(x,0)=u(x,0)-frac{1}{3}sin(x)=sin(3x)-frac{1}{3}sin(x)
$$
The solutions $v$ is
$$
v(x,y)=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}
$$
So,
begin{align}
u(x,y)&=v(x,y)+frac{1}{3}e^{-2y}sin(x) \ &=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}+frac{1}{3}sin(x)e^{-2y}.
end{align}
And,
$$
u_y(x,0)=-3sin(3x)+frac{1}{3}sin(x)-frac{2}{3}sin(x) \
=-3sin(3x)-frac{1}{3}sin(x).
$$
answered Nov 20 at 2:58
DisintegratingByParts
58.3k42478
answered Nov 20 at 2:58
DisintegratingByParts
58.3k42478
answered Nov 20 at 2:58
DisintegratingByParts
58.3k42478
answered Nov 20 at 2:58
DisintegratingByParts
58.3k42478
58.3k42478
add a comment |
add a comment |
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