Mixing
What is meant by homogeneous boundary conditions?
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I am sorry if this is basic knowledge for differential equations but it has been a long time since I took the class, I probably learnt it and forgot about it. I would appreciate the explanation. Thank you!
ordinary-differential-equations analysis
asked Dec 3 '17 at 21:46
dareToDiffer07dareToDiffer07
2817
$endgroup$
add a comment |
$begingroup$
I am sorry if this is basic knowledge for differential equations but it has been a long time since I took the class, I probably learnt it and forgot about it. I would appreciate the explanation. Thank you!
ordinary-differential-equations analysis
asked Dec 3 '17 at 21:46
dareToDiffer07dareToDiffer07
2817
$endgroup$
5
$begingroup$
It means the function is zero at the boundaries
$endgroup$
– Alex
Dec 3 '17 at 21:48
$begingroup$
Makes sense, I was thinking that was the case but was not sure. Thanks!
$endgroup$
– dareToDiffer07
Dec 3 '17 at 21:51
add a comment |
$begingroup$
I am sorry if this is basic knowledge for differential equations but it has been a long time since I took the class, I probably learnt it and forgot about it. I would appreciate the explanation. Thank you!
ordinary-differential-equations analysis
asked Dec 3 '17 at 21:46
dareToDiffer07dareToDiffer07
2817
$endgroup$
I am sorry if this is basic knowledge for differential equations but it has been a long time since I took the class, I probably learnt it and forgot about it. I would appreciate the explanation. Thank you!
ordinary-differential-equations analysis
ordinary-differential-equations analysis
asked Dec 3 '17 at 21:46
dareToDiffer07dareToDiffer07
2817
asked Dec 3 '17 at 21:46
dareToDiffer07dareToDiffer07
2817
asked Dec 3 '17 at 21:46
dareToDiffer07dareToDiffer07
2817
asked Dec 3 '17 at 21:46
dareToDiffer07dareToDiffer07
2817
asked Dec 3 '17 at 21:46
dareToDiffer07dareToDiffer07
2817
2817
5
$begingroup$
It means the function is zero at the boundaries
$endgroup$
– Alex
Dec 3 '17 at 21:48
$begingroup$
Makes sense, I was thinking that was the case but was not sure. Thanks!
$endgroup$
– dareToDiffer07
Dec 3 '17 at 21:51
add a comment |
5
$begingroup$
It means the function is zero at the boundaries
$endgroup$
– Alex
Dec 3 '17 at 21:48
$begingroup$
Makes sense, I was thinking that was the case but was not sure. Thanks!
$endgroup$
– dareToDiffer07
Dec 3 '17 at 21:51
5
5
$begingroup$
It means the function is zero at the boundaries
$endgroup$
– Alex
Dec 3 '17 at 21:48
$begingroup$
It means the function is zero at the boundaries
$endgroup$
– Alex
Dec 3 '17 at 21:48
$begingroup$
Makes sense, I was thinking that was the case but was not sure. Thanks!
$endgroup$
– dareToDiffer07
Dec 3 '17 at 21:51
$begingroup$
Makes sense, I was thinking that was the case but was not sure. Thanks!
$endgroup$
– dareToDiffer07
Dec 3 '17 at 21:51
add a comment |
2 Answers
2
active
oldest
votes
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The simplest way to test whether an equation (here the equation for the boundary conditions) is homogeneous is to substitute the zero function and see whether it equals to zero.
edited Dec 13 '18 at 14:40
LutzL
58.7k42055
answered Oct 10 '18 at 23:42
James B. VidalesJames B. Vidales
191
$endgroup$
add a comment |
$begingroup$
If your differential equation is homogeneous (it is equal to zero and not some function), for instance, $$frac{d^2y}{dx^2}+4y=0$$ and you were asked to solve the equation given the boundary conditions,
$$y(x = 0) = 0$$
$$y(x = 2pi) = 0$$
Then the boundary conditions above are known as homogenous boundary conditions.
It is important to remember that when we say homogeneous (or inhomogeneous) we are saying something not only about the differential equation itself but also about the boundary conditions as well.
Formulation and motivation for this answer have been summarized from Paul's Online Notes
answered Feb 1 at 21:26
BLAZEBLAZE
6,098112756
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The simplest way to test whether an equation (here the equation for the boundary conditions) is homogeneous is to substitute the zero function and see whether it equals to zero.
edited Dec 13 '18 at 14:40
LutzL
58.7k42055
answered Oct 10 '18 at 23:42
James B. VidalesJames B. Vidales
191
$endgroup$
add a comment |
$begingroup$
The simplest way to test whether an equation (here the equation for the boundary conditions) is homogeneous is to substitute the zero function and see whether it equals to zero.
edited Dec 13 '18 at 14:40
LutzL
58.7k42055
answered Oct 10 '18 at 23:42
James B. VidalesJames B. Vidales
191
$endgroup$
add a comment |
$begingroup$
The simplest way to test whether an equation (here the equation for the boundary conditions) is homogeneous is to substitute the zero function and see whether it equals to zero.
edited Dec 13 '18 at 14:40
LutzL
58.7k42055
answered Oct 10 '18 at 23:42
James B. VidalesJames B. Vidales
191
$endgroup$
The simplest way to test whether an equation (here the equation for the boundary conditions) is homogeneous is to substitute the zero function and see whether it equals to zero.
edited Dec 13 '18 at 14:40
LutzL
58.7k42055
answered Oct 10 '18 at 23:42
James B. VidalesJames B. Vidales
191
edited Dec 13 '18 at 14:40
LutzL
58.7k42055
edited Dec 13 '18 at 14:40
LutzL
58.7k42055
edited Dec 13 '18 at 14:40
LutzL
58.7k42055
58.7k42055
answered Oct 10 '18 at 23:42
James B. VidalesJames B. Vidales
191
answered Oct 10 '18 at 23:42
James B. VidalesJames B. Vidales
191
answered Oct 10 '18 at 23:42
James B. VidalesJames B. Vidales
191
191
add a comment |
add a comment |
$begingroup$
If your differential equation is homogeneous (it is equal to zero and not some function), for instance, $$frac{d^2y}{dx^2}+4y=0$$ and you were asked to solve the equation given the boundary conditions,
$$y(x = 0) = 0$$
$$y(x = 2pi) = 0$$
Then the boundary conditions above are known as homogenous boundary conditions.
It is important to remember that when we say homogeneous (or inhomogeneous) we are saying something not only about the differential equation itself but also about the boundary conditions as well.
Formulation and motivation for this answer have been summarized from Paul's Online Notes
answered Feb 1 at 21:26
BLAZEBLAZE
6,098112756
$endgroup$
add a comment |
$begingroup$
If your differential equation is homogeneous (it is equal to zero and not some function), for instance, $$frac{d^2y}{dx^2}+4y=0$$ and you were asked to solve the equation given the boundary conditions,
$$y(x = 0) = 0$$
$$y(x = 2pi) = 0$$
Then the boundary conditions above are known as homogenous boundary conditions.
It is important to remember that when we say homogeneous (or inhomogeneous) we are saying something not only about the differential equation itself but also about the boundary conditions as well.
Formulation and motivation for this answer have been summarized from Paul's Online Notes
answered Feb 1 at 21:26
BLAZEBLAZE
6,098112756
$endgroup$
add a comment |
$begingroup$
If your differential equation is homogeneous (it is equal to zero and not some function), for instance, $$frac{d^2y}{dx^2}+4y=0$$ and you were asked to solve the equation given the boundary conditions,
$$y(x = 0) = 0$$
$$y(x = 2pi) = 0$$
Then the boundary conditions above are known as homogenous boundary conditions.
It is important to remember that when we say homogeneous (or inhomogeneous) we are saying something not only about the differential equation itself but also about the boundary conditions as well.
Formulation and motivation for this answer have been summarized from Paul's Online Notes
answered Feb 1 at 21:26
BLAZEBLAZE
6,098112756
$endgroup$
If your differential equation is homogeneous (it is equal to zero and not some function), for instance, $$frac{d^2y}{dx^2}+4y=0$$ and you were asked to solve the equation given the boundary conditions,
$$y(x = 0) = 0$$
$$y(x = 2pi) = 0$$
Then the boundary conditions above are known as homogenous boundary conditions.
It is important to remember that when we say homogeneous (or inhomogeneous) we are saying something not only about the differential equation itself but also about the boundary conditions as well.
Formulation and motivation for this answer have been summarized from Paul's Online Notes
answered Feb 1 at 21:26
BLAZEBLAZE
6,098112756
answered Feb 1 at 21:26
BLAZEBLAZE
6,098112756
answered Feb 1 at 21:26
BLAZEBLAZE
6,098112756
answered Feb 1 at 21:26
BLAZEBLAZE
6,098112756
6,098112756
add a comment |
add a comment |
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5
$begingroup$
It means the function is zero at the boundaries
$endgroup$
– Alex
Dec 3 '17 at 21:48
$begingroup$
Makes sense, I was thinking that was the case but was not sure. Thanks!
$endgroup$
– dareToDiffer07
Dec 3 '17 at 21:51