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Separation of variables and non-constant boundaries?
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I am not sure why I am having a difficult time answering this seemingly simple question. (Perhaps I learned this topic only very superficially.)
Let us assume that we have a wave equation defined for some time $tin [0,infty)$ on some domain $D$:
$partial_t^2 phi = nabla^2 phi$,
with $phi(x,y,t)$.
The go-to method to solve this system is to use separation of variables: $phi = X(x)Y(y)T(t)$.
This makes sense if the boundaries of the domain are constants, but what if they are allowed to vary (let's only vary in space for now)?
For example, let $D$ be a box with vertical sides, and some curve $y_1(x)$ for the top, and another curve $y_2(x)$ for the bottom.
This seems like a silly question. Of course we cannot separate the solutions. We will find that $Y(y)$ is somehow dependent on $x$.
But then, we assume spatially dependent initial values in the wave equation all the time...
What am I overlooking here? I feel like there is something subtle happening and I really can't put my finger on it.
Is there a proof that anybody can provide which goes one way or another?
Thanks.
boundary-value-problem wave-equation linear-pde
asked Jul 26 at 19:36
user109527
657
add a comment |Â
up vote
0
down vote
favorite
I am not sure why I am having a difficult time answering this seemingly simple question. (Perhaps I learned this topic only very superficially.)
Let us assume that we have a wave equation defined for some time $tin [0,infty)$ on some domain $D$:
$partial_t^2 phi = nabla^2 phi$,
with $phi(x,y,t)$.
The go-to method to solve this system is to use separation of variables: $phi = X(x)Y(y)T(t)$.
This makes sense if the boundaries of the domain are constants, but what if they are allowed to vary (let's only vary in space for now)?
For example, let $D$ be a box with vertical sides, and some curve $y_1(x)$ for the top, and another curve $y_2(x)$ for the bottom.
This seems like a silly question. Of course we cannot separate the solutions. We will find that $Y(y)$ is somehow dependent on $x$.
But then, we assume spatially dependent initial values in the wave equation all the time...
What am I overlooking here? I feel like there is something subtle happening and I really can't put my finger on it.
Is there a proof that anybody can provide which goes one way or another?
Thanks.
boundary-value-problem wave-equation linear-pde
asked Jul 26 at 19:36
user109527
657
Separation of variables does not work here exactly because of what you said. The boundary conditions are coupling $x,y$ together, so how can the spatial part of the solution be of the form $X(x)Y(y)$, as in they are independent.
– Hamed
Jul 26 at 20:01
In more physical terms, the usual boundary conditions give you a standing wave with endpoints fixed on the boundary. In your case, a similar solution is a standing wave terminating on your boundary. Obviously, this a very complicated wave squeezing, streching and twisting as $x$ varies. You can probably make a homemade experiment to see this. Get a thin piece of metal which can be bent. Soak it in soap water and create a bubble. Give the metal an irregular shape and slowly oscillate up and down (careful not to break the bubble). What you'll see is a very complicated motion.
– Hamed
Jul 26 at 20:07
Yes, this was sort of what I was looking for, but not quite. Regarding the initial conditions being a function of space: This works precisely because initial conditions are defined at a t = constant value. Regarding boundaries which vary in space: I think that we need to define the boundary for a constant value of y (as in the example above). Otherwise, we need to find a new coordinate system in which we do have a boundary at a constant value of one of the coordinate variables. Separation of variables is then possible if the Laplacian in the new coordinates is diagonalizable.
– user109527
Jul 27 at 19:25
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am not sure why I am having a difficult time answering this seemingly simple question. (Perhaps I learned this topic only very superficially.)
Let us assume that we have a wave equation defined for some time $tin [0,infty)$ on some domain $D$:
$partial_t^2 phi = nabla^2 phi$,
with $phi(x,y,t)$.
The go-to method to solve this system is to use separation of variables: $phi = X(x)Y(y)T(t)$.
This makes sense if the boundaries of the domain are constants, but what if they are allowed to vary (let's only vary in space for now)?
For example, let $D$ be a box with vertical sides, and some curve $y_1(x)$ for the top, and another curve $y_2(x)$ for the bottom.
This seems like a silly question. Of course we cannot separate the solutions. We will find that $Y(y)$ is somehow dependent on $x$.
But then, we assume spatially dependent initial values in the wave equation all the time...
What am I overlooking here? I feel like there is something subtle happening and I really can't put my finger on it.
Is there a proof that anybody can provide which goes one way or another?
Thanks.
boundary-value-problem wave-equation linear-pde
asked Jul 26 at 19:36
user109527
657
I am not sure why I am having a difficult time answering this seemingly simple question. (Perhaps I learned this topic only very superficially.)
Let us assume that we have a wave equation defined for some time $tin [0,infty)$ on some domain $D$:
$partial_t^2 phi = nabla^2 phi$,
with $phi(x,y,t)$.
The go-to method to solve this system is to use separation of variables: $phi = X(x)Y(y)T(t)$.
This makes sense if the boundaries of the domain are constants, but what if they are allowed to vary (let's only vary in space for now)?
For example, let $D$ be a box with vertical sides, and some curve $y_1(x)$ for the top, and another curve $y_2(x)$ for the bottom.
This seems like a silly question. Of course we cannot separate the solutions. We will find that $Y(y)$ is somehow dependent on $x$.
But then, we assume spatially dependent initial values in the wave equation all the time...
What am I overlooking here? I feel like there is something subtle happening and I really can't put my finger on it.
Is there a proof that anybody can provide which goes one way or another?
Thanks.
boundary-value-problem wave-equation linear-pde
asked Jul 26 at 19:36
user109527
657
asked Jul 26 at 19:36
user109527
657
asked Jul 26 at 19:36
user109527
657
asked Jul 26 at 19:36
user109527
657
657
Separation of variables does not work here exactly because of what you said. The boundary conditions are coupling $x,y$ together, so how can the spatial part of the solution be of the form $X(x)Y(y)$, as in they are independent.
– Hamed
Jul 26 at 20:01
In more physical terms, the usual boundary conditions give you a standing wave with endpoints fixed on the boundary. In your case, a similar solution is a standing wave terminating on your boundary. Obviously, this a very complicated wave squeezing, streching and twisting as $x$ varies. You can probably make a homemade experiment to see this. Get a thin piece of metal which can be bent. Soak it in soap water and create a bubble. Give the metal an irregular shape and slowly oscillate up and down (careful not to break the bubble). What you'll see is a very complicated motion.
– Hamed
Jul 26 at 20:07
Yes, this was sort of what I was looking for, but not quite. Regarding the initial conditions being a function of space: This works precisely because initial conditions are defined at a t = constant value. Regarding boundaries which vary in space: I think that we need to define the boundary for a constant value of y (as in the example above). Otherwise, we need to find a new coordinate system in which we do have a boundary at a constant value of one of the coordinate variables. Separation of variables is then possible if the Laplacian in the new coordinates is diagonalizable.
– user109527
Jul 27 at 19:25
add a comment |Â
Separation of variables does not work here exactly because of what you said. The boundary conditions are coupling $x,y$ together, so how can the spatial part of the solution be of the form $X(x)Y(y)$, as in they are independent.
– Hamed
Jul 26 at 20:01
In more physical terms, the usual boundary conditions give you a standing wave with endpoints fixed on the boundary. In your case, a similar solution is a standing wave terminating on your boundary. Obviously, this a very complicated wave squeezing, streching and twisting as $x$ varies. You can probably make a homemade experiment to see this. Get a thin piece of metal which can be bent. Soak it in soap water and create a bubble. Give the metal an irregular shape and slowly oscillate up and down (careful not to break the bubble). What you'll see is a very complicated motion.
– Hamed
Jul 26 at 20:07
Yes, this was sort of what I was looking for, but not quite. Regarding the initial conditions being a function of space: This works precisely because initial conditions are defined at a t = constant value. Regarding boundaries which vary in space: I think that we need to define the boundary for a constant value of y (as in the example above). Otherwise, we need to find a new coordinate system in which we do have a boundary at a constant value of one of the coordinate variables. Separation of variables is then possible if the Laplacian in the new coordinates is diagonalizable.
– user109527
Jul 27 at 19:25
Separation of variables does not work here exactly because of what you said. The boundary conditions are coupling $x,y$ together, so how can the spatial part of the solution be of the form $X(x)Y(y)$, as in they are independent.
– Hamed
Jul 26 at 20:01
Separation of variables does not work here exactly because of what you said. The boundary conditions are coupling $x,y$ together, so how can the spatial part of the solution be of the form $X(x)Y(y)$, as in they are independent.
– Hamed
Jul 26 at 20:01
In more physical terms, the usual boundary conditions give you a standing wave with endpoints fixed on the boundary. In your case, a similar solution is a standing wave terminating on your boundary. Obviously, this a very complicated wave squeezing, streching and twisting as $x$ varies. You can probably make a homemade experiment to see this. Get a thin piece of metal which can be bent. Soak it in soap water and create a bubble. Give the metal an irregular shape and slowly oscillate up and down (careful not to break the bubble). What you'll see is a very complicated motion.
– Hamed
Jul 26 at 20:07
In more physical terms, the usual boundary conditions give you a standing wave with endpoints fixed on the boundary. In your case, a similar solution is a standing wave terminating on your boundary. Obviously, this a very complicated wave squeezing, streching and twisting as $x$ varies. You can probably make a homemade experiment to see this. Get a thin piece of metal which can be bent. Soak it in soap water and create a bubble. Give the metal an irregular shape and slowly oscillate up and down (careful not to break the bubble). What you'll see is a very complicated motion.
– Hamed
Jul 26 at 20:07
Yes, this was sort of what I was looking for, but not quite. Regarding the initial conditions being a function of space: This works precisely because initial conditions are defined at a t = constant value. Regarding boundaries which vary in space: I think that we need to define the boundary for a constant value of y (as in the example above). Otherwise, we need to find a new coordinate system in which we do have a boundary at a constant value of one of the coordinate variables. Separation of variables is then possible if the Laplacian in the new coordinates is diagonalizable.
– user109527
Jul 27 at 19:25
Yes, this was sort of what I was looking for, but not quite. Regarding the initial conditions being a function of space: This works precisely because initial conditions are defined at a t = constant value. Regarding boundaries which vary in space: I think that we need to define the boundary for a constant value of y (as in the example above). Otherwise, we need to find a new coordinate system in which we do have a boundary at a constant value of one of the coordinate variables. Separation of variables is then possible if the Laplacian in the new coordinates is diagonalizable.
– user109527
Jul 27 at 19:25
add a comment |Â
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Separation of variables does not work here exactly because of what you said. The boundary conditions are coupling $x,y$ together, so how can the spatial part of the solution be of the form $X(x)Y(y)$, as in they are independent.
– Hamed
Jul 26 at 20:01
In more physical terms, the usual boundary conditions give you a standing wave with endpoints fixed on the boundary. In your case, a similar solution is a standing wave terminating on your boundary. Obviously, this a very complicated wave squeezing, streching and twisting as $x$ varies. You can probably make a homemade experiment to see this. Get a thin piece of metal which can be bent. Soak it in soap water and create a bubble. Give the metal an irregular shape and slowly oscillate up and down (careful not to break the bubble). What you'll see is a very complicated motion.
– Hamed
Jul 26 at 20:07
Yes, this was sort of what I was looking for, but not quite. Regarding the initial conditions being a function of space: This works precisely because initial conditions are defined at a t = constant value. Regarding boundaries which vary in space: I think that we need to define the boundary for a constant value of y (as in the example above). Otherwise, we need to find a new coordinate system in which we do have a boundary at a constant value of one of the coordinate variables. Separation of variables is then possible if the Laplacian in the new coordinates is diagonalizable.
– user109527
Jul 27 at 19:25