Partial Differential Equations : Separation of Variables in Second Order Linear Equations
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I know how to solve this homogeneous problem using separation of variables .
Here, $u$ is from $mathbb{R}^2 rightarrow mathbb{R}$
$$u_{xx}+u_{yy} = 0$$
But, How to solve this non homogeneous equation using separation of variables
$$ u_ {xx} + u_{yy} = 1 $$
There are no boundary conditions given in question.
I am new in solving PDE, so please explain the answer as well.
pde
asked Nov 17 at 12:31
Doraemon
579
|
show 3 more comments
up vote
0
down vote
favorite
I know how to solve this homogeneous problem using separation of variables .
Here, $u$ is from $mathbb{R}^2 rightarrow mathbb{R}$
$$u_{xx}+u_{yy} = 0$$
But, How to solve this non homogeneous equation using separation of variables
$$ u_ {xx} + u_{yy} = 1 $$
There are no boundary conditions given in question.
I am new in solving PDE, so please explain the answer as well.
pde
asked Nov 17 at 12:31
Doraemon
579
Can you make a change of variables that gets rid of the $1$? Maybe something like $v(x,t) = u(x,t) + x^{2}/2$, though you could equally take $v(x,t) = u(x,t) + y^{2}/2$. It usually depends on what the boundary conditions are.
– Mattos
Nov 17 at 12:38
Why you have used $(x, t)$ ? $u$ is a function of $(x, y) $
– Doraemon
Nov 17 at 12:52
1
It was a typo. Just replace $t$ with $y$ in what I wrote in the previous comment.
– Mattos
Nov 17 at 13:02
There was literally a problem just like this one posted a couple of days ago. Duplicate?
– DaveNine
Nov 17 at 18:21
@DaveNine, Can you please share link of that. So, that I can know if there is one more method to solve it.
– Doraemon
Nov 18 at 1:58
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know how to solve this homogeneous problem using separation of variables .
Here, $u$ is from $mathbb{R}^2 rightarrow mathbb{R}$
$$u_{xx}+u_{yy} = 0$$
But, How to solve this non homogeneous equation using separation of variables
$$ u_ {xx} + u_{yy} = 1 $$
There are no boundary conditions given in question.
I am new in solving PDE, so please explain the answer as well.
pde
asked Nov 17 at 12:31
Doraemon
579
I know how to solve this homogeneous problem using separation of variables .
Here, $u$ is from $mathbb{R}^2 rightarrow mathbb{R}$
$$u_{xx}+u_{yy} = 0$$
But, How to solve this non homogeneous equation using separation of variables
$$ u_ {xx} + u_{yy} = 1 $$
There are no boundary conditions given in question.
I am new in solving PDE, so please explain the answer as well.
pde
pde
asked Nov 17 at 12:31
Doraemon
579
asked Nov 17 at 12:31
Doraemon
579
edited Nov 18 at 11:43
asked Nov 17 at 12:31
Doraemon
579
asked Nov 17 at 12:31
Doraemon
579
asked Nov 17 at 12:31
Doraemon
579
579
Can you make a change of variables that gets rid of the $1$? Maybe something like $v(x,t) = u(x,t) + x^{2}/2$, though you could equally take $v(x,t) = u(x,t) + y^{2}/2$. It usually depends on what the boundary conditions are.
– Mattos
Nov 17 at 12:38
Why you have used $(x, t)$ ? $u$ is a function of $(x, y) $
– Doraemon
Nov 17 at 12:52
1
It was a typo. Just replace $t$ with $y$ in what I wrote in the previous comment.
– Mattos
Nov 17 at 13:02
There was literally a problem just like this one posted a couple of days ago. Duplicate?
– DaveNine
Nov 17 at 18:21
@DaveNine, Can you please share link of that. So, that I can know if there is one more method to solve it.
– Doraemon
Nov 18 at 1:58
|
show 3 more comments
Can you make a change of variables that gets rid of the $1$? Maybe something like $v(x,t) = u(x,t) + x^{2}/2$, though you could equally take $v(x,t) = u(x,t) + y^{2}/2$. It usually depends on what the boundary conditions are.
– Mattos
Nov 17 at 12:38
Why you have used $(x, t)$ ? $u$ is a function of $(x, y) $
– Doraemon
Nov 17 at 12:52
1
It was a typo. Just replace $t$ with $y$ in what I wrote in the previous comment.
– Mattos
Nov 17 at 13:02
There was literally a problem just like this one posted a couple of days ago. Duplicate?
– DaveNine
Nov 17 at 18:21
@DaveNine, Can you please share link of that. So, that I can know if there is one more method to solve it.
– Doraemon
Nov 18 at 1:58
Can you make a change of variables that gets rid of the $1$? Maybe something like $v(x,t) = u(x,t) + x^{2}/2$, though you could equally take $v(x,t) = u(x,t) + y^{2}/2$. It usually depends on what the boundary conditions are.
– Mattos
Nov 17 at 12:38
Can you make a change of variables that gets rid of the $1$? Maybe something like $v(x,t) = u(x,t) + x^{2}/2$, though you could equally take $v(x,t) = u(x,t) + y^{2}/2$. It usually depends on what the boundary conditions are.
– Mattos
Nov 17 at 12:38
Why you have used $(x, t)$ ? $u$ is a function of $(x, y) $
– Doraemon
Nov 17 at 12:52
Why you have used $(x, t)$ ? $u$ is a function of $(x, y) $
– Doraemon
Nov 17 at 12:52
1
1
It was a typo. Just replace $t$ with $y$ in what I wrote in the previous comment.
– Mattos
Nov 17 at 13:02
It was a typo. Just replace $t$ with $y$ in what I wrote in the previous comment.
– Mattos
Nov 17 at 13:02
There was literally a problem just like this one posted a couple of days ago. Duplicate?
– DaveNine
Nov 17 at 18:21
There was literally a problem just like this one posted a couple of days ago. Duplicate?
– DaveNine
Nov 17 at 18:21
@DaveNine, Can you please share link of that. So, that I can know if there is one more method to solve it.
– Doraemon
Nov 18 at 1:58
@DaveNine, Can you please share link of that. So, that I can know if there is one more method to solve it.
– Doraemon
Nov 18 at 1:58
|
show 3 more comments
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Can you make a change of variables that gets rid of the $1$? Maybe something like $v(x,t) = u(x,t) + x^{2}/2$, though you could equally take $v(x,t) = u(x,t) + y^{2}/2$. It usually depends on what the boundary conditions are.
– Mattos
Nov 17 at 12:38
Why you have used $(x, t)$ ? $u$ is a function of $(x, y) $
– Doraemon
Nov 17 at 12:52
1
It was a typo. Just replace $t$ with $y$ in what I wrote in the previous comment.
– Mattos
Nov 17 at 13:02
There was literally a problem just like this one posted a couple of days ago. Duplicate?
– DaveNine
Nov 17 at 18:21
@DaveNine, Can you please share link of that. So, that I can know if there is one more method to solve it.
– Doraemon
Nov 18 at 1:58