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Confused about the laplace operator when used in partial differential equations.
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I have a function $u(r,t)$ that satisfies
$$fracpartial^2upartial t^2 - c^2 Delta u = 0.$$
I'm looking at a solution of a problem with this setup and it states that the above equation can be written as
$$fracpartial^2upartial t^2 + c^2Au = 0$$
with
$$A = -Delta = -frac1rfracddr(rfracddr).$$
I have two questions:
The operator $Delta$ is defined as $sum fracpartial^2partial x_i^2$ but I never see the term $fracpartial^2partial t^2$ in $Delta u$ when looking at solutions. Why is this?
I get $$Au = -frac1rfracddr(rfracdudr) = -frac1r(ru')' = -u'' - fracu'r$$ which is NOT equal to
$-Delta u = fracpartial partial r^2$ or $-Delta u = fracpartial partial r^2 + fracpartial partial r^2.$
What is going on?
pde operator-theory differential-operators
asked Aug 6 at 16:17
Heuristics
415111
add a comment |Â
up vote
1
down vote
favorite
I have a function $u(r,t)$ that satisfies
$$fracpartial^2upartial t^2 - c^2 Delta u = 0.$$
I'm looking at a solution of a problem with this setup and it states that the above equation can be written as
$$fracpartial^2upartial t^2 + c^2Au = 0$$
with
$$A = -Delta = -frac1rfracddr(rfracddr).$$
I have two questions:
The operator $Delta$ is defined as $sum fracpartial^2partial x_i^2$ but I never see the term $fracpartial^2partial t^2$ in $Delta u$ when looking at solutions. Why is this?
I get $$Au = -frac1rfracddr(rfracdudr) = -frac1r(ru')' = -u'' - fracu'r$$ which is NOT equal to
$-Delta u = fracpartial partial r^2$ or $-Delta u = fracpartial partial r^2 + fracpartial partial r^2.$
What is going on?
pde operator-theory differential-operators
asked Aug 6 at 16:17
Heuristics
415111
3
1. The Laplacian is a sum of second order derivatives on the spatial dimensions only 2. The Laplacian $A$ that you have is in polar coordinates, not Cartesian. See the polar form of the Laplacian here. Note that in your problem, $u = u(r,t)$ is not dependent on $theta$, so the second order derivatives in $theta$ in the polar Laplacian don't appear, which means $$Delta u = u_rr + fracu_rr$$ which is what you have written.
– Mattos
Aug 6 at 16:36
Okay thanks. Just one question: so the laplacian is always on spatial dimensions? If my function is dependent on time, it's not in the laplacian? @Mattos
– Heuristics
Aug 6 at 19:31
$fracpartial^2 upartial t^2$ is the Laplacian in time.
– DisintegratingByParts
Aug 10 at 3:50
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a function $u(r,t)$ that satisfies
$$fracpartial^2upartial t^2 - c^2 Delta u = 0.$$
I'm looking at a solution of a problem with this setup and it states that the above equation can be written as
$$fracpartial^2upartial t^2 + c^2Au = 0$$
with
$$A = -Delta = -frac1rfracddr(rfracddr).$$
I have two questions:
The operator $Delta$ is defined as $sum fracpartial^2partial x_i^2$ but I never see the term $fracpartial^2partial t^2$ in $Delta u$ when looking at solutions. Why is this?
I get $$Au = -frac1rfracddr(rfracdudr) = -frac1r(ru')' = -u'' - fracu'r$$ which is NOT equal to
$-Delta u = fracpartial partial r^2$ or $-Delta u = fracpartial partial r^2 + fracpartial partial r^2.$
What is going on?
pde operator-theory differential-operators
asked Aug 6 at 16:17
Heuristics
415111
I have a function $u(r,t)$ that satisfies
$$fracpartial^2upartial t^2 - c^2 Delta u = 0.$$
I'm looking at a solution of a problem with this setup and it states that the above equation can be written as
$$fracpartial^2upartial t^2 + c^2Au = 0$$
with
$$A = -Delta = -frac1rfracddr(rfracddr).$$
I have two questions:
The operator $Delta$ is defined as $sum fracpartial^2partial x_i^2$ but I never see the term $fracpartial^2partial t^2$ in $Delta u$ when looking at solutions. Why is this?
I get $$Au = -frac1rfracddr(rfracdudr) = -frac1r(ru')' = -u'' - fracu'r$$ which is NOT equal to
$-Delta u = fracpartial partial r^2$ or $-Delta u = fracpartial partial r^2 + fracpartial partial r^2.$
What is going on?
pde operator-theory differential-operators
asked Aug 6 at 16:17
Heuristics
415111
asked Aug 6 at 16:17
Heuristics
415111
asked Aug 6 at 16:17
Heuristics
415111
asked Aug 6 at 16:17
Heuristics
415111
415111
3
1. The Laplacian is a sum of second order derivatives on the spatial dimensions only 2. The Laplacian $A$ that you have is in polar coordinates, not Cartesian. See the polar form of the Laplacian here. Note that in your problem, $u = u(r,t)$ is not dependent on $theta$, so the second order derivatives in $theta$ in the polar Laplacian don't appear, which means $$Delta u = u_rr + fracu_rr$$ which is what you have written.
– Mattos
Aug 6 at 16:36
Okay thanks. Just one question: so the laplacian is always on spatial dimensions? If my function is dependent on time, it's not in the laplacian? @Mattos
– Heuristics
Aug 6 at 19:31
$fracpartial^2 upartial t^2$ is the Laplacian in time.
– DisintegratingByParts
Aug 10 at 3:50
add a comment |Â
3
1. The Laplacian is a sum of second order derivatives on the spatial dimensions only 2. The Laplacian $A$ that you have is in polar coordinates, not Cartesian. See the polar form of the Laplacian here. Note that in your problem, $u = u(r,t)$ is not dependent on $theta$, so the second order derivatives in $theta$ in the polar Laplacian don't appear, which means $$Delta u = u_rr + fracu_rr$$ which is what you have written.
– Mattos
Aug 6 at 16:36
Okay thanks. Just one question: so the laplacian is always on spatial dimensions? If my function is dependent on time, it's not in the laplacian? @Mattos
– Heuristics
Aug 6 at 19:31
$fracpartial^2 upartial t^2$ is the Laplacian in time.
– DisintegratingByParts
Aug 10 at 3:50
3
3
1. The Laplacian is a sum of second order derivatives on the spatial dimensions only 2. The Laplacian $A$ that you have is in polar coordinates, not Cartesian. See the polar form of the Laplacian here. Note that in your problem, $u = u(r,t)$ is not dependent on $theta$, so the second order derivatives in $theta$ in the polar Laplacian don't appear, which means $$Delta u = u_rr + fracu_rr$$ which is what you have written.
– Mattos
Aug 6 at 16:36
1. The Laplacian is a sum of second order derivatives on the spatial dimensions only 2. The Laplacian $A$ that you have is in polar coordinates, not Cartesian. See the polar form of the Laplacian here. Note that in your problem, $u = u(r,t)$ is not dependent on $theta$, so the second order derivatives in $theta$ in the polar Laplacian don't appear, which means $$Delta u = u_rr + fracu_rr$$ which is what you have written.
– Mattos
Aug 6 at 16:36
Okay thanks. Just one question: so the laplacian is always on spatial dimensions? If my function is dependent on time, it's not in the laplacian? @Mattos
– Heuristics
Aug 6 at 19:31
Okay thanks. Just one question: so the laplacian is always on spatial dimensions? If my function is dependent on time, it's not in the laplacian? @Mattos
– Heuristics
Aug 6 at 19:31
$fracpartial^2 upartial t^2$ is the Laplacian in time.
– DisintegratingByParts
Aug 10 at 3:50
$fracpartial^2 upartial t^2$ is the Laplacian in time.
– DisintegratingByParts
Aug 10 at 3:50
add a comment |Â
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3
1. The Laplacian is a sum of second order derivatives on the spatial dimensions only 2. The Laplacian $A$ that you have is in polar coordinates, not Cartesian. See the polar form of the Laplacian here. Note that in your problem, $u = u(r,t)$ is not dependent on $theta$, so the second order derivatives in $theta$ in the polar Laplacian don't appear, which means $$Delta u = u_rr + fracu_rr$$ which is what you have written.
– Mattos
Aug 6 at 16:36
Okay thanks. Just one question: so the laplacian is always on spatial dimensions? If my function is dependent on time, it's not in the laplacian? @Mattos
– Heuristics
Aug 6 at 19:31
$fracpartial^2 upartial t^2$ is the Laplacian in time.
– DisintegratingByParts
Aug 10 at 3:50