How does one use Green's function of the operator to get the solution of the arbitrary boundary value...
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Assume I've been given an operator $L$ and its Green's function $G(s, s')$. This is the function that solves the following:
$$ L G(s, s') = delta(s-s'), G(a,s') = G(b, s') = 0 $$
I know how to get a solution of the:
$$ L u = f, u(a) = u(b) = 0 $$
But if the task was:
$$ L u = f , u'(a) + D u(a) = A, u'(b) + C u(b) = B $$
Is there a way to even represent a solution with $G$? Or one would need to solve separately:
$$ L u = 0 , u'(a) + D u(a) = A, u'(b) + C u(b) = B $$
and
$$ L u = f , u'(a) + D u(a) = 0, u'(b) + C u(b) = 0 $$
Where the last one may be solved invoking new Green's function $N(s, s')$:
$$ L N = 0 , N_s(a, s') + D N(a, s') = 0, N_s(b, s') + C N(b, s') = 0 $$
ordinary-differential-equations pde mathematical-physics greens-function
asked Dec 25 '18 at 14:26
Al PrihodkoAl Prihodko
536
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add a comment |
$begingroup$
Assume I've been given an operator $L$ and its Green's function $G(s, s')$. This is the function that solves the following:
$$ L G(s, s') = delta(s-s'), G(a,s') = G(b, s') = 0 $$
I know how to get a solution of the:
$$ L u = f, u(a) = u(b) = 0 $$
But if the task was:
$$ L u = f , u'(a) + D u(a) = A, u'(b) + C u(b) = B $$
Is there a way to even represent a solution with $G$? Or one would need to solve separately:
$$ L u = 0 , u'(a) + D u(a) = A, u'(b) + C u(b) = B $$
and
$$ L u = f , u'(a) + D u(a) = 0, u'(b) + C u(b) = 0 $$
Where the last one may be solved invoking new Green's function $N(s, s')$:
$$ L N = 0 , N_s(a, s') + D N(a, s') = 0, N_s(b, s') + C N(b, s') = 0 $$
ordinary-differential-equations pde mathematical-physics greens-function
asked Dec 25 '18 at 14:26
Al PrihodkoAl Prihodko
536
$endgroup$
$begingroup$
Green's function depends on boundary conditions. So you can solve the usual equation with point source and your particular boundary conditions.
$endgroup$
– Yuriy S
Dec 28 '18 at 9:39
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@yuriy-s you mean, having the Gf for one conditions won't help with the others?
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– Al Prihodko
Dec 28 '18 at 10:06
1
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Yes, at least as far as I know. The standard method is finding it for each case of conditions
$endgroup$
– Yuriy S
Dec 28 '18 at 10:19
add a comment |
$begingroup$
Assume I've been given an operator $L$ and its Green's function $G(s, s')$. This is the function that solves the following:
$$ L G(s, s') = delta(s-s'), G(a,s') = G(b, s') = 0 $$
I know how to get a solution of the:
$$ L u = f, u(a) = u(b) = 0 $$
But if the task was:
$$ L u = f , u'(a) + D u(a) = A, u'(b) + C u(b) = B $$
Is there a way to even represent a solution with $G$? Or one would need to solve separately:
$$ L u = 0 , u'(a) + D u(a) = A, u'(b) + C u(b) = B $$
and
$$ L u = f , u'(a) + D u(a) = 0, u'(b) + C u(b) = 0 $$
Where the last one may be solved invoking new Green's function $N(s, s')$:
$$ L N = 0 , N_s(a, s') + D N(a, s') = 0, N_s(b, s') + C N(b, s') = 0 $$
ordinary-differential-equations pde mathematical-physics greens-function
asked Dec 25 '18 at 14:26
Al PrihodkoAl Prihodko
536
$endgroup$
Assume I've been given an operator $L$ and its Green's function $G(s, s')$. This is the function that solves the following:
$$ L G(s, s') = delta(s-s'), G(a,s') = G(b, s') = 0 $$
I know how to get a solution of the:
$$ L u = f, u(a) = u(b) = 0 $$
But if the task was:
$$ L u = f , u'(a) + D u(a) = A, u'(b) + C u(b) = B $$
Is there a way to even represent a solution with $G$? Or one would need to solve separately:
$$ L u = 0 , u'(a) + D u(a) = A, u'(b) + C u(b) = B $$
and
$$ L u = f , u'(a) + D u(a) = 0, u'(b) + C u(b) = 0 $$
Where the last one may be solved invoking new Green's function $N(s, s')$:
$$ L N = 0 , N_s(a, s') + D N(a, s') = 0, N_s(b, s') + C N(b, s') = 0 $$
ordinary-differential-equations pde mathematical-physics greens-function
ordinary-differential-equations pde mathematical-physics greens-function
asked Dec 25 '18 at 14:26
Al PrihodkoAl Prihodko
536
asked Dec 25 '18 at 14:26
Al PrihodkoAl Prihodko
536
edited Dec 28 '18 at 9:34
Al Prihodko
asked Dec 25 '18 at 14:26
Al PrihodkoAl Prihodko
536
asked Dec 25 '18 at 14:26
Al PrihodkoAl Prihodko
536
asked Dec 25 '18 at 14:26
Al PrihodkoAl Prihodko
536
536
$begingroup$
Green's function depends on boundary conditions. So you can solve the usual equation with point source and your particular boundary conditions.
$endgroup$
– Yuriy S
Dec 28 '18 at 9:39
$begingroup$
@yuriy-s you mean, having the Gf for one conditions won't help with the others?
$endgroup$
– Al Prihodko
Dec 28 '18 at 10:06
1
$begingroup$
Yes, at least as far as I know. The standard method is finding it for each case of conditions
$endgroup$
– Yuriy S
Dec 28 '18 at 10:19
add a comment |
$begingroup$
Green's function depends on boundary conditions. So you can solve the usual equation with point source and your particular boundary conditions.
$endgroup$
– Yuriy S
Dec 28 '18 at 9:39
$begingroup$
@yuriy-s you mean, having the Gf for one conditions won't help with the others?
$endgroup$
– Al Prihodko
Dec 28 '18 at 10:06
1
$begingroup$
Yes, at least as far as I know. The standard method is finding it for each case of conditions
$endgroup$
– Yuriy S
Dec 28 '18 at 10:19
$begingroup$
Green's function depends on boundary conditions. So you can solve the usual equation with point source and your particular boundary conditions.
$endgroup$
– Yuriy S
Dec 28 '18 at 9:39
$begingroup$
Green's function depends on boundary conditions. So you can solve the usual equation with point source and your particular boundary conditions.
$endgroup$
– Yuriy S
Dec 28 '18 at 9:39
$begingroup$
@yuriy-s you mean, having the Gf for one conditions won't help with the others?
$endgroup$
– Al Prihodko
Dec 28 '18 at 10:06
$begingroup$
@yuriy-s you mean, having the Gf for one conditions won't help with the others?
$endgroup$
– Al Prihodko
Dec 28 '18 at 10:06
1
1
$begingroup$
Yes, at least as far as I know. The standard method is finding it for each case of conditions
$endgroup$
– Yuriy S
Dec 28 '18 at 10:19
$begingroup$
Yes, at least as far as I know. The standard method is finding it for each case of conditions
$endgroup$
– Yuriy S
Dec 28 '18 at 10:19
add a comment |
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$begingroup$
Green's function depends on boundary conditions. So you can solve the usual equation with point source and your particular boundary conditions.
$endgroup$
– Yuriy S
Dec 28 '18 at 9:39
$begingroup$
@yuriy-s you mean, having the Gf for one conditions won't help with the others?
$endgroup$
– Al Prihodko
Dec 28 '18 at 10:06
1
$begingroup$
Yes, at least as far as I know. The standard method is finding it for each case of conditions
$endgroup$
– Yuriy S
Dec 28 '18 at 10:19