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What does the syntax C[1][x][y] mean?
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I am examining an integrability condition,
$$u_3,122=u_3,123.$$
Typing
DSolve[D[u3[x1,x2,x3],x1,x2,x2]==D[u3[x1,x2,x3],x1,x2,x3],u3,x1,x2,x3]
it into MMA gives the following output:
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[1][x1][x3] + x1 C[2][x1][x2 + x3]]
I understand that the integration 'constants' C[1] and C[2] are functions of two variables, so the result C[1][x2, x3] is clear. But what does C[1][x1][x3] mean? Specifically, why is there a concatenation of argument brackets [x1][x3] instead of an argument?
differential-equations bugs symbols
edited Aug 9 at 13:00
Michael E2
140k11190456
asked Aug 9 at 10:29
Rainer Glüge
22415
add a comment |Â
up vote
6
down vote
favorite
I am examining an integrability condition,
$$u_3,122=u_3,123.$$
Typing
DSolve[D[u3[x1,x2,x3],x1,x2,x2]==D[u3[x1,x2,x3],x1,x2,x3],u3,x1,x2,x3]
it into MMA gives the following output:
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[1][x1][x3] + x1 C[2][x1][x2 + x3]]
I understand that the integration 'constants' C[1] and C[2] are functions of two variables, so the result C[1][x2, x3] is clear. But what does C[1][x1][x3] mean? Specifically, why is there a concatenation of argument brackets [x1][x3] instead of an argument?
differential-equations bugs symbols
edited Aug 9 at 13:00
Michael E2
140k11190456
asked Aug 9 at 10:29
Rainer Glüge
22415
2
I'm adding the bugs tag because there should be three independent parameters.
– Michael E2
Aug 9 at 13:00
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I am examining an integrability condition,
$$u_3,122=u_3,123.$$
Typing
DSolve[D[u3[x1,x2,x3],x1,x2,x2]==D[u3[x1,x2,x3],x1,x2,x3],u3,x1,x2,x3]
it into MMA gives the following output:
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[1][x1][x3] + x1 C[2][x1][x2 + x3]]
I understand that the integration 'constants' C[1] and C[2] are functions of two variables, so the result C[1][x2, x3] is clear. But what does C[1][x1][x3] mean? Specifically, why is there a concatenation of argument brackets [x1][x3] instead of an argument?
differential-equations bugs symbols
edited Aug 9 at 13:00
Michael E2
140k11190456
asked Aug 9 at 10:29
Rainer Glüge
22415
I am examining an integrability condition,
$$u_3,122=u_3,123.$$
Typing
DSolve[D[u3[x1,x2,x3],x1,x2,x2]==D[u3[x1,x2,x3],x1,x2,x3],u3,x1,x2,x3]
it into MMA gives the following output:
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[1][x1][x3] + x1 C[2][x1][x2 + x3]]
I understand that the integration 'constants' C[1] and C[2] are functions of two variables, so the result C[1][x2, x3] is clear. But what does C[1][x1][x3] mean? Specifically, why is there a concatenation of argument brackets [x1][x3] instead of an argument?
differential-equations bugs symbols
edited Aug 9 at 13:00
Michael E2
140k11190456
asked Aug 9 at 10:29
Rainer Glüge
22415
edited Aug 9 at 13:00
Michael E2
140k11190456
edited Aug 9 at 13:00
Michael E2
140k11190456
edited Aug 9 at 13:00
Michael E2
140k11190456
140k11190456
asked Aug 9 at 10:29
Rainer Glüge
22415
asked Aug 9 at 10:29
Rainer Glüge
22415
asked Aug 9 at 10:29
Rainer Glüge
22415
22415
2
I'm adding the bugs tag because there should be three independent parameters.
– Michael E2
Aug 9 at 13:00
add a comment |Â
2
I'm adding the bugs tag because there should be three independent parameters.
– Michael E2
Aug 9 at 13:00
2
2
I'm adding the bugs tag because there should be three independent parameters.
– Michael E2
Aug 9 at 13:00
I'm adding the bugs tag because there should be three independent parameters.
– Michael E2
Aug 9 at 13:00
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
It looks like an error. For C[1][x][y] to make sense C[1][x] must represent a single-variable function (and C[1] should represent a function that produces a function). However, in the solution returned by DSolve, C[1] has to have a different meaning in the first term. Probably the second two parameters should be C[2] and C[3]. Another point of confusion is that in classical math, C[1][x][y] would automatically be uncurried to form C[1][x, y]. See also Curry.
We can see that the more general expression with three functions C[1], C[2], C[3] is a solution:
D[u3[x1, x2, x3], x1, x2, x2] == D[u3[x1, x2, x3], x1, x2, x3] /.
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[3][x1][x3] + x1 C[2][x1][x2 + x3]]
(* True *)
The uncurried form is also a solution:
D[u3[x1, x2, x3], x1, x2, x2] == D[u3[x1, x2, x3], x1, x2, x3] /.
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[3][x1, x3] + x1 C[2][x1, x2 + x3]]
(* True *)
Speculation: Probably the parameter generator checks the heads for unique C[n] expressions, but the head of C[1][x1][x3] is C[1][x1] and doesn't match C[1]. One can see that the order of creation with the following:
Trace[
DSolve[D[u3[x1, x2, x3], x1, x2, x2] ==
D[u3[x1, x2, x3], x1, x2, x3], u3, x1, x2, x3],
C[_][__],
TraceInternal -> True]
Remark: Note that DSolve gives an unexplainable error (without examining internals):
Last::nolast: has zero length and no last element.
answered Aug 9 at 13:14
Michael E2
140k11190456
Just to compare. Maple 2018 answers $$ it u3 left( it x1,it x2,it x3 right) =F_1 left( it x3,it x2 right) +F_2 left( it x3,it x1 right) +F_3 left( it x1,it x3+it x2 right) . $$
– user64494
Aug 9 at 14:06
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
It looks like an error. For C[1][x][y] to make sense C[1][x] must represent a single-variable function (and C[1] should represent a function that produces a function). However, in the solution returned by DSolve, C[1] has to have a different meaning in the first term. Probably the second two parameters should be C[2] and C[3]. Another point of confusion is that in classical math, C[1][x][y] would automatically be uncurried to form C[1][x, y]. See also Curry.
We can see that the more general expression with three functions C[1], C[2], C[3] is a solution:
D[u3[x1, x2, x3], x1, x2, x2] == D[u3[x1, x2, x3], x1, x2, x3] /.
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[3][x1][x3] + x1 C[2][x1][x2 + x3]]
(* True *)
The uncurried form is also a solution:
D[u3[x1, x2, x3], x1, x2, x2] == D[u3[x1, x2, x3], x1, x2, x3] /.
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[3][x1, x3] + x1 C[2][x1, x2 + x3]]
(* True *)
Speculation: Probably the parameter generator checks the heads for unique C[n] expressions, but the head of C[1][x1][x3] is C[1][x1] and doesn't match C[1]. One can see that the order of creation with the following:
Trace[
DSolve[D[u3[x1, x2, x3], x1, x2, x2] ==
D[u3[x1, x2, x3], x1, x2, x3], u3, x1, x2, x3],
C[_][__],
TraceInternal -> True]
Remark: Note that DSolve gives an unexplainable error (without examining internals):
Last::nolast: has zero length and no last element.
answered Aug 9 at 13:14
Michael E2
140k11190456
Just to compare. Maple 2018 answers $$ it u3 left( it x1,it x2,it x3 right) =F_1 left( it x3,it x2 right) +F_2 left( it x3,it x1 right) +F_3 left( it x1,it x3+it x2 right) . $$
– user64494
Aug 9 at 14:06
add a comment |Â
up vote
6
down vote
accepted
It looks like an error. For C[1][x][y] to make sense C[1][x] must represent a single-variable function (and C[1] should represent a function that produces a function). However, in the solution returned by DSolve, C[1] has to have a different meaning in the first term. Probably the second two parameters should be C[2] and C[3]. Another point of confusion is that in classical math, C[1][x][y] would automatically be uncurried to form C[1][x, y]. See also Curry.
We can see that the more general expression with three functions C[1], C[2], C[3] is a solution:
D[u3[x1, x2, x3], x1, x2, x2] == D[u3[x1, x2, x3], x1, x2, x3] /.
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[3][x1][x3] + x1 C[2][x1][x2 + x3]]
(* True *)
The uncurried form is also a solution:
D[u3[x1, x2, x3], x1, x2, x2] == D[u3[x1, x2, x3], x1, x2, x3] /.
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[3][x1, x3] + x1 C[2][x1, x2 + x3]]
(* True *)
Speculation: Probably the parameter generator checks the heads for unique C[n] expressions, but the head of C[1][x1][x3] is C[1][x1] and doesn't match C[1]. One can see that the order of creation with the following:
Trace[
DSolve[D[u3[x1, x2, x3], x1, x2, x2] ==
D[u3[x1, x2, x3], x1, x2, x3], u3, x1, x2, x3],
C[_][__],
TraceInternal -> True]
Remark: Note that DSolve gives an unexplainable error (without examining internals):
Last::nolast: has zero length and no last element.
answered Aug 9 at 13:14
Michael E2
140k11190456
Just to compare. Maple 2018 answers $$ it u3 left( it x1,it x2,it x3 right) =F_1 left( it x3,it x2 right) +F_2 left( it x3,it x1 right) +F_3 left( it x1,it x3+it x2 right) . $$
– user64494
Aug 9 at 14:06
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
It looks like an error. For C[1][x][y] to make sense C[1][x] must represent a single-variable function (and C[1] should represent a function that produces a function). However, in the solution returned by DSolve, C[1] has to have a different meaning in the first term. Probably the second two parameters should be C[2] and C[3]. Another point of confusion is that in classical math, C[1][x][y] would automatically be uncurried to form C[1][x, y]. See also Curry.
We can see that the more general expression with three functions C[1], C[2], C[3] is a solution:
D[u3[x1, x2, x3], x1, x2, x2] == D[u3[x1, x2, x3], x1, x2, x3] /.
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[3][x1][x3] + x1 C[2][x1][x2 + x3]]
(* True *)
The uncurried form is also a solution:
D[u3[x1, x2, x3], x1, x2, x2] == D[u3[x1, x2, x3], x1, x2, x3] /.
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[3][x1, x3] + x1 C[2][x1, x2 + x3]]
(* True *)
Speculation: Probably the parameter generator checks the heads for unique C[n] expressions, but the head of C[1][x1][x3] is C[1][x1] and doesn't match C[1]. One can see that the order of creation with the following:
Trace[
DSolve[D[u3[x1, x2, x3], x1, x2, x2] ==
D[u3[x1, x2, x3], x1, x2, x3], u3, x1, x2, x3],
C[_][__],
TraceInternal -> True]
Remark: Note that DSolve gives an unexplainable error (without examining internals):
Last::nolast: has zero length and no last element.
answered Aug 9 at 13:14
Michael E2
140k11190456
It looks like an error. For C[1][x][y] to make sense C[1][x] must represent a single-variable function (and C[1] should represent a function that produces a function). However, in the solution returned by DSolve, C[1] has to have a different meaning in the first term. Probably the second two parameters should be C[2] and C[3]. Another point of confusion is that in classical math, C[1][x][y] would automatically be uncurried to form C[1][x, y]. See also Curry.
We can see that the more general expression with three functions C[1], C[2], C[3] is a solution:
D[u3[x1, x2, x3], x1, x2, x2] == D[u3[x1, x2, x3], x1, x2, x3] /.
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[3][x1][x3] + x1 C[2][x1][x2 + x3]]
(* True *)
The uncurried form is also a solution:
D[u3[x1, x2, x3], x1, x2, x2] == D[u3[x1, x2, x3], x1, x2, x3] /.
u3 -> Function[x1, x2, x3,
C[1][x2, x3] + x1 C[3][x1, x3] + x1 C[2][x1, x2 + x3]]
(* True *)
Speculation: Probably the parameter generator checks the heads for unique C[n] expressions, but the head of C[1][x1][x3] is C[1][x1] and doesn't match C[1]. One can see that the order of creation with the following:
Trace[
DSolve[D[u3[x1, x2, x3], x1, x2, x2] ==
D[u3[x1, x2, x3], x1, x2, x3], u3, x1, x2, x3],
C[_][__],
TraceInternal -> True]
Remark: Note that DSolve gives an unexplainable error (without examining internals):
Last::nolast: has zero length and no last element.
answered Aug 9 at 13:14
Michael E2
140k11190456
edited Aug 9 at 14:12
answered Aug 9 at 13:14
Michael E2
140k11190456
answered Aug 9 at 13:14
Michael E2
140k11190456
answered Aug 9 at 13:14
Michael E2
140k11190456
140k11190456
Just to compare. Maple 2018 answers $$ it u3 left( it x1,it x2,it x3 right) =F_1 left( it x3,it x2 right) +F_2 left( it x3,it x1 right) +F_3 left( it x1,it x3+it x2 right) . $$
– user64494
Aug 9 at 14:06
add a comment |Â
Just to compare. Maple 2018 answers $$ it u3 left( it x1,it x2,it x3 right) =F_1 left( it x3,it x2 right) +F_2 left( it x3,it x1 right) +F_3 left( it x1,it x3+it x2 right) . $$
– user64494
Aug 9 at 14:06
Just to compare. Maple 2018 answers $$ it u3 left( it x1,it x2,it x3 right) =F_1 left( it x3,it x2 right) +F_2 left( it x3,it x1 right) +F_3 left( it x1,it x3+it x2 right) . $$
– user64494
Aug 9 at 14:06
Just to compare. Maple 2018 answers $$ it u3 left( it x1,it x2,it x3 right) =F_1 left( it x3,it x2 right) +F_2 left( it x3,it x1 right) +F_3 left( it x1,it x3+it x2 right) . $$
– user64494
Aug 9 at 14:06
add a comment |Â
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2
I'm adding the bugs tag because there should be three independent parameters.
– Michael E2
Aug 9 at 13:00