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




share|improve this question


















  • 2




    I'm adding the bugs tag because there should be three independent parameters.
    – Michael E2
    Aug 9 at 13:00














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




share|improve this question


















  • 2




    I'm adding the bugs tag because there should be three independent parameters.
    – Michael E2
    Aug 9 at 13:00












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




share|improve this question














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





share|improve this question













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share|improve this question








edited Aug 9 at 13:00









Michael E2

140k11190456




140k11190456










asked Aug 9 at 10:29









Rainer Glüge

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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












  • 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










1 Answer
1






active

oldest

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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.







share|improve this answer






















  • 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











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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.







share|improve this answer






















  • 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















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.







share|improve this answer






















  • 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













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.







share|improve this answer














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.








share|improve this answer














share|improve this answer



share|improve this answer








edited Aug 9 at 14:12

























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

















  • 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


















 

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