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0 0 0 0 0 0 0 }{CSTYLE "" -1 344 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 345 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 346 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 347 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 348 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Tim es" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Outpu t" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 } 1 3 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 13 "Linear ODE 's." }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 257 37 "ODE = Ordi nary Differential Equation." }}{PARA 0 "" 0 "" {TEXT 263 8 "ordinary" }{TEXT -1 138 " means: there is only 1 independent variable (we'll usu ally use the name x) and 1 dependent variable, which will usually be c alled y=y(x)." }{TEXT 258 6 "linear" }{TEXT -1 165 " means: If y=y(x) \+ is the dependent variable, and y'(x), y''(x), ... are the derivatives, then there are no products among y(x),y'(x),y''(x),... in the equatio n. The " }{TEXT 265 5 "order" }{TEXT -1 116 " of an ordinary different ial equation is the highest derivative that appears in the equation. T he general form of a " }{TEXT 264 11 "homogeneous" }{TEXT -1 53 " line ar ordinary differential equation of order n is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "n:=3; # take order n=3, but can also take oth er values" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"nG\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "eqn := add( a[i](x) * diff(y(x),[x$ i]) , i=0..n) = 0; # homogeneous means righthand side is 0" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eqnG/,**&-&%\"aG6#\"\"!6#%\"xG\"\"\"-%\"y GF-F/F/*&-&F*6#F/F-F/-%%diffG6$F0F.F/F/*&-&F*6#\"\"#F-F/-F76$F0-%\"$G6 $F.F=F/F/*&-&F*6#\"\"$F-F/-F76$F0-FA6$F.FGF/F/F," }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 50 "where a0(x), a1(x),...,an(x) are elements of some \+ " }{TEXT 266 20 "differential field K" }{TEXT -1 67 ", and where an(x) <> 0 (otherwise the order would be less than n)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 269 11 "Definition." }{TEXT -1 3 " A " }{TEXT 267 18 "differential field" }{TEXT -1 34 " is a field K with the properties:" }}{PARA 0 "" 0 "" {TEXT -1 15 "*) K is a field " }}{PARA 0 "" 0 "" {TEXT -1 82 "*) There is a map, denoted by ', from K to K, satisfying the following properties:" }}{PARA 0 "" 0 "" {TEXT -1 33 " (a+b)'=a'+b' for all a,b in K" }}{PARA 0 "" 0 "" {TEXT -1 65 " (a*b)' = a'*b + a*(b') for all a,b in K (the Leibni z rule)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 270 10 "Definition" }{TEXT -1 6 ". The " }{TEXT 268 18 "field of constants " }{TEXT -1 9 " of K is:" }}{PARA 0 "" 0 "" {TEXT -1 21 "C_K = \{a in \+ K | a'=0\}" }}{PARA 0 "" 0 "" {TEXT -1 61 "It is easy to prove that C_ K is a differential subfield of K." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 271 39 "We will make the following assumptions: " }}{PARA 0 "" 0 "" {TEXT -1 57 "*) The field K (and hence also C_K) h as characteristic 0." }}{PARA 0 "" 0 "" {TEXT -1 41 "*) The field C_K \+ is algebraically closed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 65 "In most of the examples we will take the different ial field K as " }{TEXT 278 1 "C" }{TEXT -1 14 "(x), so C_K = " } {TEXT 279 1 "C" }{TEXT -1 31 ", the field of complex numbers." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 272 12 "Definiti on: " }{TEXT -1 2 "a " }{TEXT 273 28 "differential field extension" } {TEXT -1 29 " of K is a field L such that:" }}{PARA 0 "" 0 "" {TEXT -1 50 "L is a differential field with a differentiation '" }}{PARA 0 " " 0 "" {TEXT -1 20 "K is a subfield of L" }}{PARA 0 "" 0 "" {TEXT -1 58 "' restricted to K coincides with the differentiation on K." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 505 "For exam ple, if K=Qbar(x, exp(x)) where Qbar is the algebraic closure of the r ational numbers Q, then L=K(exp(x+1)) is a differential field extensio n of K. But in some sense it's not a good field extension, because all we did is just add some more constants: L = K(exp(x+1)) = K(exp(x+1)/ exp(x)) = K(exp(1)), so all we did is enlarge the field of constants: \+ C_L = Qbar(exp(1)) = C_K(exp(1)). For technical reasons we'll have to \+ avoid such \"fake\" differential field extensions, so from now on we'l l assume:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 274 12 "Assumption: " }{TEXT -1 90 "The only differential field extens ions of K that we will consider are those for which the " }{TEXT 275 30 "field of constants is the same" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 315 42 "Examples \+ of differential field extensions:" }}{PARA 0 "" 0 "" {TEXT 317 9 "Exam ple1)" }{TEXT -1 263 " Let K be a differential field, let a be in K, a nd suppose that there is no b in K for which b'=a. Now take the differ ential field extension L=K(A) where A is a new variable (so A is trans cendental over K) and the differentiation is defined by A'=a. Then C_L =C_K." }}{PARA 0 "" 0 "" {TEXT -1 28 "For each c in C_K one has a " } {TEXT 316 25 "differential automorphism" }{TEXT -1 139 " g_c of L over K, i.e. an automorphism of L that commutes with ' and that is the ide ntity on K. This automorphism is defined by g_c(A)=A+c." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 318 10 "Example2) " }{TEXT -1 252 "Let K be a differential field, let a be in K, and suppose that there is no b in K for which b'=a*b. Now take the differential field \+ extension L=K(A) where A is a new variable (so A is transcendental ove r K) and the differentiation is defined by A'=a*A." }}{PARA 0 "" 0 "" {TEXT 322 10 "Example2a)" }{TEXT -1 516 " If there exists a positive i nteger n and a non-zero element b in K for which b'=n*a*b then take S= A^n/b. Then S'/S=(A^n)'/A^n-b'/b=n*A'/A-b'/b=n*a-n*a=0 so S'=0. So the n S in C_L but not in C_K, so C_L is not equal to C_K. We will always \+ avoid such differential field extensions of K. Instead take the field \+ M=K( b^(1/n) ), which is an algebraic extension of K. Now b^(1/n)' / b ^(1/n) = 1/n * b'/b = a. Every automorphism (as a field) of M over K a s a field is automatically an automorphism as a differential field." } }{PARA 0 "" 0 "" {TEXT 323 10 "Example2b)" }{TEXT -1 142 " If there is no such integer n, then C_L=C_K. Then for every c in C_L-\{0\} there \+ is a differential automorphism of L over K, namely g_c(A)=c*A." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 320 4 "Note" } {TEXT -1 54 ": For convenience of notation we'll always take C_K = " } {TEXT 319 1 "C" }{TEXT -1 190 ", the field of complex numbers (althoug h any other algebraically closed field of characteristic 0 would work \+ equally well). But this way, the previous assumption simply means the \+ following: " }{TEXT 321 103 "The fields of constants C_K and C_L shoul d always be C for all differential fields that are considered." } {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "eqn;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/,**&-&%\"aG6#\"\"!6#%\"xG\"\"\"-%\"yG F+F-F-*&-&F(6#F-F+F--%%diffG6$F.F,F-F-*&-&F(6#\"\"#F+F--F56$F.-%\"$G6$ F,F;F-F-*&-&F(6#\"\"$F+F--F56$F.-F?6$F,FEF-F-F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 243 "We will often not write the part \"= 0\" in such an equation. We will always assume that the coefficients a.i(x) are in a differential field K, and that we search for solutions of eqn in some differential field extension L of K. Often K will be " }{TEXT 280 1 " C" }{TEXT -1 4 "(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 103 "Denote by V_L(eqn) the set of solutions of the equatio n in L. Now it is easy to see that V_L(eqn) is a " }{TEXT 259 1 "C" } {TEXT -1 64 "-vector space. The dimension of V_L(eqn) depends on the f ield L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "The elements of V_K(eqn) are called " }{TEXT 281 18 "rational solu tions" }{TEXT -1 12 ". Usually K=" }{TEXT 282 1 "C" }{TEXT -1 223 "(x) , in which case this name makes sense, but even when K is some other d ifferential field then solutions in K will be called rational solution s in order to distinguish them from solutions in a differential field \+ extension." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "If K=" }{TEXT 260 1 "C" }{TEXT -1 82 "(x) then we can use ratsol s from the DEtools package to find a basis for V_K(eqn)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "eqn:=diff(diff(diff(y(x),x),x),x)+2 /x*diff(diff(y(x),x),x)-1/x^2*diff(y(x),x)+1/x^3*y(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eqnG,*-%%diffG6$-%\"yG6#%\"xG-%\"$G6$F,\"\"$\" \"\"*&*&\"\"#F1-F'6$F)-F.6$F,F4F1F1F,!\"\"F1*&-F'6$F)F,F1*$)F,F4F1F9F9 *&F)F1*$)F,F0F1F9F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "with (DEtools): # load the DEtools package" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Note: everything after # is a comment, and will be ignore d by Maple." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "ratsols(eqn, y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$*&\"\"\"F%%\"xG!\"\"F&" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "y1,y2 := op(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%#y1G%#y2G6$*&\"\"\"F)%\"xG!\"\"F*" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "Now V_K(eqn) = " }{TEXT 276 1 "C" }{TEXT -1 6 "*y1 + " }{TEXT 277 1 "C" }{TEXT -1 4 "*y2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "Now lets determine t he remaining solutions" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "d solve(eqn); # dsolve = differential solver" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,(*&%$_C1G\"\"\"F'F+F+*&%$_C2GF+F'!\"\"F+ *(%$_C3GF+F'F+-%#lnGF&F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "We \+ see that if L=" }{TEXT 261 1 "C" }{TEXT -1 99 "(x,ln(x)) then we have \+ a 3-dimensional solution space V_L(eqn) with basis y1=1/x, y2=x, y3=x* ln(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "However, we'll see below that no matter how large the field L is, " } {TEXT 262 68 "the dimension of V_L(eqn) can never be larger than the o rder of eqn." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 418 "Furthermore: If the coefficients a0(x)..an(x) are in the diffe rential field K that satisfies the assumptions (characteristic 0, and \+ alg. closed field of constants), then one can always find a differenti al field L that contains K such that the dimension of V_L(eqn) equals \+ the order of eqn. And L satisfies the assumption as well (C_L = C_K). \+ Given such a field L, and a basis y1,..,yn of V_L(eqn), we can define \+ the a " }{TEXT 283 24 "smallest field extension" }{TEXT -1 119 " of K \+ that contains a basis of solutions, namely: K(y1,..,yn). However, in g eneral this is not a differential field. A " }{TEXT 284 37 "smallest d ifferential field extension" }{TEXT -1 20 " of K that contains " } {TEXT 285 43 "a basis of n linearly independent solutions" }{TEXT -1 316 " is: PV=K(y1,..,yn,y1',..,yn',...,y1^(n-1),...,yn^(n-1)). One do es not need derivatives higher than the n-1'th derivative because the \+ n'th derivative of y.i can be expressed in terms of lower derivatives \+ using the linear relation given by the equation eqn. Such a field PV i s called a Picard-Vessiot extension of K." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 286 11 "Definition:" }{TEXT -1 115 " If e qn is a differential equation of order n with coefficients in K, then \+ a differential field PV that contains n " }{TEXT 344 1 "C" }{TEXT -1 124 "-linearly independent solutions y1,..,yn, of eqn, and is generate d (as a differential field) over K by y1,..,yn is called a " }{TEXT 287 24 "Picard-Vessiot extension" }{TEXT -1 72 " of K for eqn. As alwa ys, PV must have the same field of constants as K." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 314 "Now one can prove that f or every differential field K (at this point we need the assumptions, \+ i.e. C_K is algebraically closed and has characteristic 0) and every l inear homogeneous differential equation eqn, there exists a Picard-Ves siot extension, and this PV extension is unique up to differential iso morphisms." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 212 "We will see that there are several ways to construct such PV e xtensions. And although the theory says that then a differential isomo rphism exists, it is often not easy in concrete examples to find one e xplicitly." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 288 11 "Definition:" }{TEXT -1 5 " The " }{TEXT 289 25 "differential G alois group" }{TEXT -1 99 " G of eqn over K is the group of differenti al automorphisms of the Picard-Vessiot extension over K." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 172 "A differential au tomorphism of PV over K is an automorphism g:PV-->PV such that g(a)=a \+ for all a in K, and g commutes with the differentiation, g(a')=g(a)' f or all a in PV." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 345 9 "Notation:" }{TEXT -1 159 " V(eqn) is short for: V_PV(eqn) . This notation is slightly ambiguous because the PV extension is not \+ unique (it's only unique up to differential isomorphisms)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 232 "If g in G then g acts trivially on the coefficients of eqn (which are in K) and commut es with the differentiation, so g(y) is a solution of eqn for every so lution y of eqn, and every g in G. So the group G acts on the n-dimens ional " }{TEXT 290 1 "C" }{TEXT -1 237 "-vector space V(eqn). This act ion is faithful because PV is generated by V(eqn), so we may consider \+ G as a subgroup of GL(V(eqn)). It turns out that G is always an algebr aic group, which means that G is an algebraic subset of GL(V(eqn))." } }{PARA 0 "" 0 "" {TEXT -1 100 "After the choice of a basis y1,..,yn of V(eqn), one may think of G as an algebraic subgroup of GL_n(" }{TEXT 291 1 "C" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT 292 28 "Differential Galois theory: " }{TEXT -1 331 " The re is a Galois correspondence between differential subfields of PV and algebraic subgroups of G, just like in the usual Galois theory. There are many analogies. We'll study relations between properties of the g roup and properties of solutions, like \"connected component of G is s olvable\" <===> \"The solutions are Liouvillian\"." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 324 39 "Examples of differenti al Galois groups:" }}{PARA 0 "" 0 "" {TEXT -1 57 "Go back to the examp les of differential field extensions." }}{PARA 0 "" 0 "" {TEXT -1 14 " First look at " }{TEXT 327 9 "example 1" }{TEXT -1 80 ". y'=a. This eq uation is not homogeneous. We can make it homogeneous as follows:" }} {PARA 0 "" 0 "" {TEXT -1 35 "L1(y)=a where L1=D=differentiation." }} {PARA 0 "" 0 "" {TEXT -1 131 "L2(a)=0 where L2=D-a'/a, which means L2 is an operator that sends y to D(y)-(a'/a)*y = y' - (a'/a)*y. Clearly L2(a)=a'-(a'/a)*a=0." }}{PARA 0 "" 0 "" {TEXT -1 76 "Now L1(L2(y))=L1 (a)=0, but L1(L2(y))=L1(y')=D(y')-(a'/a)*y' = y''-(a'/a)*y'." }}{PARA 0 "" 0 "" {TEXT -1 109 "So we can make example 1 homogeneous by taking the following equation: y''-(a'/a)*y' = 0. The solutions are: " } {TEXT 325 1 "C" }{TEXT -1 3 "*1+" }{TEXT 326 1 "C" }{TEXT -1 107 "*A w here A'=a. So the PV extension is PV=K(1,1',A,A')=K(A). The differenti al Galois group is G=\{g_c | c in " }{TEXT 328 1 "C" }{TEXT -1 68 "\} \+ where g_c(A)=A+c. The group G is isomorphic to the additive group " } {TEXT 329 1 "C" }{TEXT -1 156 ". The matrix representation of g_c is a s follows: Take the following basis y1=1 y2=A of V(eqn). Then g_c(y1)= y1 and g_c(y2)=y2+c*y1, which gives the matrix:" }}{PARA 0 "" 0 "" {TEXT -1 6 "(1 c)" }}{PARA 0 "" 0 "" {TEXT -1 67 "(0 1) and G can \+ be identified with the set of all such matrices." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 330 12 "Example 2a. " }{TEXT -1 537 "If all solut ions can be found in the algebraic closure of K, then PV is an algebra ic extension of K. The differential Galois group is a finite set if an d only if PV is an algebraic extension of K (more generally, the dimen sion of G as an algebraic set equals the transcendence degree of PV ov er K). If PV is an algebraic extension of K, then the usual Galois gro up of PV as a field over K coincides with the differential Galois grou p, because an automorphism of an algebraic extension over K is automat ically a differential automorphism." }}{PARA 0 "" 0 "" {TEXT 331 12 "E xample 2b. " }{TEXT -1 46 "In this case the differential Galois group \+ is " }{TEXT 332 1 "C" }{TEXT -1 5 "^* = " }{TEXT 333 1 "C" }{TEXT -1 12 "-\{0\} = Gl_1(" }{TEXT 334 1 "C" }{TEXT -1 2 ")." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 293 33 "Constru ction of a PV when K=C(x)." }}{PARA 0 "" 0 "" {TEXT -1 104 "If a0(x).. an(x) are functions, and they are analytic at a point x=c and an(c)<>0 (note: when a0..an in " }{TEXT 294 1 "C" }{TEXT -1 48 "(x) then all \+ but a finite number of points c in " }{TEXT 295 1 "C" }{TEXT -1 66 " u nion \{infinity\} satisfy this condition. These points are called " } {TEXT 296 14 "regular points" }{TEXT -1 17 ", or also called " }{TEXT 297 19 "non-singular points" }{TEXT -1 403 ".) then by Cauchy's theore m there are n linearly independent functions y1..yn that are analytic \+ at x=c and that are solutions of eqn. So the we can take L as the fiel d of all functions that are meromorphic in a small open set that conta ins x=c, and we can find n linearly independent solutions y1..yn in L. Then we can take PV as the subfield of L generated (as a differential field) by y1,..,yn over K." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 293 "If we take an explicit equation, and compute t hese solutions y1,..,yn at two regular points, then for each regular p oint we have a PV extension, and then there must be a differential aut omorphism between these two differential fields. However, it may be di fficult to find such an automorphism." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT 298 14 "The Wronskian." }}{PARA 0 "" 0 "" {TEXT -1 95 "To each y in a differential field extension L of K, we as sociate the vector of all derivatives:" }}{PARA 0 "" 0 "" {TEXT -1 41 "Y=transpose of (y,y',y'',y''',...). Then:" }}{PARA 0 "" 0 "" {TEXT 305 13 "y1,..,yk are " }{TEXT -1 1 "C" }{TEXT 299 119 "-linearly indep endent elements of L, if and only if the corresponding vectors Y1,..,Y k are linearly independent over L." }}{PARA 0 "" 0 "" {TEXT 300 6 "Pro of:" }}{PARA 0 "" 0 "" {TEXT -1 325 "Suppose Y1,..,Yk are linearly dep endent over L. Without loss of generality we may assume that no proper subset of Y1,..,Yk is dependent over L (if there is such a set then r eplace Y1,..,Yk by that set). Let a1*Y1+...+ak*Yk=0 where a.i in L, no t all a.i=0. Then all a.i are not 0. After dividing by a1 we may assum e that a1=1." }}{PARA 0 "" 0 "" {TEXT -1 520 "Then the derivative of 0 =a1*Y1+...+ak*Yk is 0 = 0' = (a1*Y1'+...+ak*Yk') + (a1'*Y1+...+ak'*Yk) . Now the i'th entry of the vector (a1*Y1'+...+ak*Yk') is the i+1'th e ntry of the vector a1*Y1+...+ak*Yk=0, so that's 0. So we find that a1' *Y1+...+ak'*Yk=0, which is a new linear relation. However, a1'=1'=0 so we have a relation between a proper subset of Y1,..,Yk, namely a2'Y2+ ...ak'*Yk. Since we assumed that Y2,..,Yk is linearly independent, it \+ follows that this relation is trivial, i.e. a2',,.,ak'=0, so all a.i a re in " }{TEXT 301 1 "C" }{TEXT -1 42 ". So Y1,..,Yk are linearly depe ndent over " }{TEXT 335 1 "C" }{TEXT -1 41 ", and so y1,..,yk is also \+ dependent over " }{TEXT 302 1 "C" }{TEXT -1 44 ". Conversely, if y1,.. ,yk is dependent over " }{TEXT 304 1 "C" }{TEXT -1 55 " then it's easy to see that Y1,..,Yk is dependent over " }{TEXT 303 1 "C" }{TEXT -1 68 " as well, and hence also linearly dependent over the larger field \+ L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 468 "If now y1,..,yk are solutions of some equation eqn of order n, with coef ficients a.i in some differential subfield K of L, then the above stat ement still holds if we replace the vector of all derivatives (y,y',.. .) by just the first n entries: Y1=transpose of (y1, y1', y1'',...,y1^ (n-1)). This is true because if you use the relation given by eqn, the n the (>=n)'th derivatives of y1 can be expressed as K-linear combinat ions of the 0,1,2,..,n-1'th derivatives of y1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 306 11 "Conclusion:" }{TEXT -1 78 " If eqn is an equation of order n with coefficients in K, and if y 1,..,yk are " }{TEXT 307 1 "C" }{TEXT -1 101 "-linear independent elem ents of V_L(eqn) (as always L is some differential field extension of \+ K with " }{TEXT 308 1 "C" }{TEXT -1 118 " as field of constants) then \+ the corresponding vectors Y1,..,Yk are linear independent elements of \+ L^n. In particular:" }}{PARA 0 "" 0 "" {TEXT -1 11 "*) k <= n." }} {PARA 0 "" 0 "" {TEXT -1 148 "*) If k=n then the matrix W=(Y1 .. Yn) \+ has non-zero determinant. This matrix W, whose i-j'th entry is the (i- 1)'th derivative of y.j is called the " }{TEXT 309 9 "Wronskian" } {TEXT -1 37 " of y1,..,yn, denoted by W(y1,..,yn)." }}{PARA 0 "" 0 "" {TEXT -1 100 "Notation: w(y1,..,yn)=det(W(y1,..,yn)). So w(y1,..,yn)=0 <==> y1,..,yn are linearly dependent over " }{TEXT 346 1 "C" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 310 41 "General \"Construction\" of a PV ex tension." }}{PARA 0 "" 0 "" {TEXT -1 45 "Let K be a differential field with constants " }{TEXT 311 1 "C" }{TEXT -1 163 ", and let eqn be a l inear homogeneous differential eqation of order n over K. Now we can c onstruct a differential ring that contains n solutions of eqn as follo ws:" }}{PARA 0 "" 0 "" {TEXT -1 33 "R = K[y_\{i,j\}] i =1..n, j=0..n- 1" }}{PARA 0 "" 0 "" {TEXT -1 477 "The differentiation is defined as f ollows. Denote y_i = y_\{i,0\}. Then let y_\{i,j\} be the j'th derivat ive of y_i. This determines the derivative for all y_\{i,j\} except wh en j=n-1. When j=n-1 then define the derivative in such a way that y_i is a solution of eqn. This way R is a differential ring, that contain s n linearly independent solutions y_1,..,y_n of eqn. However, we can \+ not just take PV as the field of fractions of R, because R may have mu ch more constants than just " }{TEXT 312 1 "C" }{TEXT -1 319 ". Indeed , the transcendence degree of R over K is n^2, but the transcendence d egree of a true PV extension is in \{0,1,...,n^2\}, and is not necessa rily n^2. So we have to introduce relations in R in order to end up wi th the correct transcendence degree and the correct set of constants. \+ We have to take R/I where I is a " }{TEXT 313 18 "differential ideal" }{TEXT -1 70 ". But we have to be careful, we have to prevent that y1 ,..,yn become " }{TEXT 336 1 "C" }{TEXT -1 326 "-linearly dependent in R/I, in other words we have to prevent that w(y1,..,yn) becomes 0 in \+ R/I. This is done by the following trick, which is a standard trick fo r computations with ideals to make sure something is non-zero: Introdu ce one more variable z and one relation as follows: Take the following ring instead of with R:" }}{PARA 0 "" 0 "" {TEXT 337 1 "R" }{TEXT -1 32 " = K[y_\{ij\}, z]/(1-z*w(y1..yn))." }}{PARA 0 "" 0 "" {TEXT -1 103 "There is only one way to turn this into a differential ring (i.e. there's only one way to define z' in " }{TEXT 338 1 "R" }{TEXT -1 148 ") because the Leibniz rule implies the quotient rule, and we can \+ differentiate z=1/w(..) by the quotient rule, z' = -w'/w^2 = -w' * z^2 which is in " }{TEXT 339 1 "R" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 15 "Now let I be a " }{TEXT 314 26 "maximal differential idea l" }{TEXT -1 302 ". It is easy to prove that such ideal exists with Zo rn's lemma, but at this point the construction is no longer really a c onstruction but merely an existence proof, because Zorn's lemma is not constructive. A maximal differential ideal I is not necessarily a max imal ideal, but it is a prime ideal. So " }{TEXT 340 1 "R" }{TEXT -1 129 "/I is an integral domain. The field of quotients is a Picard-Vess iot extension (we skip the proof that the field of constants is " } {TEXT 347 1 "C" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 33 "EXAMPLES OF SOLUTIONS AT A POINT:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "eqn:=(x-1)^2*diff(y(x),x$2) \+ - (x-2)/x^3 *y(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eqnG,&*&),&% \"xG\"\"\"F*!\"\"\"\"#F*-%%diffG6$-%\"yG6#F)-%\"$G6$F)F,F*F**&*&,&F)F* F,F+F*F0F*F**$)F)\"\"$F*F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "S ingular points:" }}{PARA 0 "" 0 "" {TEXT -1 38 "x=0 (a coefficient ha s a pole at x=0)" }}{PARA 0 "" 0 "" {TEXT -1 46 "x=1 (the highest coe fficient vanishes at x=1)" }}{PARA 0 "" 0 "" {TEXT -1 10 "x=infinity" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 341 21 "Regula r singularities" }{TEXT -1 280 " (p is a regular singularity means tha t for any open set S for which p is in the closure of S, and for every holomorphic function y on S that satisfies the differential equation, there exists a rational function f, such abs(y(x)) " 0 "" {MPLTEXT 1 0 44 "sol := seri es(add(a[i]*(x-2)^i,i=0..6),x=2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %$solG+1,&%\"xG\"\"\"\"\"#!\"\"&%\"aG6#\"\"!F.&F,6#F(F(&F,6#F)F)&F,6# \"\"$F5&F,6#\"\"%F8&F,6#\"\"&F;-%\"OGF0\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "subs(y(x)=sol,eqn): series(%,x=2);" }}{PARA 12 " " 1 "" {XPPMATH 20 "6#+-,&%\"xG\"\"\"\"\"#!\"\",$&%\"aG6#F'F'\"\"!,(&F +6#\"\"$\"\"'*&\"\"%F&F*F&F&*&#F&\"\")F&&F+6#F-F&F(F&,,&F+6#F4\"#7*&F= F&F/F&F&*&F'F&F*F&F&*&#F&F7F&&F+6#F&F&F(*&#F1\"#;F&F8F&F&F',.&F+6#\"\" &\"#?*&\"#CF&F;F&F&*&F2F&F/F&F&*&FEF&FBF&F&*&#F&F7F&F*F&F(*&#F1FFF&F8F &F(F1-%\"OGFCF4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "subs(x=t +2, convert(%,polynom));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,*&%\"aG6# \"\"#F'*&,(&F%6#\"\"$\"\"'*&\"\"%\"\"\"F$F0F0*&#F0\"\")F0&F%6#\"\"!F0! \"\"F0%\"tGF0F0*&,,&F%6#F/\"#7*&F=F0F*F0F0*&F'F0F$F0F0*&#F0F3F0&F%6#F0 F0F7*&#F,\"#;F0F4F0F0F0)F8F'F0F0*&,.&F%6#\"\"&\"#?*&\"#CF0F;F0F0*&F-F0 F*F0F0*&FEF0FBF0F0*&#F0F3F0F$F0F7*&#F,FFF0F4F0F7F0)F8F,F0F0" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "solve(\{coeffs(%,t)\},\{seq( a[i],i=2..5)\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<&/&%\"aG6#\"\"$,$ &F&6#\"\"!#\"\"\"\"#[/&F&6#\"\"#F,/&F&6#\"\"&,&&F&6#F.#!\"(\"$?$*&#F( \"#kF.F*F.F./&F&6#\"\"%,&F9#F.\"#'**&#\"\"(\"$#>F.F*F.!\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "sol:=subs(%,sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG+/,&%\"xG\"\"\"\"\"#!\"\"&%\"aG6#\"\" !F.&F,6#F(F(,$F+#F(\"#[\"\"$,&F/#F(\"#'**&#\"\"(\"$#>F(F+F(F*\"\"%,&F/ #!\"(\"$?$*&#F4\"#kF(F+F(F(\"\"&-%\"OGF0\"\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 179 "Since a0,a1 are arbitrary constants, we see that we h ave a 2-dimensional space of solutions at x=2. It is not hard to prove that these are convergent. A basis of the solutions is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "y1, y2 := subs(a[0]=1,a[1]=0,sol), \+ subs(a[0]=0,a[1]=1,sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%#y1G%# y2G6$+-,&%\"xG\"\"\"\"\"#!\"\"F+\"\"!#F+\"#[\"\"$#!\"(\"$#>\"\"%#F1\"# k\"\"&-%\"OG6#F+\"\"'++F)F+F+#F+\"#'*F5#F3\"$?$F8F9F<" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "So a PV extension of eqn over " }{TEXT 348 1 "C" }{TEXT -1 8 "(x) is::" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "C(x, y1, y2, diff(y1,x), diff(y2,x));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#-%\"CG6'%\"xG+-,&F&\"\"\"\"\"#!\"\"F)\"\"!#F)\"#[\"\"$# !\"(\"$#>\"\"%#F/\"#k\"\"&-%\"OG6#F)\"\"'++F(F)F)#F)\"#'*F3#F1\"$?$F6F 7F:++F(#F)\"#;F*#F1F.F/#\"#:F5F3F7F6++F(F)F,#F)\"#CF/#F1F5F3F7F6" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "Now at x=3:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 171 "sol := series(add(b[i]*(x-3)^i,i=0..6),x=3); \+ subs(y(x)=sol,eqn): series(%,x=3): subs(x=t+3, convert(%,polynom)): so lve(\{coeffs(%,t)\},\{seq(b[i],i=2..5)\}): sol:=subs(%,sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG+1,&%\"xG\"\"\"\"\"$!\"\"&%\"bG6#\"\" !F.&F,6#F(F(&F,6#\"\"#F3&F,6#F)F)&F,6#\"\"%F8&F,6#\"\"&F;-%\"OGF0\"\"' " }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%$solG+1,&%\"xG\"\"\"\"\"$!\"\"&% \"bG6#\"\"!F.&F,6#F(F(,$F+#F(\"$;#\"\"#,&F/#F(\"$['*&#F(F7F(F+F(F*F),& F+#\"#\"*\"'O*z#*&#F(\"%'H\"F(F/F(F*\"\"%,&F/#\"$r#\"(!o*R\"*&#F(\"&' \\ " 0 "" {MPLTEXT 1 0 67 "with(DEtools): # Load the DEtools package which contains formal_ sol" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "formal_sol(eqn,y(x), T,x=3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7%+/%\"TG!\"\"\"\"!#F'\"$ ;#\"\"##\"\"\"\"$['\"\"$#!#\"*\"'O*z#\"\"%#F'\"&'\\<\"\"&-%\"OG6#F-\" \"'+-F&F-F-F,F/#F'\"%'H\"F3#\"$r#\"(!o*R\"F6F7F:/F&,&%\"xGF-F/F'" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "formal_sol(eqn,y(x),T,x=2); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7%+-%\"TG!\"\"\"\"!#F'\"#[\"\"$# \"\"(\"$#>\"\"%#!\"$\"#k\"\"&-%\"OG6#\"\"\"\"\"'++F&F7F7#F7\"#'*F/#!\" (\"$?$F3F4F8/F&,&%\"xGF7\"\"#F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "The formal_sol command computes the formal solutions at a point, a nd expresses them in terms of a " }{TEXT 343 15 "local parameter" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 110 "If p is a finite point, then the local parameter is T=x-p, and if p=infinity then the local p arameter is T=1/x" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "formal _sol(eqn,y(x),T,x=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7%*& +-%\"TG!\"\"\"\"!#F(\"\"#F+#\"\"\"\"#C\"\"%#F-\"#5\"\"&-%\"OG6#F-\"\"' F-F'F(*&++F'F-F-#F-F6\"\"$#F(F.F2F3F6F-F'F(/F'*&F-F-%\"xGF(" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 211 "Here you can see that x=infinity \+ is not a regular point, because one of the two solutions has a pole at T=0 (i.e. at x=infinity). But it is regular singular because the pole is no worse than a rational function." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "formal_sol(eqn,y(x),T,x=1);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7#7$*&)%\"TG-%'RootOfG6#,(\"\"\"F,%#_ZG!\"\"*$)F-\"\"#F ,F,F,+1F'F,\"\"!,&F1F,*&F1F,F(F,F.F,,&#!#>\"#9F,*&#\"\"$\"\"(F,F(F,F.F 1,&#\"\"&\"#jF,*&#\"#$)FAF,F(F,F,F<,&#\"%Tl\"%w&*F,*&#\"%'p\"\"%(>\"F, F(F,F.\"\"%,&#!&6e#\"&SR#F,*&#\"&Z!HFQF,F(F,F,F@-%\"OG6#F,\"\"'F,/F',& %\"xGF,F,F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 300 "This looks like j ust one solution, but in fact it's two solutions because RootOf(1-_Z+Z ^2) stands for two complex numbers (the two solutions of 1-Z+Z^2=0). I f e is one of those roots, then T^e*(1+..) does not go to infinity fas ter than say T^(-100) when T goes to 0 so this is still regular singul ar." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "formal_sol(eqn,y(x), T,x=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7$*()%\"TG#\"\"$\"\"#\"\" \"-%$expG6#,$*&F+F+F'!\"\"!\"#F++1F'F+\"\"!#!#X\"#;F+#\"%&Q#\"$7&F*#!& H!**\"&wX#F)#\"'z27\"')GC&\"\"%#\")Pu%f#\"(3')Q)\"\"&-%\"OG6#F+\"\"'F+ /,$*$)F'F*F+F2%\"xG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 545 "Now x=0 i s irregular singular, because exp(-2/T) can (depending on which path y ou take) go to infinity faster than any rational function when T-->0, \+ i.e. when x-->0. Note that we now have fractional powers of x because \+ T=sqrt(-x/2). When fractional powers occur, then at x=p the local para meter T is not necessarily x-p but it is (c*(x-p))^(1/d) where c is so me constant and d is a positive integer. Of course we could just take \+ (x-p)^(1/d), but if we did that in this example we'd have not just a s qrt(x) but also a sqrt(-2) in the computation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "The generalized exponents at x=0 are:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "gen_exp(eqn ,y(x),T,x=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7$,&*&\"\"\"F'%\"TG !\"\"F'#\"\"$\"\"%F'/,$*$)F(\"\"#F'!\"#%\"xG" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 139 "Now if e=1/T+3/4 then exp(int(e/x))=T^(3/2)*exp(-2/T) \+ like in formal_sol. Hint: the 3/4 becomes a 3/2 because x is not T but T^2*something." }}{PARA 0 "" 0 "" {TEXT -1 161 "It looks like there i s only one generalized exponent, but there are two (=order of equation ) generalized exponents, because there are two T's for which -2*T^2=x. " }}{PARA 0 "" 0 "" {TEXT -1 109 "For more information on generalized \+ exponents and formal solutions see the following two help pages in Map le." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "?gen_exp" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "?formal_sol" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 334 "On Wednesday I'll explain how one can calculate the formal solutions and the generalized exponents. Because this is fairl y technical (Newton Polygon, Newton Polynomial, etc.), it will be quit e time consuming to calculate these things by hand, so I would recomme nd to always compute them with the Maple software gen_exp and formal_s ol." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "44" 0 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }