{VERSION 2 3 "APPLE_PPC_MAC" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Headi ng 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }1 0 0 0 6 6 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 2" 3 4 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 4 4 0 0 0 0 0 0 -1 0 } {PSTYLE "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "" 4 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 18 "" 0 "" {TEXT -1 37 "Linear Methods of Applied Mathem atics" }}{PARA 256 "" 0 "" {TEXT -1 37 "Linear equations and linear op erators" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 162 " To set the pattern for what is to be expected for linear equations and linear operators, we recall t he situation for matrices. We examine a matrix operation " }{XPPEDIT 18 0 "Mx" "I#MxG6\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 156 " Recall, that Maple saves space in your computer by allowing the u ser to load in only the packages needed. Here, we will use the linear \+ algebra package." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(lin alg);" }}}{PARA 0 "" 0 "" {TEXT -1 67 " Defining the matrix of equ ation (1.2) can be done as follows. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "M:=matrix([[1,-2,1],[1,-2,1],[1,-2,1]]);" }}}{PARA 0 "" 0 "" {TEXT -1 174 " To see how this matrix operates on a vector , we evaluate M at a generic point [x,y,z] and refer to this matrix mu ltiplication as \"M at [x,y,z],\" or \"M of [x,y,z.,\"" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "M&*[x,y,z];\nevalm(\");" }}} {PARA 0 "" 0 "" {TEXT -1 48 " Compute M at [1,2,3], [1,3,5], and [ 2,2,2]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 " We now define a " }{TEXT 256 21 "differential operator" } {TEXT -1 189 ". As with any linear function, this operator will be def ined by saying how it operates on a member in its domain. Here, we def ine the operator A as acting on functions with two derivatives:" }} {PARA 0 "" 0 "" {TEXT -1 54 " \+ " }{XPPEDIT 18 0 "A(u) = diff(u(x),x,x) + diff(u(x),x) + \+ 2*u(x" "/-%\"AG6#%\"uG,(-%%diffG6%-F&6#%\"xGF-F-\"\"\"-F)6$-F&6#F-F-F. *&\"\"#F.-F&6#F-F.F." }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 80 "W e make this definition in the same pattern as we did that for the matr ix above:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "A:=u->diff(u(x) ,x$2)+diff(u(x),x)+2*u(x):;A(u);" }}}{PARA 0 "" 0 "" {TEXT -1 16 "For \+ example, if " }{XPPEDIT 18 0 "u(x) = x^2" "/-%\"uG6#%\"xG*$F&\"\"#" } {TEXT -1 37 " we compute exactly what you expect." }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 17 "u:=x->x^2; \nA(u);" }}}{PARA 0 "" 0 "" {TEXT -1 39 "Compute A of several functions such as " }{XPPEDIT 18 0 " x*sin(x)" "*&%\"xG\"\"\"-%$sinG6#F#F$" }{TEXT -1 3 ", " }{XPPEDIT 18 0 "x^3*exp(x)" "*&%\"xG\"\"$-%$expG6#F#\"\"\"" }{TEXT -1 9 ", and \+ " }{XPPEDIT 18 0 "exp(-x/2)*sin(sqrt(7)*x/2" "*&-%$expG6#,$*&%\"xG\"\" \"\"\"#!\"\"F+F)-%$sinG6#*(-%%sqrtG6#\"\"(F)F(F)F*F+F)" }{TEXT -1 1 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 " \+ We next address how to find the basis for the null space of these two operators. Recall that the " }{TEXT 258 10 "null space" }{TEXT -1 114 " for an operator is the collection of vectors, or objects, in the domain of the operator that get mapped to zero. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 141 " We find the null sp ace for the matrix M above. We can think of doing this in two ways. Th e first illustration is to solve the equation " }}{PARA 0 "" 0 "" {TEXT -1 57 " \+ " }{XPPEDIT 18 0 "M*x = 0" "/*&%\"MG\"\"\"%\"xGF%\"\"!" }}{PARA 0 "" 0 "" {TEXT -1 6 "for x." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "linsolve(M,[0,0,0]);" }}}{PARA 0 " " 0 "" {TEXT -1 68 "An understanding of the answer is that there are t wo constants: say " }{XPPEDIT 18 0 "t[1]" "&%\"tG6#\"\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "t[2]" "&%\"tG6#\"\"#" }{TEXT -1 72 " and vec tors in the nullspace can be written as the combination " } {XPPEDIT 18 0 "t[1]*[0,1,2]+t[2]*[1,0,-1]" ",&*&&%\"tG6#\"\"\"F'7%\"\" !F'\"\"#F'F'*&&F%6#F*F'7%F'F),$F'!\"\"F'F'" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 51 " As an alternate, we could compute the pair o f " }{TEXT 259 14 "basis elements" }{TEXT -1 20 " for the null space. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "nullspace(M);" }}}{PARA 0 "" 0 "" {TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 241 " We no w find the null space for the differential operator A defined above. A gain, there are several ways to do this. A first is to recall from sop homore differential equations that solutions of the equation Au = 0 ar e likely multiples of" }}{PARA 0 "" 0 "" {TEXT -1 46 " \+ " }{XPPEDIT 18 0 "u(x) = exp(m*x)" "/-% \"uG6#%\"xG-%$expG6#*&%\"mG\"\"\"F&F," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 23 "We have only to find m." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "u:=x->exp(m*x);\nsolve(A(u)=0,m);" }}}{PARA 0 "" 0 " " {TEXT -1 87 "An alternate is to solve differential equations as in s ophomore differential equations." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "dsolve(\{A(v)=0\},v(x));" }}}{PARA 0 "" 0 "" {TEXT -1 114 "To \+ see that this is the same result as we had with the first technique, w e convert the answer to exponential form." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "convert(\",exp);\nsimplify(\");" }}}{PARA 0 "" 0 "" {TEXT -1 140 "Finally, just as Maple computed a basis for the null spa ce of the matrix M above, so it will compute a basis for this differen tial operator." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "dsolve(\{A (v)=0\},v(x),output=basis);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 " Finally, we solve a nonhomogeneous equatio n for the two problems." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "l insolve(M,[1,1,1]);" }}}{PARA 0 "" 0 "" {TEXT -1 78 "The technique for solving the nonhomogeneous differential equation is familar." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "dsolve(\{A(v)=exp(2*x)\},v(x ));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "4 0" 5 } {VIEWOPTS 1 1 0 1 1 1803 }