P-Set 1 Solution Set. Solutions created by Melissa Diskin. Part 1 (3 points) Part 2 (4 points) ê H1 ê 2 + xl^hx ê 2L, xdd

Transcription

1 PSet Solution Set Solutions created by Melissa Diskin I (Handworked, 5 points) Simplified version of the derivative with respect to x. In[80]:= Simplify@D@ ê H ê + xl^hx ê L, xdd Out[80]= + x xê x + x LogB + xf I (5 points) Part (3 points) Defines the function. In[8]:= gfunc = H ê HHoxy + hydll^hoxy hydll Out[8]= hyd oxy Hhyd + oxyl The first derivative, followed by the second derivative. In[8]:= dgdh = D@gfunc, hydd êê Simplify dgdh = D@gfunc, 8oxy, <D êê Simplify Out[8]= oxy Hhyd + oxyl hyd oxy Hhyd + Hhyd + oxyl Log@hyd + oxydl Out[83]= hyd oxy hyd Hhyd + oxyl Ioxy + hyd I + oxy M + hyd oxy Hhyd + oxyl Log@hyd + oxyd + hyd Hhyd + oxyl Log@hyd + oxyd M Part (4 points) Plots using a table for a given range of oxygen, plotted as a function of hydgrogen.

2 HW009Sols.nb In[84]:= 8oxy,.05,,.<DD, 8hyd, 0, 0<, PlotRange Ø All, ImageSize Ø Large, BaseStyle Ø Large, AxesLabel Ø 8"Oxygen", "Hydrogen"<D Hydrogen Out[84]= Oxyge Part 3 (4 points) Creates a function to find the numerical value of the minimum around, looking from 0^6 to 0. In[85]:= minfunc@aval_d := FindMinimum@N@dgdh ê. oxy Ø avald, 8hyd,, 0^H6L, 0<D Finds the numerical minimum at oxygen = /Pi. In[86]:= minfunc@ ê PiD Out[86]= , 8hyd Ø.5083<< Creates a table of those minimum values from 0.00 to. The warning comes from the unusual behavior around 0 and can be ignored. In[87]:= Table@First@minFunc@oxDD, 8ox, 0.00,,.05<D FindMinimum::reged : The point 80.< is at the edge of the search region 9.µ0 6, 0.= in coordinate and the computed search direction points outside the region. à Out[87]= , , 0.504, , , , , , , , , , , , , , , , , <

3 HW009Sols.nb 3 Part 4 (4 points) Plots the minimum values of the function (created in the previous step). (again, warning can be ignored) In[88]:= Plot@First@minFunc@aDD, 8a,.0, <D Plot::exclul : 8Im@a + hydd 0, Im@a + hydd 0< must be a list of equalities or realvalued functions. à Out[88]= I3 (5 points) Part (3 points) Creates the matrix and displays it in matrix form. In[89]:= mat = Out[90]//MatrixForm= :: 4,, 7, >, :0,, 6, 4 5 >, : 5, 3 4,, >, : 3, 7, 3, >>; MatrixForm@ matd Calculates the inverse.

4 4 HW009Sols.nb In[9]:= êê MatrixForm Out[9]//MatrixForm= Product of the inverse and b. In[9]:= Inverse@matD.8,,, < Out[9]= : 7 767, ,, > Part ( points) Series of steps to input the new values of the matrix using operations on the rows. In[93]:= mat@@4dd = 3 mat@@dd mat@@dd + 4 mat@@3dd Out[93]= : 3 0,, 79 4, 3 0 > In[94]:= mat@@3dd = 3 mat@@dd mat@@dd Out[94]= : 3 4, 5, 4, 7 0 > In[95]:= mat@@dd = 3 mat@@dd mat@@dd Out[95]= : 4,, 5 4, 9 0 > In[96]:= MatrixForm@matD Out[96]//MatrixForm=

5 HW009Sols.nb 5 In[97]:= NullSpace@matD H*just curiousity*l Out[97]= :: , , , >> Creates x, to be solved for in following step, and b, a vector of zeros. In[98]:= x = 8x, x, x3, x4< bzero = 80, 0, 0, 0<; Out[98]= 8x, x, x3, x4< The conditions which much be true for a solution to exist, using the matrix product with x. In[300]:= eqs = Table@mat@@iDD.x ã bzero@@idd, 8i, 4<D Out[300]= : x 4 + x + x3 7 + x4 ã 0, x 4 + x + 5 x3 4 3 x x x x4 3 x ã 0, 0 0 x + 9 x4 0 ã 0, + 79 x3 4 3 x4 0 ã 0> Solves for x, x, x3 in terms of x4. In[30]:= Solve@eqs, 8x, x, x3, x4<d êê Framed Solve::svars : Equations may not give solutions for all "solve" variables. à Out[30]= 086 x x4 338 x4 ::x Ø, x Ø, x3 Ø >> Note that, while {0, 0, 0, 0} is a solution, it is not the only solution. The solutions lie on the boundary described by these equations. (0.5/ points provided if said 0, 0, 0, 0 was only solution) Plug in to verify that these solutions work. In[30]:= eqs ê. Solve@eqs, 8x, x, x3, x4<d Solve::svars : Equations may not give solutions for all "solve" variables. à Out[30]= 88True, True, True, True<< I4 (0 points) Part (5 points) Inputs the matrix and displays in matrix form.

6 6 HW009Sols.nb In[303]:= mat = ::,, 0, 0,, 0, 3, >, :,,,,,,, >, :,,, 0,,,, >, : 3,,, 0,,,, 0>, : 3, 0,,,, 0,, 0>, :0, 0,, 0,,,, 3 >, :, 3, 3,,, 0,, >, Out[304]//MatrixForm= :,,,,, 3,, >>; MatrixForm@ matd Solve with the vector b composed of placeholder variables a through h. In[305]:= LinearSolve@mat, CharacterRange@"a", "h"dd êê Simplify êê TableForm Out[305]//TableForm= 3 H50 a958 b+940 c9874 d95 e+ 747 f7 838 g+406 hl a b c58 6 d3 49 e f05 58 g h a5 090 b+53 c7 54 d+4433 e f89 80 g h a+7 b865 c654 d e+377 f6 878 g9930 h a5 86 b+336 c7 874 d7 053 e f6 598 g h H8 66 a b+940 c 84 d+39 e f5 030 g hl H 86 a+566 b 780 c4398 d6747 e+45 f47 g3606 hl H88 a3074 b+500 c966 d+573 e+50 f+54 g54 hl Part (5 points) Using several functions, finds the solution for any b, chosen from RandomReal.

7 HW009Sols.nb 7 In[306]:= matinverse = Inverse@matD; In[307]:= thesol@bvec_d := matinverse.bvec In[308]:= arandomsol := thesol@randomreal@80, <, 88<DD In[309]:= arandomsol Out[309]= ,.5958,.57343, ,.848,.657, , 0.575< This function takes advantage of Norm which finds the magnitude. In[30]:= arandommag := Norm@aRandomSolD Plots a histogram for 500 magnitudes (each of which is 8 elements wide). This is useful for looking at the magnitude distribution. In[3]:= Histogram@Table@aRandomMag, 8500<DD Out[3]= I5 (Handworked, 5 points) and I6 (0 points) This specifically is for I6; I5 follows the same process. Uses solve to find the x and y points where the ellipses intersect for generic ellipses with any coefficients a, b, c, and d.

8 8 HW009Sols.nb In[3]:= sol = Solve@8aa xx^ + bb yy^ ã, cc xx^ + dd yy^ ã <, 8xx, yy<d êê Simplify Out[3]= ::xx Ø cc HbbddL bb ccaa dd cc, yy Ø aa + cc bb cc aa dd >, :xx Ø cc HbbddL bb ccaa dd cc, yy Ø aa + cc bb cc aa dd >, :xx Ø cc HbbddL bb ccaa dd cc, yy Ø aa + cc bb cc aa dd >, :xx Ø cc HbbddL bb ccaa dd cc, yy Ø aa + cc bb cc aa dd >> 0. Tests with selected coefficients. In[33]:= sol ê. 8aa Ø ê, bb Ø ê 4, cc Ø ê 8, dd Ø ê 3< Out[33]= ::xx Ø 3, yy Ø 6 3 >, :xx Ø 3, yy Ø 6 3 >, :xx Ø 3, yy Ø 6 3 >, :xx Ø 3, yy Ø 6 3 >>. Finds the solution with coefficients given in the problem. In[34]:= sol = Solve@8aa xx^ + yy^ ã, xx^ + dd yy^ ã <, 8xx, yy<d Out[34]= ::xx Ø H + aal dd, yy Ø + aa + aa dd + aa dd >, :xx Ø H + aal dd, yy Ø + aa + aa dd + aa dd >, :xx Ø H + aal dd, yy Ø + aa + aa dd + aa dd >, :xx Ø H + aal dd, yy Ø + aa + aa dd + aa dd >> In[35]:= sol êê Simplify Out[35]= ::xx Ø + dd + aa dd, yy Ø + aa + aa dd >, :xx Ø + dd + aa dd, yy Ø + aa + aa dd >, :xx Ø + dd + aa dd, yy Ø + aa + aa dd >, :xx Ø + dd + aa dd, yy Ø + aa + aa dd >> Reduce finds the conditions on a and d which much be true. Remember that the radicands must be positive.

9 HW009Sols.nb 9 In[36]:= ReduceB + dd + aa dd > 0 && + aa > 0, 8aa, dd<, RealsF Out[36]= aa > && dd <»» dd > aa In[37]:= 8xx, yy< ê. sol Out[37]= :: H + aal dd, + aa + aa dd + aa dd >, : H + aal dd, + aa + aa dd + aa dd >, : H + aal dd, + aa + aa dd + aa dd >, : H + aal dd, + aa + aa dd + aa dd >> Your professor having fun: Finds what the intersections approach as coefficient a approaches coefficient d. In[38]:= asol = H 8xx, yy< ê. soll ê. aa Ø dd H*wcc playing*l Out[38]= :: : H + ddl dd + dd, H + ddl dd + dd, + dd + dd >, : + dd + dd >, : H + ddl dd + dd, H + ddl dd + dd, + dd >, + dd + dd >> + dd Your professor having more fun: The orbit of the intersections for changing d.

10 0 HW009Sols.nb In[39]:= 8dd, 0, 3<D H*more play*l Out[39]= I7 (Handworked, 5 points). Sets the derivative equal to Sin(t) In[30]:= dsol7 = xxx'@td ã Sin@tD Out[30]= ã Sin@tD DSolve uses the given boundary conditions to find the function of t which solves the ODE. In[3]:= DSolve@8dsol7, xxx@0d ã <, xxx@td, td Out[3]= 88xxx@tD Ø Cos@tD<< I8 (0 points) Uses the same process as I7. In[3]:= dsol8 = D@zzz@tD, 8t, <D + zzz@td ã Sin@tD Out[3]= zzz@td + ã Sin@tD

11 HW009Sols.nb In[33]:= ã bca, ã bcb<, td Out[33]= Ø 4 I4 bca Cos@tD t Cos@tD + Sin@tD + 4 bcb Sin@tD Cos@tD Sin@tD + Cos@tD Sin@ tdm>> I9 (0 points) This shows that there is no closed form solution available. In[34]:= DSolve@8rrr''@tD ã H rrr@tdl ê H + Sin@tDL, rrr@0d ã 0, rrr'@0d ã <, rrr@td, td Out[34]= ã rrr@td + Sin@tD, rrr@0d ã 0, ã >, rrr@td, tf Uses Numerical to solve. In[35]:= dsol9 = NDSolve@8rrr''@tD ã H rrr@tdl ê H + Sin@tDL, rrr@0d ã 0, rrr'@0d ã <, rrr@td, 8t, 0, 50<D Out[35]= 88rrr@tD Ø InterpolatingFunction@880., 50.<<, <>D@tD<< Plots the solution found using Numerical. In[36]:= Plot@rrr@tD ê. dsol9, 8t, 0, 50<D 3 Out[36]= I0 (5 points) Part (4 points) The key to solving I0 is to start by nondimensionalizing. If you do not understand this, review the notebook WCC posted on the 3.06 site. Defines xa in terms of xb.

12 HW009Sols.nb In[37]:= xa = xb Out[37]= xb The function: In[38]:= deltagbar = W xa xb + R T HxA Log@xAD + xb Log@xBDL Out[38]= H xbl xb W + R T HH xbl Log@ xbd + xb Log@xBDL For the "two hump" shape, the two second derivates going to zero converge a xa= xb=/ at the critical temperature (found using solve): In[39]:= critsol = Solve@D@deltaGBar, 8xB, <D ã 0, TD Out[39]= H + xbl xb W ::T Ø >> R Knowing that Tcrit will occur at /: In[330]:= Tcrit = HH T ê. critsoll ê. xb Ø ê L@@DD Out[330]= W R Full credit given if you provided a numerical answer that was correct with whatever value of W you picked. Bonus point assigned if you got this generic answer. Part 3 (4 points) Nondimensionalizing by using t = T/Tcrit. In[33]:= DimensionlessDG = SimplifyB deltagbar R Tcrit ê. T Ø t TcritF Out[33]= Ht xb tl Log@ xbd + xb H xb + t Log@xBDL Note that part of the grade for this part was based on whether or not you did the steps show in Part of this key. W should not be present in your answer, as it has units! Parts and 4 (4 points, 3 points, respectively) To plot in part, you had to use placeholder values; in part 4, with the equation nondimensionalized, you don't need placeholder values. The full range of solution shapes is now apparent using the dimensionless deltag.

13 HW009Sols.nb 3 In[33]:= Plot@ Evaluate@Table@DimensionlessDG, 8t, 0.5,.5, 0.<DD, 8xB, 0, <D Out[33]= The transition is also apparent when looking at the plot using Manipulate. In[333]:= Plot@DimensionlessDG, 8xB, 0, <D Out[333]=

14 4 HW009Sols.nb In[334]:= ê. t Ø temperature, 8xB, 0, <, PlotRange Ø 80.5, <D, 88temperature, 0.5<, 0.,.5<D temperature Out[334]=