mathematica syntax, matrix notation

*To*: mathgroup at smc.vnet.net*Subject*: [mg2127] mathematica syntax, matrix notation*From*: itcook at peseta.ucdavis.edu (Poor Little Rich )*Date*: Wed, 4 Oct 1995 01:55:24 -0400*Organization*: University of California, Davis

Hello, I am a poor foolish undergrad in chemical engineering,and am working on a project which requires that I understand the syntax of the following expression. The context is that I am trying to solve a system of ode's of the 2 point BV type. That's probably not important to know. What I would like to know is, what significance do the bracketed and double-bracketed expressions have in the following mathematica command: yend[#189#_]:=(soln=NDSolve[system[#189#],{y1,y2}, {x,-1,1}])[[1,1,2]][1] It has been suggested to me (and I am paraphrasing) that the double brackets indicate that the variable "soln" will be of a nested index function form and that the double brackets reference a part of this function, e.g: soln={{{a,b},{a2,b2}},{{a3,b3},{a4,b4}}} and thus passes a value of 1 to the function I have called "b" here,then passes b[1] to yend[#189#_], as in: yend[#189#_]:=(soln)[b[1]]. Does that seem correct? A related question is that when I enter the following command: system[#189#_]={y1'[x]==y2[x], y2'[x]==-Sin[y1[x]]+Cos[5 x], y1[-1]==0,y2[-1]==#189#} mathematica echoes back the same thing, just as you would expect, except that it inexplicably adds a #1 right before the last #189#, like this: system[#189#_]={y1'[x]==y2[x], y2'[x]==-Sin[y1[x]]+Cos[5 x], y1[-1]==0,y2[-1]==#1 #189#} Does anyone know why this is? Thank you for your help in this matter. If it helps, I am enclosing the source of these statements, a discussion on using mathematica to solve 2-point BVODE's: Boundary Value Problems Introduction In this tutorial we will discuss how to use Mathematica to solve n-th order boundary value problems (BVP). Since we can always write a n-th order BVP as a system of n first-order equations of the form y'(x)=F(y,x,u), where y={y1,y2,..yn}, and F={F1,F2, F3,..Fn} and u={u1,u2,...} , our discussion will assume that the BVP to be solved can be written in this manner. Mathematica has a numerical differential equation solver called NDSolve[] which solves initial value problems. Thus in order to use this routine for BVPs, we must implement some iteration procedure that is based on a guess for the unknown (missing) initial condition. One way to accomplish this is to use a shooting method. Suppose we have a 2nd-order BVP for y(x) subject to the BCs y1(a)=b1, y2(b)=b2. To integrate this system from x=a to x=b, we need to guess the value of y2(a). Since our guess is unlikely to be correct, the numerical solution y(x) will not satisfy the BC y2(b)=b2. Thus we must update our guess for y2(a), and integrate once again. Since the numerical solution y(x) depends parametrically on the initial guess y2(a)=#189#, we have an additional equation for #189#, viz., y2(x; #189#)=b2. Note that we have expressed the dependence on #189# as y(x; #189#). To find #189# we augment the system of ODEs with the additional equation, and solve iteratively for #189#. We now demonstate this procedure in the next section Example 1: We consider the following BVP defined by the differential equation y''(x)+sin(y(x))=cos(5x) with boundary conditions y(-1)=0, y(1)=0 This equation can be expressed as a system of two first order ODEs by making the substitutions: y1(x)=y(x) y1'(x)=y2(x) y2'(x)=-sin(y(x)+cos(5x) with boundary conditions y1(-1)=0, y2(-1)=#189# Next we assign the system of equations to a variable system: system[#189#_]={y1'[x]==y2[x], y2'[x]==-Sin[y1[x]]+Cos[5 x], y1[-1]==0,y2[-1]==#189#} Note that the variable system is a function of the unknown initial condition #189#. Next, we define a function yend[#189#_] which is equal to the solution of the ODEs at the end point x=1.Refer to the tutorial on the use of NDSolve for an explanation of the format used in this statement yend[#189#_]:=(soln=NDSolve[system[#189#],{y1,y2}, {x,-1,1}])[[1,1,2]][1] We use the Plot[] function to show how the end point yend[#189#_] varies with the initial condition Plot[yend[#189#],{#189#,-2,2}] The value of #189# that we are seeking is the value that satisfies the equation yend[#189#]==0. We make use of the FindRoot[] function to determine #189# such that yend[#189#]= y1(1)=0: FindRoot[yend[#189#]==0,{#189#,-2,2}][[1,2]] Finally, we use the Plot[] routine to plot out the solution y1[x]. Plot[Evaluate[y1[x]/.soln],{x,-1,1}] Example 2: In this example we show how to use more advanced programming in Mathematica to combine all the steps in Example 1 so as to have an efficient algorithm. The first several ines of the code are similar except that we make use of "pure functions" defined using the symbols # and &: system[#189#_]={y1'[x]==y2[x], y2'[x]==-Sin[y1[x]]+Cos[5 x], y1[-1]==0,y2[-1]==#189#} yend:=(soln=NDSolve[system[#],{y1,y2}, {x,-1,1}])[[1,1,2]][1]& FindRoot[yend[#189#]==0,{#189#,-2,2}][[1,2]] Plot[Evaluate[y1[x]/.soln],{x,-1,1}] Eample 3: In this example we show how the ideas we have developed in the previous sections can be applied to more difficult BVPs. The first BVP we solve is the 3rd-order Blasius equation that describes the velocity profile f(n) in a laminar boundary layer over a flat plate: ff''+2f'''=0, f(0)=0, f'(0)=0, f'(#176#)=1 As before, we reduce the 3rd-order equation to a system of 3 first-order equation with the following substitutions: f'(n)=v(n), v'(n)=w(n) w'(n)=-f(n)*w(n)/2 with initial conditions f(0)=0, v(0)=0, w(0)=#189# and the auxillary equation to find #189# v(#176#; #189#)=1 In the calculations we truncate the infinite domain and apply the BC at infinity at a finite location given by n=6 system[#189#_]={2 w'[n]+f[n] w[n]==0,w[n]==v'[n],v[n]==f'[n], w[0]==#189#,v[0]==0,f[0]==0} gun:=(soln=NDSolve[system[#],{w,v,f}, {n,0,10}])[[1,2,2]][10.]& FindRoot[gun[#189#]==1,{#189#,1,0.5}] Plot[Evaluate[{f[n],v[n],w[n]}/.soln],{n,0,10}] In our next BVP we consider the differential equation for stagnation in plane flow (Hiemenz flow). The similarity form for the velocity field is given by the following third-order equation: f'''+ff''-f'2+1=0 with boundary conditions f(0)=0, f'(0)=0, f'(#176#)=1 We reduce this to a system of 3 first-order equations with the following substitutions: f'(n)=v(n), v'(n)=w(n) w'(n)=v(n)2-1-f(n)w(n) with initial conditions f(0)=0, v(0)=0, w(0)=#189# and the auxillary equation to find #189# v(#176#; #189#)=1 system[#189#_]={w'[n]==v[n]^2-1-f[n] w[n],f'[n]==v[n], v'[n]==w[n],f[0]==0,v[0]==0,w[0]==#189#} gun:=(sol=NDSolve[system[#],{f,v,w},{n,0,5}])[[1,2,2]][5]& FindRoot[gun[#189#]==1,{#189#,-2,2}] Plot[Evaluate[{f[n],v[n],w[n]}/.sol],{n,0,5}] 2 {w'[n] == -1 + v[n] - f[n] w[n], f'[n] == v[n], v'[n] == w[n], f[0] == 0, v[0] == 0, w[0] == #189#} {#189# -> 1.23259} -Graphics- ^*) ---Sincerely, Rich Cook |"The volume of a body of rotation is equal to the rdcook at ucdavis.edu |generating area times the distance traveled by the 1-916-758-6897 |centroid of the area while the body is being Metaphor Alert |generated." -Beer, et al. -Sincerely, Rich Cook |"There are trivial truths & there are great rdcook at ucdavis.edu |truths. The opposite of a trivial truth is 1-916-758-6897 |plainly false. The opposite of a great truth Metaphor Alert | is also true." -Neils Bohr

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