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Re: Piecewise function involving matrices

  • To: mathgroup at smc.vnet.net
  • Subject: [mg91230] Re: Piecewise function involving matrices
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Sun, 10 Aug 2008 01:53:41 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <g7k0gn$ieh$1@smc.vnet.net>

kaveh wrote:

> I am trying to define a piecewise matrix function of time with
> Mathematica that I later want to use to transform another matrix as a
> function of time and integrate it. To make it simple I have
> 
> In[163]:= A={{1,.1},{.1,-1}};
> In[171]:= H0={{0,b},{b,0}};
> In[172]:= F[t_]:=Piecewise[{{exp[-I t] A,0<t<1},{exp[+I t] A,1<t<2}}]

Syntax error: Except if you have already defined the function exp 
somewhere else, exp should be written Exp, which is the standard 
exponential function in Mathematica.

> In[173]:= F[0.1]
> Out[173]= {{exp[-0.1 \[ImaginaryI]],0.1 exp[-0.1 \[ImaginaryI]]},{0.1
> exp[-0.1 \[ImaginaryI]],-exp[-0.1 \[ImaginaryI]]}}
> 
> In[181]:= H[t_]:= Inverse[F[t]].H0.F[t]+H0.H0
> 
> In[182]:= H[0.4]
> Out[182]= {{0.19802 b+b^2,-0.980198 b},{-0.980198 b,-0.19802 b+b^2}}
> 
> The problem is that when I evaluate H[t] for a given number, it
> returns the right value but if t is a symbol Mathematica naturally
> leaves it as unevaluated. This by itself is nice, however when I want
> to integrate H[t], I run into problems.
> Say
> In[183]:= Integrate[H[x],{x,0,1}]

<snip>

> Notice that the output is not a 2x2 matrix but has dimensions
> {2,2,2,2}!
> Same with:
> In[186]:= Integrate[H[x][[1,1]],{x,0,1}]
> Out[186]= {{0.00990099 (20. b+101. b^2),0.00990099 (-99. b+101. b^2)},
> {0.00990099 (-99. b+101. b^2),0.00990099 (-20. b+101. b^2)}}
> which should be just a number but is a 2x2 matrix!
> 
> Obviously there is something wrong with the treatment of piecewise as
> an array instead of a flat thing.

Since Mathematica 6.0, Integrate works well with *continuous* piecewise 
functions.

In your case, you should use *Boole* to write the piecewise function. 
For instance,

In[1]:= A = {{1, .1}, {.1, -1}};

H0 = {{0, b}, {b, 0}};

F[t_] := Boole[0 < t < 1] Exp[-I t] A + Boole[1 < t < 2] Exp[+I t] A

F[0.1]

H[t_] := Inverse[F[t]].H0.F[t] + H0.H0

H[0.4]

Out[4]= {{0.995004- 0.0998334 I,
   0.0995004- 0.00998334 I}, {0.0995004- 0.00998334 I, -0.995004 +
    0.0998334 I}}

Out[6]= {{(0.19802- 6.93889*10^-18 I) b +
    b^2, (-0.980198 - 4.33681*10^-19 I) b}, {(-0.980198 -
      4.33681*10^-19 I) b, (-0.19802 + 6.93889*10^-18 I) b + b^2}}

In[7]:= Integrate[H[x], {x, 0, 1}]

Out[7]= {{0.00990099 (20. b + 101. b^2), -0.980198 b}, {-0.980198 b,
   0.00990099 (-20. b + 101. b^2)}}

In[8]:= % // Dimensions

Out[8]= {2, 2}

In[9]:= Integrate[H[x][[1, 1]], {x, 0, 1}]

Out[9]= 0.00990099 (20. b + 101. b^2)

In[10]:= % // Dimensions

Out[10]= {2}


Regards,
-- Jean-Marc


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