Example 2

In[65]:=

  a={{0,1},{-36,-12}};
  u0={-2,-3};
  MatrixForm[a]

Out[65]=

  0     1
  
  -36   -12

You should recognize this system as the system from the Using Jordan Canonical Form subsection of the Matrix Exponential section. Let's recall some of that information.

A is not diagonalizable, so we must settle for the Jordan Canonical Form of A, namely A=P.J.Inv[P].

In[66]:=

  j={{-6,1},{0,-6}};
  MatrixForm[j]

Out[66]=

  -6   1
  
  0    -6

In[67]:=

  d={{-6,0},{0,-6}};
  z={{0,1},{0,0}};
  MatrixForm[d]
  MatrixForm[z]

Out[67]=

  -6   0
  
  0    -6

Out[68]=

  0   1
  
  0   0

In[69]:=

  p={{-1,1},{6,-7}};
  MatrixForm[a]
  MatrixForm[p.j.Inverse[p]]

Out[69]=

  0     1
  
  -36   -12

Out[70]=

  0     1
  
  -36   -12

The solution u is going to be Exp[t A].u0 or P.Exp[t J].Inverse[P].{-2,-3}

In[71]:=

  exptj=MatrixExp[t j];
  MatrixForm[exptj]

Out[71]=

           t
          ----
   -6 t    6 t
  E       E
  
  
  
           -6 t
  0       E

In[72]:=

  usol=p.exptj.Inverse[p].{-2,-3};
  MatrixForm[usol]

Out[72]=

  -3 t       7         -6 t    t
  ---- - 2 (---- - 6 (E     - ----))
   6 t       6 t               6 t
  E         E                 E
  
       -6 t   6 t        -42        -7    6 t
  -3 (E     - ----) - 2 (---- - 6 (---- + ----))
               6 t        6 t       6 t    6 t
              E          E         E      E

In[73]:=

  Simplify[%];
  MatrixForm[%]

Out[73]=

    2 + 15 t
  -(--------)
       6 t
      E
  
  3 (-1 + 30 t)
  -------------
       6 t
      E

Here u1=u (the solution to the original second order differential equation). If you recall from your differential equations class, you can solve u''+12u'+36u=0 by substituting u=Exp[mt] and solving the resulting equation (the characteristic equation) for m. In this case the characteristic equation is m^2+12m+36=0 which yields m=-6 (only one root). If you recall, in this case you had to "add" a solution of the form tExp[mt]. You can see this is taken care of above.

Where does this tExp[mt] come from?
The Exp[mt] comes from Exp[t D], the tExp[mt] comes from Exp[t Z], where D+Z=J

In[74]:=

  MatrixForm[d]
  MatrixForm[z]

Out[74]=

  -6   0
  
  0    -6

Out[75]=

  0   1
  
  0   0

In[76]:=

  MatrixForm[MatrixExp[t d]]
  MatrixForm[MatrixExp[t z]]

Out[76]=

   -6 t
  E       0
  
           -6 t
  0       E

Out[77]=

  1   t
  
  0   1

You can also see the tExp[mt] come into play from the matrix exponential of the Jordan block.

In[78]:=

  MatrixForm[MatrixExp[t j]]

Out[78]=

           t
          ----
   -6 t    6 t
  E       E
  
  
  
           -6 t
  0       E

So you would have u=c1Exp[mt]+c2tExp[mt], coming from the first row of Exp[t J].

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