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src/differential_equations_1.md

@@ -465,14 +465,147 @@ solution are:
    $\dot{**x**}(t) = e^{**A** t} {**x**}_{0}$, with initial condition 
     $**x**(0) = e^{**A** 0}{**x**}_0 = \mathbbm{1} {**x**}_{0} = {**x**}_{0}$.
 
-Next we will discuss how to determine a solution in general, going beyond the 
+Next we will discuss how to determine a solution in practice, beyond the 
 formal solution just presented. 
 
-    
+### Case 1: **A** diagonalizable ###
+
+For an $n \times n$ matrix $**A**$, denote the $n$ distinct eigenvectors as 
+$\{**v_1**, \cdots, **v**_n \}$. By definition, the eigenvectors satisfy the 
+equation
+
+$$**A** **v**_i = \lambda_i **v**_i, \forall i \epsilon \{1, \cdots, n \}. $$
+
+Here we give consideration to the case of distinct eigenvectors, in which case 
+the $n$ eigenvectors form a basis for $\mathbb{R}^{n}$. 
+
+To solve the equation $\dot{**x**}(t) = **A** **x**(t)$, define a set of scalar 
+functions $\{u_{1}(t), \cdots u_{n}(t) \}$ and make the following ansatz:
+
+$$**\phi**_{i} = u_{i}(t) **v_{i}**.$$
+
+Then, by differentiating,
+
+$$\dot{**{\phi}**_i}(t) = \dot{u_i}(t) **v**_{i}(t).$$
+
+The above equation can be combined with the differential equation for 
+$**\phi**_{i}(t)$, $\dot{**\phi**_{i}}(t)=**A** **\phi**_{i}(t)$, to derive the 
+following equations,
+
+$$\dot{u_i}(t) **v**_{i}(t) = **A** u_{i}(t) **v_{i}**$$
+$$\dot{u_i}(t) **v**_{i}(t) = u_{i}(t) \lambda_{i} **v_{i}** $$
+$$**v**_{i} (\dot{u_i}(t) - \lambda_i u_{i}(t)) = 0, $$
+
+where in the second last line we mase use of the fact that $**v**_i$ is an eigenvector
+of $**A**$. The obtained relation implies that 
+
+$$\dot{u_i}(t) = \lambda_i u_{i}(t).$$
+
+This is a simple differential equation, of the type dealt with in the third 
+example. The solution is found to be
+
+$$u_{i}(t) = c_i e^{\lambda_i t},$$
+
+with $c_i$ some constant. The general solution is found by adding all $n$ of the
+solutions $**\phi_{i}**(t)$,
+
+$$**x**(t) = c_{1} e^{\lambda_1 t} **v**_{1} + c_{2} e^{\lambda_2 t} **v**_{2} + \cdots + c_{n} e^{\lambda_n t} **v**_{n}.$$
+
+and the vectors $\{e^{\lambda_1 t} **v**_{1}, \cdots, e^{\lambda_n t} **v**_{n} \}$
+form a basis for the solution space since $det(**v**_1 | \cdots | **v_n**) \neq 0$
+(the $n$ eigenvectors are linearly independent). 
+
+!!! check "Example: Homogeneous first order linear system with diagonalizable 
+    constant coefficient matrix"
+
+    Define the matrix $**A** = \begin{bmatrix}
+    0 & -1 \\
+    1 & 0 \\
+    \end{bmatrix}$, and consider the DE 
+
+    $$\dot{**x**}(t) = **A** **x**(t), **x_0** = \begin{bmatrix}
+    1 \\
+    0 \\
+    \end{bmatrix}. $$
+ 
+    To proceed following the solution technique, we determine the eigenvalues of 
+    $**A**$, 
+ 
+    $$det {\begin{bmatrix} 
+    -\lambda & -1 \\
+    1 & - \lambda \\
+    \end{bmatrix}} = \lambda^2 + 1 = 0. $$
+    
+    By solving the characteristic polynomial, one finds the two eigenvalues 
+    $\lambda_{\pm} = \pm i$. 
+    
+    Focusing first on the positive eigenvalue, we can determine the first 
+    eigenvector,
+    
+    $$\begin{bmatrix} 
+    0 & -1 \\
+    1 & 0 \\
+    \end{bmatrix} \begin{bmatrix}
+    a \\
+    b\\
+    \end{bmatrix} = i \begin{bmatrix}
+    a \\
+    b \\
+    \end{bmatrix}.$$
+
+    A solution to this eigenvector equation is given by $a=1$, $b=-i$, altogether
+    implying that $\lambda_1=i, **v**_{1} = \begin{bmatrix} 
+    1 \\
+    -i \\
+    \end{bmatrix}$. 
+    
+    As for the second eigenvalue, $\lambda_{2} = -i$, we can solve the analagous
+    eigenvector equation to determine $**v**_{2} = \begin{bmatrix}
+    1 \\
+    i \\
+    \end{bmatrix}$
+    
+    Hence two independent solutions of the differential equation are:
+    
+    $$**\phi**_{1} = e^{i t}\begin{bmatrix}
+    1 \\
+    -i \\
+    \end{bmatrix}, **\phi**_{2}  = e^{-i t} \begin{bmatrix}
+    1 \\
+    i \\
+    \end{bmatrix}.$$
+ 
+    To obtain the general solutoin of the equation, the only thing which is 
+    missing is determination of linear combination coefficients for the two 
+    solutions which allow for satisfaction of the initial condition. To this end,
+    we must solve 
+    
+    $$c_1 **\phi**_{1}(t) + c_2 **\phi**_{2}(t) = \begin{bmatrix}
+    1 \\
+    0 \\
+    \end{bmatrix}$$
+    $$\begin{bmatrix}
+    c_1 + c_2 \\
+    -i c_1 + i c_2 \\
+    \end{bmatrix} = \begin{bmatrix} 
+    1 \\
+    0 \\
+    \end{bmatrix}.$$
+    
+    The second row of the vector equation for $c_1, c_2$ implies that $c_1=c_2$. 
+    The first row then implies that $c_1=c_2=\frac{1}{2}$. 
+    
+    Overall then, the general solution of the DE can be summarized
+    
+    $$\dot{**x**}(t) = \begin{bmatrix}
+    \frac{1}{2}(e^{i t} + e^{-i t}) \\
+    \frac{1}{2 i}(e^{i t} - e^{-i t}) \\
+    \end{bmatrix} = \begin{bmatrix}
+    cos(t) \\
+    sin(t) \\
+    \end{bmatrix}. $$
+
 
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