DifferentialEquationsLecture2

Add lecture notes from the second lecture on differential equations.

src/differential_equations_2.md

@@ -10,3 +10,142 @@ For this purpose we make the following definitions,
 
 $$y' = \frac{dy}{dx}, \ y'' = \frac{d^2 y}{dx^2}, \ \cdots, \ y^{(n)} = \frac{d^n y}{dx^n}.$$
 
+In the new notation, a linear $n$-th order differential equation with constant
+coefficients reads 
+
+$$y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0. $$
+
+!!! info "Linear combination of solutions are still solutions"
+
+    Note that as was the case for first order linear DE's, the propery of 
+    linearity once again means that if $y_{1}(x)$ and $y_{2}(x)$ are both 
+    solutions, and $a$ and $b$ are constants, 
+    
+    $$a y_{1}(x) + b y_{2}(x)$$
+    
+    then linear combination of the solutions is also a solution.
+    
+In order to solve a higher order linear DE we will present a trick that makes it
+possible to map the problem of solving a single $n$-th order linear DE into a
+related problem of solving a system of $n$ first order linear DE's. 
+
+To begin, define:
+
+$$y_{1} = y, \ y_{2} = y', \ \cdots, \ y_{n} = y^{(n-1)}.$$
+
+Then, the differential equation can be re-written as
+
+$$y_1 ' = y_2$$
+$$y_2 ' = y_3$$
+$$ \vdots $$
+$$y_{n-1} ' = y_{n}$$
+$$y_{n} ' = - a_{0} y_{1} - a_{1} y_{2} - \cdots - a_{n-1} y_{n}.$$
+
+Notice that together these $n$ equations form a linear first order system, the 
+first $n-1$ equations of which are trivial. Note that this trick can be used to 
+reduce any system of $n$-th order linear DE's to a larger system of first order 
+linear DE's. 
+
+Since we have discussed already the method of solution for first order linear 
+systems, we will outline the general solution to this system. As before, the 
+general solution will be the linear combination of $n$ linearly independent 
+solutions $f_{i}(x)$, $i \epsilon \{1, \cdots, n \}$, which make up a basis for 
+the solution space. That is the general solution has the form
+
+$$y(x) = c_1 f_1 (x) + c_2 f_2 (x) + \cdots + c_n f_{n}(x). $$
+
+To check that the $n$ solutions form a basis, it is sufficient to verify
+
+$$ det \begin{bmatrix} 
+f_1(x) & \hdots & f_{n}(x) \\
+f_1 ' (x) & \hdots & f_{n}'(x) \\
+\vdots & \vdots & \vdots \\
+f^{(n-1)}_{1} (x) & \hdots & f^{(n-1)}_{n} (x) \\
+\end{bmatrix}  \neq 0.$$
+
+The determinant in the preceding line is called the *Wronski determinant*. In 
+particular, to determine solutions, we need to find the eigenvalues of 
+
+$$**A** = \begin{bmatrix} 
+0 & 1 & 0 & \hdots & 0 \\
+0 & 0 & 1 & \hdots & 0 \\
+\vdots & \vdots & \vdots & \hdots & \vdots \\
+-a_0 & -a_1 & -a_2 & \hdots & -a_{n-1} \\
+\end{bmatrix}.$$
+
+It is possible to show that 
+
+$$det(**A** - \lambda \mathbbm{1}) = -P(\lambda),$$
+
+in which $P(\lambda)$ is the characteristic polynomial of the system matrix $**A**$,
+
+$$P(\lambda) = \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_0.$$
+
+As we demonstrate below, the proof relies on the co-factor expansion technique 
+for calculating a determinant. 
+
+$$- det(**A** - \lambda \mathbbm{1}) = \begin{bmatrix} 
+\lambda & -1 & 0 & \hdots & 0 \\
+0 & \lambda & -1 & \hdots & 0 \\
+\vdots & \vdots & \vdots & \hdots & \vdots \\
+a_0 & a_1 & a_2 & \hdots & a_{n-1} + \lambda \\
+\end{bmatrix} $$
+$$- det(**A** - \lambda \mathbbm{1}) =  \lambda det \begin{bmatrix}
+\lambda & -1 & 0 & \hdots & 0 \\
+0 & \lambda & -1 & \hdots & 0 \\
+\vdots & \vdots & \vdots & \hdots & \vdots \\
+a_1 & a_2 & a_3 & \hdots & a_{n-1} + \lambda \\
+\end{bmatrix} + (-1)^{n+1}a_0 det \begin{bmatrix} 
+-1 & 0 & 0 & \hdots & 0 \\
+\lambda & -1 & 0 & hdots & 0 \\
+\vdots & \vdots & \vdots & \hdots & \vdots \\
+0 & 0 & \hdots & \lambda & -1 \\
+\end{bmatrix}$$
+$$- det(**A** - \lambda \mathbbm{1}) = \lambda det \begin{bmatrix}
+\lambda & -1 & 0 & \hdots & 0 \\
+0 & \lambda & -1 & \hdots & 0 \\
+\vdots & \vdots & \vdots & \hdots & \vdots \\
+a_1 & a_2 & a_3 & \hdots & a_{n-1} + \lambda \\
+\end{bmatrix} + (-1)^{n+1} a_0 (-1)^{n-1}$$
+$$- det(**A** - \lambda \mathbbm{1}) = \lambda det \begin{bmatrix}
+\lambda & -1 & 0 & \hdots & 0 \\
+0 & \lambda & -1 & \hdots & 0 \\
+\vdots & \vdots & \vdots & \hdots & \vdots \\
+a_1 & a_2 & a_3 & \hdots & a_{n-1} + \lambda \\
+\end{bmatrix} + a_0$$
+$$- det(**A** - \lambda \mathbbm{1}) = \lambda (\lambda (\lambda \cdots + a_2) + a_1) + a_0$$
+$$- det(**A** - \lambda \mathbbm{1}) = P(\lambda).$$
+
+In the second last line of the proof we indicated that the method of co-factor 
+expansion demonstrated is repeated an additional $n-2$ times. This completes the
+proof. 
+
+With the characteristic polynomial, it is possible to write the differential 
+equation as 
+
+$$P(\frac{d}{dx})y(x) = 0.$$
+
+To determine solutions, we need to find $\lambda_i$ such that $P(\lambda_i) = 0$. 
+By the fundamental theorem of algebra, we know that $P(\lambda)$ can be written 
+as
+
+$$P(\lambda) = \overset{l}{\underset{k=1}{\Sigma}} (\lambda - \lambda_k)^{m_k}.$$
+
+In the previous equation $\lambda_k$ are the k roots of the equations, and $m_k$
+is the multiplicity of each root. Note that the multiplicities satisfy 
+$\overset{l}{\underset{k=1}{\Sigma}} m_k = n$. 
+
+If the multiplicity of each eigenvalue is one, then solutions which form the 
+basis are then given as:
+
+$$f(x) = e^{\lambda_1 x}, \ e^{\lambda_2 x}, \ \cdots, \ e^{\lambda_n x}.$$
+
+If there are eigenvalues with multiplicity greater than one, the the solutions
+which form the basis are given as 
+
+$$f(x) = e^{\lambda_1 x}, \ x e^{\lambda_1 x} , \ \cdots, \ x^{m_{1}-1} e^{\lambda_1 x}, \ etc.$$
+
+
+
+
+    
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