Add up to page 4.5

src/differential_equations_2.md

@@ -145,7 +145,124 @@ which form the basis are given as
 
 $$f(x) = e^{\lambda_1 x}, \ x e^{\lambda_1 x} , \ \cdots, \ x^{m_{1}-1} e^{\lambda_1 x}, \ etc.$$
 
+---ADD PROOF HERE---
 
+!!! check "Example: Second order homogeneous linear DE with constant coefficients"
 
+    Consider the equation 
+    
+    $$y'' + Ey = 0.$$ 
+    
+    Let us reduce this second order equation to a system of two first order 
+    equations. Define
+    
+    $$y_1=y$$
+    $$y_2=y'.$$
+    
+    Writing $**y**= \begin{bmatrix} 
+    y_1 \\
+    y_2 \\ 
+    \end{bmatrix}$, the DE can be written,
+    
+    $$\dot{**y**} = \begin{bmatrix} 
+    y_2 \\
+    -E y_1 \\
+    \end{bmatrix}$$
+    $$\dot{**y**} = \begin{bmatrix} 
+    0 & 1 \\
+    -E & 0 \\
+    \end{bmatrix} \begin{bmatrix} 
+    y_1 \\
+    y_2 \\
+    \end{bmatrix}.$$
+    
+    The characteristic polynomial of this equation is 
+    
+    $$P(\lambda) = \lambda^2 + E.$$
+    
+    There are three cases for the possible solutions, depending upon the value 
+    of E.
+    
+    **Case 1: $E>0$**
+    For ease of notation, define $E=k^2$ for some constant $k$. The 
+    characteristic polynomial can then be factored as
+    
+    $$P(\lambda) = (\lambda+ i k)(\lambda - i k). $$
+    
+    Following our formulation for the solution, the two basis functions for the 
+    solution space are 
+    
+    $$f_1(x) = e^{i k x}, \ f_2=e^{- i k x}.$$
+    
+    Alternatively, the trigonometric functions can serve as basis functions, 
+    since they are linear combinations of $f_1$ and $f_2$ which remain linearly
+    independent,
+    
+    $$\Tilde{f_1}(x)=cos(kx), \ \Tilde{f_2}(x)=sin{kx}.$$
+    
+    **Case 2: $E<0$**
+    This time, define $E=-k^2$, for constant $k$. The characteristic polynomial 
+    can then be factored as 
+    
+    $$P(\lambda) = (\lambda+ k)(\lambda -  k).$$
+
+    The two basis functions for this solution are then 
+    
+    $$f_1(x)=e^{k x}, \ f_2(x) = e^{-k x}.$$
+    
+    **Case 3: $E=0$**
+    In this case, there is a repeated eigenvalue (equal to $0$), since the 
+    characteristic polynomial reads
+    
+    $$P(\lambda) = (\lambda-0)^2.$$
+    
+    Hence the basis functions for the solution space read 
+    
+    $$f_1(x)=e^{0 x} = 1, \ f_{2}(x) = x e^{0 x} = x. $$
+
+# Partial differential equations
+
+A partial differential equation (PDE) is an equation involving a function of two or 
+more indepenedent variables and derivatives of said function. These equations
+are classified similarly to ordinary differential equations (the subject of
+our earlier study). For example, they are called linear if no terms such as
+
+$$\frac{\partial y(x,t)}{\partial x} \cdot \frac{d y(x,t)}{\partial t} \ or $$
+$$\frac{\partial^2 y(x,t)}{\partial x^2} y(x,t)$$ 
+
+occur. A PDE can be classified as $n$-th order accorind to the highest 
+derivative order of either variable occuring in the equation. For example, the 
+equation
+
+$$\frac{\partial^3 f(x,y)}{\partial x^3} + \frac{\partial f(x,t)}{\partial t} = 5$$
+
+is a $3$-rd order equation because of the third derivative with respect to x
+in the equation.
+
+To begin, we demonstrate that PDE's are of fundamental importance in physics, 
+especially in quantum physics. In particular, the Schr\"{o}dinger equation, 
+which is of central importance in quantum physics is a partial differential 
+equation with respect to time and space. This equation is very important 
+because it describes the evolution in time and space of the entire description
+of a quantum system $\psi(x,t)$, which is known as the wavefunction. 
+
+For a free particle in one dimension, the Schr\"{o}dinger equation is 
+
+$$i \hbar \frac{\partial \psi(x,t)}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi(x,t)}{\partial x^2}. $$
+
+When we studied ODEs, an initial condition was necessary in order to fully 
+specify a solution. Similarly, in the study of PDEs an initial condition is 
+required but now boundary conditions are also required. Going back to the 
+intuitive discussion from the lecture on ODEs, each of these conditions is 
+necassary in order to specify an integration constant that occurs in solving 
+the equation. In partial differential equations at least one such constant will
+arise from the time derivative and likewise at least one from the spatial 
+derivative. 
+
+For the Schr\"{o}dinger equation, we could supply the initial conditions
+
+$$\psi(x,0)= \psi_{0}(x) \ & \ \psi(0,t) = \psi{t, L} = 0.$$
 
+This particular set of boundary conditions corresponds to a particle in a box,
+a basic situation which comes up often in quantum physics. 
     
\ No newline at end of file