Merge branch 'DE2' into 'master'

DifferentialEquationsLecture2

See merge request !9

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@@ -13,6 +13,7 @@ nav:
   - Operators in quantum mechanics: '5_operators_QM.md'
   - Eigenvectors and eigenvalues: '6_eigenvectors_QM.md'
   - Differential equations 1: '7_differential_equations_1.md'
+  - Differential equations 2: '8_differential_equations_2.md'
 
 theme:
   name: material

src/8_differential_equations_2.md

+---
+title: Differential Equations 2
+---
+
+# Differential equations 2
+
+The lecture on differential equations consists of three parts, each with their own video:
+
+- [Higher order linear differential equations](#higher-order-linear-differential-equations)
+- [Partial differential equations: Separation of variables](#partial-differential-equations-separation-of-variables)
+- [Self-adjoint differential operators](#self-adjoint-differential-operators)
+
+**Total video length:  hour  minutes  seconds**
+
+## Higher order linear differential equations
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/ucvIiLgJ2i0?rel=0" frameborder="0" allow="accelerometer; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+In the previous lecture, we focused on first order linear differential equations
+as well as systems of such equations. In this lecture we switch focus to DE's 
+which involve higher derivatives of the function we would like to solve for. To
+f`%acilitate this change we are going to change notation. In the previous lecture
+we wrote differential equations for $x(t)$. In this lecture we will write DE's 
+of $y(x)$, where $y$ is an unknown function and $x$ is the independent variable. 
+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
+
+$$ det \begin{bmatrix} 
+f_1(x) & \cdots & f_{n}(x) \\
+f_1 ' (x) & \cdots & f_{n}'(x) \\
+\vdots & \vdots & \vdots \\
+f^{(n-1)}_{1} (x) & \cdots & f^{(n-1)}_{n} (x) \\
+\end{bmatrix}  \neq 0.$$
+
+The determinant in the preceding line is called the *Wronski determinant*. In 
+particular, to determine solutions, we need to find the eigenvalues of 
+
+$$A = \begin{bmatrix} 
+0 & 1 & 0 & \cdots & 0 \\
+0 & 0 & 1 & \cdots & 0 \\
+\vdots & \vdots & \vdots & \cdots & \vdots \\
+-a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \\
+\end{bmatrix}.$$
+
+It is possible to show that 
+
+$$det(A - \lambda I) = -P(\lambda),$$
+
+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. 
+
+$$- det(A - \lambda I) = \begin{bmatrix} 
+\lambda & -1 & 0 & \cdots & 0 \\
+0 & \lambda & -1 & \cdots & 0 \\
+\vdots & \vdots & \vdots & \cdots & \vdots \\
+a_0 & a_1 & a_2 & \cdots & a_{n-1} + \lambda \\
+\end{bmatrix} $$
+$$- det(A - \lambda I) =  \lambda det \begin{bmatrix}
+\lambda & -1 & 0 & \cdots & 0 \\
+0 & \lambda & -1 & \cdots & 0 \\
+\vdots & \vdots & \vdots & \cdots & \vdots \\
+a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
+\end{bmatrix} + (-1)^{n+1}a_0 det \begin{bmatrix} 
+-1 & 0 & 0 & \cdots & 0 \\
+\lambda & -1 & 0 & cdots & 0 \\
+\vdots & \vdots & \vdots & \cdots & \vdots \\
+0 & 0 & \cdots & \lambda & -1 \\
+\end{bmatrix}$$
+$$- det(A - \lambda I) = \lambda det \begin{bmatrix}
+\lambda & -1 & 0 & \cdots & 0 \\
+0 & \lambda & -1 & \cdots & 0 \\
+\vdots & \vdots & \vdots & \cdots & \vdots \\
+a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
+\end{bmatrix} + (-1)^{n+1} a_0 (-1)^{n-1}$$
+$$- det(A - \lambda I) = \lambda det \begin{bmatrix}
+\lambda & -1 & 0 & \cdots & 0 \\
+0 & \lambda & -1 & \cdots & 0 \\
+\vdots & \vdots & \vdots & \cdots & \vdots \\
+a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
+\end{bmatrix} + a_0$$
+$$- det(A - \lambda I) = \lambda (\lambda (\lambda \cdots + a_2) + a_1) + a_0$$
+$$- det(A - \lambda I) = P(\lambda).$$
+
+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 
+
+$$f(x) = e^{\lambda_1 x}, \ x e^{\lambda_1 x} , \ \cdots, \ x^{m_{1}-1} e^{\lambda_1 x}, \ etc.$$
+
+---ADD PROOF HERE---
+
+!!! check "Example: Second order homogeneous linear DE with constant coefficients"
+
+    Consider the equation 
+    
+    $$y'' + Ey = 0.$$ 
+    
+    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: Separation of variables
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/I4ghpYsFLFY?rel=0" frameborder="0" allow="accelerometer; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+### Definitions and examples
+
+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 Schrödinger 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 Schrödinger 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 Schrödinger 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 situation which is used as the base model for many derivations in quantum 
+physics. 
+
+Another example of a partial differential equation common in physics is the 
+Laplace equation
+
+$$\frac{\partial^2 \phi(x,y)}{\partial x^2}+\frac{\partial^2 \phi(x,y)}{\partial y^2}=0.$$
+
+In quantum physics Laplace's equation is important for the study of the hydrogen
+atom. In three dimensions and using spherical coordinates, the solutions to 
+Laplace's equation are special functions called spherical harmonics. In the 
+context of the hydrogen atom, these functions describe the wave function of the 
+system and a unique spherical harmonic function corresponds to each distinct set
+of quantum numbers.
+
+In the study of PDEs there is not a comprehensive overall treatment to the same 
+extent as there is for ODEs. There are several techniques which can be applied 
+to solving these equations, but the choice of technique must be tailored to the
+equation at hand. Hence we focus on some specific examples that are common in
+physics.
+
+### Separation of variables
+
+Let us focus on the one dimensional Schrödinger equation of a free particle
+
+$$i \hbar \frac{\partial \psi(x,t)}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi(x,t)}{\partial x^2}. $$
+
+To attempt a solution, we will make a *separation ansatz*,
+
+$$\psi(x,t)=\phi(x) f(t).$$
+
+!!! info "Separation ansatz"
+    The separation ansatz is a restrictive ansatz, not a fully general one. In
+    general, for such a treatment to be valid an equation and the boundary 
+    conditions given with it have to fulfill certain properties. In this course
+    however you will only be asked to use this technique when it is suitable.
+    
+Substituting the separation ansatz into the PDE,
+
+$$i \hbar \frac{\partial \phi(x)f(t)}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \phi(x)f(t)}{\partial x^2} $$
+$$i \hbar \dot{f}(t) \phi(x) = - \frac{\hbar^2}{2m} \phi''(x)f(t). $$
+
+Notice that in the above equation the derivatives on $f$ and $\phi$ can each be
+written as ordinary derivatives, $\dot{f}=\frac{df(t)}{dt}$, 
+$\phi''(x)=\frac{d^2 \phi}{dx^2}$. This is so because each is only a function of 
+one variable. 
+
+Next, divide both sides of the equation through by $\psi(x,t)=\phi(x) f(t)$,
+
+$$i \hbar \frac{\dot{f}(t)}{f(t)} = - \frac{\hbar^2}{2m} \frac{\phi''(x)}{\phi(x)} = constant := \lambda. $$
+
+In the previous line we concluded that each part of the equation must be equal 
+to a constant, which we defined as $\lambda$. This follows because the left hand
+side of the equation only has a dependence on the spatial coordinate $x$, whereas 
+the right hand side only has dependence on the time coordinate $t$. If we have 
+two functions $a(x)$ and $b(t)$ such that 
+$a(x)=b(t) \ \forall x, \ t \ \epsilon \mathbb{R}$, then $a(x)=b(t)=const$.
+
+The constant we defined, $lambda$, is called a *separation constant*. With it, we 
+can break the spatial and time dependent parts of the equation into two separate
+equations,
+
+$$i \hbar \dot{f}(t) = \lambda f(t)$$
+
+$$-\frac{\hbar^2}{2m} \phi''(x) = \lambda \phi(x) .$$
+
+To summarize, this process has broken one partial differential equation into two
+ordinary differential equations of different variables. In order to do this, we 
+needed to introduce a separation constant, which remains to be determined.
+
+### Boundary and eigenvalue problems
+
+Continuing on with the Schrödinger equation example from the previous 
+section, let us focus on 
+
+$$-\frac{\hbar^2}{2m} \phi''(x) = \lambda \phi(x),$$
+$$\phi(0)=\phi(L)=0.$$
+
+This has the form of an eigenvalue equation, in which $\lambda$ is the 
+eigenvalue, $- \frac{\hbar^2}{2m} \frac{d^2}{dx^2}[\cdot]$ is the linear 
+operator and $\phi(x)$ is the eigenfunction. 
+
+Notice that when stating the ordinary differential equation, it is specified 
+along with it's boundary conditions. Note that in contrast to an initial value
+problem, a boundary value problem does not always have a solution. For example, 
+in the figure below, regardless of the initial slope, the curves never reach $0$
+when $x=L$. 
+
+![image](figures/DE2_1.png)
+
+For boundary value problems like this, there are only solutions for particular 
+eigenvalues $\lambda$. Coming back to the example, it turns out that solutions
+only exist for $\lambda>0$ --this can be shown quickly, feel free to try it! 
+Define for simplicity $k^2:= \frac{2m \lambda}{\hbar^2}$. The equation then 
+reads
+
+$$\phi''(x)+k^2 \phi(x)=0.$$
+
+Two linearly independent solutions to this equation are 
+
+$$\phi_{1}(x)=sin(k x), \ \phi_{2}(x) = cos(k x).$$
+
+The solution to this homogeneous equation is then 
+
+$$\phi(x)=c_1 \phi_1(x)+c_2 \phi_2(x).$$
+
+The eigenvalue, $\lambda$ as well as one of the constant coefficients can be 
+determined using the boundary conditions. 
+
+$$\phi(0)=0 \ \Rightarrow \ \phi(x)=c_1 sin(k x), \ c_2=0.$$
+
+$$\phi(L)=0 \ \Rightarrow \ 0=c_1 sin(k L) .$$
+
+In turn, using the properties of the $sin(\cdot)$ function, it is now possible
+to find the allowed values of $k$ and hence also $\lambda$. The previous 
+equation implies, 
+
+$$k L = n \pi, \ n \ \epsilon \ \mathbb{N}$$
+
+$$\lambda_n = \big{(}\frac{n \pi \hbar}{L} \big{)}^2.$$
+
+The values $\lambda_n$ are the eigenvalues. Now that we have determined 
+$\lambda$, it enters into the time equation, $i \hbar \dot{f}(t) = \lambda f(t)$
+only as a constant. We can hence simply solve,
+
+$$\dot{f}(t) = -i \frac{\lambda}{\hbar} f(t)$$
+
+$$f(t) = A e^{\frac{-i \lambda t}{\hbar}}.$$
+
+In the previous equation, the coefficient $A$ can be determined if the original
+PDE was supplied with an initial condition. 
+
+Putting the solutions to the two ODEs together and redefining 
+$\tilde{A}=A \cdot c_1$, we arrive at the solutions for theb PDE,
+
+$\psi_n(x,t) = \tilde{A}_n e^{-i \frac{\lambda_n t}{\hbar}} sin(\frac{n \pi x}{L}).$
+
+Notice that there is one solution $\psi_{n}(x,t)$ for each natural number $n$. 
+These are still very special solutions. We will begin discussing next how to 
+obtain the general solution in our example. 
+
+
+## Self-adjoint differential operators
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/p4MHW0yMMvY?rel=0" frameborder="0" allow="accelerometer; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+
+As we hinted was possible earlier, let us re-write the previous equation by 
+defining a linear operator, $L$, acting on the space of functions which satisfy
+$\phi(0)=\phi(L)=0$:
+
+$$L[\cdot]:= \frac{- \hbar^2}{2m} \frac{d^2}{dx^2}[\cdot]. $$
+
+Then, the ODE can be writted as 
+
+$$L[\phi]=\lambda \phi.$$
+
+This equation looks exactly like, and turns out to be, an eigenvalue equation!
+
+!!! info "Connecting function spaces to Hilbert spaces"
+    
+    Recall that a space of functions can be transformed into a Hilbert space by 
+    equipping it with a inner product,
+    
+    $$\langle f, g \rangle = \int^{L}_{0} dx f*(x) g(x) $$
+    
+    Use of this inner product also has utility in demonstrating that particular 
+    operators are *Hermitian*. The term hermitian is precisely defined below.
+    Of considerable interest is that hermition operators have a set of nice 
+    properties including all real eigenvalues and orthonormal eigenfunctions. 
+    
+The nicest type of operators for many practical purposes are hermitian 
+operators. In quantum physics for example, any physical operator must be 
+hermitian. Denote a hilbert space $\mathcal{H}$. An opertor 
+$H: \mathcal{H} \mapsto \mathcal{H}$ is said to be hermitian if it satisfies
+
+$$\langle f, H g \rangle = \langle H f, g \rangle \ \forall \ f, \ g \ \epsilon \ \mathcal{H}.$$
+
+Now, we would like to investigate whether the operator we have been working with,
+$L$ satisfies the criterion of being hermitian over the function space 
+$\phi(0)=\phi(L)=0$ equipped with the above defined inner product (i.e. it is a
+Hilbert space). Denote this Hilbert space $\mathcal{H}_{0}$ and consider let 
+$f, \ g \ \epsilon \ \mathcal{H}_0$ denote two functions from the Hilbert space.
+Then, we can investigate
+
+$$\langle f, L g \rangle = \frac{- \hbar^2}{2m} \int^{L}_{0} dx f*(x) \frac{d^2}{dx^2}g(x).$$
+
+As a first step, it is possible to do integration by parts in the integral,
+
+$$\langle f, L g \rangle = \frac{+ \hbar^2}{2m} ( \int^{L}_{0} dx \frac{d f*}{dx} \frac{d g}{dx} - [f*(x)\frac{d g}{dx}] \big{|}^{L}_{0} )$$
+
+The boundary term vansishes, due to the boundary conditions $f(0)=f(L)=0$, 
+which directly imply $f*(0)=f*(L)=0$. Now, intergrate by parts a second time
+
+$$\langle f, L g \rangle = \frac{- \hbar^2}{2m} (\int^{L}_{0} dx \frac{d^2 f*}{dx^2} g(x) - [\frac{d f*}{dx} g(x)] \big{|}^{L}_{0} ).$$
+
+As before, the boundary term vanishes, due to the boundary conditions 
+$g(0)=g(L)=0$. Upon cancelling the boundary term however, the expression on 
+the right hand side, contained in the integral is simply 
+$\langle L f, g \rangle$. Therefore,
+
+$$\langle f, L g \rangle=\langle L f, g \rangle. $$
+
+We have demonstrated that $L$ is a hermitian operator on the space 
+$\mathcal{H}_0$. As a hermitian operator, $L$ has the property that it's 
+eigenfunctions form an orthonormal basis for the space $\mathcal{H}_0$. Hence it
+is possible to expand any function $f \ \epsilon \ \mathcal{H}_0$ in terms of
+the eigenfunctions of $L$. 
+
+!!! info "Connection to quantum states"
+    
+    Recall that q quantum state $\ket{\phi}$ can be written in an orthonormal 
+    basis $\{ \ket{u_n} \}$ as 
+    $$\ket{\phi} = \underset{n}{\Sigma} \bra{u_n} \ket{\phi} \ket{u_n}.$$ 
+    
+    In terms of hermitian operators and their eigenfunctions, the eigenfunctions
+    play the role of the orthonormal basis. In reference to our running example,
+    the 1D Schrödinger equation of a free particle, the eigenfunctions 
+    $sin(\frac{n \pi x}{L})$ play the role of the basis functions $\ket{u_n}$.
+    
+To close our running example, consider the initial condition 
+$\psi(x,o) = \psi_{0}(x)$. Since the eigenfunctions $sin(\frac{n \pi x}{L})$ 
+form a basis, we can now write the general solution to the problem as 
+
+$$\psi(x,t)  = \overset{\infinity}{\underset{n}{\Sigma}} c_n e^{-i \frac{\lambda_n t}{\hbar}} sin(\frac{n \pi x}{L}),$$
+
+where in the above we have defined the coefficients using the Fourier 
+coefficient,
+
+$$c_n:= \int^{L}_{0} dx sin(\frac{n \pi x}{L}) \psi_{0}(x). $$
+    
+### General recipie for seperable PDEs
+
+1. Make the separation ansatz to obtain separate ordinary differential 
+    equations.
+2. Choose which euation to treat as the eigenvalue equation. This will depend 
+    upon the boundary conditions. Additionally, verify that the linear 
+    differential operator $L$ in the eigenvalue equation is hermitian. 
+3.Solve the eigenvalue equation. Substitute the eigenvalues into the other 
+    equations and solve those too. 
+4. Use the orthonormal basis functions to write down the solution corresponding 
+    to the specified initial and boundary conditions. 
+
+One natural question is what if the operator $L$ from setp 2 is not hermitian? 
+It is possible to try and make it hermitian by working on a Hilbert space 
+equipped with a different inner product. This means one can consider 
+modifications to the definition of $\langle \cdot, \cdot \rangle$ such that $L$
+is hermitian with respect to the modified inner product. This type of technique 
+falls under the umbrella of *Sturm-Liouville Theory*, which forms the foundation
+for a lot of the analysis that can be done analytically on PDEs. 
+
+Another question is of course what if the equation is not separable? One 
+possible approach is to try working in a different coordinate system. There are 
+a few more analytic techniques available, however in many situations it becomes
+necessary to work with numerical methods of solution. 
+
+### Problems
+
+1.  [:grinning:] Which of the following equations for $y(x)$ is linear?
+
+    (a) y''' - y'' + x cos(x) y' + y - 1 = 0
+
+    (b) y''' + 4 x y' - cos(x) y = 0
+
+    (c) y'' + y y' = 0
+
+    (d) y'' + e^x y' - x y = 0
+
+2.  [:grinning:] Find the general solution to the equation 
+
+    $$y'' - 4 y' + 4 y = 0. $$
+
+    Show explicitly by computing the Wronski determinant that the 
+    basis for the solution space is actually linearly independent. 
+
+3.  [:grinning:] Find the general solution to the equation 
+
+    $$y''' - y'' + y' - y = 0.$$
+
+    Then find the solution to the initial conditions $y''(0) =0$, $y'(0)=1$, $y(0)=0$. 
+
+4.  [:smirk:] Take the Laplace equation in 2D:
+
+    $$\frac{\partial^2 \phi(x,y)}{\partial x^2} + \frac{\partial^2 \phi(x,y)}{\partial y^2} = 0.$$
+
+    (a) Make a separation ansatz $\phi(x,y) = f(x)g(y)$ and write 
+        down the resulting ordinary differential equations.
+
+    (b) Now assume that the boundary conditions $\phi(0,y) = \phi(L,y) =0$ for 
+    	all y, i.e.  f(0)=f(L)=0. Find all solutions $f(x)$ and the
+	corresponding eigenvalues.
+
+    (c) Finally, for each eigenvalue, find the general solution $g(y)$ for
+    	this eigenvalue. Combine this with all solutions $f(x)$ to write down
+	the general solution (we know from the lecture that the operator
+	$\frac{d^2}{dx^2}$ is Hermitian - you can thus directly assume that
+	the solutions form an orthogonal basis). 
+
+5.  [:smirk:] Take the partial differential equation
+
+    $$\frac{\partial h(x,y)}{\partial x} + x \frac{\partial h(x,y)}{\partial y} = 0. $$
+
+    Try to make a separation ansatz $h(x,y)=f(x)g(y)$. What do you observe?
+
+6.  [:sweat:] We consider the Hilbert space of functions $f(x)$ defined
+    for $x \ \epsilon \ [0,L]$ with $f(0)=f(L)=0$. 
+
+    Which of the following operators on this space is hermitian?
+
+    (a) Lf = A(x) \frac{d^2 f}{dx^2}
+
+    (b) Lf = \frac{d}{dx} \big{()} A(x) \frac{df}{dx} \big{)}
+