From 7efa29ae1abb73d9ac3dad0fc94c5b61eeaa6837 Mon Sep 17 00:00:00 2001
From: Scarlett Gauthier <s.s.gauthier@student.tudelft.nl>
Date: Wed, 19 Aug 2020 10:50:12 +0000
Subject: [PATCH] Finish example of page 6.
---
src/differential_equations_2.md | 56 +++++++++++++++++++++++++++++++--
1 file changed, 54 insertions(+), 2 deletions(-)
diff --git a/src/differential_equations_2.md b/src/differential_equations_2.md
index 3775e12..19d76fe 100644
--- a/src/differential_equations_2.md
+++ b/src/differential_equations_2.md
@@ -348,10 +348,62 @@ 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.
+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 $x=L$.
-<img src="figures/DE2_1.png" width="700">
+<img src="figures/DE2_1.png" width="650">
+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,
+
+$\psi_n(x,t) = \Tilde{A} e^{-i \frac{\lambda_n t}{\hbar}} sin(\frac{n \pi x}{L}).$
+
+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.
--
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