From 311ebe3ad5e7f308d9f952dfa31482e28bf77537 Mon Sep 17 00:00:00 2001
From: Michael Wimmer <m.t.wimmer@tudelft.nl>
Date: Tue, 15 Sep 2020 21:28:30 +0200
Subject: [PATCH] fix problem formatting
---
src/8_differential_equations_2.md | 64 +++++++++++++++----------------
1 file changed, 31 insertions(+), 33 deletions(-)
diff --git a/src/8_differential_equations_2.md b/src/8_differential_equations_2.md
index 7d543f5..33c4828 100644
--- a/src/8_differential_equations_2.md
+++ b/src/8_differential_equations_2.md
@@ -283,7 +283,7 @@ 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 ##
+### Separation of variables
Let us focus on the one dimensional Schrödinger equation of a free particle
@@ -332,7 +332,7 @@ 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 ###
+### Boundary and eigenvalue problems
Continuing on with the Schrödinger equation example from the previous
section, let us focus on
@@ -493,7 +493,7 @@ coefficient,
$$c_n:= \int^{L}_{0} dx sin(\frac{n \pi x}{L}) \psi_{0}(x). $$
-## General recipie for seperable PDEs ##
+### General recipie for seperable PDEs
1. Make the separation ansatz to obtain separate ordinary differential
equations.
@@ -518,62 +518,60 @@ 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
+### Problems
1. [:grinning:] Which of the following equations for $y(x)$ is linear?
- (a) y''' - y'' + x cos(x) y' + y - 1 = 0
+ (a) y''' - y'' + x cos(x) y' + y - 1 = 0
- (b) y''' + 4 x y' - cos(x) y = 0
+ (b) y''' + 4 x y' - cos(x) y = 0
- (c) y'' + y y' = 0
+ (c) y'' + y y' = 0
- (d) y'' + e^x y' - x y = 0
+ (d) y'' + e^x y' - x y = 0
2. [:grinning:] Find the general solution to the equation
- $$y'' - 4 y' + 4 y = 0. $$
+ $$y'' - 4 y' + 4 y = 0. $$
- Show explicitly by computing the Wronski determinant that the
- basis for the solution space is actually linearly independent.
+ 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
- $$y''' - y'' + y' - y = 0.$$
+ $$y''' - y'' + y' - y = 0.$$
- Then find the solution to the initial conditions $y''(0) =0$, $y'(0)=1$, $y(0)=0$.
+ 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:
- $$\frac{\partial^2 \phi(x,y)}{\partial x^2} + \frac{\partial^2 \phi(x,y)}{\partial y^2} = 0.$$
+ $$\frac{\partial^2 \phi(x,y)}{\partial x^2} + \frac{\partial^2 \phi(x,y)}{\partial y^2} = 0.$$
- (a) Make a separation ansatz $\phi(x,y) = f(x)g(y)$ and write
- down the resulting ordinary differential equations.
+ (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.
+ (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).
+ (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
- $$\frac{\partial h(x,y)}{\partial x} + x \frac{\partial h(x,y)}{\partial y} = 0. $$
+ $$\frac{\partial h(x,y)}{\partial x} + x \frac{\partial h(x,y)}{\partial y} = 0. $$
- Try to make a separation ansatz $h(x,y)=f(x)g(y)$. What do you observe?
+ Try to make a separation ansatz $h(x,y)=f(x)g(y)$. What do you observe?
-6. [:sweat:] *Bonus question - this kind of question will not be asked in the exam*
+6. [:sweat:] We consider the Hilbert space of functions $f(x)$ defined
+ for $x \ \epsilon \ [0,L]$ with $f(0)=f(L)=0$.
- 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?
- Which of the following operators on this space is hermitian?
+ (a) Lf = A(x) \frac{d^2 f}{dx^2}
- (a) Lf = A(x) \frac{d^2 f}{dx^2}
-
- (b) Lf = \frac{d}{dx} \big{()} A(x) \frac{df}{dx} \big{)}
+ (b) Lf = \frac{d}{dx} \big{()} A(x) \frac{df}{dx} \big{)}
--
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