In [6]:
%matplotlib inline
# plots graphs within the notebook
%config InlineBackend.figure_format='svg' # not sure what this does, may be default images to svg format
from IPython.display import Image
from IPython.core.display import HTML
def header(text):
raw_html = '<h4>' + str(text) + '</h4>'
return raw_html
def box(text):
raw_html = '<div style="border:1px dotted black;padding:2em;">'+str(text)+'</div>'
return HTML(raw_html)
def nobox(text):
raw_html = '<p>'+str(text)+'</p>'
return HTML(raw_html)
def addContent(raw_html):
global htmlContent
htmlContent += raw_html
class PDF(object):
def __init__(self, pdf, size=(200,200)):
self.pdf = pdf
self.size = size
def _repr_html_(self):
return '<iframe src={0} width={1[0]} height={1[1]}></iframe>'.format(self.pdf, self.size)
def _repr_latex_(self):
return r'\includegraphics[width=1.0\textwidth]{{{0}}}'.format(self.pdf)
class ListTable(list):
""" Overridden list class which takes a 2-dimensional list of
the form [[1,2,3],[4,5,6]], and renders an HTML Table in
IPython Notebook. """
def _repr_html_(self):
html = ["<table>"]
for row in self:
html.append("<tr>")
for col in row:
html.append("<td>{0}</td>".format(col))
html.append("</tr>")
html.append("</table>")
return ''.join(html)
font = {'family' : 'serif',
'color' : 'black',
'weight' : 'normal',
'size' : 18,
}
Solve the questions in green blocks. Save the file as ME249-Lecture-3-YOURNAME.ipynb and change YOURNAME in the bottom cell. Send me and the grader the html file not the ipynb file.
Consider a function $f$ periodic over a domain $0\leq x\leq 2\pi$, discretized by $N_x$ points. The longest wavelength wave that can be contained in the domain is $L_x$. A phyiscal understanding of Fourier series is the representation of a system as the sum of many waves fo wavelengths smaller or equal to $L_x$. In a discrete sense, the series of wave used to decompose the system is defined as: $$ a_n\exp\left(\hat{\jmath}\frac{2\pi n}{Lx}\right) $$ such that
$$ f(x) = \sum_{n=-\infty}^{\infty}a_n\exp\left(\hat{\jmath}\frac{2\pi nx}{Lx}\right) $$
and$$ a_n = \frac{1}{L_x}\int_Lf(x)\exp\left(-\hat{\jmath}\frac{2\pi nx}{Lx}\right)dx $$
Here $\hat{\jmath}^2=-1$.Often the reduction to wavenumber is used, where$$ k_n = \frac{2\pi n}{L_x} $$
Note that if $x$ is time instead of distance, $L_x$ is a time $T$ and the smallest frequency contained in the domain is $f_0=1/T_0$ and the wavenumber $n$ is $k_n=2\pi f_0n=2\pi f_n$ with $f_n$ for $\vert n\vert >1$ are the higher frequencies.In scientific computing we are interested in applying Fourier series on vectors or matrices, containing a integer number of samples. The DFT is the fourier series for the number of samples. DFT functions available in python or any other language only care about the number of samples, therefore the wavenumber is
$$ k_n=\frac{2\pi n}{N_x} $$
Consider a function $f$ periodic over a domain $0\leq x\leq 2\pi$, discretized by $N_x$ points. The nodal value is $f_i$ located at $x_i=(i+1)\Delta x$ with $\Delta x=L_x/Nx$. The DFT is defined as$$ \hat{f}_k=\sum_{i=0}^{N_x-1}f_i\exp\left(-2\pi\hat{\jmath}\frac{ik}{N_x}\right) $$
The inverse DFT is defined as$$ f_i=\sum_{k=0}^{N_x-1}\hat{f}_k\exp\left(2\pi\hat{\jmath}\frac{ik}{N_x}\right) $$
$$ \hat{f}_k=\sum_{i=-Nx/2+1}^{N_x/2}f_i\exp\left(-2\pi\hat{\jmath}\frac{ik}{N_x}\right) $$
Compared to the Fourier series, DFT or FFT assumes that the system can be accurately captured by a finite number of waves. It is up to the user to ensure that the number of computational points is sufficient to capture the smallest scale, or smallest wavelength or highest frequence. Remember that the function on which FT is applied must be periodic over the domain and the grid spacing must be uniform.
There are FT algorithms for unevenly space data, but this is beyond the scope of this notebook.The following provides examples of low- and high-pass filters based on Fourier transform. A ideal low-(high-) pass filter passes frequencies that are lower than a threshold without attenuation and removes frequencies that are higher than the threshold.
When applied to spatial data (function of $x$ rather than $t$-time), the FT (Fourier Transform) of a variable is function of wavenumbers $$ k_n=\frac{2\pi n}{L_x} $$ or wavelengths $$ \lambda_n=\frac{2\pi}{k_n} $$
The test function is defined as sum of $N_{wave}$ cosine function: $$ u(x)=\sum_{n=0}^{N_{wave}}A_n\cos\left(nx+\phi_n\right) $$ with the following first and second derivatives: $$ \frac{du}{dx}=\sum_{n=1}^{N_{wave}}-nA_n\sin\left(nx+\phi_n\right) $$ $$ \frac{d^2u}{dx^2}=\sum_{n=1}^{N_{wave}}-n^2A_n\cos\left(nx+\phi_n\right) $$ The python code for function u and its derivatives is written below. Here amplitudes $A_n$ and phases $\phi_n$ are chosen randomly of ranges $[0,1]$ and $[0,2\pi]$, respectively.
In [10]:
import matplotlib.pyplot as plt
import numpy as np
Lx = 2.*np.pi
Nx = 64
u = np.zeros(Nx,dtype='float64')
du = np.zeros(Nx,dtype='float64')
ddu = np.zeros(Nx,dtype='float64')
k_0 = 2.*np.pi/Lx
x = np.linspace(Lx/Nx,Lx,Nx)
Nwave = 24
uwave = np.zeros((Nx,Nwave),dtype='float64')
duwave = np.zeros((Nx,Nwave),dtype='float64')
dduwave = np.zeros((Nx,Nwave),dtype='float64')
ampwave = np.random.random(Nwave)
phasewave = np.random.random(Nwave)*2*np.pi
for iwave in range(Nwave):
uwave[:,iwave] = ampwave[iwave]*np.cos(k_0*iwave*x+phasewave[iwave])
duwave[:,iwave] = -k_0*iwave*ampwave[iwave]*np.sin(k_0*iwave*x+phasewave[iwave])
dduwave[:,iwave] = -(k_0*iwave)**2*ampwave[iwave]*np.cos(k_0*iwave*x+phasewave[iwave])
u = np.sum(uwave,axis=1)
du = np.sum(duwave,axis=1)
ddu = np.sum(dduwave,axis=1)
#print(u)
plt.plot(x,u,lw=2)
plt.xlim(0,Lx)
#plt.legend(loc=3, bbox_to_anchor=[0, 1],
# ncol=3, shadow=True, fancybox=True)
plt.xlabel('$x$', fontdict = font)
plt.ylabel('$u$', fontdict = font)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.show()
plt.plot(x,du,lw=2)
plt.xlim(0,Lx)
#plt.legend(loc=3, bbox_to_anchor=[0, 1],
# ncol=3, shadow=True, fancybox=True)
plt.xlabel('$x$', fontdict = font)
plt.ylabel('$du/dx$', fontdict = font)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.show()
plt.plot(x,ddu,lw=2)
plt.xlim(0,Lx)
#plt.legend(loc=3, bbox_to_anchor=[0, 1],
# ncol=3, shadow=True, fancybox=True)
plt.xlabel('$x$', fontdict = font)
plt.ylabel('$d^2u/dx^2$', fontdict = font)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.show()
First, let's perform a sanity check, i.e. $$ u=FT^{-1}\left[FT\left[u\right]\right] $$ where $FT$ designates the Fourier transform and $FT^{-1}$ its inverse.
In [11]:
#check FT^-1(FT(u)) - Sanity check
u_hat = np.fft.fft(u)
v = np.real(np.fft.ifft(u_hat))
plt.plot(x,u,'r-',lw=2,label='$u$')
plt.plot(x,v,'b--',lw=2,label='$FT^{-1}[FT[u]]$')
plt.xlim(0,Lx)
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.xlabel('$x$', fontdict = font)
plt.ylabel('$u$', fontdict = font)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.show()
print('error',np.linalg.norm(u-v,np.inf))
For now we will define the spectrum $\Phi_u$ as
$$ \Phi_u(k_n) = \hat{u}_n.\hat{u}_n^* $$
which can be interpreted as the energy contained in the $k_n$ wavenumber. This is helpful when searching for the most energetic scales or waves in our system. Thanks to the symmetries of the FFT, the spectrum is defined over $n=0$ to $N_x/2$
In [12]:
F = np.zeros(Nx/2+1,dtype='float64')
F = np.real(u_hat[0:Nx/2+1]*np.conj(u_hat[0:Nx/2+1]))
k = np.hstack((np.arange(0,Nx/2+1),np.arange(-Nx/2+1,0))) #This is how the FFT stores the wavenumbers
plt.loglog(k[0:Nx/2+1],F,'r-',lw=2,label='$\Phi_u$')
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.xlabel('$k$', fontdict = font)
plt.ylabel('$\Phi_u(k)$', fontdict = font)
plt.xticks(fontsize = 16)
y_ticks = np.logspace(-30,5,8)
plt.yticks(y_ticks,fontsize = 16) #Specify ticks, necessary when increasing font
plt.show()
The following code filters the original signal by half the wavenumbers using FFT and compares to exact filtered function
In [13]:
# filtering the smaller waves
def low_pass_filter_fourier(a,k,kcutoff):
N = a.shape[0]
a_hat = np.fft.fft(a)
filter_mask = np.where(np.abs(k) > kcut)
a_hat[filter_mask] = 0.0 + 0.0j
a_filter = np.real(np.fft.ifft(a_hat))
return a_filter
kcut=Nwave/2+1
k = np.hstack((np.arange(0,Nx/2+1),np.arange(-Nx/2+1,0)))
v = low_pass_filter_fourier(u,k,kcut)
u_filter_exact = np.sum(uwave[:,0:kcut+1],axis=1)
plt.plot(x,v,'r-',lw=2,label='filtered with fft')
plt.plot(x,u_filter_exact,'b--',lw=2,label='filtered (exact)')
plt.plot(x,u,'g:',lw=2,label='original')
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.xlabel('$x$', fontdict = font)
plt.ylabel('$u$', fontdict = font)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.show()
print('error:',np.linalg.norm(v-u_filter_exact,np.inf))
In [14]:
F = np.zeros(Nx/2+1,dtype='float64')
F_filter = np.zeros(Nx/2+1,dtype='float64')
u_hat = np.fft.fft(u)
F = np.real(u_hat[0:Nx/2+1]*np.conj(u_hat[0:Nx/2+1]))
v_hat = np.fft.fft(v)
F_filter = np.real(v_hat[0:Nx/2+1]*np.conj(v_hat[0:Nx/2+1]))
k = np.hstack((np.arange(0,Nx/2+1),np.arange(-Nx/2+1,0)))
plt.loglog(k[0:Nx/2+1],F,'r-',lw=2,label='$\Phi_u$')
plt.loglog(k[0:Nx/2+1],F_filter,'b-',lw=2,label='$\Phi_v$')
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.xlabel('$k$', fontdict = font)
plt.ylabel('$\Phi_u(k)$', fontdict = font)
plt.xticks(fontsize = 16)
plt.yticks(y_ticks,fontsize = 16)
plt.show()
From the example below, develop a function for a high-pass filter.
In [15]:
u_hat = np.fft.fft(u)
kfilter = Nwave/2
k = np.linspace(0,Nx-1,Nx)
filter_mask = np.where((k < kfilter) | (k > Nx-kfilter) )
u_hat[filter_mask] = 0.+0.j
v = np.real(np.fft.ifft(u_hat))
plt.plot(x,v,'r-',lw=2,label='Filtered (FT)')
plt.plot(x,np.sum(uwave[:,kfilter:Nwave+1],axis=1),'b--',lw=2,label='Filtered (exact)')
plt.plot(x,u,'g:',lw=2,label='original')
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.xlim(0,Lx)
plt.show()
In [17]:
F = np.zeros(Nx/2+1,dtype='float64')
F_filter = np.zeros(Nx/2+1,dtype='float64')
u_hat = np.fft.fft(u)
F = np.real(u_hat[0:Nx/2+1]*np.conj(u_hat[0:Nx/2+1]))
v_hat = np.fft.fft(v)
F_filter = np.real(v_hat[0:Nx/2+1]*np.conj(v_hat[0:Nx/2+1]))
k = np.hstack((np.arange(0,Nx/2+1),np.arange(-Nx/2+1,0)))
plt.loglog(k[0:Nx/2+1],F,'r-',lw=2,label=r'$\Phi_u$')
plt.loglog(k[0:Nx/2+1],F_filter,'b-',lw=2,label=r'$\Phi_v$')
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.xlabel('$k$', fontdict = font)
plt.ylabel('$\Phi(k)$', fontdict = font)
plt.xticks(fontsize = 16)
plt.yticks(y_ticks,fontsize = 16)
plt.show()
Going back to the Fourier series, $$ u(x) = \sum_{n=-\infty}^{\infty}a_n\exp\left(\hat{\jmath}\frac{2\pi nx}{L_x}\right)=\sum_{n=-\infty}^{\infty}a_n\exp\left(\hat{\jmath}k_nx\right) $$ with $$ k_n=\frac{2\pi n}{L_x}\, $$ it is obvious that any $m$ derivative of the real variable $u$ be: $$ \frac{d^mu}{dx^m} = \sum_{n=-\infty}^{\infty}\left(\hat{\jmath}k_n\right)^ma_n\exp\left(\hat{\jmath}k_nx\right) $$ In other words, if $u_n$ is defined as: $$ u_n(x) = a_n\exp\left(\hat{\jmath}k_nx\right) $$ then $$ \frac{d^mu_n}{dx^m} = \left(\hat{\jmath}k_n\right)^m u_n\,. $$ A $m$ derivativation in the Fourier space amounts to the multiplication of each Fourier coefficient $a_n$ by $\left(\hat{\jmath}k_n\right)^m$.
The following is a code for the first derivative of $u$:
In [18]:
u_hat = np.fft.fft(u)
k = np.hstack((np.arange(0,Nx/2+1),np.arange(-Nx/2+1,0)))
ik = 1j*k
v_hat = ik*u_hat
v = np.real(np.fft.ifft(v_hat))
plt.plot(x,v,'r-',lw=2,label='$FT^{-1}[\hat{\jmath}kFT[u]]$')
plt.plot(x,du,'b--',lw=2,label='exact derivative')
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.ylabel('$du/dx$',fontsize = 18)
plt.xlabel('$x$',fontsize = 18)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.xlim(0,Lx)
plt.show()
print('error:',np.linalg.norm(v-du))
In [19]:
mk2 = ik*ik
v_hat = mk2*u_hat
v = np.real(np.fft.ifft(v_hat))
plt.plot(x,v,'r-',lw=2,label='$FT^{-1}[-k^2FT[u]]$')
plt.plot(x,ddu,'b--',lw=2,label='exact derivative')
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.ylabel('$d^2u/dx^2$',fontsize = 18)
plt.xlabel('$x$',fontsize = 18)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.xlim(0,Lx)
plt.show()
print('error:',np.linalg.norm(v-ddu))
When the number of Fourier node is sufficient to capture all scales, derivatives computed in the Fourier space are essentially exact. The following code compares the exact first derivative with a first order upwind scheme: $$ \left.\frac{\delta u}{\delta x}\right\vert_i=\frac{u_i-u_{i-1}}{\Delta x} $$
For finite difference derivative, we will use the symbol $\delta/\delta x$ rather than $d/dx$.
Show that the scheme is first order and write the leading term in the truncation error.
In [20]:
du_fd = np.zeros(Nx,dtype='float64')
dx = Lx/Nx
du_fd[1:Nx] = (u[1:Nx]-u[0:Nx-1])/dx
du_fd[0] = (u[0]-u[Nx-1])/dx
plt.plot(x,du_fd,'r-',lw=2,label='FD derivative')
plt.plot(x,du,'b--',lw=2,label='exact derivative')
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.ylabel('$du/dx$',fontsize = 18)
plt.xlabel('$x$',fontsize = 18)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.xlim(0,Lx)
plt.show()
print('error:',np.linalg.norm(du_fd-du,np.inf))
The error is large which is compounded by the fact that np.linalg.norm does not normalize the norm by the number of points. Nonetheless the error is far greater than for the derivative using the Fourier space. To shed some light on the cause for errors, the following code computes the spectrum of the first derivative. The second graph shows the difference in spectral energy between the exact solution and the derivative approximated with finite difference per wavenumber.
However the best way to visualize the problem with spectra alone is to lower the resolution at the top cell and rerun the whole notebook. Finish the notebook as is and then rerun with $N_x=64$.
In [23]:
F = np.zeros(Nx/2+1,dtype='float64')
F_fd = np.zeros(Nx/2+1,dtype='float64')
du_hat = np.fft.fft(du)/Nx
F = np.real(du_hat[0:Nx/2+1]*np.conj(du_hat[0:Nx/2+1]))
dv_hat = np.fft.fft(du_fd)/Nx
F_fd = np.real(dv_hat[0:Nx/2+1]*np.conj(dv_hat[0:Nx/2+1]))
k = np.hstack((np.arange(0,Nx/2+1),np.arange(-Nx/2+1,0)))
plt.loglog(k[0:Nx/2+1],F,'r-',lw=2,label='$\Phi_{du/dx}$')
plt.loglog(k[0:Nx/2+1],F_fd,'b-',lw=2,label='$\Phi_{\delta u/\delta x}$')
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.xlabel('$k$', fontdict = font)
plt.ylabel('$\Phi(k)$', fontdict = font)
plt.xticks(fontsize = 16)
plt.ylim(1e-3,250)
plt.yticks(fontsize = 16)
plt.show()
print('error:',np.linalg.norm(F[0:Nwave/2]-F_fd[0:Nwave/2],np.inf))
print('error per wavenumber')
plt.loglog(k[0:Nx/2+1],np.abs(F-F_fd),'r-',lw=2)
plt.xlabel('$k$', fontdict = font)
plt.ylabel('$\Vert\Phi_{d u/d x}(k)-\Phi_{\delta u/\delta x}(k)\Vert$', fontdict = font)
plt.xticks(fontsize = 16)
plt.ylim(1e-5,250)
plt.yticks(fontsize = 16)
plt.show()
Which scales are the most affected by the finite difference scheme? What effect do you observe?
Starting from, $$ f(x) = \sum_{k=-N/2+1}^{N/2}\hat{f}_k\exp\left(\hat{\jmath}\frac{2\pi kx}{L_x}\right) $$ and introducing the wavenumber $$ \omega = \frac{2\pi k}{L_x} $$ allows to reduce the Fourier expansion to $$ f(x) = \sum_{\omega=-\pi}^{\pi}\hat{f}_\omega e^{\hat{\jmath}\omega x} $$ The derivative of $f$ becomes $$ f'(x) = \sum_{\omega=-\pi}^{\pi}\hat{f'}_\omega e^{\hat{\jmath}\omega x} $$ where $\hat{f'}_\omega$ is only a notation, not the derivative of the Fourier coefficient: $$ \hat{f'}_\omega=\hat{\jmath}\omega\hat{f}_\omega $$ Considering the symmetry of the FT, we restrict the study to $\omega\Delta x\in[0,\pi]$.
Now we want to express the first order finite difference scheme in terms of Fourier coefficients and wavenumbers. The scaled coordinates expression of the scheme is $$ \frac{\delta f}{\delta x}=\frac{f(x)-f(x-\Delta x)}{\Delta x}=\sum_{\omega=-\pi}^{\pi}\hat{f}_\omega\frac{e^{\hat{\jmath}\omega x}-e^{\hat{\jmath}\omega (x-\Delta x)}}{\Delta x}=\sum_{\omega=-\pi}^{\pi}\frac{1-e^{-\hat{\jmath}\omega\Delta x}}{\Delta x}\hat{f}_\omega e^{\hat{\jmath}\omega x} $$ We now define a modified wavenumber for the first order upwind scheme: $$ \hat{\jmath}\omega'=\frac{1-e^{-\hat{\jmath}\omega\Delta x}}{\Delta x}=\frac{1-\cos(-\omega\Delta x)-\hat{\jmath}\sin(-\omega\Delta x)}{\Delta x} $$ which reduces to $$ \hat{\jmath}\omega_\text{mod}\Delta x= \hat{\jmath}\sin(\omega\Delta x)+\left(\cos(1-\omega\Delta x)\right) $$ The modified wavenumber is no longer purely imaginary, and even the imaginary part is far from the exact wavenumber, as shown in the following plot.
In [24]:
L = np.pi
N = 32
dx = L/N
omega = np.linspace(0,L,N)
omegap_exact = omega
omegap_modified = np.sin(omega)
plt.plot(omega,omegap_exact,'k-',lw=2,label='exact')
plt.plot(omega,omegap_modified,'r-',lw=2,label='1st order upwind')
plt.xlim(0,L)
plt.ylim(0,L)
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.xlabel('$\omega\Delta x$', fontdict = font)
plt.ylabel('$\omega_{mod}\Delta x$', fontdict = font)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.show()
First, let's verify that the derivation of the modified wavenumber is correct.
In [25]:
u_hat = np.fft.fft(u)
k = np.hstack((np.arange(0,Nx/2+1),np.arange(-Nx/2+1,0)))
dx = Lx/Nx
ikm = 1j*np.sin(2.*np.pi/Nx*k)/dx+(1-np.cos(2.*np.pi/Nx*k))/dx
v_hat = ikm*u_hat
dum = np.real(np.fft.ifft(v_hat))
plt.plot(x,dum,'r-',lw=2,label='$FT^{-1}[\hat{\jmath}\omega_{mod}FT[u]]$')
plt.plot(x,du_fd,'b--',lw=2,label='$\delta u/\delta x$')
plt.legend(loc=3, bbox_to_anchor=[0, 1],
ncol=3, shadow=True, fancybox=True)
plt.ylabel('$du/dx$',fontsize = 18)
plt.xlabel('$x$',fontsize = 18)
plt.xticks(fontsize = 16)
plt.yticks(fontsize = 16)
plt.xlim(0,Lx)
plt.show()
print('error:',np.linalg.norm(du_fd-dum,np.inf))
- Explain why $\omega\Delta x$ can be defined from $0$ to $\pi$.
- Write a code to illustrate the effects of the imaginary part and real part on the derivative on the following function $$ f(x) = cos(nx) $$ defined for $x\in[0,2\pi]$, discretized with $N$ points. Study a few $n$, ranging from large scales to small scales. Note the two effects we seek to identify are phase change and amplitude change.
- Derive the modified wavenumber for the second order central finite difference scheme $$ \frac{\delta f}{\delta x}=\frac{f_{i+1}-f_{i-1}}{2\Delta x} $$
- Create a second order upwind scheme and derive the modified wavenumber. Compare the performance of the first order and the second order schemes. For the second order upwind scheme, find $a$, $b$ and $c$ such that $$ \frac{\delta f}{\delta x}=\frac{af_{i-2}+bf_{i-1}+cf_i}{\Delta x} $$
In [ ]:
!ipython nbconvert --to html ME249-Lecture-3-YOURNAME.ipynb