MATH3310 Note 6: The Fourier transform on $\mathbb{R}$

In many ways the Fourier transform on $ {\mathbb{R}}$

$ \displaystyle f(x)\mapsto\hat{f}(\omega):=\int_{\mathbb{R}}f(x)e^{-i\omega x}dx,\quad x,\omega\in\mathbb{R} $

is similar to the transformation that gets us the Fourier coefficient of a function on $ {[0,2\pi]}$

$ \displaystyle f(x)\mapsto\hat{f}(n)=\frac{1}{2\pi}\int_{[0,2\pi]}f(x)e^{-inx}dx,\quad x\in[0,2\pi],n\in\mathbb{Z}. $

We shall see these similarities and comparisons later on. But first of all, just like the study of Fourier series, we need to make clear what is the object of study in the context of Fourier transfrom on $ {\mathbb{R}}$:

• Functions on $ {\mathbb{R}}$ with ``reasonable'' decay near the infinity $ {\pm\infty.}$ For the sake of simplicity, we can just pretend that these functions are continuous with compact supports, i.e. the closure of the set
  $ \displaystyle \{x\in\mathbb{R}:f(x)\neq0\} $
  is compact in $ {\mathbb{R}}$, with the usual topology.

With the above agreement, then defining the integral $ {\hat{f}(\omega)}$ for each $ {\omega}$ will not be an issue. The function $ {\hat{f}(\omega)}$ is called the Fourier transform of $ {f}$, and $ {\omega}$ is often called the frequency variable.
The Fourier transform is easily seen to be a linear transform. Let us see three examples.

Example 1 $ {\:}$

1. The Fourier transform of the ``Dirac delta function'' at the origin $ {\delta_{0}(x)}$ is
   $ \displaystyle \int_{\mathbb{R}}\delta_{0}(x)e^{-i\omega x}dx=e^{0}=1. $
2. The Fourier transform of the function
   $ \displaystyle f(x)=\begin{cases} e^{-ax} & \text{if }x\geq0\\ e^{ax} & \text{if }x<0 \end{cases} $
   where $ {a>0}$, is
   $ \displaystyle \begin{array}{rcl} \hat{f}(x) & = & \int_{-\infty}^{0}e^{ax}e^{-i\omega x}dx+\int_{0}^{\infty}e^{-ax}e^{-i\omega x}dx\\ & = & \frac{1}{a-i\omega}+\frac{1}{a+i\omega}\\ & = & \frac{2a}{a^{2}+\omega^{2}}. \end{array} $
3. The Fourier transform of the Gaussian function
   $ \displaystyle f(x)=\frac{1}{\sqrt{\pi}}e^{-x^{2}} $
   is still a Gaussian
   $ \displaystyle \hat{f}(x)=2\sqrt{\pi}e^{-x^{2}}. $

Remark 1 In fact, the function

$ \displaystyle P_{a}(x)=\frac{1}{\pi}\frac{a}{a^{2}+x^{2}} $

is called the Poisson kernel on the real line $ {\mathbb{R}}$. Try to connect what you know about approximation of the delta function with the above Example.

In general, it is good to think of the function $ {\hat{f}(\omega)}$ as a measure of how smooth the function $ {f}$ is:

• If $ {\hat{f}(\omega)}$ is ``large'' beyond certain threshold $ {|\omega|>C}$, then the function is considered as rough, and oscillatory. An extreme example is the ``Dirac delta function'', whose frequency spreads out equally on the real line.
• If instead $ {\hat{f}(\omega)}$ is ``small'' when $ {|\omega|>C}$, then the function is considered as smooth. An good example is the Gaussian function.

To understand this better we make an analogy with the Fourier series. To do this, let us first come to the inversion map that recovers the function from its Fourier transform.

  The inverse Fourier transform. We have, for nice functions $ {f:\mathbb{R}\rightarrow\mathbb{C}}$

$ \displaystyle f(x)=\frac{1}{2\pi}\int\hat{f}(\omega)e^{ix\omega}d\omega. $

Now you can compare this with the Fourier series $ {f:[0,2\pi]\rightarrow\mathbb{C}}$

$ \displaystyle f(x)=\sum_{n\in\mathbb{Z}}\hat{f}(n)e^{inx}. $

We can still think of the function as a superposition of different freqeuncies, $ {e^{ix\omega}}$, each with coefficient $ {\hat{f}(\omega)}$, only that the sum is replaced by an integral. Of course, $ {e^{ix\omega}}$, with variable in $ {x}$, oscillate very much when $ {\omega}$ is large, and is smooth when $ {\omega}$ is small. And the Fourier transform evaluated at $ {\omega}$ just shows how much this frequency is present in the function.
Many intuitions about Fourier coefficient thus still hold about the Fourier transform.

Exercise 1 Show that

1. For a function $ {f:[0,2\pi]\rightarrow\mathbb{C}}$, its zero-th Fourier coefficient $ {\hat{f}(0)}$ is the average of the function $ {f}$ up to dividing by $ {2\pi}$ 
   $ \displaystyle \hat{f}(0)=\frac{1}{2\pi}\int_{[0,2\pi]}f(x)dx. $
2. For a function $ {f:\mathbb{R}\rightarrow\mathbb{C}}$, its Fourier transform evaluated at $ {0}$ is the average of the function 
   $ \displaystyle \hat{f}(0)=\int_{\mathbb{R}}f(x)dx. $ 

Interaction with convolution. We can define convolution of two functions on $ {\mathbb{R}}$,

$ \displaystyle f*g(x)=\int_{\mathbb{R}}f(y)g(x-y)dy $

whenever the right hand side makes sense (convergent). With straightforward computations we can show

Exercise 2    $ {\hat{f}(\omega)\hat{g}(\omega)=\widehat{f*g}(\omega).}$

Of course this is in spirit the same with the result about the Fourier coefficients. Here we have essentially the same interpretation: the Fourier transform on $ {\mathbb{R}}$ diagonalizes convolution on $ {\mathbb{R}}$.


The "unitarity" of the Fourier transform. Just as in the case of Fourier series, where we have, for a function $ {f:[0,2\pi]\rightarrow\mathbb{C}}$,

$ \displaystyle \int_{[0,2\pi]}|f(x)|^{2}dx=2\pi\sum_{n\in\mathbb{Z}}|\hat{f}(n)|^{2}. $

in the situation of the Fourier transform, we have

Exercise 3    $ {2\pi\int_{\mathbb{R}}|f(x)|^{2}dx=\int_{\mathbb{R}}|\hat{f}(\omega)|^{2}d\omega}$. (Hint: use Exercise 2.)

The right way to see this is to interprete the mapping

$ \displaystyle f\mapsto\frac{1}{\sqrt{2\pi}}\hat{f} $

as a unitary mapping, where the mapping is unitary means it preserves the Hermite inner product. Just like the case of matrices, the inverse of a unitary mapping is just its Hermittian adjoint.

Solving differential equations using Fourier transform. Because the in the Fourier transform, one must use the whole real line the compute the integral

$ \displaystyle \hat{f}(\omega)=\int_{\mathbb{R}}f(x)e^{-i\omega x}dx, $

the differential equation that can be solved is always on $ {\mathbb{R}}$, that means there is no boundary term. Because of our object of study is function with ``reasonable'' decay near the inifinity, the data term and the solution must also be of such kind.

Exercise 4 Solve the following heat equation on the real line by taking the Fourier transform in the spatial variable $ {x}$

$ \displaystyle \begin{cases} \frac{\partial u}{\partial t}-\frac{\partial^{2}u}{\partial x^{2}}=0 & (t,x)\in(0,\infty)\times\mathbb{R}\\ u(0,x)=u_{0}(x) & x\in\mathbb{R} \end{cases}. $

The sampling theorem. A function $ {f:\mathbb{R}\rightarrow\mathbb{C}}$ is called $ {(-\pi,\pi)}$-band-limited if its Fourier transform $ {\hat{f}:\mathbb{R}\rightarrow\mathbb{C}}$ is supported inside the open interval $ {(-\pi,\pi)}$, that is

$ \displaystyle \{\omega:\hat{f}(\omega)\neq0\}\subset(-\pi,\pi). $

Intuitively, we know that this function $ {f}$ cannot oscillate too fast. What is amazing is the fact that the function $ {f}$ is entirely determined by its value at a discrete set of points in $ {\mathbb{R}}$. In the following exercise we will derive such a reconstruction formula.

Exercise 5

1. Assume the inverse Fourier transform is valid for the function $ {f}$. Show that $ \displaystyle f(x)=\frac{1}{2\pi}\int_{[-\pi,\pi]}\hat{f}(\omega)e^{ix\omega}d\omega. $  
2. Now that $ {\hat{f}}$ can be thought of as a function defined on the interval $ {[-\pi,\pi]}$, which vanishes on the two end points, we can express $ {\hat{f}}$ in terms of its Fourier series. What are the Fourier coefficients $ {\{a_{n}\}_{n\in\mathbb{Z}}}$ of $ {\hat{f}}$?
3. By substituting the Fourier series expression of $ {\hat{f}}$ inside the Fourier inversion, i.e.$ \displaystyle f(x)=\frac{1}{2\pi}\int_{[-\pi,\pi]}\left(\sum_{n\in\mathbb{Z}}a_{n}e^{in\omega}\right)e^{ix\omega}d\omega, $
4. show that $ \displaystyle f(x)=\sum_{n\in\mathbb{Z}}f(n)\frac{\sin\pi(x-n)}{\pi(x-n)}. $

The above is often called the Shannon-Nyquist sampling theorem. We can think of $ {f(n)}$ as sampled from the original function $ {f}$. The sampling point are at the integers, where two neighboring sampling points are of distance $ {1=\frac{1}{2}\cdot\frac{\text{length}([-\pi,\pi])}{\pi}}$, which is often called the Nyquist rate of sampling. In real life signals may come with very high frequency, to the extent that the equipment at hand cannot capture at this reconstruction rate. This will make some different frequency appears identical to the equipment, producing the aliasing effect. But nevertheless, it is possible to sample certain special signals below this criticial Nyquist rate, a subject now known as compressed sensing.

MATH3310 Note 5: 1D heat equation on the interval $[0,2\pi]$

In this note we recap and clarify on what we learned about using spectral (Fourier series) method to solve a fundamental constant coefficient linear differential equation with simple initial/boundary conditions:

• 1D heat equation on $ {[0,2\pi]}$;

In particular, we will stress the role played by the initial/boundary conditions in these problems, a point alluded way back in Note 1. And we shall discuss the connections to the Fourier series. This is an important topic that will appear in midterm so you are suggested to read this note in details and solve all the exercises.


The equation is

$ \displaystyle \frac{\partial u}{\partial t}-\frac{\partial^{2}u}{\partial x^{2}}=0\quad(t,x)\in(0,\infty)\times(0,2\pi). $

Now the differential operator $ {\frac{\partial}{\partial t}-\frac{\partial^{2}}{\partial x^{2}}}$ should come with a boundary condition on

$ \displaystyle \left([0,\infty)\times\{0\}\right)\cup\left(\{0\}\times(0,2\pi)\right)\cup\left([0,\infty)\times\{2\pi\}\right) $

which I would like to call a parabolic boundary. First let us consider the homogeneous Dirichlet initial/boundary value problem.

$ \displaystyle \begin{cases} \frac{\partial u}{\partial t}-\frac{\partial^{2}u}{\partial x^{2}}=0 & (t,x)\in(0,\infty)\times(0,2\pi)\\ u(0,x)=u_{0}(x) & x\in(0,2\pi)\quad\text{initial condition}\\ u(t,0)=u(t,2\pi)=0 & t\in[0,\infty)\quad\text{boundary condition} \end{cases} $

You can think of the initial condition is given to the differential operator $ {\frac{\partial}{\partial t}}$ and the boundary condition is given to $ {-\frac{\partial^{2}}{\partial x^{2}}}$.
Before everything, let us recall the method of separation of variables. It is a genius idea to try for the solution which has the form

$ \displaystyle u(t,x)=X(x)T(t). $

We shall have a good sense of why this strategy should work later on. To begin we will make some computations.

Exercise 1 Show that if $ {\frac{\partial u}{\partial t}-\frac{\partial^{2}u}{\partial x^{2}}=0}$, and $ {u(t,x)=X(x)T(t)}$, then we can decouple the PDE into two eigenvalue problems

$ \displaystyle \begin{cases} \frac{dT}{dt}=\lambda T\\ \frac{d^{2}X}{dx^{2}}=\lambda X \end{cases} $

where $ {\lambda\in\mathbb{C}}$ is a constant.

We first look at the second order ODE above.

Exercise 2 Consider the equation

$ \displaystyle \frac{d^{2}X}{dx^{2}}=\lambda X $

with boundary condition

$ \displaystyle X(0)=X(2\pi)=0. $

Show that

1. $ {\lambda\in\mathbb{R}}$. (Hint: the differential operator is self-adjoint).
2. if $ {\lambda>0}$, then the only solution satisfies the boundary condition is the trivial solution $ {X(x)=0}$.
3. if $ {\lambda=0}$, then the only solution satisfies the boundary condition is the trivial solution $ {X(x)=0}$.
4. if $ {\lambda<0}$, then the solution is of the form $ {C\cdot\sin(\sqrt{-\lambda}x)}$ for some constant $ {C}$, where $ {2\sqrt{-\lambda}=n}$ must be a positive integer.

Now let us turn to the other differential equation. From the above exercise we know that $ {\lambda=-\frac{n^{2}}{4}}$ for some non-negative integer $ {n}$. It is a simple exercise to show that the solution is

$ \displaystyle T(t)=C'e^{-n^{2}t/4}, $

for some constant $ {C'}$ depending on the initial condition. Hence we get

$ \displaystyle u(t,x)=Ae^{-n^{2}t/4}\sin(\frac{nx}{2}) $

which corresponds to the initial condition $ {u_{0}=A\cdot\sin(\frac{nx}{2})}$.

Exercise 3 (Principle of superposition) Verify that if $ {n,m\in\mathbb{Z}}$ and $ {n,m>0}$, and

$ \displaystyle \begin{array}{rcl} u_{n}(t,x) & = & A_{1}e^{-n^{2}t/4}\sin(\frac{nx}{2})\\ u_{m}(t,x) & = & A_{2}e^{-m^{2}t/4}\sin(\frac{mx}{2}) \end{array} $

then $ {u=u_{n}+u_{m}}$ also satisfy $ {\frac{\partial u}{\partial t}-\frac{\partial^{2}u}{\partial x^{2}}=0}$, $ {u(0)=u(2\pi)=0}$.

We are now led to the ansatz that the general solution to the PDE $ {\frac{\partial u}{\partial t}-\frac{\partial^{2}u}{\partial x^{2}}=0}$ with the homogeneous Dirichlet boundary condition is of the form

$ \displaystyle \sum_{n=1}^{\infty}A_{n}e^{-n^{2}t/4}\sin(\frac{nx}{2}) \ \ \ \ \ (1)$

where the constants $ {\{A_{n}\}_{n=1}^{\infty}}$ will be determined from the initial condition $ {u(0,x)=u_{0}(x)}$. More precisely, we have

Exercise 4 Following the above notation, show that

$ \displaystyle A_{n}=\frac{1}{\pi}\int_{[0,2\pi]}u_{0}(x)\sin(\frac{nx}{2})dx. $

(Hint: use the orthogonality of the family of functions $ {\{\sin(\frac{nx}{2})\}_{n=1}^{\infty}}$.)

Using what is developed above, you can quickly solve the following more concrete problem.

Exercise 5 Solve the following initial/boundary value problem

$ \displaystyle \begin{cases} \frac{\partial u}{\partial t}-\frac{\partial^{2}u}{\partial x^{2}}=0 & (t,x)\in(0,\infty)\times(0,2\pi)\\ u(0,x)=20 & x\in(0,2\pi)\quad\text{initial condition}\\ u(t,0)=u(t,2\pi)=0 & t\in[0,\infty)\quad\text{boundary condition} \end{cases} $

I plotted an approximate solution of the above problem from time $t=0$ to time $t=10$ below

Notice that the above solution tends to zero when time goes to infinity. You can explore more on your own, I provide my Matlab code below.

A = linspace(0,2*pi,100);
figure; hold on;
T = linspace(0,10,50);
for i = 1:1000
    t = T(i);
    fA = heatsol(A,t);
    plot(A,fA);
    drawnow; pause(0.01);
end

function fA = heatsol(A,t)
f = @(x) 0;
for n = 1:2000
   f = @(x) f(x) + (40*(1-(-1)^n)/(n*pi))*sin(n*x/2)*exp((-n^2/4)*t);
end
fA = f(A);
end

Now let me make some explanation about the ansatz (1) we made.

• The family of functions $ {\{\sin(\frac{nx}{2})\}_{n=1}^{\infty}}$ is the set of (nontrivial) eigenfunctions of the differential operator $ {-\frac{\partial^{2}}{\partial x^{2}}}$ and the homogeneous boundary condition $ {u(0)=u(2\pi)=0}$, each corresponds to the eigenvalue $ {\frac{n^{2}}{4}}$. That $ {-\frac{\partial^{2}}{\partial x^{2}}}$ is linear, self-adjoint and positive semi-definite under the Hermite inner product implies that the eigenfunctions are orthogonal, with nonnegative eigenvalues;
• The family of functions $ {\{\sin(\frac{nx}{2})\}_{n=1}^{\infty}}$ in fact can be considered as a basis of the vector space
  $ \displaystyle \{f\;|\;f:[0,2\pi]\rightarrow\mathbb{C}\text{ is square-integrable, with }f(0)=f(2\pi)=0\} $
  so that in some appropriate sense,
  $ \displaystyle u_{0}(x)=\sum_{n=1}^{\infty}A_{n}\sin(\frac{nx}{2}) \ \ \ \ \ (2)$
  with $ {A_{n}}$ given in Exercise 4.
• Having the above expression for $ {u_{0}(x)}$ under the eigen-basis $ {\{\sin(\frac{nx}{2})\}_{n=1}^{\infty}}$, we can then study the evolution of each independent orthogonal component under the heat equation. This leads back to the technique of separation of variables.

How do we make a connection of the above problem to the Fourier series? It might be a little confusing about the formula (2) since the Fourier series of

$ \displaystyle u_{0}(x)=\begin{cases} 20 & 0<x<2\pi\\ 0 & x=0,2\pi \end{cases} $

should be just $ {20}$, with no other terms. Indeed, modifying the function on a finite number of points does not affect the computation of Fourier coefficient. The key point is that the Fourier series can NOT be applied to the above problem directly, because of the boundary condition. We shall see below that Fourier series can only take care of periodic boundary conditions. But there is actually some relation to the Fourier series! To understand this we must make a connection of the above homogenous boundary condition to the periodic boundary condition. Looking at the formula (2), and recall that the Fourier series contains only sine terms if and only if the function is odd. This inspires us to look at the odd extension of the function $ {u_{0}}$ into $ {[-2\pi,2\pi]}$

$ \displaystyle \tilde{u}_{0}(x)=\begin{cases} u_{0}(x) & x\in(0,2\pi]\\ 0 & x=0\\ -u_{0}(-x) & x\in[-2\pi,0) \end{cases} $

So in some sense the homogeneous Dirichlet boundary value problem on $ {[0,2\pi]}$ is one half of the periodic boundary value problem on $ {[-2\pi,2\pi]}$ (or equivalently the problem on a circle with radius $ {2}$). Finally, we give corresponding exercises on the homogeneous Neumann initial/boundary value problem and periodic boundary value problem.

Exercise 6 Consider the problem

$ \displaystyle \begin{cases} \frac{\partial u}{\partial t}-\frac{\partial^{2}u}{\partial x^{2}}=0 & (t,x)\in(0,\infty)\times(0,2\pi)\\ u(0,x)=u_{0}(x) & x\in(0,2\pi)\quad\text{initial condition}\\ \frac{\partial u}{\partial x}(t,0)=\frac{\partial u}{\partial x}(t,2\pi)=0 & t\in[0,\infty)\quad\text{boundary condition} \end{cases}. $

Using separation of variables, and mimic the argument for the Dirichlet boundary condition, show that the solution of the problem takes the form

$ \displaystyle \sum_{n=0}^{\infty}B_{n}e^{-n^{2}t/4}\cos(\frac{nx}{2}) $

where

$ \displaystyle B_{n}=\begin{cases} \frac{1}{2\pi}\int_{[0,2\pi]}f(x)dx & n=0\\ \frac{1}{\pi}\int_{[0,2\pi]}f(x)\cos(\frac{nx}{2})dx & n=1,2,3\dots \end{cases} $

Exercise 7 Consider the problem

$ \displaystyle \begin{cases} \frac{\partial u}{\partial t}-\frac{\partial^{2}u}{\partial x^{2}}=0 & (t,x)\in(0,\infty)\times(0,2\pi)\\ u(0,x)=u_{0}(x) & x\in(0,2\pi)\quad\text{initial condition}\\ \frac{\partial u}{\partial x}(t,0)=\frac{\partial u}{\partial x}(t,2\pi),\text{ and }u(t,0)=u(t,2\pi) & t\in[0,\infty)\quad\text{boundary condition} \end{cases}. $

Using separation of variables, and mimic the argument for the Dirichlet boundary condition, show that the solution of the problem takes the form

$ \displaystyle B_{0}+\sum_{n=1}^{\infty}A_{n}e^{-n^{2}t}\sin(nx)+\sum_{n=1}^{\infty}B_{n}e^{-n^{2}t}\cos(nx) $

where

$ \displaystyle A_{n}=\frac{1}{\pi}\int_{[0,2\pi]}u_{0}(x)\sin(nx)dx, $

and

$ \displaystyle B_{n}=\begin{cases} \frac{1}{2\pi}\int_{[0,2\pi]}f(x)dx & n=0\\ \frac{1}{\pi}\int_{[0,2\pi]}f(x)\cos(nx)dx & n=1,2,3\dots \end{cases} $

MATH3310 Note 4: Convolution

1. Convolution

The notion of convolution is so fundamental in Fourier analysis and in applications (guess anyone hasn't heard of the convolutional neural network?). We have seen its appearance last time: that the symmetric partial sum of the Fourier series of a (say) continuous function $ {f:[0,2\pi]\rightarrow\mathbb{C}}$ is obtained by convolving $ {f}$ with the Dirichlet kernel

$ \displaystyle \sum_{|n|\leq N}\hat{f}(n)e^{in\theta}=f*D_{N}\;(\theta). $

And the solution to the Laplace equation on the unit disk is obtained by convolving the given boundary value $ {u_{0}:S^{1}\equiv\partial B(0,1)\rightarrow\mathbb{C}}$ with the Poisson kernel

$ \displaystyle u(r,\theta)=u_{0}*P_{r}\;(\theta),\quad0<r<1. $

In this post we aim to grasp some basic properties of convolution in the context of Fourier series.
First, let us address again our objects of study. There are three basically equivalent categories

• Functions defined on $ {[0,2\pi]}$;
• Functions defined on the circle $ {S^{1}}$;
• $ {2\pi}$-periodic functions defined on $ {\mathbb{R}}$;

For the first class of functions, we have the agreement that values of the function at the endpoints are not important for computing Fourier coefficient, and thus can be made equivalent to the following two classes by possibly altering the function's value at the endpoints (which may not be continuous anymore, but remains integrable). The second class of functions can easily be $ {2\pi}$-periodically extended to the real line. In the context of Fourier series, convolution of two functions is indispensable with the periodicity of the functions. This is manifested in the following definition. 

Definition 1 (Convolution of two $ {2\pi}$-periodic functions) For two continuous, $ {2\pi}$-periodic functions $ {f,g:\mathbb{R}\rightarrow\mathbb{C}}$, their convolution is again a $ {2\pi}$-periodic function, defined to be
$$ \displaystyle f*g\;(x)=\int_{[0,2\pi]}f(y)g(x-y)\;dy $$

Exercise 1 Check that for the same $ {f,g}$ as above,

$ \displaystyle f*g\;(x)=\int_{[0,2\pi]}f(x-y)g(y)\;dy=g*f\;(x). $

One can think of convolution $ {f*g\;(x)}$ as a $ {g}$-weighted averaged version of $ {f}$, or equivalently one can think of it as an $ {f}$-weighted averaged version of $ {g}$, which interpretation is more sensible depends on the context. The weight function is usually called a kernel. For example, we would say the interior of the solution $ {u(r,\theta)}$ to the Laplace equation is obtained by averaging the boundary value $ {u_{0}}$ with the Poisson kernel $ {P_{r}}$.

We gather some basic properties of convolution below.

Exercise 2 Suppose $ {f,g,h:\mathbb{R}\rightarrow\mathbb{C}}$ are continuous and $ {2\pi}$-periodic. Show that

1. $ {f*(g+h)=(f*g)+(f*h)}$.
2. $ {(cf)*g=c(f*g)=f*(cg)}$ for any $ {c\in\mathbb{C}}$.
3. $ {(f*g)*h=f*(g*h)}$.

The relation between convolution and Fourier coefficient is contained in the next proposition.

Proposition 1 Suppose $ {f,g:\mathbb{R}\rightarrow\mathbb{C}}$ are continuous and $ {2\pi}$-periodic. Then

$ \displaystyle \widehat{f*g}(n)=2\pi\hat{f}(n)\hat{g}(n). $

Proof: The result can be proved directly by writing out

$ \displaystyle \begin{array}{rcl} \widehat{f*g}(n) & = & \frac{1}{2\pi}\int_{[0,2\pi]}\left(\int_{[0,2\pi]}f(x-y)g(y)dy\right)e^{-inx}dx\\ & = & \int_{[0,2\pi]}\left(\frac{1}{2\pi}\int_{[-y,2\pi-y]}f(x-y)e^{-in(x-y)}g(y)e^{-iny}d(x-y)\right)dy\\ & = & \int_{[0,2\pi]}\hat{f}(n)g(y)e^{-iny}dy\\ & = & 2\pi\hat{f}(n)\hat{g}(n). \end{array} $

In the second step we have interchanged the integral and made a change of variable. Using the fact that $ {f(x)}$ and $ {e^{inx}}$ are $ {2\pi}$-periodic, so that the integral gives $ {\hat{f}(n)}$. $ \Box$

Exercise 3 What are the Fourier coefficients of $ {f*D_{N}\;(\theta)}$, where $ {D_{N}}$ is the $ {N}$-th Dirichlet kernel? What are the Fourier coefficients of $ {f*P_{r}\;(\theta)}$, where $ {P_{r}}$ is the Poisson kernel with radius $ {r}$, $ {0<r<1}$? Plot the Fourier coefficients of $ {D_{N}}$ and $ {P_{r}}$, and how are they different?

The above relation can be interpreted this way: as we have shown in Exercise 2, the mapping $ {f\mapsto f*g}$ is linear, and in analogy to matrices we can think of it as

$ \displaystyle f\mapsto Gf $

for some matrix $ {G}$. Expressing these function under Fourier series, we find the very same mapping expressed in Fourier coefficients is

$ \displaystyle \hat{f}(n)\mapsto2\pi\hat{g}(n)\hat{f}(n) $

or equivalently, in matrix form

$ \displaystyle \hat{f}=\begin{pmatrix}\vdots\\ \hat{f}(1)\\ \hat{f}(2)\\ \hat{f}(3)\\ \vdots \end{pmatrix}\mapsto2\pi\begin{pmatrix}\ddots\\ & \hat{g}(1)\\ & & \hat{g}(2)\\ & & & \hat{g}(3)\\ & & & & \ddots \end{pmatrix}\begin{pmatrix}\vdots\\ \hat{f}(1)\\ \hat{f}(2)\\ \hat{f}(3)\\ \vdots \end{pmatrix} $

in other words, the under the Fourier basis, the convolution is diagonalized! This will happen exactly the same for the discrete Fourier transform.
We have seen that under the Fourier basis the differential operator is also diagonalized. For completeness we record this result.

Exercise 4 Show that if $ {f}$ is differentiable and $ {2\pi}$-periodic, then

$ \displaystyle \widehat{f'}(n)=in\widehat{f}(n). $

In matrix form, this is

$ \displaystyle \begin{pmatrix}\vdots\\ \hat{f'}(1)\\ \hat{f'}(2)\\ \hat{f'}(3)\\ \vdots \end{pmatrix} =  i\begin{pmatrix}\ddots\\ & 1\\ & & 2\\ & & & 3\\ & & & & \ddots \end{pmatrix}\begin{pmatrix}\vdots\\ \hat{f}(1)\\ \hat{f}(2)\\ \hat{f}(3)\\ \vdots \end{pmatrix}. $

Remark 1 In fact, differentiation can also be represented as a convolution! (But to describe the kernel will be out of scope). In any case, the above diagonalization result can be seen as a special instance of a generalized version of Proposition 1.

Exercise 5 It is an instructive exercise to compute the convolution of two vectors $v,w\in\mathbb{C}^3$ (how to define convolution of two vectors of the same size?). Note that the matrix corresponds to the convolution operation commutes with the translation operator $\tau_y: f(x) \mapsto f(x-y)$. What special properties do the matrix have? What are the eigenvectors of this convolution matrix?

2. Using Fourier series to solve the heat equation on the circle

Let us consider the following heat equation on the unit circle with initial and boundary condition

$ \displaystyle \begin{cases} \frac{\partial u}{\partial t}-\frac{\partial^{2}u}{\partial x^{2}}=0, & 0<x<2\pi,\,t>0\\ u(x,0)=\sin2x+\sin5x+\sin7x, & 0<x<2\pi\\ u(0,t)=u(2\pi,t)=0, & t>0 \end{cases}. $

Below I record the solution version from last year. Unfortunately the argument is wrong. You are invited to think about why the correct solution is obtained by pure luck with the method below.

Write $ {u(x,t)=a_0(t)+\sum^\infty_{k=1}a_k(t)\cos kx+b_k(t)\sin kx}$. Using the PDE $ {0=\frac{\partial u}{\partial t}-\frac{\partial^2u}{\partial x^2}}$,

$ \displaystyle 0=(a_0)_t+\sum^\infty_{k=1}\left((a_k)_t+k^2a_k\right)\cos kx+\left((b_k)_t+k^2b_k\right)\sin kx \ \ \ \ \ (1)$

Using the initial condition

$ \displaystyle \sin2x+\sin5x+\sin7x=a_0(0)+\sum^\infty_{k=1}a_k(0)\cos kx+b_k(0)\sin kx \ \ \ \ \ (2)$

Comparing the coefficients of (2),

$ \displaystyle \begin{aligned} b_k(0)&=\begin{cases} 1,&\text{ if } k=2,5,7\\ 0,&\text{otherwise} \end{cases}\\ a_k(0)&=0\quad\forall k\geq0 \end{aligned} $

Comparing the coefficients of (1), then we have

$ \displaystyle \begin{aligned} (a_0)_t&=0\\ (a_k)_t+k^2a_k&=0\\ (b_k)_t+k^2b_k&=0 \end{aligned} $

Therefore, we have

$ \displaystyle \begin{aligned} a_k&=0,\quad\forall k\geq 0\\ b_k&=0,\quad \forall k\geq 1\text{ and }k\neq 2,5,7 \end{aligned} $

and for $ {k=2,5,7}$

$ \displaystyle b_k=e^{-k^2t} $

Hence, the solution is

$ \displaystyle u(x,t)=e^{-4t}\sin2x+e^{-25t}\sin 5x+e^{-49t}\sin 7x $

We will talk about why the above argument is NOT correct in the next note.

Remark 2 Let us suppose we have a nice $2\pi$-periodic continuous function whose Fourier series converges to the function itself, i.e. $$ u_{0}(\theta) = \sum_{n\in\mathbb{Z}}\hat{u_{0}}(n)e^{in\theta}, $$ and we set it to be the initial value of the heat equation. In addition if we think of this function to be defined on the circle, then there will be no boundary conditions. Then we can find that the solution to the heat equation is then $$ u(t,\theta) = \sum_{n\in\mathbb{Z}}e^{-tn^2}\hat{u_{0}}(n)e^{in\theta}. $$ Compare this expression with the solution to the Laplace equation. You can recognize it as obtained by convolution with certain function, which is known as the heat kernel. You may also want to compare the Fourier coefficients of the heat kernel with those of the Dirichlet and Poisson kernel.

Problem 1 What if the initial condition of the original problem is changed to $ u(0,x) = 20 $ for $0 < x < 2\pi$ and the boundary condition is changed to $u(t,0)=u(t,2\pi)=0$ for $ t>0$? What if the boundary condition is $u(t,0)=u(t,2\pi)=20$ ? (Hint: you may find the answer in this note.) 

MATH3310 Note 3: Best approximation property of Fourier series; "good" and "bad" approximation to the Dirac delta

1. Best approximation property of Fourier series

Last time we considered the vector space $ {C([0,2\pi],\mathbb{C})}$ (with coefficient in $ {\mathbb{C}}$, of course) and equipped it with the Hermite inner product

$ \displaystyle \langle f,g\rangle=\int_{[0,2\pi]}f(x)\overline{g(x)}dx. $

The inner product induces a norm $ {\|f\|=\left(\langle f,f\rangle\right)^{1/2}}$. The distance between two functions $ {f,g}$ is then $ {\|f-g\|}$.

Recall that the $ {n}$-th Fourier coefficient is obtained by

$ \displaystyle \hat{f}(n)=\frac{1}{2\pi}\langle f,e^{inx}\rangle, $

and we showed that the family $ {\{e^{inx}\}_{n\in\mathbb{Z}}}$ is orthogonal, meaning that

$ \displaystyle \langle e^{inx},e^{imx}\rangle=\begin{cases} 2\pi & \text{if }n=m\\ 0 & \text{otherwise} \end{cases}. $

In other words, $ {{\displaystyle \langle f,\frac{e^{inx}}{\|e^{inx}\|}\rangle\frac{e^{inx}}{\|e^{inx}\|}=\hat{f}(n)e^{inx}}}$ is obtained by projecting $ {f}$ onto the 1-dimensional linear subspace spanned by $ {e^{inx}}$. And the Fourier series of $ {f}$ the summation of all these orthogonal pieces together

$ \displaystyle \sum_{n\in\mathbb{Z}}\hat{f}(n)e^{inx}. $

With this geometric picture in mind we can now prove quite easily that, e.g. the projected function $ {\hat{f}(n)e^{inx}}$ minimizes the distance

$ \displaystyle \|f-g\|,\quad g=ce^{inx},c\in\mathbb{C}. $

This will follow from Pythagoras theorem for inner product spaces.

Theorem 1 (Pythagoras) Let $ {(V,\langle\cdot,\cdot\rangle)}$ be an inner product space. If $ {X,Y\in V}$ and $ {\langle X,Y\rangle=0}$, then

$ \displaystyle \|X+Y\|^{2}=\|X\|^{2}+\|Y\|^{2}. $

Proof: Expanding the squre $ {\|X+Y\|^{2}}$ and using the bilinearity and symmetric property of the inner prodcut we find that

$ \displaystyle \begin{array}{rcl} \|X+Y\|^{2} & = & \langle X+Y,X+Y\rangle\\ & = & \langle X,X+Y\rangle+\langle Y,X+Y\rangle\\ & = & \langle X,X\rangle+\langle X,Y\rangle+\langle Y,X\rangle+\langle Y,Y\rangle\\ & = & \|X\|^{2}+\|Y\|^{2}. \end{array} $

$ \Box$

 Now to show the minimization property, we consider for arbitrary $ {c\in\mathbb{C}}$,

$ \displaystyle \|f-ce^{inx}\|^{2}=\|f-\hat{f}(n)e^{inx}+(\hat{f}(n)-c)e^{inx}\|. $

Note that

$ \displaystyle \langle f-\hat{f}(n)e^{inx},e^{inx}\rangle=\langle f,e^{inx}\rangle-2\pi\hat{f}(n)=0, $

so that $ {f-\hat{f}(n)e^{inx}}$ is orthogonal to $ {(\hat{f}(n)-c)e^{inx}}$. Then we have by the Pythagoras theorem,

$ \displaystyle \|f-ce^{inx}\|^{2}=\|f-\hat{f}(n)e^{inx}\|^{2}+\|(\hat{f}(n)-c)e^{inx}\|^{2}. $

Since $ {\|(\hat{f}(n)-c)e^{inx}\|^{2}\geq0}$ for any $ {c}$, this implies

$ \displaystyle \|f-\hat{f}(n)e^{inx}\|^{2}\leq\|f-ce^{inx}\|^{2},\quad\forall c\in\mathbb{C}. $

Exercise 1 Generalise the above argument to any finite sum of the terms in the Fourier series with $ {\text{span}\{e^{inx}\}}$ replaced the corresponding subspace. We thus see that the finite Fourier sum are the best approximation of the given function $ {f}$ using complex exponentials (i.e. trignometric polynomials), using the norm $ {\|\cdot\|}$ we defined before as the "error metric''.

Remark 1 The space $ {C([0,2\pi],\mathbb{C})}$ is particular nice since we can always make sense of the integrals such as $ {\int_{[0,2\pi]}f(x)e^{-inx}dx}$ whenever $ {f\in C([0,2\pi],\mathbb{C})}$ as a Riemann integral. The above argument will also work for the more general space of Riemann square-integrable functions and Lebesgue square-integrable functions.

We have one more remarkable property of Fourier coefficients, namely

$ \displaystyle \sum_{n\in\mathbb{Z}}|\hat{f}(n)|^{2}=\int_{[0,2\pi]}|f(x)|^{2}dx,\quad f\in C([0,2\pi],\mathbb{C}) $

We call this the isometry property of Fourier coefficient, and the equality is often called the Parseval's identity. This follows from the best approximation property and the Weierstrass approximation theorem, which states that continuous functions can be approximated to arbitrary accuracy by trignometric polynomials. We won't get into the details here. After we introduce the discrete Fourier transform, you will definitely recognize the similarity. Spoiler: discrete Fourier transform is a unitary linear transform of finite dimensional vector space (or unitary up to a constant, depending on the definition convention).

2. Dirichlet kernel; Poisson kernel

The Dirichlet kernel arises from the symmetric partial sum formula

$ \displaystyle \begin{array}{rcl} \sum_{|n|\leq N}\hat{f}(n)e^{in\theta} & = & \sum_{|n|\leq N}\frac{1}{2\pi}\int_{[0,2\pi]}f(x)e^{-inx}e^{in\theta}dx\\ & = & \frac{1}{2\pi}\int_{[0,2\pi]}f(x)\sum_{|n|\leq N}e^{i(\theta-x)n}dx\\ & = & f*D_{N}(\theta) \end{array} $

where $ {*}$ stands for convolution, and $ {D_{N}(x)=\frac{1}{2\pi}\sum_{|n|\leq N}e^{ixn}}$ is called the $ {n}$-th Dirichlet kernel.

Exercise 2 Show that for each $ {N\in\mathbb{N}}$,
$$ \displaystyle \int_{[0,2\pi]}\frac{1}{2\pi}\sum_{|n|\leq N}e^{inx}dx=1. $$

We showed in class that

$ \displaystyle \sum_{|n|\leq N}e^{ixn}=\frac{\sin(N+\frac{1}{2})x}{\sin\frac{1}{2}x} $

and below I plotted the graph for this kernel when $ {N=1,4,10}$ respectively.

 You can see that the kernel fluctuate more and more when $ {N}$ gets larger, peaking at origin. Notice also that the it takes both positive and negative values. This type of fluctuatation should be considered as very large. In fact, one can show

$ \displaystyle \int_{[0,2\pi]}|D_{N}(x)|dx\geq C\log N $

for some constant $ {C>0}$. This implies that the absolute integral of the Dirichlet kernels cannot be uniformly bounded. It is this fact that leads to the Gibbs phenomenon we mentioned in the last tutorial. Nevertheless, it is possible to make sense of the convergence in appropriate sense

$ \displaystyle D_{N}(x)\rightarrow\delta_{0}(x) $

where $ {\delta_{0}(x)}$ is the "Dirac delta function'' at the origin. For example, the pointwise convergence of the symmetric partial sum holds for smooth functions; also, the convergence

$ \displaystyle \sum_{|n|\leq N}\hat{f}(n)e^{inx}\rightarrow f(x)\text{ in the norm }\|\cdot\|, $

can be seen as

$ \displaystyle f*D_{N}\rightarrow f*\delta_{0}=f\text{ in the norm }\|\cdot\|. $

We only need to know the qualitative behavior here. In any case, the Dirichlet kernel can be thought of a "bad'' approximation of the "Dirac delta function''.

Now we turn to the Poisson kernel. It is closely related to the following boundary value problem

$ \displaystyle \begin{cases} \Delta u=0 & \text{in }B(0,1)\subset\mathbb{R}^{2}\\ u=u_{0} & \text{on }\partial B(0,1) \end{cases}. \ \ \ \ \ (1)$

In class we made the ansatz that the solution is of the form

$ \displaystyle u(r,\theta)=a_{0}+\sum_{k=1}^{\infty}(a_{k}\cos k\theta+b_{k}\sin k\theta)r^{k} $

where $ {(r,\theta)}$ is the polar coordinate in $ {\mathbb{R}^{2}}$. Using Exercise 4 of Tutorial note 2, we can recognize it as the so-called Abel summation of the Fourier series of a real valued function

$ \displaystyle \sum_{n\in\mathbb{Z}}r^{|n|}\hat{g}(n)e^{in\theta}. $

Note that when $ {0<r<1}$, this expression is absolutely summable if e.g. $ {\hat{g}(n)}$ are uniformly bounded!
What is the function $ {g}$ here? Suppose $ {g}$, defined on the unit circle $ {\partial B(0,1)}$, is nice and smooth. Letting $ {r\rightarrow1}$, we should have

$ \displaystyle \sum_{n\in\mathbb{Z}}\hat{g}(n)e^{in\theta}=g(\theta)=u_{0}(\theta). $

In other words,

$ \displaystyle \hat{g}(n)=\widehat{u_{0}}(n). $

Hence

$ \displaystyle \begin{array}{rcl} u(r,\theta) & = & \sum_{n\in\mathbb{Z}}r^{|n|}\widehat{u_{0}}(n)e^{in\theta}\\ & = & \sum_{n\in\mathbb{Z}}r^{|n|}\frac{1}{2\pi}\int_{[0,2\pi]}u_{0}(x)e^{-inx}dx\\ & = & \frac{1}{2\pi}\int_{[0,2\pi]}u_{0}(x)\sum_{n\in\mathbb{Z}}r^{|n|}e^{in(\theta-x)}dx \end{array} $

which again we recognize as a convolution

$ \displaystyle u(r,\theta)=u_{0}*P_{r}(\theta) $

where

$ \displaystyle \begin{array}{rcl} P_{r}(\theta) & = & \frac{1}{2\pi}\sum_{n\in\mathbb{Z}}r^{|n|}e^{in\theta}\\ & = & \frac{1}{2\pi}(\frac{re^{-i\theta}}{1-re^{-i\theta}}+1+\frac{re^{i\theta}}{1-re^{i\theta}})\\ & = &\frac{1}{2\pi} \frac{1-r^{2}}{1+r^{2}-2r\cos\theta} \end{array} $

is the Poisson kernel with parameter $ {r}$. We can check that the solution to the boundary value problem (1) is given by convolution of the boundary value with the Poisson kernel.

Exercise 3 Show that for all $ {0<r<1}$,
$$ \displaystyle \int_{[0,2\pi]}|P_{r}(\theta)|d\theta=1. $$
Note that $ {P_{r}(\theta)>0}$ for all $ {0<r<1}$ and $ {\theta\in[0,2\pi]}$.

 Now you should think of $ {u(r,\theta)}$ as an approximation of $ {u_{0}}$ here, where you have a parameter $ {r}$ to adjust how closely you want it to approximate $ {u_{0}}$ (by making $ {r}$ close to $ {1}$) or how smooth you want $ {u(r,\theta)}$ (by making $ {r}$ closer to zero). You can compare $ {r}$ with the $ {N}$ in Dirichlet kernel. I plotted the kernel for $ {r=0.3,0.5,0.8}$ below.

Notice that the kernel is always positive and its absolute integral is uniformly bounded. This stands in strong contrast to the Dirichlet kernel. And we say that the Poisson kernal is a "good'' approximation of the "Dirac delta function".

There is also a physical interpretation for the Poisson kernel. Imaging putting a ideal charge on the boundary of the disk, say at $(1,0)$. Then the Poisson kernel describes the (negative) electrical potential inside the unit disk. I plotted the figure below.

MATH3310 Note 2: Real and complex Fourier series; Gibbs phenomenon

The idea of representing a general function on e.g. the interval $ {[0,2\pi]}$ by an infinite sum of sines and cosines was invented by Joseph Fourier (1768-1830) in the context of solving the 1D wave equation and the 2D heat equation on the unit disk. For this matter you are recommended to have a look at Chapter 1 of Fourier Analysis: an introduction by Stein and Shakarchi. Our lecture also motivates the study from roughly the same perspective. Here I want to stress one more point of view from the beginning which you should remember

  The Fourier series of a function is a representation of the function under a particular "basis". 

 You have probably often heard of "spectral decomposition'' or equivalently "eigen-decomposition'' used to describe the Fourier series. These terminologies are used essentially because constant coefficent linear differential operators are diagonalized under such a representation. Of course, diagonalized systems are much easier to solve!

But at this moment we won't jump ahead of ourselves in a hurry to understand what these statements mean precisely. In this post we aim to grasp some basic properties of the Fourier series.

A technical note: the object of study here appears in basically three equivalent forms:

• functions on the interval $ {[0,2\pi]}$ (or $ [0,2\pi)$, or $(0,2\pi)$, etc)
• functions defined on the unit circle $ {S^{1}}$;
•  $ {2\pi}$-periodic functions

The last two categories can be easily seen to be equivalent, while for the first one, to be strictly correct, one has to require $f(0)=f(2\pi)$. We can safely ignore this issue since the effect of a point in the integration is negligible, but at the same time it's important to remember that pointwise convergence issue will arise once we choose to adapt such an relaxation. See Section 2.

Some Examples of the Calderón-Zygmund operators

In this post we study various examples of ``singular integral operator''. Those included here are

• The Riesz transforms, which recover the Hessian of a Schwartz function from its Laplacian;
• Regular parametrix of constant coefficient linear (hypo)elliptic operators, and their distributional derivatives, which are responsible for the regularity property of their solutions;
• Littlewood Paley projections, which is a convenient tool of multiresolution analysis.

In the above examples the kernel of the operator is ``singular'', in the sense that it's not always locally integrable, yet the operator has nice boundedness properties, established by Calderón and Zymund, and others. Roughly speaking, that they can have such properties is due to the following two properties:

• The kernels have certain cancellation condition, or when it is a convolutional operator, the closely related property that their Fourier transform (i.e. corresponding multiplier) is bounded. In general it is the $L^2$ boundedness of the operator. This in some sense makes it possible to bound those functions that have big height but low oscillation.
• The kernels, and their derivatives, have "integrable decay'', or the closely related property that their multipliers are $ {C^{\infty}}$ functions that satisfy some differential inequalities corresponding to that decay. This in some sense makes it possible to bound those functions that have low height but big oscillation.

There is a more precise procedure to implement above intuitive argument, that is the famous Calderón-Zygmund decomposition, together with real interpolation. However, it should be noted that the argument only leads to a weak type bound in the endpoint case $L^{1}$ and $L^{\infty}$, and in that case more delicate nature of the functions (cf. the $\bf{H^{1}}$ and $BMO$) have to be considered. These boundedness properties will not be studied here, instead I want to present how these examples arise naturally in various questions, then I will identify them as such operators, and explain why studying them is important.

Singular integrals: a starter

 We begin with the simpliest subject of our current theme of study, namely the Hilbert transform. It often appears in two forms. One is as a Fourier multiplier operator: in the case the underlying space is $ {\mathbb{R}}$,

$ \displaystyle H(f)(x)=\int_{\mathbb{R}}f^{\wedge}(\xi)\frac{\text{sign}(\xi)}{i}e^{2\pi ix\xi}\thinspace d\xi $

for $ {f\in L^{2}(\mathbb{R})}$, where

$ \displaystyle \text{sign}(\xi)=\begin{cases} 1 & \xi>0\\ 0 & \xi=0\\ -1 & \xi<0 \end{cases};\ f^{\wedge}(\xi):=\int_{\mathbb{R}}f(x)e^{-2\pi i\xi x}\,dx $

or in the case the underlying space is a circle $ {\mathbb{T}}$,

$ \displaystyle H(f)(\theta)=\sum_{n\in\mathbb{Z}}f^{\wedge}(n)\frac{\text{sign}(n)}{i}e^{in\theta} $

where $ {f\in L^{2}([0,2\pi])}$, and

$ \displaystyle \text{sign}(n)=\begin{cases} 1 & n>0\\ 0 & n=0\\ -1 & n<0 \end{cases};\ f^{\wedge}(n):=\frac{1}{2\pi}\int_{[0,2\pi]}f(x)e^{-inx}\,dx. $

(Note that the sign function vanishes on a measure zero set on $ {\mathbb{R}}$, but on a measure 1 set on $ {\mathbb{Z}}$. This ultimately causes the constant mode of the periodic function jumping out in various identities, compared to their analogs for functions on $ {\mathbb{R}}$. This is however not essential.) The other formulation is as a "singular integral operator'', as will be ellaborated below. It is in the 19th and earlier 20th centuries that the properties of this operator came into interest, for its fundamental connection with the following two questions that arised in Analysis.

The first one is that of convergence, in some appropriate sense, of the symmetric partial sum of the Fourier series of an integrable function $ {f:[0,2\pi]\rightarrow\mathbb{C}}$

$ \displaystyle S_{N}(f)(\theta):=\sum_{|n|\leq N}f^{\wedge}(n)e^{in\theta}\rightarrow f(\theta)\ ? $

At that time Lebesgue's integration theory had not been developed. However, that under the condition of $ {C^{1}}$ regularity of the function, the pointwise convergence was already clear to Dirichlet, in 1829, who told in private that he would soon resolve the case when $ {f}$ is continuous. This was disproved in 1873, by du Bois Reymond. Then after the work of Lebesgue, under the condition of square integrability of $ {f}$, the $ {L^{2}}$ convergence can be established using elemental Hilbert space technique, invented during the end of the 19th century. Later, achieved by Marcel Riesz in 1923, the $ {L^{p}}$ convergence was shown to be true for all $ {1<p<\infty}$. Nevertheless, the "local'' convergence behavior turns out much more intricate, and the a.e. convergence for $ {L^{2}}$ functions was eventually resolved by Carleson, in 1966.

The Hilbert transform enters the above question as follows. The symmetric partial sum can be regarded as the convolution with the Dirichlet kernel (this is no incidence: a bounded multiplier must correspond to a distributional kernel that commutes with translations)

$ \displaystyle S_{N}(f)(\theta)=f*D_{N}\thinspace(\theta), $

where

$ \displaystyle D_{N}(x)=\sum_{|n|\leq N}e^{inx}. $

The integral operator so defined is actually the orthogonal projection $ {P_{|n|\leq N}}$ to the space spanned by $ {\{e^{inx}\}_{|n|\leq N}}$, which is associated with the following multiplier on the intergers

$ \displaystyle m_{S_{N}}(n)=\chi_{|n|\leq N}=\begin{cases} 1 & |n|\leq N\\ 0 & \text{otherwise} \end{cases}. $

Denote the multiplier of the Hilber transform on the circle by $ {m_{H}(n):=\frac{1}{i}\text{sign}(n)}$. And notice that

$ \displaystyle m_{S_{N}}(n)=\frac{i}{2}\left(m_{H}(n+N)-m_{H}(n-N)\right)+\frac{1}{2}\chi_{\{\pm N\}}. $

Then, in terms of operator,

$ \displaystyle S_{N}(f)(\theta)=\frac{ie^{-iN\theta}H(e^{iN\theta}f)-ie^{iN\theta}H(e^{-iN\theta}f)}{2}+\frac{P_{\pm N}}{2}. $

Hence the study of the boundedness properties of the family of operators $ {S_{N}}$ reduces to the boundedness property of operators $ {H}$ and $ {P_{\pm N}}$.

The second question is concerned with a different summation method of Fourier series, and its connection to complex analysis. Here for each $ {\theta\in[0,2\pi]}$, the Fourier series is called Abel summable to $ {f(\theta)}$ if

$ \displaystyle \sum_{n\in\mathbb{Z}}f^{\wedge}(n)r^{|n|}e^{in\theta}\rightarrow f(\theta) $

as $ {r\rightarrow1}$. Similar to the symmetric partial sum, Abel summation can also be written as a convolution of $ {f}$ with a kernel. This time it is the Poisson kernel for the unit disk

$ \displaystyle f*\mathcal{P}_{r}\thinspace(\theta), $

where

$ \displaystyle \mathcal{P}_{r}(\theta)=\sum_{n\in\mathbb{Z}}r^{|n|}e^{in\theta}=\frac{1-r^{2}}{1-2r\cos\theta+r^{2}}. $

We remark that if for a.e. $ {\theta\in[0,2\pi]}$, a Fourier series is Abel summable and moreover, $ {f^{\wedge}(n)=0}$ for all $ {n<0}$, then $ {f}$ , originally defined on the unit circle, can be extended to a holomorphic function inside the unit disk, by

$ \displaystyle F(z)=\sum_{n\in\mathbb{Z}}f^{\wedge}(n)z^{n}. $

It is of interest that when the a.e. convergence holds. In sharp contrast to the symmetric partial sum method, it turns out that the result holds even for integrable functions. This is to be explained by the fact that the family of Poisson kernels, indexed in $ {r}$, forms an approximations to the identity, while it is not the case for the Dirichlet kernels.

Now we ask the converse, that is, given a holomorphic function $ {F}$ in the unit disk, then under what conditions $ {F}$ will have radial limit on the circle, i.e.

$ \displaystyle \lim_{r\rightarrow1^{-}}F(re^{i\theta})\ \text{ exists for a.e. $\theta\in [0,2\pi]$?} $

Suppose $ {F}$ has the power series expansion $ {\sum_{n=0}^{\infty}a_{n}z^{n}}$. Now by Parseval's identity

$ \displaystyle \sum_{n=0}^{\infty}|a_{n}|^{2}r^{2n}=\frac{1}{2\pi}\int_{[0,2\pi]}|F(re^{i\theta})|^{2}d\theta,\ 0\leq r<1. $

Thus, assume that

$ \displaystyle \sup_{0\leq r<1}\int_{[0,2\pi]}|F(re^{i\theta})|^{2}d\theta<\infty, \ \ \ \ \ (1)$

so that the series $ {\sum_{n=0}^{\infty}|a_{n}|^{2}}$ converges to the above quantity, and thus has a corresponding $ {L^{2}}$ function $ {F_{0}(e^{i\theta}).}$ Since $ {L^{2}([0,2\pi])\subset L^{1}([0,2\pi])}$, by the above paragraph we have the limit $ {\lim_{r\rightarrow1^{-}}F(re^{i\theta})}$ exists for a.e. $ {\theta}$. This result was obtained by Fatou.

The space of holomorphic functions on the unit disk satisfying the bound (1) is called the Hardy space $ {H^2(\mathbb{D})}$, with norm $ {\|F\|_{H^{2}(\mathbb{D})}}$ defined to be the quantity on the LHS of (1). It can be identified with the closed subspace $ {\ell^{2}(\mathbb{Z}_{+})}$ of the Hilbert space $ {\ell^{2}(\mathbb{Z})}$, or equivalently the space $ {\mathcal{S}}$ of square integrable functions on the cirlce whose Fourier coefficients vanish for $ {n<0}$. In view of this, define the orthogonal projection from $ {L^{2}([0,2\pi])}$ to $ {\mathcal{S}}$, and write in the corresponding element in $ {H^{2}(\mathbb{D})}$

$ \displaystyle \begin{array}{rcl} P(f)(z): & = & \sum_{n=0}^{\infty}a_{n}z^{n}.\\ & = & \sum_{n=0}^{\infty}\left(\frac{1}{2\pi}\int_{[0,2\pi]}f(e^{i\theta})e^{-in\theta}\thinspace d\theta\right)z^{n}\\ & = & \frac{1}{2\pi}\int_{[0,2\pi]}f(e^{i\theta})\sum_{n=0}^{\infty}\left(e^{-i\theta}z\right)^{n}\thinspace d\theta\\ & = & \frac{1}{2\pi i}\int_{S^{1}}\frac{f(\zeta)}{\zeta-z}\thinspace d\zeta, \end{array} $

which turns out to be the Cauchy integral on the circle. On the other hand, notice that

$ \displaystyle P(f)(e^{i\theta})=\frac{f(e^{i\theta})+iH(f)(e^{i\theta})}{2}. \ \ \ \ \ (2)$

For this reason, $ {H(f)}$ is often called conjugate to $ {f}$, and hence can be also identified the imaginary part of the Cauchy kernel of the unit circle

$ \displaystyle \frac{1}{i(\zeta-z)}d\zeta=\frac{ie^{i\psi}d\psi}{i(e^{i\psi}-re^{i\theta})}=\mathcal{P}_{r}(\theta-\psi)d\psi+\frac{d\psi}{2}+i\mathcal{Q}_{r}(\theta-\psi)d\psi. $

Here $ {Q_{r}(\theta)}$ is known as the conjugate Poisson kernel. In explicit form,

$ \displaystyle \mathcal{Q}_{r}(\theta-\psi)=\frac{r\sin(\theta-\psi)}{1-2r\cos\theta+r^{2}}\rightarrow\frac{1}{2}\cot(\frac{\theta-\psi}{2}) $

as $ {r\rightarrow1^{-}}$. Unlike Poisson's kernel, it is not absolutely integrable, thus called "singular'', near the diagonal $ {\theta=\psi}$, but note that $ {\cot(x)}$ is an odd function on $ {[-\pi,\pi]}$, which is one manifestation of the {cancellation property} for such singular integral operators.

Now we turn to the analogous results of the above for functions on the real line, where the unit disk is replaced by the upper half space $ {\mathbb{R}_{+}^{2}:=\{(x,y)\in\mathbb{R}^{2}:y>0\}}$. Starting with a function $ {F_{0}^{\wedge}\in L^{2}(0,\infty)}$, define its Laplace transform to be

$ \displaystyle F(x+iy)=\int_{[0,\infty)}F_{0}^{\wedge}(\xi)e^{2\pi i\xi z}\thinspace d\xi, \ \ \ \ \ (3)$

which can be viewed as an "integral version'' of Abel summation of the Fourier series $ {\sum_{n\geq0}f^{\wedge}(n)}$. (The analogue for $ {\sum_{n\in\mathbb{Z}}f^{\wedge}(n)}$ will be given by $ {\int_{\mathbb{R}}f^{\wedge}(\xi)e^{-2\pi|\xi|y}e^{2\pi ix\xi}\thinspace d\xi}$.) Since for any $ {\delta>0}$, if $ {y\geq\delta}$, the integral has the uniform bound

$ \displaystyle \int_{[0,\infty)}|F_{0}^{\wedge}(\xi)e^{2\pi i\xi z}|\thinspace d\xi\leq\left(\int_{[0,\infty)}|F_{0}^{\wedge}(\xi)|^{2}\thinspace d\xi\right)^{1/2}\left(\int_{[0,\infty)}e^{-4\pi\xi\delta}\thinspace d\xi\right)^{1/2}, $

so $ {F}$ is in fact holomorphic in $ {\mathbb{R}_{+}^{2}}$. Plancherel's identity implies that

$ \displaystyle \int_{\mathbb{R}}|F(x+iy)|^{2}\thinspace dx=\int_{[0,\infty)}|F_{0}^{\wedge}(\xi)|^{2}e^{-4\pi\xi y}\thinspace d\xi\leq\|F_{0}^{\wedge}\|_{L^{2}(0,\infty)}^{2} $

and by the monotone convergence theorem

$ \displaystyle \sup_{y>0}\int_{\mathbb{R}}|F(x+iy)|^{2}\thinspace dx=\|F_{0}^{\wedge}\|_{L^{2}(0,\infty)}^{2}. $

As before, the Hardy space $ {H^{2}(\mathbb{R}_{+}^{2})}$ is defined to be the space of holomorphic functions on $ {\mathbb{R}_{+}^{2}}$ satisfying the condition

$ \displaystyle \sup_{y>0}\int_{\mathbb{R}}|F(x+iy)|^{2}\thinspace dx<\infty, $

and equipped with the norm $ {\|F\|_{H^{2}(\mathbb{R}_{+}^{2})}}$ to be the above quantity.

Given such a function $ {F}$ with suitable decaying property, via a contour integration argument, its inverse Laplace transform satisfies, for any $ {y_{1},y_{2}>0}$.

$ \displaystyle \int_{\mathbb{R}}F(x+iy_{1})e^{-2\pi i\xi(x+iy_{1})}\thinspace dx=\int_{\mathbb{R}}F(x+iy_{2})e^{-2\pi i\xi(x+iy_{2})}\thinspace dx. $

In other words, there is a deck transformation rule

$ \displaystyle F_{y_{1}}^{\wedge}(\xi)e^{2\pi y_{1}\xi}=F_{y_{2}}^{\wedge}(\xi)e^{2\pi y_{2}\xi}. $

A limiting argument shows this it is true for all $ {F\in H^{2}(\mathbb{R}_{+}^{2})}$. Hence the function $ {F_{0}^{\wedge}(\xi):=F_{y}^{\wedge}(\xi)e^{2\pi y\xi}}$ is well defined, and by Plancherel's identity

$ \displaystyle \int_{\mathbb{R}}|F(x+iy)|^{2}\thinspace dx=\int_{\mathbb{R}}|F_{0}^{\wedge}(\xi)|^{2}e^{-4\pi y\xi}\thinspace dx. $

Thus,

$ \displaystyle \|F\|_{H^{2}(\mathbb{R}_{+}^{2})}=\int_{\mathbb{R}}|F_{0}^{\wedge}(\xi)|^{2}\thinspace dx $

and $ {H^{2}(\mathbb{R}_{+}^{2})}$ can be identified with the Hilbert space $ {L^{2}(0,\infty)}$ via (3).

It then follows that $ {F}$ has boundary values $ {F_{0}}$, the inverse Fourier transform of $ {F_{0}^{\wedge}}$, in the $ {L^{2}}$ sense. It also holds a.e., once we have the Poisson integral representation

$ \displaystyle F(x+iy)=\int_{[0,\infty)}F_{0}^{\wedge}(\xi)e^{2\pi i\xi z}\thinspace d\xi=\int_{\mathbb{R}}F_{0}(x-t)\mathcal{P}_{y}(t)\thinspace dt $

or more generally,

$ \displaystyle \int_{\mathbb{R}}f^{\wedge}(\xi)e^{-2\pi|\xi|y}e^{2\pi ix\xi}\thinspace d\xi=\int_{\mathbb{R}}f(x-t)\mathcal{P}_{y}(t)\thinspace dt $

where $ {\mathcal{P}_{y}(x)=\frac{1}{\pi}\frac{y}{x^{2}+y^{2}}}$ is the Poisson kernel for the upper half space. To see this, define the Cauchy kernel for $ {y>0}$

$ \displaystyle C(x+iy)=\int_{\mathbb{R}}e^{2\pi i|\xi|z}\thinspace d\xi=\int_{\mathbb{R}}2\chi_{[0,\infty)}e^{2\pi i\xi z}\thinspace d\xi=\frac{i}{\pi z}. $

Then

$ \displaystyle \begin{array}{rcl} \mathcal{P}_{y}(x) & = & \int_{\mathbb{R}}e^{-2\pi|\xi|y}e^{2\pi ix\xi}\thinspace d\xi\\ & = & \int_{[0,\infty)}e^{-2\pi\xi y}e^{2\pi ix\xi}\thinspace d\xi+\int_{[0,\infty)}e^{-2\pi\xi y}e^{-2\pi ix\xi}\thinspace d\xi\\ & = & \frac{1}{2}(C(x+iy)+C(-x+iy))\\ & = & \frac{1}{\pi}\frac{y}{x^{2}+y^{2}}. \end{array} $

Notice that for $ {y>0}$

$ \displaystyle \frac{i}{\pi z}=\mathcal{P}_{y}(x)+i\mathcal{Q}_{y}(x) $

where $ {\mathcal{Q}_{y}(x)=\frac{1}{\pi}\frac{x}{x^{2}+y^{2}}}$ is the conjugate Poisson kernel. We would like to know what happens when $ {y\rightarrow0}$. This is the clearest in terms of multipliers. We have, corresponding to the above,

$ \displaystyle 2\chi_{[0,\infty)}e^{-2\pi\xi y}=e^{-2\pi|\xi|y}+\frac{1}{i}\text{sign}(\xi)e^{-2\pi|\xi|y} $

and as $ {y\rightarrow0}$,

$ \displaystyle 2\chi_{[0,\infty)}(\xi)=1+\text{sign}(\xi)=1+im_{H}(\xi), $

where $ {m_{H}(\xi)=\frac{1}{i}\text{sign}(\xi)}$ is the Fourier multiplier of the Hilbert transform on the real line. One thus has

$ \displaystyle H(f)(x)=\lim_{y\rightarrow0}\int_{\mathbb{R}}f(x-t)\mathcal{Q}_{y}(t)\thinspace dt $

in $ {L^{2}}$. We shall now focus on distributional property of $ {\mathcal{Q}_{y}(t)}$ as $ {y\rightarrow0}$. Assume $ {\varphi}$ is a test function. For any $ {\epsilon>0}$,

$ \displaystyle \begin{array}{rcl} & & \int_{\mathbb{R}}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt\\ & = & \int_{|t|<\epsilon}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt+\int_{|t|\geq\epsilon}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt\\ & = & \int_{|t|<\epsilon}(\varphi(t)-\varphi(0))\mathcal{Q}_{y}(t)\thinspace dt+\int_{|t|\geq\epsilon}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt\\ & =: & A+B \end{array} $

Here the {cancellation property} of $ {\mathcal{Q}_{y}(x)}$ plays a role. It is clear that $ {A=O(\epsilon)}$, and so

$ \displaystyle \int_{\mathbb{R}}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt=\lim_{\epsilon\rightarrow0}\int_{|t|\geq\epsilon}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt. $

We now wish to take $ {y\rightarrow0}$. When $ {|t|\geq\epsilon}$ it is clear that $ {\mathcal{Q}_{y}(t)\rightarrow\frac{1}{\pi t}}$ as $ {y\rightarrow0}$. This amounts to estimate

$ \displaystyle \frac{1}{\pi}\int_{|t|\geq\epsilon}\varphi(t)\thinspace\frac{dt}{t}-\int_{\mathbb{R}}\varphi(t)\mathcal{Q}_{\epsilon}(t)\thinspace dt=\int_{\mathbb{R}}\varphi(t)\Delta_{\epsilon}(t)\thinspace dt, $

where

$ \displaystyle \Delta_{\epsilon}(t)=\begin{cases} -\mathcal{Q}_{\epsilon}(t) & \text{if }|t|<\epsilon\\ \frac{1}{t}-\mathcal{Q}_{\epsilon}(t) & \text{if }|t|\geq\epsilon \end{cases}, $

which again satisfies a cancellation property. It's then not hard to see that

$ \displaystyle \begin{array}{rcl} & & \left|\int_{\mathbb{R}}\varphi(t)\Delta_{\epsilon}(t)\thinspace dt\right|\\ & = & \left|\int_{|t|<\epsilon}(\varphi(t)-\varphi(0))\Delta_{\epsilon}(t)\thinspace dt\right|+\left|\int_{\epsilon\leq|t|<1}(\varphi(t)-\varphi(0))\Delta_{\epsilon}(t)\thinspace dt\right|+\left|\int_{|t|\geq1}\varphi(t)\Delta_{\epsilon}(t)\thinspace dt\right|\\ & = & O(\epsilon+\epsilon\log\epsilon+\epsilon). \end{array} $

We conclude that $ {\mathcal{Q}_{y}(x)\rightarrow\frac{1}{\pi}\text{p.v.}(\frac{1}{x})}$ in distribution as $ {y\rightarrow0}$, and the Hilbert transform ca be written as a singular integral operator

$ \displaystyle H(f)=\frac{1}{\pi}\text{p.v.}(\frac{1}{x})*f. $

Moreover, since $ {\mathcal{P}_{y}(x)\rightarrow\frac{1}{\pi}\delta_{0}}$ in distributon as $ {y\rightarrow0}$, denoting the distributional limit of $ {\frac{1}{z}}$ to be $ {\frac{1}{x+i0}}$ as $ {y\rightarrow0}$, we have the following distributional identity for kernels

$ \displaystyle \frac{1}{x+i0}=\text{p.v.}(\frac{1}{x})-i\pi\delta $

corresponding to identity for multipliers above.
Again, one can define the orthogonal projection $ {P}$ from $ {L^{2}(\mathbb{R})}$ to the closed subspace that contains functions whose Fourier transforms are supported on $ {[0,\infty)}$. It is realized by the Cauchy integral

$ \displaystyle F(z)=\int_{\mathbb{R}}\frac{f(t)}{t-z}\thinspace dt=\int_{[0,\infty)}f^{\wedge}(\xi)e^{2\pi i\xi z}\thinspace d\xi,\ z\in\mathbb{R}_{+}^{2} $

which maps $ {L^{2}(\mathbb{R})}$ onto $ {H^{2}(\mathbb{R}_{+}^{2})}$.

Finite Difference Methods for One Way Wave Equation

1.1. A prototypical example: Euler method for an ODE

Already in the 18th century, Leonhard Euler had considered the problem of finding an approximation of the integral curve associated to an initial value problem of an ordinary differential equation. The solution to the initial value problem can be represented as a parametrised curve lying in the spatial domain: one can imagine it is a particle going along the curve, the position and velocity at certain time is defined by the differential equation and the initial position and velocity of the particle. More concretely, we may consider the first order ODE with an initial condition

$ \displaystyle \frac{dx}{dt}(t)=f(t,x(t))\quad x(0)=x_{0} $

where, say, the solution lies in the plane $ {x(t)\in\mathbb{R}^{2}}$ for time interval $ {t\in(0,1)}$, and $ {f:(0,1)\times\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}}$. The famous Picard-Lindel{ö}f theorem asserts that if $ {f}$ is uniformly Lipschitz in the spatial variable $ {x}$ and continuous in the time variable $ {y}$, then there exists a unique solution $ {x(t)}$ to the initial value problem. Of course, we don't expect to get closed form solution for every given right hand side satisfying the theorem. So, as Euler did, we want to try to directly integrate the differential equation by discrete approximation. The obvious thing to try is

$ \displaystyle x(t_{i})-x(t_{i-1})\approx\Delta t\cdot f(t_{i-1},x(t_{i-1})), \ \ \ \ \ (1)$

where the time interval has been discretized $ {0=t_{0}<t_{1}<\cdots<t_{n}=1}$ with equal distance $ {\Delta t}$. Taking approximation

$ \displaystyle \tilde{x}(t_{i})-\tilde{x}(t_{i-1})=\Delta t\cdot f(t_{i-1},\tilde{x}(t_{i-1})), $

we can then compute $ {\tilde{x}(t_{i})}$ assuming $ {x(t_{i-1})}$ is known. This is called Euler method. Since we are computing $ {\tilde{x}(t_{i})}$ in terms of its values at previous times, we say it is an explicit method. In Riemann integration terms, we note that here we evaluate the integrand at the left endpoint in each subinterval. If evaluating at the right endpoint we arrive at an implicit method, called the backward Euler method, because now we have to solve a more difficult equation about $ {\tilde{x}(t_{i})}$. It is of most interest to study the error produced by the discrete approximation. Inspecting more closely, we see that (1) is actually the first order Taylor approximation of $ {x(t_{i})}$ based at $ {t_{i-1}}$: if the solution $ {x(t)}$ is twice differentiable, by Taylor's theorem, we have an estimate

$ \displaystyle x(t_{i})=x(t_{i-1})+\Delta t\cdot f(t_{i-1},x(t_{i-1}))+O(\Delta t^{2}) $

where $ {O(\Delta t^{2})}$ is a quantity bounded by $ {C\cdot\Delta t^{2}}$ for some constant $ {C}$ and for all $ {\Delta t}$ small enough. Note that here we are using the true values of the solution. Rearranging and writing $ {t=t_{i}}$, $ {\Delta t=h}$, $ {dx/dt=f}$, the above estimate turns to

$ \displaystyle \frac{x(t+h)-x(t)}{h}-\frac{dx(t)}{dt}=O(h). $

The difference in the LHS is known as the truncation error of the method. Since $ {h}$ is in its first power in the RHS $ {O(h)}$, we say that Euler method is of first order accurate. The above discussion holds when $ {h}$ is small enough. The behavior of Euler's method can be wild if $ {h}$ is sufficiently large. Indeed, consider the stiff problem:

$ \displaystyle \frac{dx}{dt}(t)=-k\cdot x(t),k>0 $

This equation has solutions of the form $ {x(t)=Ae^{-kt}+C}$. Clearly any particular solution decays to a fixed value as $ {t\rightarrow\infty}$, but one can check if the time step $ {h\gg1/k}$, then the numerical solution will oscillate or even diverge to infinity. In some sense the solution is only stable (here, we mean continuous dependence on the initial condition with fixed step size) within a region of choice of $ {h}$, or sometimes a expression in terms of h; we call a stability region the subset where the numerical solution for the stiff problem is stable. It turns out that here much of the analysis of a difference method (e.g. Euler's method) to solving the stiff problem is about the associated difference equation, which further breaks down to a linear algebraic problem of analysing eigenvalues of the associated linear system, recalling that the solution of a difference equation can be written as a linear combination of roots of the characteristic polynomial (though the case of multiple roots is slightly more complicated).

We shall see how these ideas in this case of Euler method correspond to those in the analysis of finite difference methods for hyperbolic PDEs, by examining the prototypical example, that of the one way wave equation with the initial condition

$ \displaystyle u_{t}+au_{x}=0\quad u(0,x)=u_{0}(x),{\tag{$\star$}} \ \ \ \ \ $

where $ {u:(0,T)\times\mathbb{R}\rightarrow\mathbb{R}}$, and $ {a}$ is a constant. We note that if $ {b,f=0}$, the unique solution is given by

$ \displaystyle u(t,x)=u_{0}(x-at), $

as we see that the initial state is propogating at a constant speed in the direction $ {(1,a)}$. The homogeneous equations such as $ {(\star)}$ do not have issues of approximating the RHS. For sake of including a more comprehensive picture, we will also temporarily discuss the inhomogeneous case

$ \displaystyle u_{t}+au_{x}=f\quad u(0,x)=u_{0}(x), $

where $ {f:(0,T)\times\mathbb{R}\rightarrow\mathbb{R}}$ is allowed to depend on both the time variable and spatial variable, but NOT $ {u}$, otherwise it will be nonlinear. The closed form solution can be given by the method of variation of parameter:

$ \displaystyle u(t,x)=u_{0}(x-at)+\int_{0}^{t}f(s,x-a(t-s))ds. $

We shall work with the initial value problem that is well-posed: if for any $ {t\in(0,T)}$, there is a constant $ {C_{T}}$ such that any solution $ {u(t,x)}$ will satisfy

$ \displaystyle \int_{-\infty}^{\infty}|u(t,x)|^{2}dx\leq C_{T}\int_{-\infty}^{\infty}|u(0,x)|^{2}dx. $

It is important to note that in this post we don't consider nonlinear PDEs, as the analysis below in general does not apply.