# Ngô Quốc Anh

## March 31, 2010

### Green’s function and differential equations, 2

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 22:28

So far we have the closed form for the Green function for heat equation on the half-plane in 1D, i.e., the Green function for

$\displaystyle u_t - au_{xx}=0, \quad x >0, t>0$

subject to the boundary condition

$\displaystyle u(0,t)=0$

(see this topic for details). The construction of the Green function for such problems relies on the reflection principle and this is our main aim in this entry.

Let first consider our problem for the whole domain. Having the existence of the Green function, denoted $G_{\mathbb R}(x,t;y,\tau)$, for the whole domain, the general solution can be expressed in terms of $G_{\mathbb R}$ as

$\displaystyle u(x,t) = \int_{ - \infty }^{ + \infty } {{G_\mathbb{R}}(x,t;y,0)u(y,0)dy}$.

As we are interested in the half-plane domain, once the Green function $G$ exists, the solution is of the following from

$\displaystyle u(x,t) = \int_0^{ + \infty } {G(x,t;y,0)u(y,0)dy}$.

The idea of the reflection principle comes from the fact that if $g(x)$ is an odd function, then

$\displaystyle \int_{-\infty}^{+\infty} g(x)dx=2\int_0^{+\infty} g(x)dx$.

Therefore, if we denote by $G$ the following function

$\displaystyle G(x,t;y,\tau ) = {{G_\mathbb{R}}(x,t;y,\tau ) - {G_\mathbb{R}}(x,t; - y,\tau )}$

we then see that $G$ is odd, that means for the half-plane domain, after putting

$\displaystyle u(y,0)=-u(-y,0)$

## March 18, 2010

### Green’s function for viscous Burgers type equation, 2

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 15:47

Followed by a question posted here we are now interested in how can we find the Green function for the following problem

$u_t+au_x-u_{xx}=0, \quad x>0$

where $a$ a constant and with boundary condition $u(0,t)=0$.

Note that, for heat equation, in one variable, the Green’s function is a solution of the initial value problem

$u_t-u_{xx}=0$

with boundary condition

$\displaystyle u(x,0)=\delta(x)$

where $\delta$ is the Dirac delta function and this Green function is already known as

$\displaystyle G(x,y,t) = \frac{1}{{\sqrt {4\pi at} }}{e^{ - \frac{{{{(x - y)}^2}}}{{4at}}}}$.

Note that, the Green’s function for heat equation in the whole line $\mathbb R$

$u_t-u_{xx}=0$

is

$\displaystyle G_a(x,y,t) = \frac{1}{{\sqrt {4\pi at} }}{e^{ - \frac{{{{(x - y)}^2}}}{{4at}}}}$.

However, this is not true once we restrict ourselves to the half-plan

$u_t-u_{xx}=0, \quad x>0$

with boundary condition

$\displaystyle u(x,0)=0$.

In order to find the correct one, we need to use the so-called reflection principle as follows: For each $x>0$, we define $\widetilde x =- x$, called the inverse point of $x$. It can be readily verified that the function

$\displaystyle G(x,y,t)=G_a(x,y,t)-G_a(x,\widetilde y,t)$

which implies

$\displaystyle G(x,y,t) = \frac{1}{{\sqrt {4\pi at} }}\left( {{e^{ - \frac{{{{(x - y)}^2}}}{{4at}}}} - {e^{ - \frac{{{{(x + y)}^2}}}{{4at}}}}} \right)$.

Now let us go back to the viscous Burgers type equation. By the following change of variable

$\displaystyle v = {e^{ - \frac{a}{2}x}}u$

one easily sees that

$\displaystyle\begin{gathered} {v_t} = {e^{ - \frac{a}{2}x}}{u_t} \hfill \\ {v_x} = - \frac{a}{2}{e^{ - \frac{a}{2}x}}u + {e^{ - \frac{a}{2}x}}{u_x} \hfill \\ {v_{xx}} = \frac{{{a^2}}}{4}{e^{ - \frac{a}{2}x}}u - a{e^{ - \frac{a}{2}x}}{u_x} + {e^{ - \frac{a}{2}x}}{u_{xx}} \hfill \\ \end{gathered}$

which gives

$\displaystyle\begin{gathered} {v_t} + \frac{{{a^2}}}{4}v - {v_{xx}} = {e^{ - \frac{a}{2}x}}{u_t} + \frac{{{a^2}}}{4}{e^{ - \frac{a}{2}x}}u - \frac{{{a^2}}}{4}{e^{ - \frac{a}{2}x}}u + a{e^{ - \frac{a}{2}x}}{u_x} - {e^{ - \frac{a}{2}x}}{u_{xx}} \hfill \\ \qquad\qquad= {e^{ - \frac{a}{2}x}}{u_t} + a{e^{ - \frac{a}{2}x}}{u_x} - {e^{ - \frac{a}{2}x}}{u_{xx}} \hfill \\ \qquad\qquad= 0 \hfill \\ \end{gathered}$.

In order to eliminate the term $\frac{a^2}{2}v$, we use the following change of variable

$\displaystyle w = {e^{\frac{{{a^2}}}{4}t}}v$

we obtain

$\displaystyle\begin{gathered} {w_t} = \frac{{{a^2}}}{4}{e^{\frac{{{a^2}}}{4}t}}v + {e^{\frac{{{a^2}}}{4}}}{v_t} \hfill \\ {w_x} = {e^{\frac{{{a^2}}}{4}t}}{v_x} \hfill \\ {w_{xx}} = {e^{\frac{{{a^2}}}{4}t}}{v_{xx}} \hfill \\ \end{gathered}$

which yields

$\displaystyle {w_t} - {w_{xx}} = \frac{{{a^2}}}{4}{e^{\frac{{{a^2}}}{4}t}}v + {e^{\frac{{{a^2}}}{4}t}}{v_t} - {e^{\frac{{{a^2}}}{2}t}}{v_{xx}} = {e^{\frac{{{a^2}}}{2}t}}\left( {{v_t} + \frac{{{a^2}}}{4}v - {v_{xx}}} \right) = 0$.

Hence we have the following summary

$\displaystyle {u_t} + a{u_x} - {u_{xx}} = 0\xrightarrow{{v = {e^{ - \frac{a}{2}x}}u}}{v_t} + \frac{{{a^2}}}{4}v - {v_{xx}} = 0\xrightarrow{{w = {e^{\frac{{{a^2}}}{4}t}}v}}{w_t} - {w_{xx}} = 0$.

We are now in a position to talk about the most interesting point of this entry: How to find the Green function for backward once we already have the Green function for pushforward? It is obvious to see that the boundary condition doesn’t change, i.e.

$\displaystyle w(0,t) = v(0,t) = u(0,t) = 0$.

The Green function for the last equation is already known, called

$\displaystyle {G_w}(x,y,t) = \frac{1}{{\sqrt {4\pi t} }}\left( {{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}} - {e^{ - \frac{{{{(x + y)}^2}}}{{4t}}}}} \right)$.

The Green function for the equation for $v$ can be constructed by looking at the substitution used. Precisely,

$\displaystyle {G_v}(x,y,t) = {e^{ - \frac{{{a^2}}}{4}t}}\frac{1}{{\sqrt {4\pi t} }}\left( {{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}} - {e^{ - \frac{{{{(x + y)}^2}}}{{4t}}}}} \right)$.

The Green function for the equation for $u$ is constructed a little bit crazy, what we need is the following

$\displaystyle {G_u}(x,y,t) = {e^{\frac{a}{2}(x - y)}}{e^{ - \frac{{{a^2}}}{4}t}}\frac{1}{{\sqrt {4\pi t} }}\left( {{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}} - {e^{ - \frac{{{{(x + y)}^2}}}{{4t}}}}} \right)$.

The most important property one should check is the boundary condition, roughly speaking, the Green function must satisfy

$\displaystyle G(x,0,t)=0$.

## April 13, 2009

### Green’s function for viscous Burgers type equation

Filed under: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 12:08

We now find Green function for the following equation $u_t+au_x-u_{xx}=0$. In this topic, I will not consider any boundary value condition. My approach is the Fourier transform. From the equation $u_t+au_x-u_{xx}=0$ we apply Fourier transform to the both sides with respect to $x$. We then obtain

$\displaystyle {\widehat u_t}\left( {\xi ,t} \right) + a\left( {i\xi } \right)\widehat u\left( {\xi ,t} \right) - {\left( {i\xi } \right)^2}\widehat u\left( {\xi ,t} \right) = 0$

or equivalently,

$\displaystyle {\widehat u_t}\left( {\xi ,t} \right) = - \left( {ai\xi + {\xi ^2}} \right)\widehat u\left( {\xi ,t} \right)$.

The above identity can be regarded as a first order ODE with respect to $t$ with constant coefficients, solving this ODE we obtain

$\displaystyle\widehat u\left( {\xi ,t} \right) = {e^{ - \left( {ai\xi + {\xi ^2}} \right)t}}\widehat u\left( {\xi ,0} \right)$.

Taking inverse Fourier transform gives

$\displaystyle u\left( {x,t} \right) = \frac{1}{2\pi} \int_\mathbb{R} {{e^{ix \cdot \xi }}{e^{ - \left( {ai\xi + {\xi ^2}} \right)t}}\widehat u\left( {\xi ,0} \right)d\xi }$.

Note that

$\displaystyle \widehat u\left( {\xi ,0} \right) = \int_\mathbb{R} {{e^{ - iy \cdot \xi }}u\left( {y,0} \right)dy}$.

Therefore,

$\displaystyle \begin{gathered} u\left( {x,t} \right) = \frac{1} {{2\pi }}\int_\mathbb{R} {{e^{ix\cdot\xi }}{e^{ - \left( {ai\xi + {\xi ^2}} \right)t}}\left( {\int_\mathbb{R} {{e^{ - iy\cdot\xi }}u\left( {y,0} \right)dy} } \right)d\xi } \hfill \\ \quad \qquad = \int_\mathbb{R} {\left( {\frac{1} {{2\pi }}\int_\mathbb{R} {{e^{ix\cdot\xi }}{e^{ - \left( {ai\xi + {\xi ^2}} \right)t}}{e^{ - iy\cdot\xi }}d\xi } } \right)u\left( {y,0} \right)dy} n \hfill \\ \quad \qquad = \int_\mathbb{R} {\left( {\frac{1} {{2\pi }}\int_\mathbb{R} {{e^{i\left( {x - y} \right)\cdot\xi }}{e^{ - \left( {ai\xi + {\xi ^2}} \right)t}}d\xi } } \right)u\left( {y,0} \right)dy} . \hfill \\ \end{gathered}$

Thus, the Green function for equation $u_t+au_x-u_{xx}=0$ is the following

$\displaystyle\mathbb{G}(x,t; y, \tau) = \frac{1} {{2\pi }}\int_\mathbb{R} {{e^{i (x-y) \cdot \xi }}{e^{ - \left( {ai\xi + {\xi ^2}} \right)(t-\tau)}}d\xi }$.

Clearly, the above Green function has the following properties

• $\displaystyle\mathbb{G}(x,t; y, t)=\delta(x-y)$.
• $\displaystyle{\mathbb{G}_t}(x,t; y, \tau) + a{\mathbb{G}_x}(x,t; y, \tau) - {\mathbb{G}_{xx}}(x,t; y, \tau) = 0$.

I now propose the following question: if we impose a boundary condition, then how can we find an appropriate Green function for such problem?

## March 13, 2009

### Green’s function and differential equations

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 12:31

In mathematics, a Green’s function is a type of function used to solve inhomogeneous differential equations subject to boundary conditions. The term is also used in physics, specifically in quantum field theory, electrodynamics and statistical field theory, to refer to various types of correlation functions, even those that do not fit the mathematical definition; for this sense, see Correlation function (quantum field theory) and Green’s function (many-body theory).

Green’s functions are named after the British mathematician George Green, who first developed the concept in the 1830s. In the modern study of linear partial differential equations, Green’s functions are largely studied from the point of view of fundamental solutions instead.

Definition and uses

Technically, a Green’s function, $G(x,s)$, of a linear differential operator $L=L(x)$ acting on distributions over a subset of the Euclidean space $\mathbb R^n$, at a point $s$, is any solution of

$LG(x,s)=\delta(x-s)$                         (1)

where $\delta$ is the Dirac delta function.

This technique can be used to solve differential equations of the form

$Lu(x)=f(x)$                                      (2)

If the kernel of $L$ is nontrivial, then the Green’s function is not unique. However, in practice, some combination of symmetry, boundary conditions and/or other externally imposed criteria will give a unique Green’s function. Also, Green’s functions in general are distributions, not necessarily proper functions.

Green’s functions are also a useful tool in condensed matter theory, where they allow the resolution of the diffusion equation, and in quantum mechanics, where the Green’s function of the Hamiltonian is a key concept, with important links to the concept of density of states. The Green’s functions used in those two domains are highly similar, due to the analogy in the mathematical structure of the diffusion equation and Schrödinger equation. As a side note, the Green’s function as used in Physics is usually defined with the opposite sign; that is, $LG(x,s)=-\delta(x-s)$. This definition does not significantly change any of the properties of the Green’s function.