Ngô Quốc Anh

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


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



\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.

See also: Green’s function and differential equations

1 Comment »

  1. It is nice. Thanks.

    Comment by TTT — March 18, 2010 @ 17:36

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Blog at

%d bloggers like this: