Green’s function for viscous Burgers type equation, 2

March 18, 2010

Green’s function for viscous Burgers type equation, 2

Filed under: PDEs — Tags: Green functions — 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 $\mathbb R$

$u_t-u_{xx}=0$

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

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

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

1 Comment »

It is nice. Thanks.

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

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Ngô Quốc Anh

March 18, 2010