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

one has the following fact $\displaystyle\begin{gathered} u(x,t) = \int_{ - \infty }^{ + \infty } {{G_\mathbb{R}}(x,t;y,0)u(y,0)dy} \hfill \\ \qquad= \int_{ - \infty }^0 {{G_\mathbb{R}}(x,t;y,0)u(y,0)dy} + \int_0^{ + \infty } {{G_\mathbb{R}}(x,t;y,0)u(y,0)dy} \hfill \\ \qquad= \int_0^{ + \infty } {{G_\mathbb{R}}(x,t; - y,0)u( - y,0)dy} + \int_0^{ + \infty } {{G_\mathbb{R}}(x,t;y,0)u(y,0)dy} \hfill \\ \qquad= \int_0^{ + \infty } {({G_\mathbb{R}}(x,t;y,0) - {G_\mathbb{R}}(x,t; - y,0))u(y,0)dy} \hfill \\ \qquad= \int_0^{ + \infty } {G(x,t;y,0)u(y,0)dy}. \hfill \\ \end{gathered}$

This is exact what we need. To be general, let us consider the following problem $\displaystyle u_t - au_{xx}=0, \quad x >x_0 > -\infty, t>0$

subject to the boundary condition $\displaystyle u(x_0,t)=0$.

There is not much difference even we have the presence of $x_0$. To this purpose, we try to do as above to figure out the function $G$ in terms of the function $G_{\mathbb R}$. Indeed, $\displaystyle\begin{gathered} \int_{ - \infty }^{ + \infty } {{G_\mathbb{R}}(x,t;y,0)u(y,0)dy} = \int_{ - \infty }^{{x_0}} {{G_\mathbb{R}}(x,t;y,0)u(y,0)dy} + \int_{{x_0}}^{ + \infty } {{G_\mathbb{R}}(x,t;y,0)u(y,0)dy} \hfill \\ \qquad= \int_{ - \infty }^0 {{G_\mathbb{R}}(x,t;y + {x_0},0)u(y + {x_0},0)dy} + \int_0^{ + \infty } {{G_\mathbb{R}}(x,t;y + {x_0},0)u(y + {x_0},0)dy} \hfill \\ \qquad= \int_0^{ + \infty } {{G_\mathbb{R}}(x,t; - y - {x_0},0)u( - y - {x_0},0)dy} + \int_0^{ + \infty } {{G_\mathbb{R}}(x,t;y + {x_0},0)u(y + {x_0},0)dy} \hfill \\ \qquad= \int_0^{ + \infty } {({G_\mathbb{R}}(x,t;y + {x_0},0) - {G_\mathbb{R}}(x,t; - y - {x_0},0))u(y + {x_0},0)dy} \hfill \\ \qquad= \int_{{x_0}}^{ + \infty } {({G_\mathbb{R}}(x,t;y,0) - {G_\mathbb{R}}(x,t; - y - 2{x_0},0))u(y,0)dy}. \hfill \\ \end{gathered}$

Thus the Green function we need to construct is $\displaystyle G(x,t;y,\tau) = {G_\mathbb{R}}(x,t;y,\tau) - {G_\mathbb{R}}(x,t; - y - 2{x_0},\tau)$.

We place here the formula for $G_{\mathbb R}$ $\displaystyle {G_\mathbb{R}}(x,t;y,\tau ) = \frac{1}{{\sqrt {4\pi a(t - \tau )} }}{e^{ - \frac{{{{(x - y)}^2}}}{{4at}}}}$.

Question. How can we construct a Green function if we impose the following boundary condition $u_x(x_0,t)=0$?

See also: Green’s function and differential equations and Heat equation from Wiki