0.6 Solution of the partial differential equations  (Page 4/13)

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Suppose one wished to find the solution to the Poisson equation in the semi-infinite domain, $y>0$ with the specification of either $u=0$ or $\partial u/\partial n=0$ on the boundary, $y=0$ . Denote as ${u}^{0}\left(x,y,z\right)$ the solution to the Poisson equation for a distribution of sources in the semi-infinite domain $y>0$ . The solutions for the Dirichlet or Neumann boundary conditions at $y=0$ are as follows.

$\begin{array}{c}u\left(x,y,z\right)={u}^{0}\left(x,y,z\right)-{u}^{0}\left(x,-y,z\right),\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}u=0\phantom{\rule{0.277778em}{0ex}}at\phantom{\rule{0.277778em}{0ex}}y=0\hfill \\ u\left(x,y,z\right)={u}^{0}\left(x,y,z\right)+{u}^{0}\left(x,-y,z\right),\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}du/dy=0\phantom{\rule{0.277778em}{0ex}}at\phantom{\rule{0.277778em}{0ex}}y=0\hfill \end{array}$

The first function is an odd function of $y$ and it vanishes at $y=0$ . The second is an even function of y and its normal derivative vanishes at $y=0$ .

An example of the method of images to satisfy either the Dirichlet or Neumann boundary conditions is illustrated in the following figure. The black curve is the response to a line sink at $x=1.5$ . We desire to have either the function or the derivative at $x=0$ to vanish. The red curve is a line sink at $x=-1.5$ . The sum of the two functions is symmetric about $x=0$ and has zero derivative there. The difference is anti-symmetric about $x=0$ and vanishes at $x=0$ .

Now suppose there is a second boundary that is parallel to the first, i.e. $y=a$ that also has a Dirichlet or Neumann boundary condition. The domain of the Poisson equation is now $0 . Denote as ${u}^{1}$ the solution that satisfies the $BC$ at $y=0$ . A solution that satisfies the Dirichlet or Neumann boundary conditions at $y=a$ are as follows.

$\begin{array}{c}u\left(x,y,z\right)={u}^{1}\left(x,y,z\right)-{u}^{1}\left(x,2a-y,z\right),\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}u=0\phantom{\rule{0.277778em}{0ex}}\text{at}\phantom{\rule{0.277778em}{0ex}}y=a\hfill \\ u\left(x,y,z\right)={u}^{1}\left(x,y,z\right)+{u}^{1}\left(x,2a-y,z\right),\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}du/dy=0\phantom{\rule{0.277778em}{0ex}}\text{at}\phantom{\rule{0.277778em}{0ex}}y=a\hfill \end{array}$

This solution satisfies the solution at $y=a$ , but no longer satisfies the solution at $y=0$ . Denote this solution as ${u}^{2}$ and find the solution to satisfy the $BC$ at $y=0$ . By continuing this operation, one obtains by induction a series solution that satisfies both boundary conditions. It may be more convenient to place the boundaries symmetric with respect to the axis in order to simplify the recursion formula.

Assignment 7.3

Calculate the solution for a unit line source at the origin of the $x$ , $y$ plane with zero flux boundary conditions at $y=+1$ and $y=-1$ . Prepare a contour plot of the solution for $0 . What is the limiting solution for large $x$ ? Note: The boundary conditions are conditions on the derivative. Thus the solution is arbitrary by a constant.

Existence and uniqueness of the solution to the poisson equation

If the boundary conditions for Poisson equation are the Neumann boundary conditions, there are conditions for the existence to the solution and the solution is not unique. This is illustrated as follows.

$\begin{array}{c}{\nabla }^{2}u=-\rho \phantom{\rule{1.em}{0ex}}\text{in}\phantom{\rule{0.277778em}{0ex}}V,\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\mathbf{n}•\nabla u=f\phantom{\rule{0.277778em}{0ex}}\text{on}\phantom{\rule{0.277778em}{0ex}}S\hfill \\ \phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}{\nabla }^{2}u\phantom{\rule{0.166667em}{0ex}}dV=-\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}\rho \phantom{\rule{0.166667em}{0ex}}dV\hfill \\ \int \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\int \phantom{\rule{-11.66656pt}{0ex}}◯\mathbf{n}\nabla u\phantom{\rule{0.166667em}{0ex}}dS=-\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}\rho \phantom{\rule{0.166667em}{0ex}}dV\hfill \\ \int \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\int \phantom{\rule{-11.66656pt}{0ex}}◯f\phantom{\rule{0.166667em}{0ex}}dS=-\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}\rho \phantom{\rule{0.166667em}{0ex}}dV\hfill \end{array}$

This necessary condition for the existence of a solution is equivalent to the statement that the flux leaving the system must equal the sum of sources in the system. The solution to the Poisson equation with the Neumann boundary condition is arbitrary by a constant. If a constant is added to a solution, this new solution will still satisfy the Poisson equation and the Neumann boundary condition.

Green's function for the diffusion equation

We showed above how the solution to the Poisson equation with homogeneous boundary conditions could be obtained from the Green's function by convolution and method of images. Here we will obtain the Green's function for the diffusion equation for an infinite domain in one, two, or three dimensions. The Green's function is for the parabolic PDE

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