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The equation $\frac{\partial f}{\partial t} = \frac{\partial^2f}{\partial x^2}$ has the fundamental solution (in one dimension) $f(x,t) = \frac{1}{2\sqrt{t}}\exp (-x^2/4t)$ if there are no boundary conditions.

If there's a boundary condition in the form $\frac{\partial f}{\partial x}\Big|_{x=1}=0$, we can supposedly sometimes find a solution that satisfies the boundary condition by placing a "ghost charge" in the point $x=2$. The book I'm reading has little to do with PDE:s and just mentions this as a sidenote. I don't know what it means. Is there an easy way to explain this?

Benjamin Lindqvist
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    This is called the "Method of Images". If you had asked about the case with boundary condition f(1,t) = 0, then I could answer your question - if p(x,t) is your fundamental solution, then p(x,t) - p(x-2,t) is also a solution, PLUS it satisfies the boundary condition f(1,t)=0 – osbert May 27 '16 at 08:28
  • It seems to work for this boundary condition too. Thanks. Is there any situation in which it doesn't work? – Benjamin Lindqvist May 27 '16 at 08:51
  • If you had more boundary conditions, e.g. at $x=-1$, you might have problems. Sometimes you have to add 'images' to cancel out the effects of other images, and end up with an infinite line of images (still a valid solution). If you were in more than 1 dimension, then this method still works for some special boundary shapes, e.g. planes, spheres, but not for arbitrary rough boundaries. – osbert May 27 '16 at 09:18

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As noted in a comment, the boundary condition $f(1,t)=0$ can be accommodated using a negative image charge at $x=2$, leading to the solution $p(x,t)-p(x-2,t)$ (with $p$ the fundemantal solution you gave).

To get $f'(1,t)=0$, you need a positive image charge at $x=2$, leading to the solution $f(x,t)=p(x,t)+p(x-2,t)$, with

\begin{align} \def\pa#1{\left.\frac{\partial#1}{\partial x}\right|_{x=1}} \pa f&=\pa{p(x,t)}+\pa{p(x-2,t)}\\ &=\pa{p(x,t)}+\pa{p(2-x,t)} \\ &=\pa{p(x,t)}-\pa{p(x,t)} \\ &=0\;. \end{align}

joriki
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  • Great. So just one more detail - the book said that "under some conditions, we can use a ghost charge". Any idea what they mean by that? – Benjamin Lindqvist May 27 '16 at 09:07
  • Note that we 'should' write $f(x,t) = p(x,t)+p(2-x,t)$ - we have reflected $x$ about 1, $x \to 2-x$ - but since this equation is symmetric in $x$ it makes no difference of course. – osbert May 27 '16 at 09:13
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    @BenjaminLindqvist: In one dimension, there isn't much of a condition, since you can always get the function value or its derivative to vanish at a single point by a suitable image charge. Perhaps they're referring to higher dimensions, where you can apply this method if you can get the sum of two fundamental solutions to vanish on the boundary. For instance, in electrostatics in three dimensions, you can do this if the boundary is a plane or a sphere; see Wikipedia. – joriki May 27 '16 at 09:13