In $2$ dimension, take $-\Delta u=0$ on $\{(x,y\},y\geq 0\}$ with $u(x,y=0)=f(x)$, $u_y(x,y=0)=g(x)$ where $f$ and $g$ are smooth function. I want to justify whether this problem is well posed.

My first question is, what do we mean by a problem is well-posed? In my opinion a problem is well-posed if this problem

$(1)$ has a solution

$(2)$ the solution is unique.

$(3)$ the solution continuously depends on the data, i.e., the boundary value in my problem.

But my friends tell me that I don't need uniqueness, only existence is enough for well-posed problem. I confused, do I need uniqueness or not?

Now, go back to my problem. The book states that this problem is not well-posed and gives an example such that $$ \frac{1}{n^2} e^{n\epsilon y}\sin(nx) $$ But I failed to see why this example give me the contradiction...

Any help is really welcome!

  • $\begingroup$ Made a very minor edit to your post, changing the adjective "continuous" to the adverb "continuously". Happy Boxing Day! $\endgroup$ – Robert Lewis Dec 26 '14 at 21:56
  • $\begingroup$ @RobertLewis haha thx! You too $\endgroup$ – spatially Dec 27 '14 at 20:59

My understanding of "well-posed" matches the items (1), (2), (3) you gave. Note that (3) is vague in that "continuously" is not specified. One of many possible interpretations of (3) is: evaluating the solution at any fixed point $(x_0,y_0)$ gives a continuous functional on the initial data with respect to the uniform norm. [One can also ask for the whole solution $u(x,y)$ to depend continuously in the uniform norm, which is more demanding.]

Let's take $$f(x) = \frac{1}{n} \cos(nx), \qquad g(x) = 0 $$ as initial data for this problem. Observe that both $f$ and $g$ converge to $0$ uniformly. On the other hand, the solution is
$$u(x,y) = \frac{1}{n} \cos(nx)\cosh (ny )$$ Hence $u(0,1) = \frac{1}{n} \cosh n\to\infty$ as $n\to\infty$.

Note that $1/n$ could be replaced by any other negative power of $n$ here.

The example $\frac{1}{n^2} e^{n\epsilon y}\sin(nx)$ works similarly: both $u(x,0)$ and $u_y(x,0)$ tend to $0$ uniformly.

It is also true that existence fails for this problem, but to give an example of that is a bit more difficult. Here is one: $$f(x) = \sum_{n=1}^\infty e^{-n} \cos(nx), \qquad g(x) = 0 $$ The solution is $$u(x,y) = \sum_{n=1}^\infty e^{-n} \cos(nx)\cosh ny$$ which is a harmonic function for $0<y<1$... but it has serious issues at $y=1$ and beyond.

  • $\begingroup$ Thank you for your answer. So the contradiction in my example is at the "continuous depends on data" right? Since as $f\to 0$, the solution is not going to $0$. But if I give in addition that $f$ vanish at $\infty$, this question should be well-posed, since at this time I will have maximal principle and hence the solution will depends on data. Am I correct? $\endgroup$ – spatially Dec 28 '14 at 23:42
  • $\begingroup$ If you restrict consideration to such well-behaved solutions, then continuous dependency holds. But existence fails already for $f(x)=\cos x$ and $g(x)=0$, since there is no solution for this data in your function space. (The example could be modified to make $f$ vanish at infinity too, but I'd rather not) $\endgroup$ – user147263 Dec 28 '14 at 23:52
  • $\begingroup$ I see. So if I only give data $f(x)$ and ask $f$ vanish at infinite, but ask no condition for $g$, i.e., does not have a boundary condition for $u_y$, then the problem is well-posed! $\endgroup$ – spatially Dec 29 '14 at 0:04
  • $\begingroup$ Yes, that's correct -- with only $f$ it becomes the Dirichlet problem for Laplace equation, well-posed. With both $f$ and $g$ it's the Cauchy (initial value) problem, ill-posed. $\endgroup$ – user147263 Dec 29 '14 at 0:10
  • $\begingroup$ I see I see. Thank you! $\endgroup$ – spatially Dec 29 '14 at 0:29

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