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!
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.