Ill-posedness and well-posedness Why is the backwards heat equation an ill-posed problem?
$$\frac{∂u}{∂t}=-k\frac{∂^2u}{∂x^2}$$. And what makes this heat conduction equation $$\frac{∂u}{∂t}=k\frac{∂^2u}{∂x^2}$$ well-posed?
 A: The well-posedness of a PDE in the sense of Hadamard as you said requires that:


*

*Existence: A solution exists

*Uniqueness: The solution is unique

*Stability: The solution changes continuously with the initial/boundary conditions.


To check these points you need to define your boundary and initial values, since they too determine if the problem is well-posed. For example take this set-up for the heat equation:
\begin{equation}
\begin{cases}
u_t - \alpha^2u_{xx} = 0,\quad &t \in (0,\infty),~ x\in \Omega\\
u(t,x) = 0 ,\quad &t \in (0,\infty),~ x\in \partial\Omega\\
u(0,x) = u_0(x),\quad &t =0,~ x\in \Omega\\
\end{cases}
\end{equation}
For this problem we can compute a  general solution of the shape:
$$u(t,x) = e^{-\alpha^2\lambda^2t}\left[A\sin(\lambda t)+B\cos(\lambda t)\right]$$
Having found a general shape we can insert the boundary and initial conditions and will end up having a solution purely dependent ($A,B,\lambda$ expressed in terms of those BCs and ICs) on these initial and boundary values. This hence shows existence.
Knowing that a solution exists one can go ahead and show using the maximum principle that the solution is indeed unique.
The stability condition can be proven by showing the direct dependency of the analytical solution on the BC/IC.
The backwards heat equation does posses a unique solution for a common set up of boundary and initial conditions. It can however be shown that the solution, which eventually explodes, does not depend on the initial conditions anymore. For further detail you can check the ill-posedness of the backward heat equation.
