Well-posedness of heat-equation PDE with only one initial condition Consider the PDE given by $u_t = \alpha u_{xx}$ with initial condition $u(x, 0) = f(x)$.
Now suppose we discretize the problem in the time variable, so we approximate $u_t(x, t)$ by a finite difference quotient:
$$ u_t(x, t) \approx \frac{u(x, t + \Delta t) - u(x, t)}{\Delta t}$$
Using the PDE above, we now have a recurrence $$u(x, t + \Delta t) = u(x, t) + \alpha u_{xx}(x, t)\Delta t$$
which allows us to compute an approximation for $u(x, t + \Delta t)$ given $u(x, t)$.
If we choose the number $\Delta t$ to be sufficiently small, we expect the accuracy of this approximation to increase, and we expect that dividing a time interval $[0, T]$ into increasingly many segments will improve the accuracy of the computed value for $u(x, T)$ given an initial condition $u(x, 0) = f(x)$ (this is what most finite-difference methods essentially do to compute such solutions numerically).
Now it would seem from the above argument that knowing $u(x, 0)$ is sufficient to compute $u(x, T)$ to any desired degree of accuracy, by choosing $\Delta t$ sufficiently small. Therefore the solution seems to be uniquely determined by this initial condition.
On the other hand, it seems that usually we need as many initial/boundary conditions as the highest order of partial derivatives for each variable. You might even wonder, what happens for example if there was some fixing of the temperature on a line $x = 0$ for $t \gt 0$, for example there could be some additional boundary condition like $u(0, t) = g(t), t \gt 0$ with $g(0) = f(0)$ to make sure $u(0, 0)$ is consistent with both the initial and boundary condition.
So it would seem that the method described above will "miss" the possible solutions where we add some additional arbitrary boundary condition on $x = 0$, since it can only converge to a unique solution.
I conclude that there must be an error in some of my considerations above, but I am not sure exactly what it is.
I see three possible answers:


*

*The solution is uniquely determined if I only have the initial condition $u(x, 0) = f(x)$.

*The solution is not uniquely determined, and additional boundary conditions need to be provided.

*The solution is uniquely determined, but there are additional boundary conditions that somehow appear implicitly in the initial value problem above, such as some behavior at infinity or continuity assumption that reduce the number of degrees of freedom.
I'd like to find which of these is the correct conclusion and why that is the case.
 A: There are non-zero solutions of the heat equation that have $0$ initial data $u(x,0)$. Solutions are not unique, without some growth condition at $\pm\infty$ on the initial data.
Non-uniqueness of solutions of the heat equation
See also https://mathoverflow.net/questions/72195/unconditional-nonexistence-for-the-heat-equation-with-rapidly-growing-data
A: Just to flesh out all the background, in the one dimensional heat equation:
$$ u_t = \alpha u_{xx} $$
$u$ represents Temperature, and $\alpha := \frac{k}{\rho C_p}$ is a constant that describes the thermal diffusivity of the material. Suppose we know the initial temperature profile $u(x,0) = f(x)$. Then the question simply becomes how does heat begin diffusing based on this profile? 
Physically, the way that the material interacts with its surroundings will matter. Even if the initial temperature profile is uniform at $f(x) = u_0$, does the $1$-D material have its ends fixed at a different temperature? If so, then we have a Dirichlet boundary condition. Is heat not allowed to leave out the ends? If so, then we have the Naumann boundary condition where $\frac{\partial u}{\partial x}|_{x = \text{endpoints}} = 0$. Is heat being convected at the ends? Do we have some combination of all of these? We can see that a purely physical framing of the question necessitates the need to know some sort of boundary condition.
I'm sure you already knew all of the above though, so let's try and talk about what happens when you try to solve the heat equation anyways. The problem is going to come from how you define $u_{xx}$ at your boundary.
Suppose we try to do this numerically using the method of line. Then we need $3$ points to approximate $u_{xx}$. Before we can even touch $u_{xx}$ though, we need to calculate $u_x$. The first thing I'm wondering about is how you'd even define such a thing without boundary conditions. Let the left-end of the material be given by $x=a$, and the right end by $x = b$. Let's also define $u_0 = u(a,t)$, $\Delta x = \frac{b - a}{n}$, and suppose we divide our material into $n$ nodes. Then $u_{i} = u((b-a)i/n + a, t) = u(a + \Delta x \cdot i, t)$. Say we calculate $u_i$ by a forward difference
$$ u_i = \frac{u_{i + 1} - u_i}{\Delta x}$$
Then $u_i$ is defined at $x = a$, but not $x = b$! You face a similar problem if you try to do a backward difference. 
Suppose we try to skirt this by defining a forward difference for all nodes, except the last where we do a backward difference. Then,
\begin{align}
u_x(a + n\Delta x,t) &= \frac{u_{n} - u_{n-1}}{\Delta x}\\
u_x(a + (n-1)\Delta x,t) &= \frac{u_{n} - u_{n-1}}{\Delta x}
\end{align}
Which would then force a second derivative of $0$ at $x = b$. If you try to have forward differences for the first half of your rod, and backward differences for the second half, then you'll implicitly have a symmetric condition of sorts imposed in the middle. 
In essence, since different definitions of $u_{xx}$ result in different solutions, we see that the problem isn't well formulated without boundary conditions.
