Unique Solution to the Heat Equation

Question

Consider the IVP consisting of the homogeneous heat equation \begin{equation} v_t(x,t)=\alpha v_{xx}(x,t), \quad (x,t)\in(-\infty,\infty)\times(0,\infty), \tag{1} \end{equation} subject to the initial condition \begin{equation} v(x,0)=f(x), \quad x\in (-\infty,\infty). \tag{2} \end{equation} Here $\alpha>0$ and $f\in C^\infty$. It can be shown using Fourier transform methods that a solution to the IVP is given by $$ v(x,t)=\int_{-\infty}^\infty \frac{f(s)}{\sqrt{4\pi\alpha t}}\exp\left(-\frac{(x-s)^2}{4\alpha t}\right) ds. $$ I am trying to determine if this is a unique solution. I am familiar with the contradiction method where we assume that there exist two distinct solutions, $v_1(x)$ and $v_2(x)$, to the IVP. Hence, $$v(x)=v_1(x,t)-v_2(x,t) \tag{3}$$ is a solution to $(1)$ as this equation is linear and homogeneous. By looking at $(2)$, we find that $$ v(x,0)=v_1(x,0)-v_2(x,0)=f(x)-f(x)=0. $$ Hence, this does not satisfy $(2)$ and so $(3)$ is not a solution to the IVP. What is the implication of this? Is $(3)$ a unique solution only if $f(x)=0$?

Answer 1

No, that's not a correct argument. (And what you're trying to prove is not true.)

In order to show uniqueness, you would like to conclude that $v(x,t) = v_1(x,t) - v_2(x,t)$ is identically zero. As you noticed, $v$ solves the heat equation with the initial condition $v(x,0)=0$, so the question now is whether that IVP (not your original IVP (1)+(2)) has a unique solution, namely the trivial solution which is identically zero. The fact that $v$ doesn't satisfy (2) is completely irrelevant.

However, the IVP for $v$ actually has nontrivial solutions unless you impose some extra assumptions! See, for example, this Math Overflow question: "Wild" solutions of the heat equation: how to graph them?.

So without such assumptions, the solution to the IVP (1)+(2) is not unique.