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 $f(x)=0$?


1 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.

  • $\begingroup$ Thanks for your response. I'm wondering if it's possible to impose extra conditions to ensure the IVP $(1)+(2)$ has a unique solution. What if $u(x,t)=0$ for $t<0$? $\endgroup$
    – Bell
    Feb 27 at 7:20
  • $\begingroup$ Also, if $v(x,t)$ was the unique solution to the IVP with $v(x,0)$, how does this help us to determine the uniqueness to the IVP with $(1)+(2)$? $\endgroup$
    – Bell
    Feb 27 at 7:27
  • 1
    $\begingroup$ The condition $u(x,t) = 0$ for $t<0$ doesn't make any difference. You can require some decay condition as $x \to \pm\infty$, or require that $u$ is nonnegative everywhere. There's some mention of this in the MO question that I linked to. $\endgroup$ Feb 27 at 9:02
  • $\begingroup$ Regarding your second question, I thought you said that you were familiar with this method already? If $v$ (which is defined to be $v_1 - v_2$) has to be the zero function, then $v_1$ has to be the same function as $v_2$, which is exactly what uniqueness means. $\endgroup$ Feb 27 at 9:04

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