Consider $$u_t - \Delta u + u = 0$$ $$\frac{\partial u}{\partial \nu} = 0$$ $$u(T) = u(0)$$ on a domain $\Omega$ (the BC is obviously on $\partial\Omega$. If $u$ solves this PDE, clearly as does $\lambda u $ for any constant $\lambda.$ So uniqueness is not there.

Suppose we have two solutions $u$ and $v$. Their difference $d = u-v$ satisfies $$d_t - \Delta d + d = 0$$ $$\frac{\partial d}{\partial \nu} = 0$$ $$d(T) = d(0)$$

Multiplying the equation by $d$ and integrating by parts we get $$\frac{1}{2}\frac{d}{dt}\int_{\Omega} d^2(t) + \int_{\Omega} |\nabla d(t)|^2 + \int_{\Omega} d^2(t) = 0$$ Now integrating by time $$\frac{1}{2}|d(T)|_{L^2} - \frac{1}{2}|d(0)|_{L^2} + \int_{0}^T |\nabla d(t)|_{L^2}^2 + \int_0^2 |d(t)|_{L^2}^2 = 0$$ but the first two terms cancel each other out, and we get $d=0$ in $H^1$. So this shows that there is a unique solution.

What am I doing wrong???!?!

Edit: maybe $0$ is the only solution to this problem? How to prove this if so? What happens in nonhomogenous case?

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    $\begingroup$ Intuitively, it seems like the energy of a solution has to dissipate, so my guess is that the requirement $u(T) = u(0)$ eliminates the possibility of non-zero solutions. $\endgroup$
    – BaronVT
    Jan 31, 2014 at 19:28
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    $\begingroup$ $u \equiv 0$ is the unique solution. $\endgroup$
    – gerw
    Jan 31, 2014 at 20:48
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    $\begingroup$ Oh yeah, I guess this follows trivially from the energy argument: if $\lambda \neq 1$, then $d = u - \lambda u \equiv 0$, i.e. $(1 - \lambda) u \equiv 0$ implying $u \equiv 0$. $\endgroup$
    – BaronVT
    Jan 31, 2014 at 21:06
  • $\begingroup$ @BaronVT No need for that, the energy argument says that any solution is equal to $0$, so you don't even need to bother considering the possibility of there being two solutions. $\endgroup$
    – JLA
    Feb 1, 2014 at 21:06

2 Answers 2


There exists $w_k$, $k\in\mathbb N$, an orthonormal basis of $L^2(\Omega)$, consisting of the eigenvectors of the Laplacian, i.e., $$ -\Delta w_k=\lambda_kw_k \,\,\,\text{in $\Omega$},\\ \partial_\nu w_k=0 \,\,\,\text{on $\partial\Omega$}, $$ with $\lambda\ge 0$ and $\lambda_k\to\infty$.

If $u$ is a solution of your equation, with $u(x,0)=\sum_{k\in\mathbb N}c_kw_k(x)$, then $$ u(x,t)=\sum_{k=1}^\infty\mathrm{e}^{-(\lambda_k+1)t}w_k(x). $$ Clearly $$ \|u(\cdot,t)\|\le \mathrm{e}^{-t}\|u(\cdot,0)\|. $$ So if $\|u(\cdot,t)\|= \|u(\cdot,T)\|$, then $u\equiv 0$.

Thus your solution is the identically zero.


A slightly different approach: writing $y(x,t)=e^tu(x,t)$, the system becomes equivalent to the following homogeneous heat equation: $\displaystyle \frac{dy}{dt}-\Delta y=0$, $\displaystyle\frac{\partial y}{\partial\nu}=0,\, y(x,T)=e^Ty(x,0)$. Now multiplying by $y$ and integrating by parts, we get $\displaystyle \frac{d}{dt}\int_{\Omega}\frac{y^2}{2}=-\int_{\Omega}|\nabla y|^2$, so that $\displaystyle\int_{\Omega}\frac{y^2}{2}$ is decreasing in time. Hence $\displaystyle \int_{\Omega}y^2(x,T)=e^{2T}\int_{\Omega}y^2(x,0)\ge e^{2T}\int_{\Omega}y^2(x,T),$ and this implies $y(x,0)=y(x,T)=0$, and also since $\displaystyle\int_{\Omega}\frac{y^2}{2}$ is decreasing in time but always positive and is zero initially, $y(x,t)=0$.


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