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Let $M$ be a smooth compact closed manifold. Let $u \in H^1(0,T;H^{-1}(M)) \cap L^2(0,T;H^1(M))$ be a solution of $$u_t - \Delta u - u = 0$$ $$u(0)=u(T)$$ satisfying $\int_M u(t) = 0$ for all $t$. Is there any way to show that $u$ must be zero (i.e. solutions are unique)?

The problem is the $-u$ term. We have Poincare's inequality in this but it does not help much. I would appreciate not using a method to do with eigenvalue problems since this PDE is a simplified version of what I am working on.

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    $\begingroup$ I am not sure how you discuss eigenvalue problems using a method not to do with them. The issue here is whether -1 is an eigenvalue of the Laplacian, and there are certainly examples where it is. $\endgroup$
    – Michael Renardy
    Commented Sep 4, 2014 at 18:36
  • $\begingroup$ @MichaelRenardy I thought a variational energy method might work since I work with weak solutions and this is just a nice linear PDE. $\endgroup$
    – TW.
    Commented Sep 5, 2014 at 10:40

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Your claim holds true if and only if $-1$ is not an eigenvalue of $\Delta$.

If there is a nontrivial function $v:M\to\mathbb R$ with $\Delta v=-v$ (then necessarily $\int_Mv=0$), then $u(x,t)=v(x)$ solves your PDE with all conditions but does not vanish identically.

Suppose then that $-1$ is not an eigenvalue. Then it follows that the line $-1+i\mathbb R\subset\mathbb C$ is in the resolvent set of $\Delta$. Because of the boundary condition the solution $u$ can be thought of as a periodic solution (w.r.t. time). Then it is very convenient to take the Fourier series of $u$ in $t$: $$ \hat u(x,s) = \int_0^Tu(x,t)e^{2\pi its/T}dt $$ for $s\in\mathbb Z$. Since $\widehat{u_t}=-\frac{2\pi is}{T}\hat u$, the PDE becomes $$ \left(\Delta+(1+\frac{2\pi is}{T})\right)\hat u = 0. $$ Notice that the equations for different Fourier components decouple. Since $-(1+\frac{2\pi is}{T})$ is in the resolvent for all $s$, we have $\hat u=0$ and consequently $u=0$.

Even if we do not assume that $-1$ is not an eigenvalue, we know that the spectrum is real. Therefore the above argument shows that any solutions $u$ has to be constant in time ($\hat u(x,s)=0$ when $s\neq0$) and thus an eigenfunction of the Laplacian — and the eigenvalue should come as no surprise.

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  • $\begingroup$ Thank you for this answer. I have a question: suppose we replace the $-u$ with $f(x,t)u$ for some arbitrary space-time function $f$ (so the equation reads $u_t - \Delta u + fu=0$). I wonder then if I should integrate over and space and time and consider the operator $-\Delta u + f\text{Id}$ acting on $L^2(0,T;H^1)$ to $L^2(0,T;H^{-1})$. Maybe something similar to your answer would work. $\endgroup$
    – TW.
    Commented Sep 5, 2014 at 10:55
  • $\begingroup$ @TobiasWood, if $f$ does not depend on $t$, something similar to my approach might work, but I don't want to work it out. If $f$ depends on $t$, the Fourier transform gives a convolution term and things get a whole lot messier. My approach might be worth a try, but it doesn't look simple. $\endgroup$
    – Joonas Ilmavirta
    Commented Sep 5, 2014 at 13:55

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