For the sake of concreteness, we consider the linear Schrödinger equation $$ \partial_t u = i\Delta u, \ \ \ \ u(0, x) = u_0(x). $$ The solution is typically obtained by taking the Fourier transform of both sides, giving $\widehat{\partial_t u}(t, \xi) = -i|\xi|^2 \hat{u}(t, \xi)$.

The next step is where I have questions. Assuming that everything is nice enough (for instance, in Tao's book, he assumes $u_0$ is Schwartz), a dominated convergence argument gives $\widehat{\partial_t u}(t, \xi) = \partial_t \hat{u}(t, \xi)$, and then we get an ODE that solves to $$ \hat{u}(t, \xi) =e^{-i|\xi|^2}\hat{u}_0(\xi) \implies u(t, x) = e^{it\Delta}u_0(x). $$ This is then referred to as "the solution to the Schrödinger equation, with initial data $u_0$."

My question: How do we know that there are no other solutions, that may not satisfy the right decay/smoothness criteria to justify pulling the Fourier transform into the time derivative of $u$? I agree that there are no other solutions $u$ that are "nice enough" to justify this. But how do we rule out the existence of solutions $u$ such that $\partial_t \hat{u} \neq \widehat{\partial_t u}$? For instance, I don't understand how just assuming $u_0$ Schwartz is enough to guarantee this.

Any help is much appreciated.

  • 7
    $\begingroup$ In my book I only establish uniqueness for solutions that are $C^1_t {\mathcal S}_x$ by this argument. Without some regularity in time and decay in space uniqueness can fail; see Exercise 2.24 of the book. $\endgroup$
    – Terry Tao
    Nov 13, 2019 at 5:54

1 Answer 1


In fact, in exactly the same book as you referred to, exercise 2.24 gives a counter example. Tao himself has commented above. Maybe he is too busy to give more details. Roughly speaking, the reason that this uniqueness may fail is that, the constructed counter-example grows too fast near infinity, so that it is out of the scope of the tempered distribution, and the Fourier transform etc fail to make nice sense.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.