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Sep 30 at 0:19 history edited Sergey Tikhomirov CC BY-SA 4.0
deleted 91 characters in body
Sep 30 at 0:17 vote accept Sergey Tikhomirov
Sep 28 at 22:53 answer added Michael Renardy timeline score: 5
Sep 27 at 19:58 comment added Thomas Kojar @SergeyTikhomirov it works for both, see the answer in the above link.
Sep 27 at 19:58 history edited Sergey Tikhomirov CC BY-SA 4.0
deleted 76 characters in body; edited title
Sep 27 at 19:49 comment added Sergey Tikhomirov It seems that Feymann-Kac formula helps with backward solution of $u_t = -u_{xx}$, which is a lot different.
Sep 26 at 1:12 history edited Sergey Tikhomirov CC BY-SA 4.0
added 64 characters in body
Sep 26 at 1:10 comment added Sergey Tikhomirov Yes, indeed. I have some restrictions on $f$. It is an indicator of a certain set in $(x, t)$. I am editing question.
Sep 26 at 1:00 comment added Michael Renardy As it stands, the answer is obviously yes. Just pick u to be whatever you want it to be and then choose f accordingly. But this is probably not what you intended.
S Sep 26 at 0:56 history suggested J. W. Tanner CC BY-SA 4.0
corrected spelling
Sep 26 at 0:29 review Suggested edits
S Sep 26 at 0:56
S Sep 25 at 22:57 review First questions
Sep 26 at 0:41
S Sep 25 at 22:57 history asked Sergey Tikhomirov CC BY-SA 4.0