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I would like find a solution $\{\phi(x,t);t\in\mathbb{R}^+,x\in\mathbb{R}\}$ of the following PDE:

$$ \frac{\partial\phi}{\partial t}=\frac{\partial^2\phi}{\partial x^2}-\phi f(x) , $$ with boundary conditions $$ \phi(0,x)=1,\text{ and }\frac{\partial\phi}{\partial t}(0,x)=f(x). $$ The function $f$ will ideally be as general as possible. If such generality is unhelpful, then I am willing to consider any particular choice of $f$ satisfying:

  1. $f(x)=f(-x)$ for any $x\in\mathbb{R}$,
  2. $f>0$, and
  3. $f(x)<f(y)$ for any $|x|>|y|$.

For example, $f(x)=1/(1+x^2)$ or $f(x)=1/(e^x+e^{-x})$.

This seemingly simple looking problem is completely baffling me. Does anyone have any suggestion for how to proceed? Do we even have existence and uniqueness?

user301395
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    Getting an analytical solution for some arbitrary $f(\phi)$ is a very hard problem. You'll be quite famous if you ever figure out how to do that. Occasionally, you can transform your PDE to be linear again depending on what $f(\phi)$ is. Do you have a specific $f$ that you're interesting in? – AlkaKadri Jun 21 '18 at 17:23
  • The function $f$ is a function of $x$, not $\phi$. Hopefully this is clearer now that I have slightly edited the question. – user301395 Jun 21 '18 at 17:59
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    As a parabolic equation, it can have some similarity solution... – rafa11111 Jun 21 '18 at 18:03
  • Perhaps I did not get that well, but I think you only provided the initial conditions for your problem... – rafa11111 Jun 21 '18 at 19:25
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    This is quite similar in form to the linear Schrodinger equation... for the case where $f$ is a positive constant function (and thus satisfying your 3 conditions), this can be easily solved via substitution $\psi = \exp(ft)\phi$. – Alex Jones Jun 21 '18 at 19:25
  • Thanks Alexander, that works nicely. Ideally $f$ will not be constant however. I have edited the condition 3 now. – user301395 Jun 21 '18 at 19:43
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    You're not going to find an explicit solution for any non-trivial set of functions $f$. – Disintegrating By Parts Jun 21 '18 at 23:43
  • The equation is first order in $t$. You cannot specify a condition on $\frac{\partial\phi}{\partial t}$ at $t=0$. – Disintegrating By Parts Jun 23 '18 at 01:29

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