Suppose I have a differential equation $$\frac{\partial\theta}{\partial t}=\frac{\partial^2\theta}{\partial x^2}$$ satisfied by $\theta(x,t)$ and I then had that $\theta(x,t)=k F(x,t)$, and I know $F(x,t)$ is dimensionless. What is a general method in this situation to answer the question "Show $F$ is a function of only the similarity variable $\eta=\eta(x,t)$"?

I am not really sure how to approach this as I don't recall ever encountering a problem like this. Any help apprecieted.

  • $\begingroup$ Is this what you are looking for? $\endgroup$ May 3, 2017 at 10:53
  • $\begingroup$ @Mattos I have seen a few documents like this, but they all seem to say "suggest a solution of the form $F(\xi)$ rather than proving that the solution must be of this form. Is there any way to prove that? $\endgroup$
    – John Doe
    May 3, 2017 at 11:01
  • $\begingroup$ The solution to the heat equation can take many different forms, depending on boundary conditions, initial conditions and domain type. The solution you are asking for is just one particular type of solution, named a similarity solution, which is a solution which is invariant under a stretching transformation. By taking particular products between the variables themselves and also with the original function, you can remove the scaling factor entirely which yields an invariant solution to the original PDE. $\endgroup$ May 3, 2017 at 12:44
  • $\begingroup$ @Mattos Ah yes, I do understand that. So if I gave the exact BCs and ICs, then it would be possible to prove that $F(x,t)=F(\eta)$ only? $\endgroup$
    – John Doe
    May 3, 2017 at 12:54

1 Answer 1


Let's take another example, the 1D wave equation: $$\frac{\partial^2 u}{\partial t^2} - c^2 \frac{\partial^2 u}{\partial x^2}=0.$$ Take $u(x,t) = U(x \pm ct)$, then $u$ satisfies the wave equation. Of course initial and boundary conditions also need to be taken into account, this will set $U$.

Can you work out an equivalent way for your (heat?) equation?

  • $\begingroup$ Yes, I am working with the heat equation. My question was about how to find the $\eta$ and show that the solution is a function of $\eta$ only. I can see that for the heat equation, the quantity equivalent to the $x+ct$ here is $x/\sqrt{t}$ but how can you justify/prove that $F$ is a function of this only? $\endgroup$
    – John Doe
    Apr 13, 2017 at 22:59
  • $\begingroup$ In your example, also $U(x-ct)$ would work too, so it is possible that $u(x,t)$ is given by a combination of these, and so isn't exclusively a function of one of these - so how can you be sure? (Of course it is different for the heat equation). $\endgroup$
    – John Doe
    Apr 13, 2017 at 23:01
  • 1
    $\begingroup$ If you exhibit a solution and can show uniciy, then you're done, aren't you? $\endgroup$
    – Joce
    Apr 14, 2017 at 9:12
  • $\begingroup$ Oh ok, yes you're right. So how can I show that $F(x,t)=F(\eta)$ is unique without determining $F$ explicitly? $\endgroup$
    – John Doe
    Apr 14, 2017 at 14:11
  • $\begingroup$ I have determined that if $F(x,t)$ solves the wave equation, then so does $F(\lambda x, \lambda^2 t)$. So $F(\eta)$ would account for this scaling property, where $\eta(x,t)=\frac{x}{\sqrt{t}}$. But how can I then conclude that $F$ is a function of $\eta$ only? $\endgroup$
    – John Doe
    Apr 15, 2017 at 21:55

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