I have the partial differential equation:

$\qquad U_{x} - 4 x^{2} U_{t} = x U, \qquad U(x,0)=f(x), \qquad x \in (- \infty, \infty), \qquad t \in (0, \infty)$

Where $U_{x} = \frac{\partial U}{\partial x}$, and the solution I arrived at using Method of Characteristics is:

$\qquad U(x,t) = f\Big( (x^{3}+\frac{3}{4}t)^{\frac{1}{3}} \Big) * e^{\frac{x^{2}}{2} - \frac{1}{2} (x^{3} + \frac{3}{4}t)^{\frac{2}{3}}}$

I am unsure how to define the arbitrary function $f(x,t)$ in that way when trying to plug the solution back into the differential equation to check to see if it is a solution. (This solution may not be correct)

  • 1
    $\begingroup$ Have you read the relevant part in the documentation? $\endgroup$
    – yohbs
    Sep 23, 2015 at 21:09
  • $\begingroup$ @yohbs Yeah I did but I tried it and it didn't work. I think I screwed up a bracket or something. Thank you though. $\endgroup$
    – Kvo
    Sep 25, 2015 at 14:10
  • $\begingroup$ @yohbs Could you provide the relevant part of the documentation ? The link is not working $\endgroup$
    – q than a
    Jun 20, 2020 at 15:43
  • 1
    $\begingroup$ @qthana I guess it's this though I really can't remember at this point what I meant 5 years ago $\endgroup$
    – yohbs
    Jun 21, 2020 at 5:57

1 Answer 1


In case reading the documentation, as recommended by yohbs, did not fully answer your question, try

U[x_, t] = f[(x^3 + 3 t/4)^(1/3)] Exp[x^2/2 - (x^3 + 3 t/4)^(2/3)/2];
Simplify[D[U[x, t], x] - 4 x^2 D[U[x, t], t] - x U[x, t]]
(* 0 *)

By the way, Exp[ - (x^3 + 3 t/4)^(2/3)/2] can be absorbed into f without loss of generality.

  • $\begingroup$ Ah! Thank you. I tried something and it didn't work. I must have just screwed up the brackets or something. This worked perfectly. $\endgroup$
    – Kvo
    Sep 25, 2015 at 14:09

Your Answer

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

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