Say we have the following zero-flux nonlinear boundary value problem: $$ u_{xx} + f(u) = 0, $$ $$u_x(0) = 0 = u_x(L),$$ Now if $u$ is a solution this implies that $w = u(L-x)$ is also a solution. How do they come to that implication?


1 Answer 1


Because $\frac{\partial^2}{\partial x^2} u(L-x) = - \frac{\partial}{\partial x} u_x(L-x) = u_{xx}(L-x)$. Since $u(x)$ fulfills the differential equation, so will $u(L-x)$. That $u(L-x)$ fulfills the bondary conditions is clear.

  • $\begingroup$ shouldn't we have $\frac{\partial^2}{\partial x^2} (u(L-x))$? $\endgroup$ Oct 15, 2015 at 10:21
  • $\begingroup$ Well, $\frac{\partial^2}{\partial x^2}$ is a differential operator, that acts on the function $u$, not on the result of applying $u$ to $L-x$. $\endgroup$ Oct 15, 2015 at 10:26
  • $\begingroup$ Thought it was still $u(t,x)$ so you should read $u*(L-x)$ but indeed it is only $u(x)$ and so the differential operator only applies to the function $u$ of $(L-x)$. $\endgroup$ Oct 15, 2015 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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