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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?

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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.

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  • $\begingroup$ 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

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