# One solution implies the other solution Second order Differential Equation

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?

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.
• shouldn't we have $\frac{\partial^2}{\partial x^2} (u(L-x))$? Oct 15, 2015 at 10:21
• 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 $L-x$. Oct 15, 2015 at 10:26
• 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 $(L-x)$. Oct 15, 2015 at 10:33