If I have the following PDE; $${{\partial_t^2 u(t,x)}} - {\partial_x^2 u(t,x)} = 0 \\ u(0,x) = g(x) \\ {\partial_t u(0,x)} = h(x) $$
Is is true if $g$ and $h$ are monotone increasing then $u(x,t)$ is also monotone increasing?
I started by finding the sub-equations and I found one of the solutions as $$w(t,x) = \int_{x-t}^{x+t} h({\xi}) ~ d{\xi} $$ but this is not necessarily monotone increasing, does this conclude that the statement does not hold? I couldn't find the second solution. help appreciated.
Use d'Alembert's formula to express the solution as $$ u(t,x) = \tfrac12 \left[g(x-t) + g(x+t)\right] + \tfrac12\int_{x-t}^{x+t} h(\xi)\,\text d \xi . $$ It remains to evaluate the sign of $$ u_x(t,x) = \tfrac12 \left[g'(x-t) + g'(x+t)\right] + \tfrac12 \left[h(x+t) - h(x-t)\right] $$ when $g'$ and $h'$ are positive. Using the mean value theorem, there exists $\xi$ such that $$ h(x+t) - h(x-t) = 2t\, h'(\xi) , \qquad x-t\leq \xi \leq x+t . $$ Thus, we can conclude for positive times.