Consider the wave equation on $[0,L]$ $$\frac{1}{c^2}\cdot\frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2}, \ c^2 = \frac{\tau}{\rho}$$

Suppose we have $u(x,0)=f(x), \frac{\partial u}{\partial t}(x,0) = g(x), u(0,t)=u(L,t)=0$. The key assumption here is that $f \in C^1$ and $g \in C^0$, i.e., $g$ is only assumed to be continuous.

The energy is defined as usual as $$E(t) = \frac{1}{2}\rho\int_{0}^{L}(\frac{\partial u}{\partial t})^2dx+\frac{1}{2}\tau\int_{0}^{L}(\frac{\partial u}{\partial x})^2dx$$

The problem is $\frac{\partial u}{\partial t}(x,0) = g(x)$ is only continuous, so when I find $\frac{dE(t)}{dt} = 0$, this is only valid for all $t>0$. So I only can conclude that $E(t)$ is a constant for all $t>0$. However, I want to say $E(t) = E(0),\ \forall t$.

How is this possible? What kind of argument could justify this step?

I'm not sure if the above is also related to the question: For a general function $f$ defined on $[0,\infty)$ but differentiable on $(0,\infty)$, if I get $f'>0$ on $(0, \infty)$, how can I say $f$ is an increasing function on $[0,\infty)$?

Many thanks!


If you have a function defined and continuous on $[0,+\infty)$ which is differentiable on $(0,+\infty)$ and whose derivative has a finite limit $$ L=\lim\limits_{t\to 0^+}f(t) $$ then $f$ is differentiable (from the right) at $t=0$ and its derivative is $f'(0)=L$. This follows from the mean value theorem, applied on $[0,h]$ and taking the limit $h\to 0^+$.


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