# Wave Equation - Conservation of Energy

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!

## 1 Answer

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 $$h\to 0^+$$.