1. Wave Equation Question

A string with displacement y(x,t), where $a \leq x \leq b$ is fixed at each end so that y(a,t) = 0 = y(b,t) for all time t. The string's energy is give by:

$E(t) = \int^b_a {[\frac{1}{2} T (\frac{\partial y}{\partial x})^2 + \frac{1}{2} p (\frac{\partial y}{\partial t})^2}]\,dx$ where the first term is the tensile energy and the second term represents ths string's kinetic energy. Given the wave equation holds, show that the string's energy is constant througout the motion. Hint: You may assume the integrand is sufficiently smooth to allow the use of the Leibniz integral rule.

I can't see how to start this question. So far I have got:
$(\frac{\partial y}{\partial x})^2 = (X'(x)T(t))^2$ and
$(\frac{\partial y}{\partial t})^2 = (X(x)T'(t))^2$ but I can't see how this will help me. Any thoughts on where to start?

2. Originally Posted by worc3247
A string with displacement y(x,t), where $a \leq x \leq b$ is fixed at each end so that y(a,t) = 0 = y(b,t) for all time t. The string's energy is give by:

$E(t) = \int^b_a {[\frac{1}{2} T (\frac{\partial y}{\partial x})^2 + \frac{1}{2} p (\frac{\partial y}{\partial t})^2}]\,dx$ where the first term is the tensile energy and the second term represents ths string's kinetic energy. Given the wave equation holds, show that the string's energy is constant througout the motion. Hint: You may assume the integrand is sufficiently smooth to allow the use of the Leibniz integral rule.

I can't see how to start this question. So far I have got:
$(\frac{\partial y}{\partial x})^2 = (X'(x)T(t))^2$ and
$(\frac{\partial y}{\partial t})^2 = (X(x)T'(t))^2$ but I can't see how this will help me. Any thoughts on where to start?
Start by using the hint and differentiating (with respect to t) under the integral sign:

$\displaystyle E'(t) = \frac d{dt}\int^b_a \Bigl(\tfrac{1}{2} T \Bigl(\frac{\partial y}{\partial x}\Bigr)^2 + \tfrac{1}{2} p \Bigl(\frac{\partial y}{\partial t}\Bigr)^2\Bigr)\,dx = \int^b_a \Bigl( T\frac{\partial y}{\partial x}\frac{\partial^2 y}{\partial t\partial x} + p \frac{\partial y}{\partial t}\frac{\partial^2 y}{\partial t^2}\Bigr)\,dx.$

Now use integration by parts on the first term of that integral:

$\displaystyle \int^b_a T\frac{\partial y}{\partial x}\frac{\partial^2 y}{\partial t\partial x}\,dx = \Bigl[T\frac{\partial y}{\partial x}\frac{\partial y}{\partial t}\Bigr]_a^b - \int^b_a T\frac{\partial^2 y}{\partial x^2}\frac{\partial y}{\partial t}\,dx = -\int^b_a T\frac{\partial^2 y}{\partial x^2}\frac{\partial y}{\partial t}\,dx$ (because $\frac{\partial y}{\partial t}$ vanishes at the ends of the interval).

Therefore $\displaystyle E'(t) = \int^b_a \frac{\partial y}{\partial t}\Bigl(p\frac{\partial^2 y}{\partial t^2} - T\frac{\partial^2 y}{\partial x^2}\Bigr)\,dx.$

You want to show that $E'(t) = 0$ (and therefore $E(t)$ is constant). So you want that last integral to be zero. I'm guessing that the form of the wave equation relevant for this problem is $p\frac{\partial^2 y}{\partial t^2} - T\frac{\partial^2 y}{\partial x^2} = 0$, in which case everything works nicely.

3. I realize that this thread is old, so forgive me for bringing it up again. However, I'm solving a similar problem and I have a question about the derivation that Opalg has given. I believe it to be correct (the book I'm using for reference, "Introduction to Partial Differential Equations - A Computational Approach" by Tveito and Winther does the exact same thing), but there is one step that I don't understand.

In the integration by parts, what is it that allows you to assume that the time-derivative is zero on the boundary? We have only assumed Dirichlet boundary conditions in the problem statement. Does it necessarily follow that the derivative is zero on the boundary also?

4. Originally Posted by jamesdt
In the integration by parts, what is it that allows you to assume that the time-derivative is zero on the boundary? We have only assumed Dirichlet boundary conditions in the problem statement. Does it necessarily follow that the derivative is zero on the boundary also?
You are told that the displacement is zero (and therefore constant!) at the ends of the interval. So its time-derivative must also be zero there.

5. Originally Posted by Opalg
You are told that the displacement is zero (and therefore constant!) at the ends of the interval. So its time-derivative must also be zero there.
Yes, of course! Thank you for the clarification!