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Consider the initial value problem:

\begin{cases} u_{tt} &= c^2 u_{xx} \ \ & \text{for} \ -\infty < x < \infty, \ 0 \leq t < \infty\\ u(x,0) &= \phi(x) \ \ & \text{for} \ -\infty < x < \infty\\ u_t(x,0) &= \psi(x) \ \ & \text{for} \ -\infty < x < \infty\\ \end{cases} where $\phi$ has compact support (that is, outside some bounded interval, $\phi$ is zero), and $\psi(x) = 0$. Define the kinetic energy $KE = \frac{1}{2}\int_{-\infty}^{\infty} \rho u_t^{2} dx$ and the potential energy $PE = \frac{1}{2} \int_{-\infty}^{\infty} T u_x^{2} dx$. Show not that, for large enough times $t$, each of $KE$ and $PE$ is itself constant, and they are equal to each other. Can you prove the same thing if the inital velocity $\psi$ merely has compact support, instead of being zero?

I am not sure how to start this, how am I to show that $KE$ and $PE$ are constant? I usually post some work but I am not sure how to start this. Any help would be useful.

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2 Answers 2

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Define $E_k(t) \equiv \frac{1}{2}\int_{\mathbb{R}}u_t^2(x,t){\rm d}x$ and $E_p(t) \equiv \frac{1}{2}\int_{\mathbb{R}}c^2u_x^2(x,t){\rm d}x$. Then taking the derivative and using integration by parts we get $$\frac{d}{dt}(E_k(t) + E_p(t)) = \int_{\mathbb{R}}u_t[u_{tt} - c^2u_{xx}]{\rm d}x = 0$$ so the total energy is conserved. To show that $E_p(t) = E_k(t)$ for large $t$ we need an expression for the solution. This is given by d'Alemberts formula

$$u(x,t) = \frac{\phi(x+ct) + \phi(x-ct)}{2} + \frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s){\rm d}s$$

from which you can compute $u_t$, $u_x$ and derive

$$E_k(t) - E_p(t) = \frac{1}{2}\int_{\mathbb{R}}[\psi(x+ct)+c\phi'(x+ct)][\psi(x-ct)-c\phi'(x-ct)]{\rm d}x$$

To make the algebra above simpler note that $u_t^2 - c^2u_x^2 = (u_t - cu_x)(u_t+cu_x)$. Now since $\phi$ and $\psi$ has compact support then there is a $M>0$ such that $\phi'(x) = \psi(x) = 0$ for all $|x| > M$. Now if $ct > \frac{M}{2}$ then either $|x-ct| > M$ or $|x+ct|>M$ for all $x$ so the integrand above is identical to zero and $E_p = E_k$ follows. Finally since we know that $E_k + E_p$ is constant it follows that $E_p$ and $E_k$ has to be constant for large $t$.

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We have \begin{equation} \begin{cases} u_{tt} = c^2 u_{xx} \ \ & \text{for} \ -\infty < x < \infty, \ 0 \leq t < \infty\\ u(x,0) = \phi(x) \ \ & \text{for} \ -\infty < x < \infty\\ u_t(x,0) = \psi(x) \ \ & \text{for} \ -\infty < x < \infty \end{cases} \end{equation} Therefore \begin{equation} \phi(x) = 0 \ \ \ \text{and} \ \ \ \psi(x) = 0 \end{equation} and \begin{equation} KE = \frac{1}{2}\int_{-\infty}^{\infty} \rho u_t^{2} dx \end{equation} \begin{equation} PE = \frac{1}{2} \int_{-\infty}^{\infty} T u_x^{2} dx \end{equation} From equation $(1)$ $u_{tt} = c^2 u_{xx}$ then \begin{equation} u_{tt} - c^2 u_{xx} = 0 \end{equation} Here, $A = -c^2, B = 0$, and $C = 1$. Thus \begin{equation} B^2 - 4AC = 4C^2 > 0 \end{equation} So the equation is hyperbolic. The equation characteristics are \begin{equation} \frac{dt}{dx} = \pm\frac{1}{C} \end{equation} or $$\xi = x - ct = \ \text{const} \ \ \ \text{and} \ \ \ \eta = x + ct = \ \text{const}$$ Now, in terms of new coordinates $\xi$ and $\eta$ then $$u_{xx} = u_{\xi \xi} + 2u_{\xi \eta} + u_{\eta \eta} \ \ \ u_{tt} = c^2(u_{\xi\xi} - 2 u_{\xi\eta + u_{\eta\eta}}$$ Thus, equation $(1)$ becomes \begin{equation} -4c^2 u_{\xi\eta} = 0 \end{equation} Therefore $c\neq 0$, \begin{equation} u_{\xi \eta} = 0 \end{equation} So, integrating $u_{\xi\eta} = 0$ twice we then get the solution \begin{equation} u(\xi \eta) = \phi(\xi) + \psi(\eta) \end{equation} where $\phi$ and $\psi$ are arbitrary functions. So, in terms of $x$ and $t$ \begin{equation} u(x,t) = \phi(x - ct) + \psi(x + ct) \end{equation} Now from \begin{equation} u(x,0) = \phi(x) \ \ \ \text{and} \ \ \ u_t(x,0) = \psi(x) \end{equation} So, \begin{equation} \phi(x) + \psi(x) = \phi(x) \end{equation} and \begin{equation} -c \phi^{\prime}(x) + c\psi^{\prime}(x) = \psi(x) \end{equation} Integrating equation $(14)$ \begin{equation} -c \phi(x) + c\phi(x) = \int_{x_0}^{x}\psi(\tau)d\tau \end{equation} where $x_0$ is an arbitrary constant. Now equation $(13)$ and $(15)$ becomes $$\phi(x) = \frac{1}{2}\phi(x) - \frac{1}{2c}\int_{x_0}^{x}\psi(\tau)d\tau$$ and $$\psi(x) = \frac{1}{2}\psi(x) + \frac{1}{2c}\int_{x_0}^{x}\psi(\tau)d\tau$$ Thus, equation $(11)$ gives the solution (D'Alembert solution) of the Cauchy problem as $$u(x,t) = \frac{1}{2}\left[ \phi(x - ct) + \phi(x + ct) \right] + \frac{1}{2c}\int_{x-ct}^{x+ct}\psi(\tau)d\tau$$ So we have $$u_t(x,t) = \frac{1}{2}\frac{\partial}{\partial t}\left[ \phi(x - ct) + \phi(x + ct) \right] + \frac{1}{2c}\frac{\partial}{\partial t}\left[ \int_{x-ct}^{x+ct} \psi(\tau)d\tau \right]$$ and $$u_{x}(x,t) = \frac{1}{2}\frac{\partial}{\partial x}\left[ \phi(x-ct) + \phi(x+ct) \right] + \frac{1}{2c}\frac{\partial}{\partial x}\left[ \int_{x - ct}^{x+ct}\psi(\tau)d\tau \right]$$ therefore $$KE = \frac{1}{2}\int_{-\infty}^{\infty}\rho u_{t}^{2}dx$$ So, $$KE = \frac{1}{2}\int_{-\infty}^{\infty}\rho\left[ \frac{1}{2}\frac{\partial}{\partial t} \left[ \phi(x-ct) + \phi(x + ct) \right] \right] + \frac{1}{2c}\frac{\partial}{\partial t}\left[\int_{x - ct}^{x+ct}\psi(\tau)d\tau \right]^2dx$$ and $$PE = \frac{1}{2} \int_{-\infty}^{\infty} T u_x^{2} dx$$ So, $$PE = \frac{1}{2}\int_{-\infty}^{\infty}T\left[ \frac{1}{2}\frac{\partial}{\partial x}\left[ \phi(x - ct) + \phi(x + ct) \right] + \frac{1}{2c}\frac{\partial}{\partial x}\left[ \int_{x-ct}^{x+ct} \psi(\tau)d\tau \right] \right]^2dx$$ Thus from above, we can see that for $t > 0$, $KE$ and $PE$ is constant thus $KE = PE$ at some stage.

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  • $\begingroup$ 1) $T$ and $\rho$ not defined anywhere (I see OP uses it, but it's not in the PDE so it would be good to simply note that here $\rho = 1$ and $T = c^2$) 2) "Thus from above, we can see that" I don't see how this follows. If it did follow it seems it holds for all $\phi$ as you don't seem to use the critical assumption that $\phi$ has compact support. 3) "KE and PE is constant thus KE=PE at some stage." How does KE and PE being constant imply that they are equal? $\endgroup$
    – Winther
    Jul 19, 2017 at 15:34
  • $\begingroup$ @Winther The OP (class mate) should have noted that we have proved that $KE + PE$ is constant. Let me think of how I can respond to your points, will get back to you. $\endgroup$
    – Wolfy
    Jul 19, 2017 at 15:59

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