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I'm reading this book. In Ch. 3.4, it studies the wave equation $u_{tt}=c^2u_{xx}$ with BCs $u_x(0,t)=0,\,u_x(L,t)=0$, and ICs $u(x,0)=f(x),\,u_t(x,0)=g(x)$.

The total energy of a string is the summation of the kinetic energy and the potential energy: $E=E_k+E_p=\frac{1}{2}\int_0^Lu_t^2dx+\frac{c^2}{2}\int_0^Lu_x^2dx$. I know that $E_k=mv^2/2$ and $E_p=kx^2/2$.

However, in the integral, $E_k$ term does not contain mass and the $E_p$ term has $u_x^2$ in stead of the square of the displacement. How can I figure out the unit and the dimension?

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  • $\begingroup$ It is a bit hard to answer without having the book. Anyway, in this kind of mathematical works it is very common to normalize constants to $1$, thus losing the possibility of doing dimensional analysis but simplifying formulas a bit. $\endgroup$ – Giuseppe Negro Feb 27 '16 at 16:18
  • $\begingroup$ Anyway, this might be useful. Note that the term with $u_x^2$ accounts for the deformation of an element of the string, not its displacement. The string gains potential energy by deforming itself. $\endgroup$ – Giuseppe Negro Feb 27 '16 at 16:19
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I have wondered similar things myself many times in the past (here an example). Mathematical books usually simplify physical constants to $1$, and moreover, they occasionally abuse of language. (Example. The word "energy" is one of the most common in mathematics, but it does not always match its physical meaning).

However, here the "energy" you mentioned really is the same energy a physicist would talk about. At this link I gave my 2 cents on this fact. This is explained in a (IMHO) clear way in the classic physics book by Halliday (first volume, section 18.4 of the Italian translation - I hope that the numbering does not change from the original version of the book). That's where I took the formulas in the linked post.

Note that the "dimensionally correct" equation for a vibrating string is

$$ \mu \frac{\partial^2 u}{\partial t^2} - T\frac{\partial^2 u}{\partial x^2}=0,$$

with the notations of the linked post (i.e. $\mu=$ mass density, $T=$ tension). This is dimensionally consistent: $u$ is a displacement, so $$ [u]=L, $$ ("dimension of $u$ is Length"), $\mu$ is a mass density, so $$ [\mu]=ML^{-1}$$ ("mass times length$^{-1}$"), and $T$ is tension, which is a force, so $$ [T]=MLt^{-2}$$ ("mass times length times time$^{-2}$"). Therefore $$ \left[ \mu \frac{\partial^2 u}{\partial t^2} \right] = Mt^{-2} = \left[ T\frac{\partial^2 u}{\partial x^2} \right].$$

The equation you wrote, that is $$ u_{tt} - c^2 u_{xx}=0$$ corresponds to this one with $$ c=\sqrt{\frac{T}{\mu}}.$$ You can check that this $c$ has the dimensions of speed: $[c]=Lt^{-1}$.

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