Author Topic: question to determine the general solution of wave equation  (Read 368 times)

Wanying Zhang

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question to determine the general solution of wave equation
« on: January 28, 2019, 03:58:31 PM »
I have difficulties obtaining the general solution of the equation $u_{tt} - c^2u_{xx} = 0$. From the online textbook Section2.3, it mentions $v = u_t +cu_x$, and it gets the result $v_t - cv_x = 0$ by chain rule. But when I expand $v_t - cv_x = 0$, I get an extra term $x'(t)$ when applies chain rule to $v_t$ because I think $u_{t}$ in $v$ can be two separated into to parts which are $u_t(t)$ and $u_t(x(t))$. Then applying chain rule, it becomes $v_t = u_{tt} + x'(t) +c(u_{xt} + x'(t))$ where the term $x'(t)$ can not be cancelled. I wonder where is the problem of my thought.

Victor Ivrii

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Re: question to determine the general solution of wave equation
« Reply #1 on: January 29, 2019, 03:51:02 AM »
I have difficulties obtaining the general solution of the equation $u_{tt} - c^2u_{xx} = 0$. From the online textbook Section2.3, it mentions $v = u_t +cu_x$, and it gets the result $v_t - cv_x = 0$ by chain rule.
Not by a chain rule. Just from equation
Quote
But when I expand $v_t - cv_x = 0$, I get an extra term $x'(t)$ when applies chain rule to $v_t$ because I think $u_{t}$ in $v$ can be two separated into to parts which are $u_t(t)$ and $u_t(x(t))$. Then applying chain rule, it becomes $v_t = u_{tt} + x'(t) +c(u_{xt} + x'(t))$ where the term $x'(t)$ can not be cancelled. I wonder where is the problem of my thought.
Several days ago it was rewritten to make it more clear. Read online version