Toronto Math Forum

APM346--2019 => APM346--Lectures & Home Assignments => Home Assignment 3 => Topic started by: Wanying Zhang on January 28, 2019, 03:58:31 PM

Title: question to determine the general solution of wave equation
Post by: Wanying Zhang 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.

Title: Re: question to determine the general solution of wave equation
Post by: Victor Ivrii on January 29, 2019, 03:51:02 AM

Quote from: Wanying Zhang 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.

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