d'Alembert's Solution to 1-D wave equation

Question

I am supposed to be using d'Alembert's solution to solve an IVP. I am generally confused with how the book is getting the correct solution. The problem is as follows:

d'Alembert's solution: $u(x,t) = \frac{1}{2}[f(x + at) + f(x-at) + \frac{1}{2a}\int_{x-at}^{x+at}g(s)ds$

Problem: $a^2u_{xx} = u_{tt}$ $; -\infty < x < \infty$ $, t>0$

Initial Conditions: $u(x,0) =f(x) , u_{t=0} = g(x)$

Specific to this problem: $f(x) = \sin(x) , g(x) = 1$

For the first initial condition $u(x,0) = \sin(x)$, I got the following: $$ \sin(x) = \frac{1}{2}[\sin(x) + \sin(x)] + \frac{1}{2a}[\int_x^xg(s)ds]$$ $$::\sin(x) = \sin(x)$$

For the second initial condition $u_{t=0} = 1$, I got the following: $$ u_t = \frac{1}{2}[a\cos(x+at) -a\cos(x-at)] + \frac{\partial}{\partial t}[\frac{1}{2a}\int_{x-at}^{x+at}g(s)ds]$$ $$u_{t=0} =1 : $$ $$ 1 = \frac{1}{2}[(a\cos(x) -a\cos(x)] + \frac{\partial}{\partial t}[\frac{1}{2a}\int_x^xg(s)ds]$$ $$1 = a\cos(x)$$ $$a = \frac{1}{\cos(x)}$$

I think I am missing something because I don't understand how I am supposed to get the solution from here. Am I supposed to plug in what I got for 'a' back into $u(x,t)$ and solve from there? The book has the solution as $u(x,t) = \sin(x)\cos(at) + t$. Can anyone provide a hint or show me where I am misunderstanding something? Thank you!

Answer 1

Your problem stems from the extremely poor notation $$u_{t=0}$$ What you meant to write was $$u_t(x,0)$$ I personally abhor subscript derivative notation, so I will write this as $$\partial_t u(x,0)$$ We can see that $$\partial_tu(x,t)=\frac{c}{2}\big(f'(x+ct)-f'(x-ct)\big)+\frac{\partial}{\partial t}\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\mathrm ds$$ Using the fundamental theorem of calculus, this is just $$\partial_tu(x,t)=\frac{c}{2}\big(f'(x+ct)-f'(x-ct)\big)+\frac{1}{2}\big(g(x+ct)+g(x-ct)\big)$$ And using the $f,g$ given in your question and substituting $t\to 0$ we get $$\partial_t u(x,0)=1=\frac{c}{2}\big(\cos(x)-\cos(x)\big)+\frac{1}{2}(1+1)=1$$ As expected.