Let $u(x,t)$ be a function that satisfies the PDE $$u_{xx}-u_{t t} = e^x+6t,$$ $x\in \mathbb{R}, t>0,$ and the initial conditions $$u(x,0)=\sin(x) ,u_{t}(x,0)=0$$ for every $x\in \mathbb{R}.$ Then what is the value of $u(\pi/2,\pi/2)?$
I'm not getting how to solve an inhomogeneous wave equation; can anyone give a hint?
On
Let $\begin{cases}p=x+t\\q=x-t\end{cases}$ ,
Then $u_x=u_pp_x+u_qq_x=u_p+u_q$
$u_{xx}=(u_p+u_q)_x=(u_p+u_q)_pp_x+(u_p+u_q)_qq_x=u_{pp}+u_{pq}+u_{pq}+u_{qq}=u_{pp}+2u_{pq}+u_{qq}$
$u_t=u_pp_t+u_qq_t=u_p-u_q$
$u_{tt}=(u_p-u_q)_t=(u_p-u_q)_pp_t+(u_p-u_q)_qq_t=u_{pp}-u_{pq}-u_{pq}+u_{qq}=u_{pp}-2u_{pq}+u_{qq}$
$\therefore u_{pp}+2u_{pq}+u_{qq}-(u_{pp}-2u_{pq}+u_{qq})=e^\frac{p+q}{2}+6\times\dfrac{p-q}{2}$
$4u_{pq}=e^\frac{p+q}{2}+3(p-q)$
$u_{pq}=\dfrac{e^\frac{p+q}{2}}{4}+\dfrac{3(p-q)}{4}$
$u(p,q)=f(p)+g(q)+e^\frac{p+q}{2}+\dfrac{3(p^2q-pq^2)}{8}$
$u(p,q)=f(p)+g(q)+e^\frac{p+q}{2}+\dfrac{3pq(p-q)}{8}$
$u(x,t)=f(x+t)+g(x-t)+e^x+\dfrac{3(x^2-t^2)t}{4}$
$u(x,0)=\sin x$ :
$f(x)+g(x)+e^x=\sin x$
$f(x)+g(x)=\sin x-e^x......(1)$
$u_t(x,t)=f_t(x+t)+g_t(x-t)+\dfrac{3x^2}{4}-\dfrac{9t^2}{4}=f_x(x+t)-g_x(x-t)+\dfrac{3x^2}{4}-\dfrac{9t^2}{4}$
$u_t(x,0)=0$ :
$f_x(x)-g_x(x)+\dfrac{3x^2}{4}=0$
$f(x)-g(x)=c-\dfrac{x^3}{4}......(2)$
$\therefore f(x)=\dfrac{\sin x-e^x}{2}-\dfrac{x^3}{8}+\dfrac{c}{2}$ , $g(x)=\dfrac{\sin x-e^x}{2}+\dfrac{x^3}{8}-\dfrac{c}{2}$
Hence $u(x,t)=\dfrac{\sin(x+t)-e^{x+t}}{2}-\dfrac{(x+t)^3}{8}+\dfrac{c}{2}+\dfrac{\sin(x-t)-e^{x-t}}{2}+\dfrac{(x-t)^3}{8}-\dfrac{c}{2}+e^x+\dfrac{3(x^2-t^2)t}{4}$
$u(x,t)=\dfrac{\sin(x+t)+\sin(x-t)+2e^x-e^{x+t}-e^{x-t}}{2}+\dfrac{(x-t)^3-(x+t)^3}{8}+\dfrac{3(x^2-t^2)t}{4}$
$u\left(\dfrac{\pi}{2},\dfrac{\pi}{2}\right)=\dfrac{2e^\frac{\pi}{2}-e^\pi-1}{2}-\dfrac{\pi^3}{8}$
Hint
Define$$w(x,t)\triangleq u(x,t)-e^x+t^3$$therefore $$u_{xx}-u_{tt}=e^x+6t\implies w_{xx}-w_{tt}=0$$with the following initial conditions$$w(x,0)=u(x,0)-e^x=\sin x-e^x\\w_t(x,0)=u_t(x,0)=0$$Now, how is a homogeneous wave equation solved in terms of $w$?