Given the 2nd order linear PDE
\begin{align} u_{tt}&=\frac{1}{9}u_{xx}, & &t>0,\ x \in \mathbb{R} \tag 1 \\ u(x,0)&=\frac{1}{x^2+9},&& x \in \mathbb{R} \tag 2\\ u_t&(x,0)=0, && x \in \mathbb{R} \tag 3\\ \lim_{|x|\to+\infty} u(x,t)&=\lim_{|x|\to+\infty} u_t(x,t)=0,&& t\geq0 \tag4 \end{align}
Solution Let $c=\frac{1}{3}$ We calculate the discriminant $$D=4c^2>0$$ So it is hyperbolic.
Now we have to find the characteristics $$\frac{dx}{dt}=\frac{b\pm\sqrt{D}}{a}=\pm c$$ Thus, \begin{align} k_1&=x+ct\\ k_2&=x-ct \end{align}
I used the following substitutions \begin{align} \xi&=x-ct\\ \eta&=x+ct \end{align} and $v(\xi(x, y), \eta(x,y)) = u(x, y)$.
We calculate the second-order partial derivatives and we have
- $u_{tt}=c^2 u_{\xi \xi}-2 c^2 u_{\xi \eta}+ c^2 u_{\eta \eta}$
- $u_{xx}=u_{\xi \xi}+2u_{\xi \eta}+u_{\eta \eta} $
So I got the canonical form $$c^2 u_{\xi \xi}-2 c^2 u_{\xi \eta}+ c^2 u_{\eta \eta}= c^2(u_{\xi \xi}+2u_{\xi \eta}+u_{\eta \eta})$$
So,$4u_{\xi \eta}=0\tag 5$ then we integrate with $(5)$ with respect of $\eta$ and then we integrate the result with respect of $\xi$ and we will have $$u(\xi,\eta)=f(\xi)+g(\eta)\Leftrightarrow u(x,t)=f(t-cx)+g(t+cx) \tag 6$$ where $f,g$ arbitrary functions.
We will follow the D'Alembert Method to Solve the problem for the following initial conditions
\begin{align} u(x,0)&=\varphi(x)=\frac{1}{x^2+9},&& x \in \mathbb{R} \\ u_t(x,0)&=h(x)=0, && x \in \mathbb{R} \\ \lim_{|x|\to+\infty} u(x,t)&=\lim_{|x|\to+\infty} u_t(x,t)=0,&& t\geq0 \end{align}
- The solution, for $t=0$, could be written $$u(x,0)=f(x)+g(x) \Leftrightarrow f(x)+g(x)=\varphi(x)=\frac{1}{x^2+9} \tag 7$$
We differentiate $(6)$ with respect of $t$ and we get $$u_t(x,t)= c(g'(x+ct)-f'(x-ct))$$ Then for t=0 $$u_t(x,0)= c(g'(x)-f(x))\Leftrightarrow c(g'(x)-f'(x)) =h(x)=0 \Leftrightarrow g'(x)=f'(x) \tag 8$$
Now we differentiate $(7)$ with respect of $x$ and we get $$f'(x)+g'(x)=\varphi'(x)=-\frac{2x}{(x^2+9)^2} \tag 9$$
From $(8)$ and $(9)$ we have $$2f'(x)=-\frac{2x}{(x^2+9)^2}$$
So $$\int 2 f'(x)dx=\int -\frac{2x}{(x^2+9)^2}dx \Leftrightarrow f(x)= \frac{1}{2(x^2+9)}+f_0 \Leftrightarrow f(x)= \frac{\varphi(x)}{2}+f_0\tag{10}$$ and $$g(x)=\frac{\varphi(x)}{2}+f_0 \tag{11}$$ Where $f_0$ arbitrary constant. Then the solution to the initial condition problem is
$$u(x,t)=f(−)+g(+)=\frac{\varphi(−)}{2}+\frac{\varphi(+)}{2}$$
But it does not seem right to me. Any advice?