- #1
Markov2
- 149
- 0
Solve
$\begin{aligned} & {{u}_{tt}}={{u}_{xx}}+t,\text{ }t>0,\text{ }x\in \mathbb R, \\
& u(x,0)=x \\
& {{u}_{t}}(x,0)=1.
\end{aligned}
$
Okay first I should set $v(x,t)=u(x,t)-\dfrac16 t^3,$ then $u(x,t)=v(x,t)+\dfrac16 t^3$ so $u_{tt}=v_{tt}+t$ and $u_{xx}=v_{xx}$ so $v_{tt}+t=v_{xx}+t\implies v_{tt}=v_{xx},$ and $u(x,0)=v(x,0)$ and $u_t(x,0)=v_t(x,0),$ so I need to solve
$\begin{aligned} & {{v}_{tt}}={{v}_{xx}},\text{ }t>0,\text{ }x\in \mathbb R, \\
& v(x,0)=x \\
& {{v}_{t}}(x,0)=1.
\end{aligned}$
which is a simple application of the formula and then once found $v$ the problem is solved! Is it correct?
$\begin{aligned} & {{u}_{tt}}={{u}_{xx}}+t,\text{ }t>0,\text{ }x\in \mathbb R, \\
& u(x,0)=x \\
& {{u}_{t}}(x,0)=1.
\end{aligned}
$
Okay first I should set $v(x,t)=u(x,t)-\dfrac16 t^3,$ then $u(x,t)=v(x,t)+\dfrac16 t^3$ so $u_{tt}=v_{tt}+t$ and $u_{xx}=v_{xx}$ so $v_{tt}+t=v_{xx}+t\implies v_{tt}=v_{xx},$ and $u(x,0)=v(x,0)$ and $u_t(x,0)=v_t(x,0),$ so I need to solve
$\begin{aligned} & {{v}_{tt}}={{v}_{xx}},\text{ }t>0,\text{ }x\in \mathbb R, \\
& v(x,0)=x \\
& {{v}_{t}}(x,0)=1.
\end{aligned}$
which is a simple application of the formula and then once found $v$ the problem is solved! Is it correct?
