- #1
Markov2
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Solve
$\begin{aligned} & {{u}_{tt}}={{u}_{xx}}+1+x,\text{ }0<x<1,\text{ }t>0 \\
& u(x,0)=\frac{1}{6}{{x}^{3}}-\frac{1}{2}{{x}^{2}}+\frac{1}{3},\text{ }{{u}_{t}}(x,0)=0,\text{ }0<x<1, \\
& {{u}_{x}}(0,t)=0=u(1,t),\text{ }t>0.
\end{aligned}
$
Here's something new for me, the boundary condition $u_x.$ I've always seen the $u_t$ condition, but what to do in this case?
$\begin{aligned} & {{u}_{tt}}={{u}_{xx}}+1+x,\text{ }0<x<1,\text{ }t>0 \\
& u(x,0)=\frac{1}{6}{{x}^{3}}-\frac{1}{2}{{x}^{2}}+\frac{1}{3},\text{ }{{u}_{t}}(x,0)=0,\text{ }0<x<1, \\
& {{u}_{x}}(0,t)=0=u(1,t),\text{ }t>0.
\end{aligned}
$
Here's something new for me, the boundary condition $u_x.$ I've always seen the $u_t$ condition, but what to do in this case?