klipbodok6

2022-07-02

Solve the following initial value problem:
${u}_{t}+{u}_{x}=v,\phantom{\rule{0ex}{0ex}}{v}_{t}+{v}_{x}=-u,\phantom{\rule{0ex}{0ex}}u\left(0,x\right)={u}_{0}\left(x\right),\phantom{\rule{0ex}{0ex}}v\left(0,x\right)={v}_{0}\left(x\right).$

trantegisis

Expert

${u}_{tt}+{u}_{xt}={v}_{t}$
${u}_{xt}+{u}_{xx}={v}_{x}$
$\therefore {v}_{t}+{v}_{x}={u}_{tt}+2{u}_{xt}+{u}_{xx}=-u$
${u}_{tt}+2{u}_{xt}+{u}_{xx}+u=0$
Let $\left\{\begin{array}{l}p=x+t\\ q=x-t\end{array}$
Then $\frac{\mathrm{\partial }u}{\mathrm{\partial }x}=\frac{\mathrm{\partial }u}{\mathrm{\partial }p}\frac{\mathrm{\partial }p}{\mathrm{\partial }x}+\frac{\mathrm{\partial }u}{\mathrm{\partial }q}\frac{\mathrm{\partial }q}{\mathrm{\partial }x}=\frac{\mathrm{\partial }u}{\mathrm{\partial }p}+\frac{\mathrm{\partial }u}{\mathrm{\partial }q}$
$\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{x}^{2}}=\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(\frac{\mathrm{\partial }u}{\mathrm{\partial }p}+\frac{\mathrm{\partial }u}{\mathrm{\partial }q}\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }p}\left(\frac{\mathrm{\partial }u}{\mathrm{\partial }p}+\frac{\mathrm{\partial }u}{\mathrm{\partial }q}\right)\frac{\mathrm{\partial }p}{\mathrm{\partial }x}+\frac{\mathrm{\partial }}{\mathrm{\partial }q}\left(\frac{\mathrm{\partial }u}{\mathrm{\partial }p}+\frac{\mathrm{\partial }u}{\mathrm{\partial }q}\right)\frac{\mathrm{\partial }q}{\mathrm{\partial }x}=\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{p}^{2}}+\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }pq}+\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }pq}+\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{q}^{2}}=\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{p}^{2}}+2\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }pq}+\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{q}^{2}}$
$\frac{\mathrm{\partial }u}{\mathrm{\partial }t}=\frac{\mathrm{\partial }u}{\mathrm{\partial }p}\frac{\mathrm{\partial }p}{\mathrm{\partial }t}+\frac{\mathrm{\partial }u}{\mathrm{\partial }q}\frac{\mathrm{\partial }q}{\mathrm{\partial }t}=\frac{\mathrm{\partial }u}{\mathrm{\partial }p}-\frac{\mathrm{\partial }u}{\mathrm{\partial }q}$
$\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }x\mathrm{\partial }t}=\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(\frac{\mathrm{\partial }u}{\mathrm{\partial }p}-\frac{\mathrm{\partial }u}{\mathrm{\partial }q}\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }p}\left(\frac{\mathrm{\partial }u}{\mathrm{\partial }p}-\frac{\mathrm{\partial }u}{\mathrm{\partial }q}\right)\frac{\mathrm{\partial }p}{\mathrm{\partial }x}+\frac{\mathrm{\partial }}{\mathrm{\partial }q}\left(\frac{\mathrm{\partial }u}{\mathrm{\partial }p}-\frac{\mathrm{\partial }u}{\mathrm{\partial }q}\right)\frac{\mathrm{\partial }q}{\mathrm{\partial }x}=\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{p}^{2}}-\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }pq}+\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }pq}-\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{q}^{2}}=\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{p}^{2}}-\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{q}^{2}}$
$\therefore \frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{p}^{2}}-2\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }pq}+\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{q}^{2}}+2\left(\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{p}^{2}}-\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{q}^{2}}\right)+\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{p}^{2}}+2\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }pq}+\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{q}^{2}}+u=0$
$4\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{p}^{2}}+u=0$
$u\left(p,q\right)=f\left(q\right)\mathrm{sin}\frac{p}{2}+g\left(q\right)\mathrm{cos}\frac{p}{2}$
$u\left(t,x\right)=f\left(x-t\right)\mathrm{sin}\frac{x+t}{2}+g\left(x-t\right)\mathrm{cos}\frac{x+t}{2}$
${u}_{t}\left(t,x\right)=-{f}_{t}\left(x-t\right)\mathrm{sin}\frac{x+t}{2}+\frac{f\left(x-t\right)}{2}\mathrm{cos}\frac{x+t}{2}-{g}_{t}\left(x-t\right)\mathrm{cos}\frac{x+t}{2}-\frac{g\left(x-t\right)}{2}\mathrm{sin}\frac{x+t}{2}$
${u}_{x}\left(t,x\right)={f}_{x}\left(x-t\right)\mathrm{sin}\frac{x+t}{2}+\frac{f\left(x-t\right)}{2}\mathrm{cos}\frac{x+t}{2}+{g}_{x}\left(x-t\right)\mathrm{cos}\frac{x+t}{2}-\frac{g\left(x-t\right)}{2}\mathrm{sin}\frac{x+t}{2}$
$\therefore v\left(t,x\right)={f}_{x-t}\left(x-t\right)\mathrm{sin}\frac{x+t}{2}+{g}_{x-t}\left(x-t\right)\mathrm{cos}\frac{x+t}{2}+f\left(x-t\right)\mathrm{cos}\frac{x+t}{2}-g\left(x-t\right)\mathrm{sin}\frac{x+t}{2}$
Hence
$\left\{\begin{array}{l}u\left(t,x\right)=f\left(x-t\right)\mathrm{sin}\frac{x+t}{2}+g\left(x-t\right)\mathrm{cos}\frac{x+t}{2}\\ v\left(t,x\right)={f}_{x-t}\left(x-t\right)\mathrm{sin}\frac{x+t}{2}+{g}_{x-t}\left(x-t\right)\mathrm{cos}\frac{x+t}{2}+f\left(x-t\right)\mathrm{cos}\frac{x+t}{2}-g\left(x-t\right)\mathrm{sin}\frac{x+t}{2}\end{array}$

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