# Using implicit differentiation, verify that u=f(x−tu) satisfies (del u(x,t))/(del t)+u(x,t)(del u(x,t))/(del x). Could someone explain how implicit differentiation works with a pde?

Using implicit differentiation, verify that $u=f\left(x-tu\right)$ satisfies $\frac{\mathrm{\partial }u\left(x,t\right)}{\mathrm{\partial }t}+u\left(x,t\right)\frac{\mathrm{\partial }u\left(x,t\right)}{\mathrm{\partial }x}$.
Could someone explain how implicit differentiation works with a pde?
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geoforoiunpwd
Differentiating $u=f\left(x-t\phantom{\rule{thinmathspace}{0ex}}u\right)$ first with respect to $t$ and then with respect to $x$ we get:
${u}_{t}={f}^{\prime }\left(x-t\phantom{\rule{thinmathspace}{0ex}}u\right)\left(-u-t\phantom{\rule{thinmathspace}{0ex}}{u}_{t}\right),$
${u}_{x}={f}^{\prime }\left(x-t\phantom{\rule{thinmathspace}{0ex}}u\right)\left(1-t\phantom{\rule{thinmathspace}{0ex}}{u}_{x}\right).$
Multiplying the second equation by u and adding both equations we get
${u}_{t}+u\phantom{\rule{thinmathspace}{0ex}}{u}_{x}=-t\phantom{\rule{thinmathspace}{0ex}}{f}^{\prime }\left(x-t\phantom{\rule{thinmathspace}{0ex}}u\right)\left({u}_{t}+u\phantom{\rule{thinmathspace}{0ex}}{u}_{x}\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(1+t\phantom{\rule{thinmathspace}{0ex}}{f}^{\prime }\right)\left({u}_{t}+u\phantom{\rule{thinmathspace}{0ex}}{u}_{x}\right)=0.$
If ${f}^{\prime }\ge 0$ (i.e. $f$ is increasing) then $1+t\phantom{\rule{thinmathspace}{0ex}}{f}^{\prime }>1$ for all $t\ge 0$ and ${u}_{t}+u\phantom{\rule{thinmathspace}{0ex}}{u}_{x}=0$ follows.
If ${f}^{\prime }<0$ at some point, there is a ${T}_{\text{max}}$ such that $1+t\phantom{\rule{thinmathspace}{0ex}}{f}^{\prime }>0$ on $\left[0,{T}_{\text{max}}\right)$,