Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 107941 by mohammad17 last updated on 13/Aug/20

Commented by kaivan.ahmadi last updated on 13/Aug/20

(∂θ/∂t)=nt^(n−1) e^((−r^2 )/(2t)) +t^n (r^2 /(2t^2 ))e^((−r^2 )/(2t)) =(nt^(n−1) +(r^2 /2)t^(n−2) )e^((−r^2 )/(2t))  (∗)  and  (∂θ/∂r)=t^n  ((−2r)/(2t))e^((−r^2 )/(2t)) =−rt^(n−1) e^((−r^2 )/(2t))   and  (1/r^2 )(∂/∂r)(r^2 (∂θ/∂r))=(1/r^2 ) (∂/∂r)(−r^3 t^(n−1) e^((−r^2 )/(2t)) )=  (1/r^2 )(−3r^2 t^(n−1) e^((−r^2 )/(2t)) +r^4 t^(n−2) e^((−r^2 )/(2t)) )=  (−3t^(n−1) +r^2 t^(n−2) )e^((−r^2 )/(2t))    (∗∗)  and from (∗)=(∗∗)  we have  −3t^(n−1) +r^2 t^(n−2) =nt^(n−1) +(r^2 /2)t^(n−2) ⇒  (n+3)t^(n−1) +((r^2 /2)−r^2 )t^(n−2) =0⇒  (n+3)t^(n−1) −(r^2 /2)t^(n−2) =0⇒  (n+3)t−(r^2 /2)=0⇒n+3=(r^2 /(2t))⇒n=−3+(r^2 /(2t))

$$\frac{\partial\theta}{\partial{t}}={nt}^{{n}−\mathrm{1}} {e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} +{t}^{{n}} \frac{{r}^{\mathrm{2}} }{\mathrm{2}{t}^{\mathrm{2}} }{e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} =\left({nt}^{{n}−\mathrm{1}} +\frac{{r}^{\mathrm{2}} }{\mathrm{2}}{t}^{{n}−\mathrm{2}} \right){e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} \:\left(\ast\right) \\ $$$${and} \\ $$$$\frac{\partial\theta}{\partial{r}}={t}^{{n}} \:\frac{−\mathrm{2}{r}}{\mathrm{2}{t}}{e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} =−{rt}^{{n}−\mathrm{1}} {e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} \\ $$$${and} \\ $$$$\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\frac{\partial}{\partial{r}}\left({r}^{\mathrm{2}} \frac{\partial\theta}{\partial{r}}\right)=\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{\partial}{\partial{r}}\left(−{r}^{\mathrm{3}} {t}^{{n}−\mathrm{1}} {e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} \right)= \\ $$$$\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\left(−\mathrm{3}{r}^{\mathrm{2}} {t}^{{n}−\mathrm{1}} {e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} +{r}^{\mathrm{4}} {t}^{{n}−\mathrm{2}} {e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} \right)= \\ $$$$\left(−\mathrm{3}{t}^{{n}−\mathrm{1}} +{r}^{\mathrm{2}} {t}^{{n}−\mathrm{2}} \right){e}^{\frac{−{r}^{\mathrm{2}} }{\mathrm{2}{t}}} \:\:\:\left(\ast\ast\right) \\ $$$${and}\:{from}\:\left(\ast\right)=\left(\ast\ast\right)\:\:{we}\:{have} \\ $$$$−\mathrm{3}{t}^{{n}−\mathrm{1}} +{r}^{\mathrm{2}} {t}^{{n}−\mathrm{2}} ={nt}^{{n}−\mathrm{1}} +\frac{{r}^{\mathrm{2}} }{\mathrm{2}}{t}^{{n}−\mathrm{2}} \Rightarrow \\ $$$$\left({n}+\mathrm{3}\right){t}^{{n}−\mathrm{1}} +\left(\frac{{r}^{\mathrm{2}} }{\mathrm{2}}−{r}^{\mathrm{2}} \right){t}^{{n}−\mathrm{2}} =\mathrm{0}\Rightarrow \\ $$$$\left({n}+\mathrm{3}\right){t}^{{n}−\mathrm{1}} −\frac{{r}^{\mathrm{2}} }{\mathrm{2}}{t}^{{n}−\mathrm{2}} =\mathrm{0}\Rightarrow \\ $$$$\left({n}+\mathrm{3}\right){t}−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0}\Rightarrow{n}+\mathrm{3}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}{t}}\Rightarrow{n}=−\mathrm{3}+\frac{{r}^{\mathrm{2}} }{\mathrm{2}{t}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com