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Question Number 143920 by mohammad17 last updated on 19/Jun/21
Answered by Dwaipayan Shikari last updated on 19/Jun/21
a^2 u=Ce^(±ax+t)
$${a}^{\mathrm{2}} {u}={Ce}^{\pm{ax}+{t}} \\ $$
Commented by mohammad17 last updated on 19/Jun/21
can you give stebs sir
$${can}\:{you}\:{give}\:{stebs}\:{sir} \\ $$
Commented by Dwaipayan Shikari last updated on 19/Jun/21
I don′t know any systematic way to solve P.D.E . But i observed that One can check (∂u/∂t)=Ce^(±ax+t) and (∂^2 u/∂x^2 )=(±a)^2 Ce^(±ax+t) ⇒(∂^2 u/∂x^2 )=a^2 (∂u/∂t)
$${I}\:{don}'{t}\:{know}\:{any}\:{systematic}\:{way}\:{to}\:{solve} \\ $$$${P}.{D}.{E}\:.\:{But}\:{i}\:{observed}\:{that}\: \\ $$$${One}\:{can}\:{check}\:\frac{\partial{u}}{\partial{t}}={Ce}^{\pm{ax}+{t}} \:\:\:{and}\:\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }=\left(\pm{a}\right)^{\mathrm{2}} {Ce}^{\pm{ax}+{t}} \\ $$$$\Rightarrow\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }={a}^{\mathrm{2}} \frac{\partial{u}}{\partial{t}} \\ $$
Answered by Olaf_Thorendsen last updated on 19/Jun/21
(∂u/∂t) = a^2 (∂^2 u/∂x^2 ) (1) Usually, we try to find such a solution u(x,t) = A(x)B(t) (1) : A(x)B′(t) = a^2 A′′(x)B(t) a^2 ((A′′(x))/(A(x))) = ((B′(t))/(B(t))) (2) Necessarily (2) = constant (2) : ((B′(t))/(B(t))) = cst = −α^2 B(t) = B_0 e^(−α^2 t) (3) (2) : a^2 ((A′′(x))/(A(x))) = cst = −α^2 A(x) = λcos(((αx)/a))+μsin(((αx)/a)) (4) Limit conditions : u(0,t) = 0 = λB_0 e^(−α^2 t) ⇒ λ = 0 u(x,t) = μsin(((αx)/a))B_0 e^(−α^2 t) = U_0 sin(((αx)/a))e^(−α^2 t) u(l,t) = 0 = U_0 sin(((αl)/a))e^(−α^2 t) ⇒ ((αl)/a) = π u(x,t) = U_0 sin(((πx)/l))e^(−((π^2 a^2 )/l^2 )t) u(x,0) = 3sin(((πx)/l)) = U_0 sin(((πx)/l))e^0 ⇒ U_0 = 3 Finally u(x,t) = 3sin(((πx)/l))e^(−((π^2 a^2 t)/l^2 ))
$$\frac{\partial{u}}{\partial{t}}\:=\:{a}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Usually},\:\mathrm{we}\:\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{such}\:\mathrm{a}\:\mathrm{solution} \\ $$$${u}\left({x},{t}\right)\:=\:\mathrm{A}\left({x}\right)\mathrm{B}\left({t}\right) \\ $$$$\left(\mathrm{1}\right)\::\:\mathrm{A}\left({x}\right)\mathrm{B}'\left({t}\right)\:=\:{a}^{\mathrm{2}} \mathrm{A}''\left({x}\right)\mathrm{B}\left({t}\right) \\ $$$${a}^{\mathrm{2}} \frac{\mathrm{A}''\left({x}\right)}{\mathrm{A}\left({x}\right)}\:=\:\frac{\mathrm{B}'\left({t}\right)}{\mathrm{B}\left({t}\right)}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{Necessarily}\:\left(\mathrm{2}\right)\:=\:\mathrm{constant} \\ $$$$\left(\mathrm{2}\right)\::\:\frac{\mathrm{B}'\left({t}\right)}{\mathrm{B}\left({t}\right)}\:\:=\:\mathrm{cst}\:=\:−\alpha^{\mathrm{2}} \\ $$$$\mathrm{B}\left({t}\right)\:\:=\:\mathrm{B}_{\mathrm{0}} {e}^{−\alpha^{\mathrm{2}} {t}} \:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)\::\:{a}^{\mathrm{2}} \frac{\mathrm{A}''\left({x}\right)}{\mathrm{A}\left({x}\right)}\:=\:\mathrm{cst}\:=\:−\alpha^{\mathrm{2}} \\ $$$$\mathrm{A}\left({x}\right)\:=\:\lambda\mathrm{cos}\left(\frac{\alpha{x}}{{a}}\right)+\mu\mathrm{sin}\left(\frac{\alpha{x}}{{a}}\right)\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$ \\ $$$$\mathrm{Limit}\:\mathrm{conditions}\:: \\ $$$$ \\ $$$${u}\left(\mathrm{0},{t}\right)\:=\:\mathrm{0}\:=\:\lambda\mathrm{B}_{\mathrm{0}} {e}^{−\alpha^{\mathrm{2}} {t}} \\ $$$$\Rightarrow\:\lambda\:=\:\mathrm{0} \\ $$$${u}\left({x},{t}\right)\:=\:\mu\mathrm{sin}\left(\frac{\alpha{x}}{{a}}\right)\mathrm{B}_{\mathrm{0}} {e}^{−\alpha^{\mathrm{2}} {t}} \:=\:\mathrm{U}_{\mathrm{0}} \mathrm{sin}\left(\frac{\alpha{x}}{{a}}\right){e}^{−\alpha^{\mathrm{2}} {t}} \\ $$$$ \\ $$$${u}\left({l},{t}\right)\:=\:\mathrm{0}\:=\:\mathrm{U}_{\mathrm{0}} \mathrm{sin}\left(\frac{\alpha{l}}{{a}}\right){e}^{−\alpha^{\mathrm{2}} {t}} \\ $$$$\Rightarrow\:\frac{\alpha{l}}{{a}}\:=\:\pi \\ $$$${u}\left({x},{t}\right)\:=\:\mathrm{U}_{\mathrm{0}} \mathrm{sin}\left(\frac{\pi{x}}{{l}}\right){e}^{−\frac{\pi^{\mathrm{2}} {a}^{\mathrm{2}} }{{l}^{\mathrm{2}} }{t}} \\ $$$$ \\ $$$${u}\left({x},\mathrm{0}\right)\:=\:\mathrm{3sin}\left(\frac{\pi{x}}{{l}}\right)\:=\:\mathrm{U}_{\mathrm{0}} \mathrm{sin}\left(\frac{\pi{x}}{{l}}\right){e}^{\mathrm{0}} \\ $$$$\Rightarrow\:\mathrm{U}_{\mathrm{0}} \:=\:\mathrm{3} \\ $$$$ \\ $$$$\mathrm{Finally}\:{u}\left({x},{t}\right)\:=\:\mathrm{3sin}\left(\frac{\pi{x}}{{l}}\right){e}^{−\frac{\pi^{\mathrm{2}} {a}^{\mathrm{2}} {t}}{{l}^{\mathrm{2}} }} \\ $$
Commented by mohammad17 last updated on 20/Jun/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by Dwaipayan Shikari last updated on 20/Jun/21
Thanks sir
$${Thanks}\:{sir} \\ $$

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