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Question Number 146761 by tabata last updated on 15/Jul/21
Solve the partial defferintial equation u_t =a^2 u_(xx) ,0<x<L ,t>0 u(0,t)=0 and u(L,t)=0 and u_x (x,0)=f(x)
$${Solve}\:{the}\:{partial}\:{defferintial}\:{equation} \\ $$$${u}_{{t}} ={a}^{\mathrm{2}} {u}_{{xx}} \:\:\:,\mathrm{0}<{x}<{L}\:,{t}>\mathrm{0} \\ $$$$ \\ $$$${u}\left(\mathrm{0},{t}\right)=\mathrm{0}\:\:{and}\:{u}\left({L},{t}\right)=\mathrm{0}\:\:{and}\:{u}_{{x}} \left({x},\mathrm{0}\right)={f}\left({x}\right) \\ $$
Commented by tabata last updated on 15/Jul/21
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Answered by Olaf_Thorendsen last updated on 15/Jul/21
u_t = a^2 u_(xx) (1) In physics, the two variables t (the time) and x (the distance) are always separated : u(x) = A(t)B(x) (1) : A′(t)B(x) = a^2 A(t)B′′(x) ⇒ ((A′(t))/(A(t))) = a^2 ((B′′(x))/(B(x))) A function of t (left) is equal to a function of x (right). As t and x are independent, necessarily it′s a constant ((A′(t))/(A(t))) = a^2 ((B′′(x))/(B(x))) = constant C_1 (2) (2) : ((A′(t))/(A(t))) = C_1 ⇒ A(t) = C_2 e^(C_1 t) In the real life, C_1 <0 because there is a dissipation of energy. Let C_1 = −w^2 A(t) = C_2 e^(−w^2 t) (2) : ((B′′(x))/(B(x))) = (C_1 /a^2 ) = −(w^2 /a^2 ) B′′(x)+(ω^2 /a^2 )B(x) = 0 B(x) = αcos((w/a)x)+βsin((w/a)x) u(0,t) = 0 ⇒ B(0) = 0 ⇒ α = 0 u(L,t) = 0 ⇒ B(L) = 0 ⇒ βsin(((wL)/a)) = 0 ⇒ ((wL)/a) = 2π w = ((2πa)/L) B(x) = βsin(((2πx)/L)) and A(t) = C_2 e^(−((4π^2 a^2 )/L^2 )t) u(x,t) = A(t)B(x) u(x,t) = C_2 e^(−((4π^2 a^2 )/L^2 )t) βsin(((2πx)/L)) u(x,t) = C_3 e^(−((4π^2 a^2 )/L^2 )t) sin(((2πx)/L)) u_x (x,t) = ((2π)/L)C_3 e^(−((4π^2 a^2 )/L^2 )t) cos(((2πx)/L)) u_x (x,0) = f(x) ⇒ ((2π)/L)C_3 cos(((2πx)/L)) = f(x) ⇒ C_3 = (L/(2π)).((f(x))/(cos(((2πx)/L)))) Finally : u(x,t) = (L/(2π))e^(−((4π^2 a^2 )/L^2 )t) f(x)tan(((2πx)/L))
$${u}_{{t}} \:=\:{a}^{\mathrm{2}} {u}_{{xx}} \:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{In}\:\mathrm{physics},\:\mathrm{the}\:\mathrm{two}\:\mathrm{variables}\:{t} \\ $$$$\left(\mathrm{the}\:\mathrm{time}\right)\:\mathrm{and}\:{x}\:\left(\mathrm{the}\:\mathrm{distance}\right) \\ $$$$\mathrm{are}\:\mathrm{always}\:\mathrm{separated}\:: \\ $$$${u}\left({x}\right)\:=\:\mathrm{A}\left({t}\right)\mathrm{B}\left({x}\right) \\ $$$$\left(\mathrm{1}\right)\::\:\mathrm{A}'\left({t}\right)\mathrm{B}\left({x}\right)\:=\:{a}^{\mathrm{2}} \mathrm{A}\left({t}\right)\mathrm{B}''\left({x}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{A}'\left({t}\right)}{\mathrm{A}\left({t}\right)}\:=\:{a}^{\mathrm{2}} \frac{\mathrm{B}''\left({x}\right)}{\mathrm{B}\left({x}\right)} \\ $$$$\mathrm{A}\:\mathrm{function}\:\mathrm{of}\:{t}\:\left(\mathrm{left}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{of}\:{x}\:\left(\mathrm{right}\right).\:\mathrm{As}\:{t}\:\mathrm{and}\:{x}\:\mathrm{are} \\ $$$$\mathrm{independent},\:\mathrm{necessarily}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\:\frac{\mathrm{A}'\left({t}\right)}{\mathrm{A}\left({t}\right)}\:=\:{a}^{\mathrm{2}} \frac{\mathrm{B}''\left({x}\right)}{\mathrm{B}\left({x}\right)}\:=\:\mathrm{constant}\:\mathrm{C}_{\mathrm{1}} \:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\::\:\:\frac{\mathrm{A}'\left({t}\right)}{\mathrm{A}\left({t}\right)}\:=\:\mathrm{C}_{\mathrm{1}} \:\Rightarrow\:\mathrm{A}\left({t}\right)\:=\:\mathrm{C}_{\mathrm{2}} {e}^{\mathrm{C}_{\mathrm{1}} {t}} \\ $$$$\mathrm{In}\:\mathrm{the}\:\mathrm{real}\:\mathrm{life},\:\mathrm{C}_{\mathrm{1}} <\mathrm{0}\:\mathrm{because}\:\mathrm{there}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{dissipation}\:\mathrm{of}\:\mathrm{energy}.\:\mathrm{Let}\:\mathrm{C}_{\mathrm{1}} \:=\:−{w}^{\mathrm{2}} \\ $$$$\mathrm{A}\left({t}\right)\:=\:\mathrm{C}_{\mathrm{2}} {e}^{−{w}^{\mathrm{2}} {t}} \\ $$$$\left(\mathrm{2}\right)\::\:\frac{\mathrm{B}''\left({x}\right)}{\mathrm{B}\left({x}\right)}\:=\:\frac{\mathrm{C}_{\mathrm{1}} }{{a}^{\mathrm{2}} }\:=\:−\frac{{w}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\mathrm{B}''\left({x}\right)+\frac{\omega^{\mathrm{2}} }{{a}^{\mathrm{2}} }\mathrm{B}\left({x}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{B}\left({x}\right)\:=\:\alpha\mathrm{cos}\left(\frac{{w}}{{a}}{x}\right)+\beta\mathrm{sin}\left(\frac{{w}}{{a}}{x}\right) \\ $$$${u}\left(\mathrm{0},{t}\right)\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{B}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:\Rightarrow\:\alpha\:=\:\mathrm{0} \\ $$$${u}\left(\mathrm{L},{t}\right)\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{B}\left(\mathrm{L}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\beta\mathrm{sin}\left(\frac{{w}\mathrm{L}}{{a}}\right)\:=\:\mathrm{0}\:\Rightarrow\:\frac{{w}\mathrm{L}}{{a}}\:=\:\mathrm{2}\pi \\ $$$${w}\:=\:\frac{\mathrm{2}\pi{a}}{\mathrm{L}} \\ $$$$\mathrm{B}\left({x}\right)\:=\:\beta\mathrm{sin}\left(\frac{\mathrm{2}\pi{x}}{\mathrm{L}}\right)\:\mathrm{and}\:\mathrm{A}\left({t}\right)\:=\:\mathrm{C}_{\mathrm{2}} {e}^{−\frac{\mathrm{4}\pi^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{L}^{\mathrm{2}} }{t}} \\ $$$${u}\left({x},{t}\right)\:=\:\mathrm{A}\left({t}\right)\mathrm{B}\left({x}\right) \\ $$$${u}\left({x},{t}\right)\:=\:\mathrm{C}_{\mathrm{2}} {e}^{−\frac{\mathrm{4}\pi^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{L}^{\mathrm{2}} }{t}} \beta\mathrm{sin}\left(\frac{\mathrm{2}\pi{x}}{\mathrm{L}}\right) \\ $$$${u}\left({x},{t}\right)\:=\:\mathrm{C}_{\mathrm{3}} {e}^{−\frac{\mathrm{4}\pi^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{L}^{\mathrm{2}} }{t}} \mathrm{sin}\left(\frac{\mathrm{2}\pi{x}}{\mathrm{L}}\right) \\ $$$${u}_{{x}} \left({x},{t}\right)\:=\:\frac{\mathrm{2}\pi}{\mathrm{L}}\mathrm{C}_{\mathrm{3}} {e}^{−\frac{\mathrm{4}\pi^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{L}^{\mathrm{2}} }{t}} \mathrm{cos}\left(\frac{\mathrm{2}\pi{x}}{\mathrm{L}}\right) \\ $$$${u}_{{x}} \left({x},\mathrm{0}\right)\:=\:{f}\left({x}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{2}\pi}{\mathrm{L}}\mathrm{C}_{\mathrm{3}} \mathrm{cos}\left(\frac{\mathrm{2}\pi{x}}{\mathrm{L}}\right)\:=\:{f}\left({x}\right) \\ $$$$\Rightarrow\:\mathrm{C}_{\mathrm{3}} \:=\:\frac{\mathrm{L}}{\mathrm{2}\pi}.\frac{{f}\left({x}\right)}{\mathrm{cos}\left(\frac{\mathrm{2}\pi{x}}{\mathrm{L}}\right)} \\ $$$$\mathrm{Finally}\:: \\ $$$${u}\left({x},{t}\right)\:=\:\frac{\mathrm{L}}{\mathrm{2}\pi}{e}^{−\frac{\mathrm{4}\pi^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{L}^{\mathrm{2}} }{t}} {f}\left({x}\right)\mathrm{tan}\left(\frac{\mathrm{2}\pi{x}}{\mathrm{L}}\right) \\ $$

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