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