Question Number 100891 by bemath last updated on 29/Jun/20
$$\mathrm{u}_{\mathrm{tt}} \:=\:\mathrm{u}_{\mathrm{xx}} \:−\:\mathrm{6x}\:;\:\mathrm{0}\leqslant\mathrm{x}<\pi\:,\:\mathrm{t}>\mathrm{0} \\ $$$$\mathrm{u}_{\left(\mathrm{0},\mathrm{t}\right)} \:=\:\mathrm{0}\:;\:\mathrm{u}_{\left(\pi,\mathrm{t}\right)} \:=\:\pi^{\mathrm{3}} +\mathrm{3}\pi \\ $$$$\mathrm{u}_{\left(\mathrm{x},\mathrm{0}\right)} \:=\:\mathrm{x}^{\mathrm{3}} +\mathrm{3x}+\mathrm{3sin}\:\mathrm{x} \\ $$$$\mathrm{u}_{\mathrm{t}} \left(\mathrm{x},\mathrm{0}\right)\:=\:\mathrm{0}\: \\ $$
Answered by bramlex last updated on 29/Jun/20
$${u}_{{xx}} −{u}_{{tt}} \:=\:\mathrm{6}{x}\:\Rightarrow\left({D}_{{x}} ^{\mathrm{2}} −{D}_{{t}} ^{\mathrm{2}} \right){u}\:=\:\mathrm{6}{x} \\ $$$${u}\:=\:\frac{\mathrm{1}}{{D}_{{x}} ^{\mathrm{2}} }\left(\mathrm{1}−\frac{{D}_{{t}} ^{\mathrm{2}} }{{D}_{{x}} ^{\mathrm{2}} }\right)\left(\mathrm{6}{x}\right) \\ $$$$\rightarrow{Taylor}\:{series}\:\left(\mathrm{1}−{x}\right)^{−\mathrm{1}} =\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +… \\ $$$${u}\:=\:\frac{\mathrm{1}}{{D}_{{x}} ^{\mathrm{2}} }\left(\mathrm{1}+\frac{{D}_{{t}} ^{\mathrm{2}} }{{D}_{{x}} ^{\mathrm{2}} }+…\right)\left(\mathrm{6}{x}\right) \\ $$$${u}=\:\frac{\mathrm{1}}{{D}_{{x}} ^{\mathrm{2}} }\left(\mathrm{6}{x}+\mathrm{0}+…\right)\: \\ $$$${u}\:=\:{x}^{\mathrm{3}} +{g}_{\mathrm{1}} \left({t}\right){x}+{g}_{\mathrm{2}} \left({t}\right) \\ $$$${u}\left(\mathrm{0},{t}\right)\:\&\:{u}\left(\pi,{t}\right)\:\rightarrow\:{g}_{\mathrm{1}} \left({t}\right)=\mathrm{3}\:,\:{g}_{\mathrm{2}} \left({t}\right)=\mathrm{0} \\ $$$${u}_{\mathrm{1}} \left({x},{t}\right)\:=\:{x}^{\mathrm{3}} +\mathrm{3}{x}\: \\ $$$${u}_{{xx}} −{u}_{{tt}} =\:\mathrm{0}\:,\:{set}\:{u}=\emptyset\left({x}+{at}\right) \\ $$$$\emptyset''\left({x}+{at}\right)×{a}^{\mathrm{2}} =\mathrm{0}\rightarrow\mathrm{1}−{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}\:=\:\pm\mathrm{1}\:\rightarrow\:{u}_{\mathrm{2}} \left({x},{t}\right)=\emptyset\left({x}+{t}\right)+\emptyset\left({x}−{t}\right) \\ $$$${u}={u}_{\mathrm{1}} +{u}_{\mathrm{2}} \:\because\:{u}\left({x},{t}\right)=\emptyset\left({x}+{t}\right)+\emptyset\left({x}−{t}\right)+{x}^{\mathrm{3}} +\mathrm{3}{x} \\ $$$${so}\:{u}\left({x},\mathrm{0}\right)=\:{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{3sin}\:{x} \\ $$$$\mathrm{2}\emptyset\left({x}\right)=\:\mathrm{3sin}\:{x}\:\rightarrow\:\emptyset\left({x}\pm{t}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:\left({x}\pm{t}\right) \\ $$$${u}\left({x},{t}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:\left({x}+{t}\right)+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:\left({x}−{t}\right)+\:{x}^{\mathrm{3}} +\mathrm{3}{x}\:\bigstar \\ $$$$ \\ $$