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Question Number 100891 by bemath last updated on 29/Jun/20

u_(tt) = u_(xx) − 6x ; 0≤x<π , t>0 u_((0,t)) = 0 ; u_((π,t)) = π^3 +3π u_((x,0)) = x^3 +3x+3sin x u_t (x,0) = 0

$$\mathrm{u}_{\mathrm{tt}} \:=\:\mathrm{u}_{\mathrm{xx}} \:−\:\mathrm{6x}\:;\:\mathrm{0}\leqslant\mathrm{x}<\pi\:,\:\mathrm{t}>\mathrm{0} \\ $$ $$\mathrm{u}_{\left(\mathrm{0},\mathrm{t}\right)} \:=\:\mathrm{0}\:;\:\mathrm{u}_{\left(\pi,\mathrm{t}\right)} \:=\:\pi^{\mathrm{3}} +\mathrm{3}\pi \\ $$ $$\mathrm{u}_{\left(\mathrm{x},\mathrm{0}\right)} \:=\:\mathrm{x}^{\mathrm{3}} +\mathrm{3x}+\mathrm{3sin}\:\mathrm{x} \\ $$ $$\mathrm{u}_{\mathrm{t}} \left(\mathrm{x},\mathrm{0}\right)\:=\:\mathrm{0}\: \\ $$

Answered by bramlex last updated on 29/Jun/20

u_(xx) −u_(tt) = 6x ⇒(D_x ^2 −D_t ^2 )u = 6x u = (1/D_x ^2 )(1−(D_t ^2 /D_x ^2 ))(6x) →Taylor series (1−x)^(−1) =1+x+x^2 +x^3 +... u = (1/D_x ^2 )(1+(D_t ^2 /D_x ^2 )+...)(6x) u= (1/D_x ^2 )(6x+0+...) u = x^3 +g_1 (t)x+g_2 (t) u(0,t) & u(π,t) → g_1 (t)=3 , g_2 (t)=0 u_1 (x,t) = x^3 +3x u_(xx) −u_(tt) = 0 , set u=∅(x+at) ∅′′(x+at)×a^2 =0→1−a^2 =0 a = ±1 → u_2 (x,t)=∅(x+t)+∅(x−t) u=u_1 +u_2 ∵ u(x,t)=∅(x+t)+∅(x−t)+x^3 +3x so u(x,0)= x^3 +3x+3sin x 2∅(x)= 3sin x → ∅(x±t) = (3/2)sin (x±t) u(x,t) = (3/2)sin (x+t)+(3/2)sin (x−t)+ x^3 +3x ★

$${u}_{{xx}} −{u}_{{tt}} \:=\:\mathrm{6}{x}\:\Rightarrow\left({D}_{{x}} ^{\mathrm{2}} −{D}_{{t}} ^{\mathrm{2}} \right){u}\:=\:\mathrm{6}{x} \\ $$ $${u}\:=\:\frac{\mathrm{1}}{{D}_{{x}} ^{\mathrm{2}} }\left(\mathrm{1}−\frac{{D}_{{t}} ^{\mathrm{2}} }{{D}_{{x}} ^{\mathrm{2}} }\right)\left(\mathrm{6}{x}\right) \\ $$ $$\rightarrow{Taylor}\:{series}\:\left(\mathrm{1}−{x}\right)^{−\mathrm{1}} =\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +... \\ $$ $${u}\:=\:\frac{\mathrm{1}}{{D}_{{x}} ^{\mathrm{2}} }\left(\mathrm{1}+\frac{{D}_{{t}} ^{\mathrm{2}} }{{D}_{{x}} ^{\mathrm{2}} }+...\right)\left(\mathrm{6}{x}\right) \\ $$ $${u}=\:\frac{\mathrm{1}}{{D}_{{x}} ^{\mathrm{2}} }\left(\mathrm{6}{x}+\mathrm{0}+...\right)\: \\ $$ $${u}\:=\:{x}^{\mathrm{3}} +{g}_{\mathrm{1}} \left({t}\right){x}+{g}_{\mathrm{2}} \left({t}\right) \\ $$ $${u}\left(\mathrm{0},{t}\right)\:\&\:{u}\left(\pi,{t}\right)\:\rightarrow\:{g}_{\mathrm{1}} \left({t}\right)=\mathrm{3}\:,\:{g}_{\mathrm{2}} \left({t}\right)=\mathrm{0} \\ $$ $${u}_{\mathrm{1}} \left({x},{t}\right)\:=\:{x}^{\mathrm{3}} +\mathrm{3}{x}\: \\ $$ $${u}_{{xx}} −{u}_{{tt}} =\:\mathrm{0}\:,\:{set}\:{u}=\emptyset\left({x}+{at}\right) \\ $$ $$\emptyset''\left({x}+{at}\right)×{a}^{\mathrm{2}} =\mathrm{0}\rightarrow\mathrm{1}−{a}^{\mathrm{2}} =\mathrm{0} \\ $$ $${a}\:=\:\pm\mathrm{1}\:\rightarrow\:{u}_{\mathrm{2}} \left({x},{t}\right)=\emptyset\left({x}+{t}\right)+\emptyset\left({x}−{t}\right) \\ $$ $${u}={u}_{\mathrm{1}} +{u}_{\mathrm{2}} \:\because\:{u}\left({x},{t}\right)=\emptyset\left({x}+{t}\right)+\emptyset\left({x}−{t}\right)+{x}^{\mathrm{3}} +\mathrm{3}{x} \\ $$ $${so}\:{u}\left({x},\mathrm{0}\right)=\:{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{3sin}\:{x} \\ $$ $$\mathrm{2}\emptyset\left({x}\right)=\:\mathrm{3sin}\:{x}\:\rightarrow\:\emptyset\left({x}\pm{t}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:\left({x}\pm{t}\right) \\ $$ $${u}\left({x},{t}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:\left({x}+{t}\right)+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:\left({x}−{t}\right)+\:{x}^{\mathrm{3}} +\mathrm{3}{x}\:\bigstar \\ $$ $$ \\ $$

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