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Question Number 146761 by tabata last updated on 15/Jul/21

$$Solvethepartialdefferintialequation$$$$u_t=a^2{u}_{xx},0<x<L,t>0$$$$$$$$u(0,t)=0andu(L,t)=0and{u}_x(x,0)=f\left(x\right)$$

Commented by tabata last updated on 15/Jul/21

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Answered by Olaf_Thorendsen last updated on 15/Jul/21

$$u_t=a^2{u}_{xx}\left(1\right)$$$$\mathrm{In}\mathrm{physics},\mathrm{the}\mathrm{two}\mathrm{variables}t$$$$\left(\mathrm{the}\mathrm{time}\right)\mathrm{and}x\left(\mathrm{the}\mathrm{distance}\right)$$$$\mathrm{are}\mathrm{always}\mathrm{separated}:$$$$u\left(x\right)=\mathrm{A}\left(t\right)\mathrm{B}\left(x\right)$$$$\left(1\right):{\mathrm{A}}^{\prime }\left(t\right)\mathrm{B}\left(x\right)=a^2\mathrm{A}\left(t\right){\mathrm{B}}^{″}\left(x\right)$$$$\Rightarrow \frac{{\mathrm{A}}^{\prime }\left(t\right)}{\mathrm{A}\left(t\right)}=a^2\frac{{\mathrm{B}}^{″}\left(x\right)}{\mathrm{B}\left(x\right)}$$$$\mathrm{A}\mathrm{function}\mathrm{of}t\left(\mathrm{left}\right)\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{a}$$$$\mathrm{function}\mathrm{of}x\left(\mathrm{right}\right).\mathrm{As}t\mathrm{and}x\mathrm{are}$$$$\mathrm{independent},\mathrm{necessarily}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{constant}$$$$\frac{{\mathrm{A}}^{\prime }\left(t\right)}{\mathrm{A}\left(t\right)}=a^2\frac{{\mathrm{B}}^{″}\left(x\right)}{\mathrm{B}\left(x\right)}=\mathrm{constant}{\mathrm{C}}_1\left(2\right)$$$$\left(2\right):\frac{{\mathrm{A}}^{\prime }\left(t\right)}{\mathrm{A}\left(t\right)}={\mathrm{C}}_1\Rightarrow \mathrm{A}\left(t\right)={\mathrm{C}}_2{e}^{{\mathrm{C}}_{1}t}$$$$\mathrm{In}\mathrm{the}\mathrm{real}\mathrm{life},{\mathrm{C}}_1<0\mathrm{because}\mathrm{there}\mathrm{is}$$$$\mathrm{a}\mathrm{dissipation}\mathrm{of}\mathrm{energy}.\mathrm{Let}{\mathrm{C}}_1=-w^2$$$$\mathrm{A}\left(t\right)={\mathrm{C}}_2{e}^{-w^{2}t}$$$$\left(2\right):\frac{{\mathrm{B}}^{″}\left(x\right)}{\mathrm{B}\left(x\right)}=\frac{{\mathrm{C}}_1}{a^2}=-\frac{w^2}{a^2}$$$${\mathrm{B}}^{″}\left(x\right)+\frac{{\omega }^2}{a^2}\mathrm{B}\left(x\right)=0$$$$\mathrm{B}\left(x\right)=\alpha \cos \left(\frac{w}{a}x\right)+\beta \sin \left(\frac{w}{a}x\right)$$$$u(0,t)=0\Rightarrow \mathrm{B}\left(0\right)=0\Rightarrow \alpha =0$$$$u(\mathrm{L},t)=0\Rightarrow \mathrm{B}\left(\mathrm{L}\right)=0$$$$\Rightarrow \beta \sin \left(\frac{w\mathrm{L}}{a}\right)=0\Rightarrow \frac{w\mathrm{L}}{a}=2\pi$$$$w=\frac{2\pi a}{\mathrm{L}}$$$$\mathrm{B}\left(x\right)=\beta \sin \left(\frac{2\pi x}{\mathrm{L}}\right)\mathrm{and}\mathrm{A}\left(t\right)={\mathrm{C}}_2{e}^{-\frac{4{\pi }^2{a}^2}{{\mathrm{L}}^2}t}$$$$u(x,t)=\mathrm{A}\left(t\right)\mathrm{B}\left(x\right)$$$$u(x,t)={\mathrm{C}}_2{e}^{-\frac{4{\pi }^2{a}^2}{{\mathrm{L}}^2}t}\beta \sin \left(\frac{2\pi x}{\mathrm{L}}\right)$$$$u(x,t)={\mathrm{C}}_3{e}^{-\frac{4{\pi }^2{a}^2}{{\mathrm{L}}^2}t}\sin \left(\frac{2\pi x}{\mathrm{L}}\right)$$$$u_x(x,t)=\frac{2\pi }{\mathrm{L}}{\mathrm{C}}_3{e}^{-\frac{4{\pi }^2{a}^2}{{\mathrm{L}}^2}t}\cos \left(\frac{2\pi x}{\mathrm{L}}\right)$$$$u_x(x,0)=f\left(x\right)$$$$\Rightarrow \frac{2\pi }{\mathrm{L}}{\mathrm{C}}_3\cos \left(\frac{2\pi x}{\mathrm{L}}\right)=f\left(x\right)$$$$\Rightarrow {\mathrm{C}}_3=\frac{\mathrm{L}}{2\pi }.\frac{f\left(x\right)}{\cos \left(\frac{2\pi x}{\mathrm{L}}\right)}$$$$\mathrm{Finally}:$$$$u(x,t)=\frac{\mathrm{L}}{2\pi }{e}^{-\frac{4{\pi }^2{a}^2}{{\mathrm{L}}^2}t}f\left(x\right)\tan \left(\frac{2\pi x}{\mathrm{L}}\right)$$

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