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Question Number 215550 by MrGaster last updated on 10/Jan/25
Let u^((1)) ,u^((2)) s.t. { ((u_(tt) ^((1)) =((βˆ‚^2 /βˆ‚x_1 ^2 )+(βˆ‚^2 /βˆ‚x_i ^2 ))u^((1)) )),((u^((1)) (x_1 ,x_2 ,0)=𝛙(x_1 ,x_2 ))),((u^((1)) (x_1 ,x_2 ,0)=0)) :}, { ((u_(tt) ^((2)) =((βˆ‚^2 /βˆ‚x_1 ^2 )+(βˆ‚^2 /βˆ‚x_2 ^2 )+c^2 )u^((2)) )),((u^((2)) (x_1 x_2 ,0)=0)),((u_t ^((2)) (x_1 ,x_2 ,0)=𝛙(x_1 ,x_2 ))) :} prove:u^((2)) (x_1 ,x_2 ,t)=(1/(2𝛑))∫∫_(𝛏_1 ^2 +𝛏_2 ^2 ≀t^2 ) ((e^(𝛏_2 c) u^((1)) (x_1 ,x_2 ,𝛏_1 )d𝛏_1 d𝛏_2 )/( (√(t^2 βˆ’π›_1 ^2 βˆ’π›_2 ^2 ))))
$$\boldsymbol{\mathrm{Let}}\:\boldsymbol{{u}}^{\left(\mathrm{1}\right)} ,\boldsymbol{{u}}^{\left(\mathrm{2}\right)} \boldsymbol{\mathrm{s}}.\boldsymbol{\mathrm{t}}.\begin{cases}{\boldsymbol{{u}}_{\boldsymbol{{tt}}} ^{\left(\mathrm{1}\right)} =\left(\frac{\partial^{\mathrm{2}} }{\partial\boldsymbol{{x}}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} }{\partial\boldsymbol{{x}}_{{i}} ^{\mathrm{2}} }\right)\boldsymbol{{u}}^{\left(\mathrm{1}\right)} }\\{\boldsymbol{{u}}^{\left(\mathrm{1}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\mathrm{0}\right)=\boldsymbol{\psi}\left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} \right)}\\{\boldsymbol{{u}}^{\left(\mathrm{1}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\mathrm{0}\right)=\mathrm{0}}\end{cases},\begin{cases}{\boldsymbol{{u}}_{\boldsymbol{{tt}}} ^{\left(\mathrm{2}\right)} =\left(\frac{\partial^{\mathrm{2}} }{\partial\boldsymbol{{x}}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} }{\partial\boldsymbol{{x}}_{\mathrm{2}} ^{\mathrm{2}} }+\boldsymbol{{c}}^{\mathrm{2}} \right)\boldsymbol{{u}}^{\left(\mathrm{2}\right)} }\\{\boldsymbol{{u}}^{\left(\mathrm{2}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} \boldsymbol{{x}}_{\mathrm{2}} ,\mathrm{0}\right)=\mathrm{0}}\\{\boldsymbol{{u}}_{\boldsymbol{{t}}} ^{\left(\mathrm{2}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\mathrm{0}\right)=\boldsymbol{\psi}\left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} \right)}\end{cases} \\ $$$$\mathrm{prove}:\boldsymbol{{u}}^{\left(\mathrm{2}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\boldsymbol{{t}}\right)=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\pi}}\int\int_{\boldsymbol{\xi}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\xi}_{\mathrm{2}} ^{\mathrm{2}} \leq\boldsymbol{{t}}^{\mathrm{2}} } \frac{\boldsymbol{{e}}^{\boldsymbol{\xi}_{\mathrm{2}} \boldsymbol{{c}}} \boldsymbol{{u}}^{\left(\mathrm{1}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\boldsymbol{\xi}_{\mathrm{1}} \right)\boldsymbol{{d}\xi}_{\mathrm{1}} \boldsymbol{{d}\xi}_{\mathrm{2}} }{\:\sqrt{\boldsymbol{{t}}^{\mathrm{2}} βˆ’\boldsymbol{\xi}_{\mathrm{1}} ^{\mathrm{2}} βˆ’\boldsymbol{\xi}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$
Answered by MrGaster last updated on 03/Feb/25
(x,t)=(1/(2Ο€))∫∫_(ΞΎ_1 ^2 +ΞΎ_2 ^2 ≀t^2 ) ((e^(ΞΎ2c) Ο†(ΞΎ_1 ,ΞΎ_2 )dΞΎ_1 dΞΎ_2 )/( (√(t^2 βˆ’ΞΎ_1 ^2 βˆ’ΞΎ_2 ^2 )))) Then,u^((2)) (x_1 ,x_2 ,t)=(βˆ‚/βˆ‚t)(tG(x,t)) By Duhamel^, s principle,u^((2)) (x_1 ,x_2 ,t)=∫_0 ^t G(x_1 ,x_2 ,Ο„)dΟ„ Thus,u^((2)) (x_1 ,x_2 ,t)=(1/(2Ο€))∫_0 ^t ∫∫_(ΞΎ_1 ^2 +ΞΎ_2 ^2 ≀(tβˆ’Ο„)^2 ) ((e^(ΞΎ_2 c) ψ(x_1 ,x_2 ,Ο„)dΞΎ_1 dΞΎ_2 )/( (√((tβˆ’Ο„)^2 βˆ’ΞΎ_1 ^2 βˆ’ΞΎ_2 ^2 ))))dΟ„ u^((2)) (x_1 ,x_2 ,t)=(1/(2𝛑))∫∫_(𝛏_1 ^2 +𝛏_2 ^2 ≀t^2 ) ((e^(𝛏_2 c) u^((1)) (x_1 ,x_2 ,𝛏_1 )d𝛏_1 d𝛏_2 )/( (√(t^2 βˆ’π›_1 ^2 βˆ’π›_2 ^2 )))) [Q.E.D]
$$\left({x},{t}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\int\int_{\xi_{\mathrm{1}} ^{\mathrm{2}} +\xi_{\mathrm{2}} ^{\mathrm{2}} \leq{t}^{\mathrm{2}} } \frac{{e}^{\xi\mathrm{2}{c}} \phi\left(\xi_{\mathrm{1}} ,\xi_{\mathrm{2}} \right){d}\xi_{\mathrm{1}} {d}\xi_{\mathrm{2}} }{\:\sqrt{{t}^{\mathrm{2}} βˆ’\xi_{\mathrm{1}} ^{\mathrm{2}} βˆ’\xi_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$$\mathrm{Then},{u}^{\left(\mathrm{2}\right)} \left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{t}\right)=\frac{\partial}{\partial{t}}\left({tG}\left({x},{t}\right)\right) \\ $$$$\mathrm{By}\:\mathrm{Duhamel}^{,} \mathrm{s}\:\mathrm{principle},{u}^{\left(\mathrm{2}\right)} \left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{t}\right)=\int_{\mathrm{0}} ^{{t}} {G}\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,\tau\right){d}\tau \\ $$$$\mathrm{Thus},{u}^{\left(\mathrm{2}\right)} \left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{t}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{{t}} \int\int_{\xi_{\mathrm{1}} ^{\mathrm{2}} +\xi_{\mathrm{2}} ^{\mathrm{2}} \leq\left({t}βˆ’\tau\right)^{\mathrm{2}} } \frac{{e}^{\xi_{\mathrm{2}} {c}} \psi\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,\tau\right){d}\xi_{\mathrm{1}} {d}\xi_{\mathrm{2}} }{\:\sqrt{\left({t}βˆ’\tau\right)^{\mathrm{2}} βˆ’\xi_{\mathrm{1}} ^{\mathrm{2}} βˆ’\xi_{\mathrm{2}} ^{\mathrm{2}} }}{d}\tau \\ $$$$\boldsymbol{{u}}^{\left(\mathrm{2}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\boldsymbol{{t}}\right)=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\pi}}\int\int_{\boldsymbol{\xi}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\xi}_{\mathrm{2}} ^{\mathrm{2}} \leq\boldsymbol{{t}}^{\mathrm{2}} } \frac{\boldsymbol{{e}}^{\boldsymbol{\xi}_{\mathrm{2}} \boldsymbol{{c}}} \boldsymbol{{u}}^{\left(\mathrm{1}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\boldsymbol{\xi}_{\mathrm{1}} \right)\boldsymbol{{d}\xi}_{\mathrm{1}} \boldsymbol{{d}\xi}_{\mathrm{2}} }{\:\sqrt{\boldsymbol{{t}}^{\mathrm{2}} βˆ’\boldsymbol{\xi}_{\mathrm{1}} ^{\mathrm{2}} βˆ’\boldsymbol{\xi}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$$\left[\mathrm{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$

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