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How can we solve the set of two coupled non-homogeneous pdes? \begin{align} K_1\frac{\partial^4 u_1}{\partial x^4}+m_1\frac{\partial ^2u_1(x,t)}{\partial t^2}+k(u_1-u_2)=F_1(t) \delta(x-x_1), \end{align} \begin{align} K_2\frac{\partial^4 u_2}{\partial x^4}+m_2\frac{\partial ^2u_2(x,t)}{\partial t^2}+k(u_2-u_1)=F_2(t)\delta(x-x_2). \end{align} In here, $K_i, m_i, x_i$ and $k$ are constants where $i=1,2$.

Boundary conditions:

$u_1(0,t)=\frac{\partial^2 u_1}{\partial x^2}(0,t)=u_1(l,t)=\frac{\partial^2 u_1}{\partial x^2}(l,t)=0$,

$u_2(0,t)=u_2(l,t)=0$

Initial conditions:

$u_i(x,0)=w_{i0}(x),$

$\frac{\partial u_i}{\partial x}(x,0)=y_{i0}(x)$ for $i=1,2.$

Since $F_1(t)$ and $F_2(t)$ are unspecified (ungiven) functions, solutions $u_1,u_2$ which we seek will be depended on $F_1(t)$ and $F_2(t)$.

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Let $\frac{m_{i}}{K_{i}}=\lambda_{i}$, $\frac{k}{K_{i}}=\psi_{i}$ and $\frac{F_{i}(t)}{K_{i}}=f_{i}(t)$. When written in the vector form, the equations take the form $$\frac{\partial^{4}}{\partial{x}^{4}}U(x, t)+\Lambda\frac{\partial^{2}}{\partial{t}^{2}}U(x, t)+\Psi{U}(x, t)=\Delta(x)\mathcal{F}(t)$$ With $\Lambda_{ij}=\delta_{ij}\lambda_{i}$, $\Psi_{11}=k_{1}$, $\Psi_{12}=-k_{1}$, $\Psi_{21}=-k_{2}$, $\Psi_{22}=k_{2}$, $\Delta_{ij}=\delta_{ij}\delta(x-x_{i})$ being $2\times2$ matricies and $U=[u_{1}, u_{2}]$ and $\mathcal{F}(t)=[f_{1}(t), f_{2}(t)]$. The functions satisfying your boundary conditions are the Dirichlet functions, thus it is natural to do the expansion $$U(x, t)=\sum_{k\in\mathbb{Z}}\hat{U}_{k}(t)\sin\Big(\frac{\pi{k}x}{l}\Big)$$ Where $\hat{U}_{k}(t)$'s are the vector coefficients depending only on time. This leads to $$\sum_{k\in\mathbb{Z}}\Big[\Big(\frac{\pi{k}}{l}\Big)^{4}+\Lambda\frac{\partial^{2}}{\partial{t}^{2}}+\Psi\Big]\hat{U}_{k}(t)\sin\Big(\frac{\pi{k}x}{l}\Big)=\Delta(x)\mathcal{F}(t)$$ Now, we multiply both sides by $\sin\Big(\frac{\pi{m}x}{l}\Big)$ and integrate on $[0, l]$. Here I hake an assumption that $x_{1}, x_{2}$ are within the interval $[0, l]$, otherwise the lhs is just zero and the equation is trivial. Thus we have $$\frac{2}{l}\Big[\Big(\frac{\pi{m}}{l}\Big)^{4}+\Lambda\frac{\partial^{2}}{\partial{t}^{2}}+\Psi\Big]\hat{U}_{m}(t)=\hat{\Delta}\mathcal{F}(t)$$ Where $$\hat{\Delta}_{ij}=\delta_{ij}\sin\Big(\frac{\pi{m}x_{i}}{l}\Big)$$ Now it is reasonable to define the matrix $$\tilde{\Psi}=\Lambda^{-1}\Big[\Psi+I\Big(\frac{\pi{m}}{l}\Big)^{4}\Big]$$ and the vector $$\mathcal{G}(t)=\frac{l}{2}\Lambda^{-1}\hat{\Delta}(m)\mathcal{F}(t)$$ With this symbolism you have $$\frac{\partial^{2}}{\partial{t}^{2}}\hat{U}_{m}(t)+\tilde{\Psi}\hat{U}_{m}(t)=\mathcal{G}(t)$$ This equation may be solved using the method of Green's functions (which in this case are matrix valued). Thus $$\hat{U}_{m}(t)=\int_{\mathbb{R}}G(t, t')\mathcal{G}(t')dt'$$ With $$\frac{\partial^{2}}{\partial{t}^{2}}G(t, t')+\tilde{\Psi}G(t, t')=I\delta(t-t')$$ Now you perform the Fourier trnasformatrion to give $$[-I\omega^{2}+\tilde{\Psi}]\tilde{G}(\omega, t')=\frac{1}{\sqrt{2\pi}}Ie^{-i\omega{t}'}$$ Thus $$\tilde{G}(\omega, t')=\frac{e^{-i\omega{t}'}}{\sqrt{2\pi}}[-I\omega^{2}+\tilde{\Psi}]^{-1}$$ Hence $$G(t, t')=\frac{1}{2\pi}\int_{\mathbb{R}}[-I\omega^{2}+\tilde{\Psi}]^{-1}e^{i\omega{(t-t')}}d\omega$$ So, youare left with inverting matricies and evaluating integrals, I hope you can do that.

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