Find the general solution to $3u_{tt}+10u_{xt}+3u_{xx}=\sin(x+t)$

Question

Part of my homework is to solve: $$3u_{tt}+10u_{xt}+3u_{xx}=\sin(x+t)$$

My workings: I made a change of variable $\zeta = x+t$ and $\eta = x-t$. This turns the PDE into (after some application of the chain rule and simplification) $$16u_{\zeta\zeta}-4u_{\eta \eta}=\left(4\frac{\delta}{\delta \zeta} -2 \frac{\delta}{\delta \eta}\right)\left(4\frac{\delta}{\delta \zeta} +2 \frac{\delta}{\delta \eta}\right)u=\sin(\zeta)$$ Now let $v=4u_{\zeta}+2u_{\eta}$ and we are left with solving the following two equations: \begin{eqnarray*} 4v_{\zeta}-2v_{\eta}&=&\sin(\zeta)\\ 4u_{\zeta}+2u_{\eta}&=&v\\ \end{eqnarray*} I was hoping that after changing to the variables $\zeta$ and $\eta$ I would be left with either a $u_{\zeta \zeta}$ term or a $u_{\eta \eta}$ term which would make it easy to solve (or possible a $u_{\zeta \eta}$ term which would still be OK). However I am now stuck here. Any help would be appreciated!

Answer 1

$\because3\dfrac{\partial^2}{\partial t^2}+10\dfrac{\partial^2}{\partial x\partial t}+3\dfrac{\partial^2}{\partial x^2}=\left(3\dfrac{\partial}{\partial t}+\dfrac{\partial}{\partial x}\right)\left(\dfrac{\partial}{\partial t}+3\dfrac{\partial}{\partial x}\right)$

$\therefore$ Let $\begin{cases}\zeta=x-3t\\\eta=t-3x\end{cases}$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial\zeta}\dfrac{\partial\zeta}{\partial x}+\dfrac{\partial u}{\partial\eta}\dfrac{\partial\eta}{\partial x}=\dfrac{\partial u}{\partial\zeta}-3\dfrac{\partial u}{\partial\eta}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial\zeta}-3\dfrac{\partial u}{\partial\eta}\right)=\dfrac{\partial}{\partial\zeta}\left(\dfrac{\partial u}{\partial\zeta}-3\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\zeta}{\partial x}+\dfrac{\partial}{\partial\eta}\left(\dfrac{\partial u}{\partial\zeta}-3\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\eta}{\partial x}=\dfrac{\partial^2u}{\partial\zeta^2}-3\dfrac{\partial^2u}{\partial\zeta\partial\eta}-3\left(\dfrac{\partial^2u}{\partial\zeta\partial\eta}-3\dfrac{\partial^2u}{\partial\eta^2}\right)=\dfrac{\partial^2u}{\partial\zeta^2}-6\dfrac{\partial^2u}{\partial\zeta\partial\eta}+9\dfrac{\partial^2u}{\partial\eta^2}$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial\eta}\dfrac{\partial\eta}{\partial t}+\dfrac{\partial u}{\partial\zeta}\dfrac{\partial\zeta}{\partial t}=\dfrac{\partial u}{\partial\eta}-3\dfrac{\partial u}{\partial\zeta}$

$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial}{\partial t}\left(\dfrac{\partial u}{\partial\eta}-3\dfrac{\partial u}{\partial\zeta}\right)=\dfrac{\partial}{\partial\eta}\left(\dfrac{\partial u}{\partial\eta}-3\dfrac{\partial u}{\partial\zeta}\right)\dfrac{\partial\eta}{\partial t}+\dfrac{\partial}{\partial\zeta}\left(\dfrac{\partial u}{\partial\eta}-3\dfrac{\partial u}{\partial\zeta}\right)\dfrac{\partial\zeta}{\partial t}=\dfrac{\partial^2u}{\partial\eta^2}-3\dfrac{\partial^2u}{\partial\zeta\partial\eta}-3\left(\dfrac{\partial^2u}{\partial\zeta\partial\eta}-3\dfrac{\partial^2u}{\partial\zeta^2}\right)=\dfrac{\partial^2u}{\partial\eta^2}-6\dfrac{\partial^2u}{\partial\zeta\partial\eta}+9\dfrac{\partial^2u}{\partial\zeta^2}$

$\dfrac{\partial^2u}{\partial x\partial t}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial\eta}-3\dfrac{\partial u}{\partial\zeta}\right)=\dfrac{\partial}{\partial\zeta}\left(\dfrac{\partial u}{\partial\eta}-3\dfrac{\partial u}{\partial\zeta}\right)\dfrac{\partial\zeta}{\partial x}+\dfrac{\partial}{\partial\eta}\left(\dfrac{\partial u}{\partial\eta}-3\dfrac{\partial u}{\partial\zeta}\right)\dfrac{\partial\eta}{\partial x}=\dfrac{\partial^2u}{\partial\zeta\partial\eta}-3\dfrac{\partial^2u}{\partial\zeta^2}-3\left(\dfrac{\partial^2u}{\partial\eta^2}-3\dfrac{\partial^2u}{\partial\zeta\partial\eta}\right)=-3\dfrac{\partial^2u}{\partial\zeta^2}+10\dfrac{\partial^2u}{\partial\zeta\partial\eta}-3\dfrac{\partial^2u}{\partial\eta^2}$

$\therefore3\left(\dfrac{\partial^2u}{\partial\eta^2}-6\dfrac{\partial^2u}{\partial\zeta\partial\eta}+9\dfrac{\partial^2u}{\partial\zeta^2}\right)+10\left(-3\dfrac{\partial^2u}{\partial\zeta^2}+10\dfrac{\partial^2u}{\partial\zeta\partial\eta}-3\dfrac{\partial^2u}{\partial\eta^2}\right)+3\left(\dfrac{\partial^2u}{\partial\zeta^2}-6\dfrac{\partial^2u}{\partial\zeta\partial\eta}+9\dfrac{\partial^2u}{\partial\eta^2}\right)=\sin\dfrac{\zeta+\eta}{-2}$

$3\dfrac{\partial^2u}{\partial\eta^2}-18\dfrac{\partial^2u}{\partial\zeta\partial\eta}+27\dfrac{\partial^2u}{\partial\zeta^2}-30\dfrac{\partial^2u}{\partial\zeta^2}+100\dfrac{\partial^2u}{\partial\zeta\partial\eta}-30\dfrac{\partial^2u}{\partial\eta^2}+3\dfrac{\partial^2u}{\partial\zeta^2}-18\dfrac{\partial^2u}{\partial\zeta\partial\eta}+27\dfrac{\partial^2u}{\partial\eta^2}=-\sin\dfrac{\zeta+\eta}{2}$

$64\dfrac{\partial^2u}{\partial\zeta\partial\eta}=-\sin\dfrac{\zeta+\eta}{2}$

$\dfrac{\partial^2u}{\partial\zeta\partial\eta}=-\dfrac{1}{64}\sin\dfrac{\zeta+\eta}{2}$

$u=F(\zeta)+G(\eta)+\dfrac{1}{16}\sin\dfrac{\zeta+\eta}{2}$

$u=F(x-3t)+G(t-3x)-\dfrac{\sin(x+t)}{16}$

Answer 2

Hint: $16 u_{\zeta \zeta} - 4 u_{\eta \eta} = \sin(\zeta)$ is a wave equation with forcing. Find a "steady-state" solution that depends only on $\zeta$, and then solve the homogeneous equation.