2.2.1 \(u_{xx} + u_{xt} - 6 u_{tt} = 0\)

problem number 79

Added May 23, 2019.

From Math 5587 midterm I, Fall 2016, practice exam, problem 8.

Solve for \(u(x,t)\) with \(u(x,0)=x\) and \(u_t(x,0)=0\) by factoring the PDE into two transport PDE \begin {align*} u_{xx} + u_{xt} - 6 u_{tt} = 0 \end {align*}

Mathematica


\[\{\{u(x,t)\to x\}\}\]

Maple


\[u \left (x , t\right ) = x\]

Hand solution

Solve \(u_{xx}+u_{xt}-6u_{tt}=0\) with \(u\left ( x,0\right ) =x\) and \(u_{t}\left ( x,0\right ) =0\). Writing the PDE as \[ \left ( \frac {\partial }{\partial x}+3\frac {\partial }{\partial t}\right ) \left ( \frac {\partial }{\partial x}-2\frac {\partial }{\partial t}\right ) u=0 \] Let \begin {equation} \frac {\partial u}{\partial x}-2\frac {\partial u}{\partial t}=w\left ( x,t\right ) \tag {1} \end {equation} Then the PDE becomes\begin {equation} \frac {\partial w}{\partial x}+3\frac {\partial w}{\partial t}=0\tag {2} \end {equation} From (1), \[ w\left ( x,0\right ) =u_{x}\left ( x,0\right ) -2u_{t}\left ( x,0\right ) \] But \(u\left ( x,0\right ) =x\) hence \(u_{x}\left ( x,0\right ) =1\), and \(u_{t}\left ( x,0\right ) =0\). Therefore the above gives \[ w\left ( x,0\right ) =1 \] Hence we need to solve (2) for \(w\left ( x,t\right ) \) with the above initial condition. The characteristics for (2) are\begin {align*} \frac {dx}{ds} & =1\\ \frac {dt}{ds} & =3\\ \frac {dw}{ds} & =0 \end {align*}

With \(x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,w\left ( 0\right ) =1\).  The above equations give\begin {align*} x & =s+x\left ( 0\right ) =s+\xi \\ t & =3s\\ w & =u\left ( 0\right ) =1 \end {align*}

Hence \[ w\left ( x,y\right ) =1 \] From (1)\begin {equation} \frac {\partial u}{\partial x}-2\frac {\partial u}{\partial t}=1\tag {3} \end {equation} with \(u\left ( x,0\right ) =x\).  The characteristics for (3) are\begin {align*} \frac {dx}{ds} & =1\\ \frac {dt}{ds} & =-2\\ \frac {du}{ds} & =1 \end {align*}

With \(x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,u\left ( 0\right ) =\xi \).  The above equations give\begin {align*} x & =s+x\left ( 0\right ) =s+\xi \\ t & =-2s\\ u & =s+u\left ( 0\right ) =s+\xi \end {align*}

But \(s=-\frac {t}{2}\) and \(\xi =x-s=x-\left ( -\frac {t}{2}\right ) =x+\frac {t}{2}\), hence \begin {align*} u\left ( x,t\right ) & =-\frac {t}{2}+x+\frac {t}{2}\\ & =x \end {align*}

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