2.2.3 \(u_{xx} - 3 u_{xt} - 4 u_{tt} = 0\)

problem number 81

Added May 25, 2019.

From HW 3, UMN Math 5587, Fall 2016, problem 2.

Solve for \(u(x,t)\) with \(u(x,0)=e^x\) and \(u_t(x,0)=0\) by factoring the PDE into two transport PDE \begin {align*} u_{xx} - 3 u_{xt} - 4 u_{tt} = 0 \end {align*}

Mathematica


\[\left \{\left \{u(x,t)\to \frac {1}{5} \left (4 e^{5 t/4}+1\right ) e^{x-t}\right \}\right \}\]

Maple


\[u \left (x , t\right ) = \frac {{\mathrm e}^{-t +x}}{5}+\frac {4 \,{\mathrm e}^{\frac {t}{4}+x}}{5}\]

Hand solution

Solve \(u_{xx}-3u_{xt}-4u_{tt}=0\) with \(u\left ( x,0\right ) =e^{x}\) and \(u_{t}\left ( x,0\right ) =x\). Writing the PDE as \[ \left ( \frac {\partial }{\partial x}+\frac {\partial }{\partial t}\right ) \left ( \frac {\partial }{\partial x}-4\frac {\partial }{\partial t}\right ) u=0 \] Let \begin {equation} \frac {\partial u}{\partial x}-4\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}+\frac {\partial w}{\partial t}=0\tag {2} \end {equation} From (1), \[ w\left ( x,0\right ) =u_{x}\left ( x,0\right ) -4u_{t}\left ( x,0\right ) \] But \(u\left ( x,0\right ) =e^{x}\) hence \(u_{x}\left ( x,0\right ) =e^{x}\), and \(u_{t}\left ( x,0\right ) =0\). Therefore the above gives \[ w\left ( x,0\right ) =e^{x}\] 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} & =1\\ \frac {dw}{ds} & =0 \end {align*}

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

Since \(\xi =x-s=x-t\), then the last equation above gives\[ w\left ( x,y\right ) =e^{x-t}\] Using the above into (1) gives\begin {equation} \frac {\partial u}{\partial x}-4\frac {\partial u}{\partial t}=e^{x-t}\tag {3} \end {equation} with \(u\left ( x,0\right ) =e^{x}\).  The characteristics for (3) are\begin {align*} \frac {dx}{ds} & =1\\ \frac {dt}{ds} & =-4\\ \frac {du}{ds} & =e^{x-t} \end {align*}

With \(x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,u\left ( 0\right ) =e^{\xi }\).  The above two equations give\begin {align*} x & =s+x\left ( 0\right ) =s+\xi \\ t & =-4s \end {align*}

Solving the above for \(s,\xi \) gives \(\xi =x-s=x+\frac {t}{4}\). Therefore \(\frac {du}{ds}=e^{x-t}\) becomes \begin {align*} \frac {du}{ds} & =e^{s+\xi -\left ( -4s\right ) }\\ & =e^{5s+\xi } \end {align*}

Solving the above gives\begin {align*} u & =\frac {1}{5}e^{5s+\xi }+u\left ( 0\right ) \\ & =\frac {1}{5}e^{5s+\xi }+e^{\xi } \end {align*}

Converting to \(x,t\), using \(s=\frac {-t}{4}\) and \(\xi =x+\frac {t}{4}\) gives\begin {align*} u\left ( x,y\right ) & =\frac {1}{5}e^{5\left ( \frac {-t}{4}\right ) +\left ( x+\frac {t}{4}\right ) }+e^{\left ( x+\frac {t}{4}\right ) }\\ & =\frac {1}{5}e^{x-t}+e^{\left ( x+\frac {t}{4}\right ) } \end {align*}

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