#### 2.2.3 $$u_{xx} - 3 u_{xt} - 4 u_{tt} = 0$$

problem number 81

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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