#### 2.2.2 $$u_{xx} - u_{xt} - 12 u_{tt} = 0$$

problem number 80

From Math 5587 midterm I, Fall 2016, practice exam, problem 9.

Solve for $$u(x,t)$$ with $$u(x,0)=0$$ and $$u_t(x,0)=x$$ by factoring the PDE into two transport PDE \begin {align*} u_{xx} - u_{xt} - 12 u_{tt} = 0 \end {align*}

Mathematica

$\left \{\left \{u(x,t)\to t x-\frac {t^2}{24}\right \}\right \}$

Maple

$u \left (x , t\right ) = -\frac {1}{24} t^{2}+t x$

Hand solution

Solve $$u_{xx}-u_{xt}-12u_{tt}=0$$ with $$u\left ( x,0\right ) =x$$ and $$u_{t}\left ( x,0\right ) =x$$. Writing the PDE as $\left ( \frac {\partial }{\partial x}+3\frac {\partial }{\partial t}\right ) \left ( \frac {\partial }{\partial x}-4\frac {\partial }{\partial t}\right ) u=0$ 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}+3\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 $$u\left ( x,0\right ) =0$$ hence $$u_{x}\left ( x,0\right ) =0$$, and $$u_{t}\left ( x,0\right ) =x$$. Therefore the above gives $w\left ( x,0\right ) =-4x$ 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 $$x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,w\left ( 0\right ) =-4\xi$$. The above equations give\begin {align*} x & =s+x\left ( 0\right ) =s+\xi \\ t & =3s\\ w & =u\left ( 0\right ) =-4\xi \end {align*}

Solving for $$\xi ,s$$ from the ﬁrst 2 equations. $$\xi =x-s=x-\frac {t}{3}$$. From the last equation above$w\left ( x,y\right ) =-4\left ( x-\frac {t}{3}\right )$ Using the above into (1) gives\begin {equation} \frac {\partial u}{\partial x}-4\frac {\partial u}{\partial t}=-4\left ( x-\frac {t}{3}\right ) \tag {3} \end {equation} with $$u\left ( x,0\right ) =0$$.  The characteristics for (3) are\begin {align*} \frac {dx}{ds} & =1\\ \frac {dt}{ds} & =-4\\ \frac {du}{ds} & =-4\left ( x-\frac {t}{3}\right ) \end {align*}

With $$x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,u\left ( 0\right ) =0$$.  The above 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 $$\frac {du}{ds}=-4\left ( x-\frac {t}{3}\right )$$ becomes \begin {align*} \frac {du}{ds} & =-4\left ( s+\xi -\frac {\left ( -4s\right ) }{3}\right ) \\ & =-4\left ( s+\xi +\frac {4s}{3}\right ) \\ & =-\frac {28}{3}s-4\xi \end {align*}

Solving the above gives\begin {align*} u & =-\frac {28}{3}\frac {s^{2}}{2}-4\xi s+u\left ( 0\right ) \\ & =-\frac {28}{3}\frac {s^{2}}{2}-4\xi s \end {align*}

Converting to $$x,t$$\begin {align*} u\left ( x,y\right ) & =-\frac {28}{3}\frac {1}{2}\left ( -\frac {t}{4}\right ) ^{2}-4\left ( x+\frac {t}{4}\right ) \left ( -\frac {t}{4}\right ) \\ & =-\frac {1}{24}\left ( t^{2}-24xt\right ) \\ & =-\frac {1}{24}t^{2}+xt \end {align*}

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