#### 2.1.73 $$u_t + t u_x = 0$$ with $$u(x,0)=e^x$$

problem number 73

From Math 5587 midterm I, Fall 2016, practice exam, problem 4.

Solve for $$u(x,t)$$ \begin {align*} u_t + t u_x = 0 \end {align*}

With with $$u(x,0)=e^x$$

Mathematica

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

Maple

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

Hand solution

Solve $$u_{t}+xu_{x}=0$$ with $$u\left ( x,0\right ) =e^{x}$$. Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin {align} \frac {dt}{ds} & =1\tag {1}\\ \frac {dx}{ds} & =t\tag {2}\\ \frac {du}{ds} & =0\tag {3} \end {align}

With initial conditions at $$s=0$$$t\left ( 0\right ) =0,x\left ( 0\right ) =\xi ,u\left ( 0\right ) =e^{\xi }$ Equation (1) gives\begin {align} t & =s+t\left ( 0\right ) \nonumber \\ & =s\tag {5} \end {align}

Equation (2) now becomes $$\frac {dx}{ds}=s$$, whose solution is \begin {align} x & =\frac {s^{2}}{2}+x\left ( 0\right ) \nonumber \\ x & =\frac {s^{2}}{2}+\xi \tag {6} \end {align}

From (5,6) solving for $$\xi$$ gives\begin {align} \xi & =x-\frac {s^{2}}{2}\nonumber \\ & =x-\frac {t^{2}}{2}\tag {7} \end {align}

Equation (3) gives\begin {align*} u & =u\left ( 0\right ) \\ & =e^{\xi }\\ & =e^{\left ( x-\frac {t^{2}}{2}\right ) } \end {align*}

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