#### 2.1.10 Transport equation $$u_t+\frac {1}{x^2+4} u_x= 0$$ IC $$u(x,0)=e^{x^3+12 x}$$

problem number 10

Exam problem. Math 5587, fall 2019. UMN

solve for $$u(x,t)$$ the PDE $$u_t+\frac {1}{x^2+4} u_x= 0$$ IC $$u(x,0)=e^{x^3+12 x}$$

Mathematica

$\left \{\left \{u(x,t)\to e^{-3 t+x^3+12 x}\right \}\right \}$

Maple

$u \left (x , t\right ) = {\mathrm e}^{x^{3}-3 t +12 x}$

Hand solution

Solve $u_{t}+\frac {1}{x^{2}+4}u_{x}=0$ with initial conditions $$u\left ( x,0\right ) =e^{x^{3}+12x}$$.

Solution

Let $$u=u\left ( x\left ( t\right ) ,t\right )$$. Then \begin {equation} \frac {du}{dt}=\frac {\partial u}{\partial x}\frac {dx}{dt}+\frac {\partial u}{\partial t}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dt} & =0\tag {3}\\ \frac {dx}{dt} & =\frac {1}{x^{2}+4}\tag {4} \end {align}

Solving (3) gives\begin {align} u & =u\left ( x\left ( 0\right ) \right ) \nonumber \\ & =e^{x\left ( 0\right ) ^{3}+12x\left ( 0\right ) }\tag {5} \end {align}

We just need to ﬁnd $$x\left ( 0\right )$$ to ﬁnish the solution. From (4)\begin {equation} \frac {x^{3}}{3}+4x=t+C\tag {6} \end {equation}

At $$t=0$$

$\frac {x\left ( 0\right ) ^{3}}{3}+4x\left ( 0\right ) =C$

Hence (6) becomes

\begin {align*} \frac {x^{3}}{3}+4x & =t+\frac {x\left ( 0\right ) ^{3}}{3}+4x\left ( 0\right ) \\ x\left ( 0\right ) ^{3} & =3\left ( \frac {x^{3}}{3}+4x-t-4x\left ( 0\right ) \right ) \\ & =x^{3}+12x-3t-12x\left ( 0\right ) \end {align*}

Substituting this back into (5) gives\begin {align*} u\left ( x\left ( t\right ) ,t\right ) & =\exp \left ( x^{3}+12x-3t-12x\left ( 0\right ) +12x\left ( 0\right ) \right ) \\ & =e^{x^{3}+12x-3t} \end {align*}

The following is an animation of the solution

Source code used for the above

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