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

Added Oct 8, 2019.

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 ✓

ClearAll["Global`*"]; 
pde =  D[u[x,t], t] +1/(x^2+4)*D[u[x,t], x]== 0; 
ic = u[x,0]==Exp[x^3+12*x]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde,ic}, u[x,t], {x,t}], 60*10]];
 

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

Maple ✓

restart; 
pde := diff(u(x,t), t) +1/(x^2+4)*diff(u(x,t),x) =0; 
ic:=u(x,0)=exp(x^3+12*x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,t))),output='realtime')); 
sol:=simplify(expand(sol));
 

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

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

Figure 2.13: Source code 2D

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