#### 2.1.6 Transport equation $$u_t+2 u_x= \sin x$$ IC $$u(0,x)=\sin x$$. Peter Olver textbook, 2.2.5

problem number 6

Taken from Peter Olver textbook, Introduction to Partial diﬀerential equations.

Solve $$u_t+2 u_x= \sin x$$ with IC $$u(0,x)=\sin x$$.

Mathematica

$\left \{\left \{u(t,x)\to \frac {1}{2} (-2 \sin (2 t-x)+\cos (2 t-x)-\cos (x))\right \}\right \}$

Maple

$u \left (t , x\right ) = -\frac {\cos (x )}{2}+\frac {\cos \left (2 t -x \right )}{2}-\sin \left (2 t -x \right )$

Hand solution

Solve \begin {equation} u_{t}+2u_{x}=\sin x\tag {1} \end {equation} With initial conditions$$\ u\left ( x,0\right ) =\sin x\,$$.

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} & =\sin x\left ( t\right ) \tag {3}\\ \frac {dx}{dt} & =2\tag {4} \end {align}

Solving (3) gives\begin {equation} \int du=\int \sin x\left ( t\right ) dt\tag {3A} \end {equation} From (4)$x=2t+x\left ( 0\right )$ Substituting the above into (3A) gives\begin {align} \int du & =\int \sin \left ( 2t+x\left ( 0\right ) \right ) dt\nonumber \\ u & =\frac {-\cos \left ( 2t+x\left ( 0\right ) \right ) }{2}+C\tag {3B} \end {align}

At $$t=0$$ the above becomes\begin {align*} \sin \left ( x\left ( 0\right ) \right ) & =\frac {-\cos \left ( x\left ( 0\right ) \right ) }{2}+C\\ C & =\sin \left ( x\left ( 0\right ) \right ) +\frac {\cos \left ( x\left ( 0\right ) \right ) }{2} \end {align*}

Hence (3B) becomes$u=\frac {-\cos \left ( 2t+x\left ( 0\right ) \right ) }{2}+\sin \left ( x\left ( 0\right ) \right ) +\frac {\cos \left ( x\left ( 0\right ) \right ) }{2}$ But $$x\left ( 0\right ) =x-2t$$, therefore\begin {align*} u\left ( x,t\right ) & =\frac {-\cos \left ( 2t+x-2t\right ) }{2}+\sin \left ( x-2t\right ) +\frac {\cos \left ( x-2t\right ) }{2}\\ & =\frac {-\cos \left ( x\right ) }{2}+\sin \left ( x-2t\right ) +\frac {\cos \left ( x-2t\right ) }{2} \end {align*}

An alternative approach to solve transport PDE is by using Lagrange-charpit method

$\frac {dt}{1}=\frac {dx}{2}=\frac {du}{\sin x}$ $$\frac {dt}{1}=\frac {dx}{2}$$ gives $$\frac {dx}{dt}=2$$ or $$x=2t+C_{1}$$. Hence $C_{1}=x-2t$ And $$\frac {dx}{2}=\frac {du}{\sin x}$$ gives $$\frac {du}{dx}=\frac {1}{2}\sin x$$. Integrating gives $$u=\frac {-1}{2}\cos x+C_{2}$$. Therefore $C_{2}=u+\frac {1}{2}\cos x$ But $$C_{2}=F\left ( C_{1}\right )$$ where $$F$$ is arbitrary function. Therefore\begin {align} u+\frac {1}{2}\cos x & =F\left ( x-2t\right ) \nonumber \\ u\left ( t,x\right ) & =F\left ( x-2t\right ) -\frac {1}{2}\cos x \tag {1} \end {align}

When $$t=0$$, $$u\left ( 0,x\right ) =\sin x$$, therefore the above becomes\begin {align*} \sin x & =F\left ( x\right ) -\frac {1}{2}\cos x\\ F\left ( x\right ) & =\sin x+\frac {1}{2}\cos x\\ F\left ( z\right ) & =\sin z+\frac {1}{2}\cos z \end {align*}

Therefore the solution (1) can now be written as\begin {align*} u\left ( t,x\right ) & =\left ( \sin \left ( x-2t\right ) +\frac {1}{2}\cos \left ( x-2t\right ) \right ) -\frac {1}{2}\cos x\\ & =\sin \left ( x-2t\right ) +\frac {1}{2}\cos \left ( x-2t\right ) -\frac {1}{2}\cos x \end {align*}

The following is an animation of the solution

 3D 2D

Source code used for the above

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