2.1.7 Transport equation \(u_t+\frac {1}{1+x^2} u_x= 0\) IC \(u(x,0)=\frac {1}{1+(3+x)^2}\). Peter Olver textbook, page 27

problem number 7

Added Sept 12, 2019.

Taken from Peter Olver textbook, Introduction to Partial differential equations. Example 2.4, page 27.

Solve \(u_t+\frac {1}{1+x^2} u_x= 0\) with IC \(u(x,0)=\frac {1}{1+(3+x)^2}\)

Mathematica


\[\left \{\left \{u(x,t)\to \frac {2 \left (\sqrt {\left (-3 t+x^3+3 x\right )^2+4}-3 t+x^3+3 x\right )^{2/3}}{\sqrt [3]{2} x^3 \sqrt [3]{\sqrt {\left (-3 t+x^3+3 x\right )^2+4}-3 t+x^3+3 x}+3 \sqrt [3]{2} x \sqrt [3]{\sqrt {\left (-3 t+x^3+3 x\right )^2+4}-3 t+x^3+3 x}+16 \left (\sqrt {\left (-3 t+x^3+3 x\right )^2+4}-3 t+x^3+3 x\right )^{2/3}+6\ 2^{2/3} \sqrt {\left (-3 t+x^3+3 x\right )^2+4}-3 \sqrt [3]{2} t \sqrt [3]{\sqrt {\left (-3 t+x^3+3 x\right )^2+4}-3 t+x^3+3 x}+\sqrt [3]{2} \sqrt {\left (-3 t+x^3+3 x\right )^2+4} \sqrt [3]{\sqrt {\left (-3 t+x^3+3 x\right )^2+4}-3 t+x^3+3 x}-12 \sqrt [3]{2} \sqrt [3]{\sqrt {\left (-3 t+x^3+3 x\right )^2+4}-3 t+x^3+3 x}-18\ 2^{2/3} t+6\ 2^{2/3} x^3+18\ 2^{2/3} x+2\ 2^{2/3}}\right \}\right \}\]

Maple


\[u \left (x , t\right ) = \frac {1}{\left (\frac {\left (4 x^{3}-12 t +12 x +4 \sqrt {9 \left (-\frac {1}{3} x^{3}+t -x \right )^{2}+4}\right )^{\frac {1}{3}}}{2}-\frac {2}{\left (4 x^{3}-12 t +12 x +4 \sqrt {9 \left (-\frac {1}{3} x^{3}+t -x \right )^{2}+4}\right )^{\frac {1}{3}}}\right )^{2}+3 \left (4 x^{3}-12 t +12 x +4 \sqrt {9 \left (-\frac {1}{3} x^{3}+t -x \right )^{2}+4}\right )^{\frac {1}{3}}-\frac {12}{\left (4 x^{3}-12 t +12 x +4 \sqrt {9 \left (-\frac {1}{3} x^{3}+t -x \right )^{2}+4}\right )^{\frac {1}{3}}}+10}\]

Hand solution

Solve \[ u_{t}+\frac {1}{x^{2}+1}u_{x}=0 \] With initial conditiomns \(u\left ( 0,x\right ) =\frac {1}{1+\left ( x+3\right ) ^{2}}\)

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}+1}\tag {4} \end {align}

Solving (3) gives\begin {align} u & =u\left ( x\left ( 0\right ) \right ) \nonumber \\ & =\frac {1}{1+\left ( x\left ( 0\right ) +3\right ) ^{2}}\tag {5} \end {align}

We just need to find \(x\left ( 0\right ) \) to finish the solution. From (4)\begin {align} \left ( x^{2}+1\right ) dx & =dt\nonumber \\ \frac {x^{3}}{3}+x & =t+C\tag {6} \end {align}

At \(t=0\)\[ \frac {x\left ( 0\right ) ^{3}}{3}+x\left ( 0\right ) =C \] Hence (6) becomes\[ \frac {x^{3}}{3}+x-t=\frac {x\left ( 0\right ) ^{3}}{3}+x\left ( 0\right ) \] This is Cubic in \(x\left ( 0\right ) \). The solution is complicated and will not be given. All what is left is to substitute this solution back in (5) and this is what the computer did above.

The following is an animation of the solution obtained from CAS. Animation agrees with textbook screen shots.

Source code used for the above

pict
Figure 2.10:Source code

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