These commands work only on first order PDEs, and return the given PDEs characteristic system depending on a parameter _s. Also, when the first argument is a characteristic strip, by passing the keyword reverse, charstrip returns the PDE behind it.

The splitstrip command returns the characteristic system divided into subsets, each one containing equations uncoupled with the ODEs of the other subsets. Depending on the PDE, this splitting of the characteristic system may strongly simplify its solving.

The Maple command usually used for solving coupled systems of ODEs (that is, the subsets returned by splitstrip) is dsolve (see dsolve/system).

Furthermore, it is possible to use the characteristic strip method for solving first order PDEs, by calling pdsolve with the option HINT = strip (see pdsolve).

This function is part of the PDEtools package, and so it can be used in the form charstrip(..) only after executing the command with(PDEtools). However, it can always be accessed through the long form of the command by using PDEtools[charstrip](..).

$\mathrm{with}\left(\mathrm{PDEtools}\right)\:$

$\mathrm{PDE}≔x\mathrm{diff}\left(f\left(x\,y\,z\right)\,z\right)-f\left(x\,y\,z\right)+y^2\mathrm{diff}\left(f\left(x\,y\,z\right)\,y\right)=0$

$\mathrm{PDE}≔x\left(\frac{\partial }{\partial z}f\left(x\,y\,z\right)\right)-f\left(x\,y\,z\right)+y^2\left(\frac{\partial }{\partial y}f\left(x\,y\,z\right)\right)=0$

$\mathrm{sys0}≔\mathrm{charstrip}\left(\mathrm{PDE}\,f\left(x\,y\,z\right)\right)$

$\mathrm{sys0}≔\left\{\frac{\ⅆ}{\ⅆ\mathrm{\_s}}f\left(\mathrm{\_s}\right)=f\left(\mathrm{\_s}\right)\,\frac{\ⅆ}{\ⅆ\mathrm{\_s}}x\left(\mathrm{\_s}\right)=0\,\frac{\ⅆ}{\ⅆ\mathrm{\_s}}y\left(\mathrm{\_s}\right)={y\left(\mathrm{\_s}\right)}^2\,\frac{\ⅆ}{\ⅆ\mathrm{\_s}}z\left(\mathrm{\_s}\right)=x\left(\mathrm{\_s}\right)\right\}$

$\mathrm{dsolve}\left(\mathrm{sys0}\,\left\{f\left(\mathrm{\_s}\right)\,x\left(\mathrm{\_s}\right)\,y\left(\mathrm{\_s}\right)\,z\left(\mathrm{\_s}\right)\right\}\,\mathrm{explicit}\right)$

$\left\{f\left(\mathrm{\_s}\right)=\mathrm{c\_\_4}{\ⅇ}^{\mathrm{\_s}}\,x\left(\mathrm{\_s}\right)=\mathrm{c\_\_2}\,y\left(\mathrm{\_s}\right)=\frac{1}{-\mathrm{\_s}+\mathrm{c\_\_3}}\,z\left(\mathrm{\_s}\right)=\mathrm{\_s}\mathrm{c\_\_2}+\mathrm{c\_\_1}\right\}$

$\mathrm{sys1}≔\mathrm{splitstrip}\left(\mathrm{PDE}\,f\left(x\,y\,z\right)\right)$

$\mathrm{sys1}≔\left\{\left\{\frac{\ⅆ}{\ⅆ\mathrm{\_s}}f\left(\mathrm{\_s}\right)=f\left(\mathrm{\_s}\right)\right\}\,\left\{\frac{\ⅆ}{\ⅆ\mathrm{\_s}}y\left(\mathrm{\_s}\right)={y\left(\mathrm{\_s}\right)}^2\right\}\,\left\{\frac{\ⅆ}{\ⅆ\mathrm{\_s}}x\left(\mathrm{\_s}\right)=0\,\frac{\ⅆ}{\ⅆ\mathrm{\_s}}z\left(\mathrm{\_s}\right)=x\left(\mathrm{\_s}\right)\right\}\right\}$

$\mathrm{ans}≔\mathrm{map}\left(u\mapsto \mathrm{dsolve}\left(u\,\mathrm{indets}\left(u\,\mathrm{Function}\right)\,\mathrm{explicit}\right)\,\mathrm{sys1}\right)$

$\mathrm{ans}≔\left\{\left\{f\left(\mathrm{\_s}\right)=\mathrm{c\_\_1}{\ⅇ}^{\mathrm{\_s}}\right\}\,\left\{y\left(\mathrm{\_s}\right)=\frac{1}{-\mathrm{\_s}+\mathrm{c\_\_1}}\right\}\,\left\{x\left(\mathrm{\_s}\right)=\mathrm{c\_\_2}\,z\left(\mathrm{\_s}\right)=\mathrm{\_s}\mathrm{c\_\_2}+\mathrm{c\_\_1}\right\}\right\}$

$\mathrm{PDE}≔\mathrm{diff}\left(z\left(x\,y\right)\,x\right)\mathrm{diff}\left(z\left(x\,y\right)\,y\right)-z\left(x\,y\right)=0$

$\mathrm{PDE}≔\left(\frac{\partial }{\partial x}z\left(x\,y\right)\right)\left(\frac{\partial }{\partial y}z\left(x\,y\right)\right)-z\left(x\,y\right)=0$

$\mathrm{charstrip}\left(\mathrm{PDE}\,z\left(x\,y\right)\,\mathrm{simplifyusingpde}=\mathrm{false}\right)$

$\left\{\frac{\ⅆ}{\ⅆ\mathrm{\_s}}x\left(\mathrm{\_s}\right)={\mathrm{\_p}}_2\left(\mathrm{\_s}\right)\,\frac{\ⅆ}{\ⅆ\mathrm{\_s}}y\left(\mathrm{\_s}\right)={\mathrm{\_p}}_1\left(\mathrm{\_s}\right)\,\frac{\ⅆ}{\ⅆ\mathrm{\_s}}z\left(\mathrm{\_s}\right)=2{\mathrm{\_p}}_1\left(\mathrm{\_s}\right){\mathrm{\_p}}_2\left(\mathrm{\_s}\right)\,\frac{\ⅆ}{\ⅆ\mathrm{\_s}}{\mathrm{\_p}}_1\left(\mathrm{\_s}\right)={\mathrm{\_p}}_1\left(\mathrm{\_s}\right)\,\frac{\ⅆ}{\ⅆ\mathrm{\_s}}{\mathrm{\_p}}_2\left(\mathrm{\_s}\right)={\mathrm{\_p}}_2\left(\mathrm{\_s}\right)\right\}\mathbf{where}\left\{{\mathrm{\_p}}_1=\frac{\partial }{\partial x}z\left(x\,y\right)\,{\mathrm{\_p}}_2=\frac{\partial }{\partial y}z\left(x\,y\right)\right\}$

$\mathrm{charstrip}\left(\,z\left(x\,y\right)\,\mathrm{reverse}\right)$

$\left(\frac{\partial }{\partial x}z\left(x\,y\right)\right)\left(\frac{\partial }{\partial y}z\left(x\,y\right)\right)+\mathrm{c\_\_1}-z\left(x\,y\right)$