Given a PDE, mapde's main goal is to map it into another PDE with a different format, perhaps easier to solve. (Of the options mentioned above, homo and ccoeff are not yet implemented.)

Description of options

The options canom and canop only work with linear second order PDEs with only two differentiation variables. The mapde command always succeeds as long as the coefficients are constant. With variable coefficients, mapde succeeds only sometimes; in those cases, the mapping is realized through a nonlinear transformation which appears as the algebraic inversion of the solution of a pair of auxiliary first order PDEs. However, the system does not always succeed in solving that auxiliary problem.

The option noF, at present, only works when the received PDE is a homogeneous function of the indeterminate function and its derivatives. In these cases, mapde introduces the indeterminate function as equal to the exponential of another (auxiliary) function.

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

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

$\mathrm{PDE}≔{\mathrm{diff}\left(f\left(x\,y\,z\right)\,x\right)}^3=f\left(x\,y\,z\right)\mathrm{diff}\left(f\left(x\,y\,z\right)\,y\,y\right)\mathrm{diff}\left(f\left(x\,y\,z\right)\,z\right)$

$\mathrm{PDE}≔{\left(\frac{\partial }{\partial x}f\left(x\,y\,z\right)\right)}^3=f\left(x\,y\,z\right)\left(\frac{{\partial }^2}{\partial y^2}f\left(x\,y\,z\right)\right)\left(\frac{\partial }{\partial z}f\left(x\,y\,z\right)\right)$

$\mathrm{pdsolve}\left(\mathrm{PDE}\right)$

$f\left(x\,y\,z\right)=\mathrm{f\_\_1}\left(x\right)\mathrm{f\_\_2}\left(y\right)\mathrm{f\_\_3}\left(z\right)\mathbf{where}\left[\left\{{\left(\frac{\ⅆ}{\ⅆx}\mathrm{f\_\_1}\left(x\right)\right)}^3={\mathrm{\_c}}_1{\mathrm{f\_\_1}\left(x\right)}^3\,\frac{\ⅆ}{\ⅆz}\mathrm{f\_\_3}\left(z\right)=\frac{{\mathrm{\_c}}_1\mathrm{f\_\_3}\left(z\right)}{{\mathrm{\_c}}_2}\,\frac{{\ⅆ}^2}{\ⅆy^2}\mathrm{f\_\_2}\left(y\right)={\mathrm{\_c}}_2\mathrm{f\_\_2}\left(y\right)\right\}\right]$

$\mathrm{PDE1}≔\mathrm{mapde}\left(\mathrm{PDE}\,\mathrm{noF}\right)$

$\mathrm{PDE1}≔{\left(\frac{\partial }{\partial x}\mathrm{f\_\_1}\left(x\,y\,z\right)\right)}^3-\left(\frac{\partial }{\partial z}\mathrm{f\_\_1}\left(x\,y\,z\right)\right){\left(\frac{\partial }{\partial y}\mathrm{f\_\_1}\left(x\,y\,z\right)\right)}^2-\left(\frac{\partial }{\partial z}\mathrm{f\_\_1}\left(x\,y\,z\right)\right)\left(\frac{{\partial }^2}{\partial y^2}\mathrm{f\_\_1}\left(x\,y\,z\right)\right)=0\mathbf{where}\left\{\mathrm{f\_\_1}\left(x\,y\,z\right)=\ln \left(f\left(x\,y\,z\right)\right)\right\}$

$\mathrm{ans}≔\mathrm{pdsolve}\left(\mathrm{op}\left(1\,\mathrm{PDE1}\right)\right)$

$\mathrm{ans}≔\mathrm{f\_\_1}\left(x\,y\,z\right)=\mathrm{f\_\_2}\left(x\right)+\mathrm{f\_\_3}\left(y\right)+\mathrm{f\_\_4}\left(z\right)\mathbf{where}\left[\left\{\frac{\ⅆ}{\ⅆx}\mathrm{f\_\_2}\left(x\right)={\mathrm{\_c}}_1\,\frac{\ⅆ}{\ⅆz}\mathrm{f\_\_4}\left(z\right)={\mathrm{\_c}}_3\,\frac{{\ⅆ}^2}{\ⅆy^2}\mathrm{f\_\_3}\left(y\right)=\frac{{\mathrm{\_c}}_1^3}{{\mathrm{\_c}}_3}-{\left(\frac{\ⅆ}{\ⅆy}\mathrm{f\_\_3}\left(y\right)\right)}^2\right\}\right]$

$\mathrm{build}\left(\mathrm{ans}\right)$

$\mathrm{f\_\_1}\left(x\,y\,z\right)=-\frac{{\mathrm{\_c}}_1^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}y}{\sqrt{{\mathrm{\_c}}_3}}+{\mathrm{\_c}}_{3}z+\frac{\ln \left(\frac{{\mathrm{\_c}}_3{\left({\left({\ⅇ}^{\frac{{\mathrm{\_c}}_1^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}y}{\sqrt{{\mathrm{\_c}}_3}}}\right)}^2\mathrm{c\_\_2}-\mathrm{c\_\_3}\right)}^2}{{\mathrm{\_c}}_1^3}\right)}{2}-\ln \left(2\right)+{\mathrm{\_c}}_{1}x+\mathrm{c\_\_1}+\mathrm{c\_\_4}$

$\mathrm{PDE}≔a\mathrm{diff}\left(f\left(x\,y\right)\,x\,x\right)+2b\mathrm{diff}\left(f\left(x\,y\right)\,y\,x\right)+c\mathrm{diff}\left(f\left(x\,y\right)\,y\,y\right)$

$\mathrm{PDE}≔a\left(\frac{{\partial }^2}{\partial x^2}f\left(x\,y\right)\right)+2b\left(\frac{{\partial }^2}{\partial x\partial y}f\left(x\,y\right)\right)+c\left(\frac{{\partial }^2}{\partial y^2}f\left(x\,y\right)\right)$

$\mathrm{mapde}\left(\mathrm{PDE}\,\mathrm{canop}\right)$

$\left(-\frac{b^2}{a}+c\right)\left(\frac{{\partial }^2}{\partial {\mathrm{\_\ξ1}}^2}f\left(\mathrm{\_\ξ1}\,\mathrm{\_\ξ2}\right)\right)+a\left(\frac{{\partial }^2}{\partial {\mathrm{\_\ξ2}}^2}f\left(\mathrm{\_\ξ1}\,\mathrm{\_\ξ2}\right)\right)\mathbf{where}\left\{\mathrm{\_\ξ1}=y-\frac{bx}{a}\,\mathrm{\_\ξ2}=x\right\}$

$\mathrm{mapde}\left(\mathrm{PDE}\,\mathrm{canom}\right)$

$-\frac{2\left(2ca-2b^2\right)\left(\frac{{\partial }^2}{\partial \mathrm{\_\ξ1}\partial \mathrm{\_\ξ2}}f\left(\mathrm{\_\ξ1}\,\mathrm{\_\ξ2}\right)\right)}{\sqrt{-4ca+4b^2}}\mathbf{where}\left\{\mathrm{\_\ξ1}=\frac{\left(2\sqrt{-ca+b^2}-2b\right)x}{2a}+y\,\mathrm{\_\ξ2}=\frac{\left(\frac{4\sqrt{-ca+b^2}b}{-4ca+4b^2}+1\right)x}{2}-\frac{ya}{2\sqrt{-ca+b^2}}\right\}$

$\mathrm{PDE1}≔\mathrm{subs}\left(\left\{a=x^2\,b=yx\,c=y^2\right\}\,\mathrm{PDE}\right)$

$\mathrm{PDE1}≔x^2\left(\frac{{\partial }^2}{\partial x^2}f\left(x\,y\right)\right)+2yx\left(\frac{{\partial }^2}{\partial x\partial y}f\left(x\,y\right)\right)+y^2\left(\frac{{\partial }^2}{\partial y^2}f\left(x\,y\right)\right)$

$\mathrm{PDE2}≔\mathrm{mapde}\left(\mathrm{PDE1}\,\mathrm{canop}\right)$

$\mathrm{PDE2}≔{\mathrm{\_\ξ1}}^2\left(\frac{{\partial }^2}{\partial {\mathrm{\_\ξ1}}^2}f\left(\mathrm{\_\ξ1}\,\mathrm{\_\ξ2}\right)\right)\mathbf{where}\left\{\mathrm{\_\ξ1}=x\,\mathrm{\_\ξ2}=\frac{y}{x}\right\}$

$\mathrm{pdsolve}\left(\mathrm{op}\left(1\,\mathrm{PDE2}\right)\right)$

$f\left(\mathrm{\_\ξ1}\,\mathrm{\_\ξ2}\right)=\mathrm{f\_\_1}\left(\mathrm{\_\ξ2}\right)\mathrm{\_\ξ1}+\mathrm{f\_\_2}\left(\mathrm{\_\ξ2}\right)$