The pdetest command returns either 0 (when the PDE is annulled by the solution sol), indicating that the solution is correct, or a remaining algebraic expression (obtained after simplifying the PDE with respect to the proposed solution), indicating that the solution might be wrong.

When PDE is a system, given as a set or list, possibly including boundary conditions, for each of the elements in the set/list pdetest will return a 0 or the remaining algebraic expression; the advantage of giving PDE as a list is that you can thus determine which element (if any) is not satisfied by the solution.

The pdetest command can also be used to reduce a PDE to a simpler problem by giving an "ansatz", instead of an explicit solution, since it will return the nonzero remaining part.

$\mathrm{PDE}≔{\ⅇ}^{\frac{{\partial }^5}{\partial x^5}f\left(x\,y\,z\,t\right)}\+\left(\frac{{\partial }^4}{\partial y^4}f\left(x\,y\,z\,t\right)\right)g\left(x\right)h\left(y\right)\=0$

$\mathrm{PDE}≔{\ⅇ}^{\frac{{\partial }^5}{\partial x^5}f\left(x\,y\,z\,t\right)}+\left(\frac{{\partial }^4}{\partial y^4}f\left(x\,y\,z\,t\right)\right)g\left(x\right)h\left(y\right)=0$

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

$\mathrm{ans}≔\left(f\left(x\,y\,z\,t\right)=\mathrm{f\_\_1}\left(x\right)+\mathrm{f\_\_2}\left(y\right)+\mathrm{f\_\_5}\left(z\,t\right)\right)\&where\left[\left\{\frac{{\ⅆ}^4}{\ⅆy^4}\mathrm{f\_\_2}\left(y\right)=-\frac{{\mathrm{\_c}}_1}{h\left(y\right)}\,\frac{{\ⅆ}^5}{\ⅆx^5}\mathrm{f\_\_1}\left(x\right)=\ln \left({\mathrm{\_c}}_{1}g\left(x\right)\right)\right\}\,\left(\mathrm{f\_\_5}\left(z\,t\right)\,\mathrm{are\; arbitrary\; functions.}\right)\right]$

$\mathrm{pdetest}\left(\mathrm{ans}\,\mathrm{PDE}\right)$

$0$

$\mathrm{PDE}≔\left(\frac{\partial }{\partial y}f\left(x\,y\right)\right)\left(\frac{\partial }{\partial y}\arctan \left(x^{\frac{1}{2}}y\right)\right)\+\left(\frac{\partial }{\partial x}f\left(x\,y\right)\right)\left(\frac{\partial }{\partial x}\arctan \left(x^{\frac{1}{2}}y\right)\right)\=0$

$\mathrm{PDE}≔\left(\frac{\partial }{\partial y}f\left(x\,y\right)\right)\left(\frac{\partial }{\partial y}\arctan \left(\sqrt{x}y\right)\right)+\left(\frac{\partial }{\partial x}f\left(x\,y\right)\right)\left(\frac{\partial }{\partial x}\arctan \left(\sqrt{x}y\right)\right)=0$

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

$\mathrm{ans}≔f\left(x\,y\right)=\mathrm{f\_\_1}\left(-2x^2+y^2\right)$

$\mathrm{pdetest}\left(\mathrm{ans}\,\mathrm{PDE}\right)$

$0$

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

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

$\mathrm{ans}≔\mathrm{pdsolve}\left(\mathrm{PDE}\,\mathrm{HINT}\=\mathrm{strip}\right)$

$\mathrm{ans}≔\left(x{\left(\frac{\partial }{\partial y}f\left(x\,y\right)\right)}^2-\frac{\partial }{\partial x}f\left(x\,y\right)-f\left(x\,y\right)=0\right)\&where\left[\left\{\left\{f\left(\mathrm{\_s}\right)=\left(-\mathrm{\_s}{\mathrm{c\_\_4}}^2-{\ⅇ}^{-\mathrm{\_s}}\mathrm{c\_\_2}+\mathrm{c\_\_1}\right){\ⅇ}^{2\mathrm{\_s}}\,x\left(\mathrm{\_s}\right)=-\mathrm{\_s}+\mathrm{c\_\_5}\,y\left(\mathrm{\_s}\right)=2\left(-\mathrm{\_s}+\mathrm{c\_\_5}+1\right)\mathrm{c\_\_4}{\ⅇ}^{\mathrm{\_s}}+\mathrm{c\_\_3}\,{\mathrm{\_p}}_1\left(\mathrm{\_s}\right)=\left(-{\mathrm{c\_\_4}}^2{\ⅇ}^{\mathrm{\_s}}+\mathrm{c\_\_2}\right){\ⅇ}^{\mathrm{\_s}}\,{\mathrm{\_p}}_2\left(\mathrm{\_s}\right)=\mathrm{c\_\_4}{\ⅇ}^{\mathrm{\_s}}\right\}\right\}\,\&and\left(\left\{{\mathrm{\_p}}_1=\frac{\partial }{\partial x}f\left(x\,y\right)\,{\mathrm{\_p}}_2=\frac{\partial }{\partial y}f\left(x\,y\right)\right\}\right)\right]$

$\mathrm{pdetest}\left(\mathrm{ans}\,\mathrm{PDE}\right)$

$0$

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

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

$\mathrm{ansatz}≔f\left(x\,y\right)\=F\left(x\right){\ⅇ}^y$

$\mathrm{ansatz}≔f\left(x\,y\right)=F\left(x\right){\ⅇ}^y$

$\mathrm{ans\_1}≔\mathrm{pdetest}\left(\mathrm{ansatz}\,\mathrm{PDE}\right)$

$\mathrm{ans\_1}≔{\ⅇ}^y\left(F\left(x\right)x-\frac{\ⅆ}{\ⅆx}F\left(x\right)-F\left(x\right)\right)$

$\mathrm{factor}\left(\mathrm{ans\_1}\right)$

${\ⅇ}^y\left(F\left(x\right)x-\frac{\ⅆ}{\ⅆx}F\left(x\right)-F\left(x\right)\right)$

$\mathrm{ans\_F}≔\mathrm{dsolve}\left(\mathrm{ans\_1}\,F\left(x\right)\right)$

$\mathrm{ans\_F}≔F\left(x\right)=\mathrm{c\_\_1}{\ⅇ}^{\frac{x\left(x-2\right)}{2}}$

$\mathrm{ans}≔\mathrm{subs}\left(\mathrm{ans\_F}\,\mathrm{ansatz}\right)$

$\mathrm{ans}≔f\left(x\,y\right)=\mathrm{c\_\_1}{\ⅇ}^{\frac{x\left(x-2\right)}{2}}{\ⅇ}^y$

$\mathrm{pdetest}\left(\mathrm{ans}\,\mathrm{PDE}\right)$

$0$

$\mathrm{sys}≔\left[\frac{\partial }{\partial t}u\left(x\,t\right)\=\frac{{\partial }^2}{\partial x^2}u\left(x\,t\right)-v\left(x\,t\right)\,\frac{\partial }{\partial t}v\left(x\,t\right)\=\frac{{\partial }^2}{\partial x^2}v\left(x\,t\right)-u\left(x\,t\right)\right]$

$\mathrm{sys}≔\left[\frac{\partial }{\partial t}u\left(x\,t\right)=\frac{{\partial }^2}{\partial x^2}u\left(x\,t\right)-v\left(x\,t\right)\,\frac{\partial }{\partial t}v\left(x\,t\right)=\frac{{\partial }^2}{\partial x^2}v\left(x\,t\right)-u\left(x\,t\right)\right]$

$\mathrm{sol}≔\mathrm{pdsolve}\left(\mathrm{sys}\,\left\{u\left(x\,t\right)\,v\left(x\,t\right)\right\}\right)$

$\mathrm{sol}≔\left\{u\left(x\,t\right)=\mathrm{c\_\_1}\cos \left(x\right)+\mathrm{c\_\_2}{\ⅇ}^x+\mathrm{c\_\_3}\sin \left(x\right)+\frac{\mathrm{c\_\_4}}{{\ⅇ}^x}+{\ⅇ}^t\mathrm{c\_\_6}+\frac{\mathrm{c\_\_5}}{{\ⅇ}^t}\,v\left(x\,t\right)=-{\ⅇ}^t\mathrm{c\_\_6}+\frac{\mathrm{c\_\_5}}{{\ⅇ}^t}-\mathrm{c\_\_1}\cos \left(x\right)+\mathrm{c\_\_2}{\ⅇ}^x-\mathrm{c\_\_3}\sin \left(x\right)+\frac{\mathrm{c\_\_4}}{{\ⅇ}^x}\right\}$

$\mathrm{pdetest}\left(\mathrm{sol}\,\mathrm{sys}\right)$

$\left[0\,0\right]$

$\mathrm{pde}≔\frac{\partial }{\partial t}u\left(x\,t\right)\=k\left(\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}u\left(x\,t\right)\right)\right)\+Q$

$\mathrm{pde}≔\frac{\partial }{\partial t}u\left(x\,t\right)=k\left(\frac{{\partial }^2}{\partial x^2}u\left(x\,t\right)\right)+Q$

${\mathrm{bc}}_1≔u\left(0\,t\right)\=2{\ⅇ}^{kt}-\frac{1Q}{k}$

${\mathrm{bc}}_1≔u\left(0\,t\right)=2{\ⅇ}^{kt}-\frac{Q}{k}$

$\mathrm{sol}≔u\left(x\,t\right)\={\mathrm{\_C1}}^2{\ⅇ}^{x\+kt}-\left({\mathrm{\_C1}}^2-2\right){\ⅇ}^{-x\+kt}-\frac{1Q{x}^2}{2k}\+\frac{1Q\left({\mathrm{\_C1}}^2-2\right)x}{{\mathrm{\_C1}}^{2}k}-\frac{1Q}{k}$

$\mathrm{sol}≔u\left(x\,t\right)={\mathrm{c\_\_1}}^2{\ⅇ}^{kt+x}-\left({\mathrm{c\_\_1}}^2-2\right){\ⅇ}^{kt-x}-\frac{Q{x}^2}{2k}+\frac{Q\left({\mathrm{c\_\_1}}^2-2\right)x}{{\mathrm{c\_\_1}}^{2}k}-\frac{Q}{k}$

$\mathrm{pdetest}\left(\mathrm{sol}\,\left[\mathrm{pde}\,{\mathrm{bc}}_1\right]\right)$

$\left[0\,0\right]$

${\mathrm{bc}}_2≔{\mathrm{D}}_{1\,1}\left(u\right)\left(0\,t\right)\=2{\ⅇ}^{kt}-\frac{1Q}{k}$

${\mathrm{bc}}_2≔{\mathrm{D}}_{1,1}\left(u\right)\left(0\,t\right)=2{\ⅇ}^{kt}-\frac{Q}{k}$

$\mathrm{pdetest}\left(\mathrm{sol}\,\left[\mathrm{pde}\,{\mathrm{bc}}_2\right]\right)$

$\left[0\,0\right]$