dsubs substitutes the given sequence of equations into expr from left to right, the same way subs does. However, unlike subs, dsubs will substitute the left hand side of each substitution equation until this derivative is eliminated from the target (when this is possible). When many substitution equations are given, they are applied sequentially, as in subs. dsubs also works with anticommutative variables set using the Physics package.

The exception to this rule occurs when a substitution cannot be applied successively, like the algebraic substitution $x=x^2$. In this case the result returned is the one obtained by performing a single substitution.

All substitution equations must be in solved form.

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

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

$\mathrm{eq1}≔\mathrm{diff}\left(f\left(x\right)\,x\right)=f\left(x\right)$

$\mathrm{eq1}≔\frac{\ⅆ}{\ⅆx}f\left(x\right)=f\left(x\right)$

$\mathrm{expr}≔\mathrm{diff}\left(f\left(x\right)\,x\,x\right)-f\left(x\right)$

$\mathrm{expr}≔\frac{{\ⅆ}^2}{\ⅆx^2}f\left(x\right)-f\left(x\right)$

$\mathrm{subs}\left(\mathrm{eq1}\,\mathrm{expr}\right)$

$\frac{\ⅆ}{\ⅆx}f\left(x\right)-f\left(x\right)$

$\mathrm{dsubs}\left(\mathrm{eq1}\,\mathrm{expr}\right)$

$0$

$\mathrm{eq2}≔\mathrm{diff}\left(f\left(x\,y\,z\right)\,x\,z\right)=\mathrm{diff}\left(f\left(x\,y\,z\right)\,y\right),\mathrm{diff}\left(f\left(x\,y\,z\right)\,y\,y\right)=\mathrm{diff}\left(f\left(x\,y\,z\right)\,x\,x\right)$

$\mathrm{eq2}≔\frac{{\partial }^2}{\partial x\partial z}f\left(x\,y\,z\right)=\frac{\partial }{\partial y}f\left(x\,y\,z\right),\frac{{\partial }^2}{\partial y^2}f\left(x\,y\,z\right)=\frac{{\partial }^2}{\partial x^2}f\left(x\,y\,z\right)$

$\mathrm{dsubs}\left(\mathrm{eq2}\,\mathrm{diff}\left(f\left(x\,y\,z\right)\,x\,y\,z\right)\right)$

$\frac{{\partial }^2}{\partial x^2}f\left(x\,y\,z\right)$

$\mathrm{with}\left(\mathrm{Physics}\right)$

$\left[\mathrm{`*`}\,\mathrm{`.`}\,\mathrm{Annihilation}\,\mathrm{AntiCommutator}\,\mathrm{Antisymmetrize}\,\mathrm{Assume}\,\mathrm{Bra}\,\mathrm{Bracket}\,\mathrm{Check}\,\mathrm{Christoffel}\,\mathrm{Coefficients}\,\mathrm{Commutator}\,\mathrm{CompactDisplay}\,\mathrm{Coordinates}\,\mathrm{Creation}\,\mathrm{D\_}\,\mathrm{Dagger}\,\mathrm{Decompose}\,\mathrm{Define}\,\mathrm{D\γ}\,\mathrm{DiracConjugate}\,\mathrm{Einstein}\,\mathrm{EnergyMomentum}\,\mathrm{Expand}\,\mathrm{ExteriorDerivative}\,\mathrm{Factor}\,\mathrm{FeynmanDiagrams}\,\mathrm{FeynmanIntegral}\,\mathrm{Fundiff}\,\mathrm{Geodesics}\,\mathrm{GrassmannParity}\,\mathrm{Gtaylor}\,\mathrm{Intc}\,\mathrm{Inverse}\,\mathrm{Ket}\,\mathrm{KillingVectors}\,\mathrm{KroneckerDelta}\,\mathrm{LagrangeEquations}\,\mathrm{LeviCivita}\,\mathrm{Library}\,\mathrm{LieBracket}\,\mathrm{LieDerivative}\,\mathrm{Normal}\,\mathrm{NumericalRelativity}\,\mathrm{Parameters}\,\mathrm{PerformOnAnticommutativeSystem}\,\mathrm{Projector}\,\mathrm{Psigma}\,\mathrm{Redefine}\,\mathrm{Ricci}\,\mathrm{Riemann}\,\mathrm{Setup}\,\mathrm{Simplify}\,\mathrm{SortProducts}\,\mathrm{SpaceTimeVector}\,\mathrm{StandardModel}\,\mathrm{Substitute}\,\mathrm{SubstituteTensor}\,\mathrm{SubstituteTensorIndices}\,\mathrm{SumOverRepeatedIndices}\,\mathrm{Symmetrize}\,\mathrm{TensorArray}\,\mathrm{Tetrads}\,\mathrm{ThreePlusOne}\,\mathrm{ToContravariant}\,\mathrm{ToCovariant}\,\mathrm{ToFieldComponents}\,\mathrm{ToSuperfields}\,\mathrm{Trace}\,\mathrm{TransformCoordinates}\,\mathrm{Vectors}\,\mathrm{Weyl}\,\mathrm{`^`}\,\mathrm{dAlembertian}\,\mathrm{d\_}\,\mathrm{diff}\,\mathrm{g\_}\,\mathrm{gamma\_}\right]$

$\mathrm{Setup}\left(\mathrm{anticommutativepre}=\left\{Q\,\mathrm{\theta }\right\}\right)$

$\mathrm{*\; Partial\; match\; of\; \text{'}}\mathrm{anticommutativepre}\mathrm{\text{'}\; against\; keyword\; \text{'}}\mathrm{anticommutativeprefix}\text{'}$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\left[\mathrm{anticommutativeprefix}=\left\{Q\,\mathrm{\theta }\right\}\right]$

$\mathrm{PDEtools}:-\mathrm{declare}\left(Q\left(x\,y\,\mathrm{\theta }\left[1\right]\,\mathrm{\theta }\left[2\right]\right)\right)$

$Q\left(x\,y\,{\mathrm{\theta }}_1\,{\mathrm{\theta }}_2\right)\mathrm{will\; now\; be\; displayed\; as}Q$

$q≔\mathrm{PDEtools}:-\mathrm{diff\_table}\left(Q\left(x\,y\,\mathrm{\theta }\left[1\right]\,\mathrm{\theta }\left[2\right]\right)\right)\:$

$\mathrm{pde}\left[1\right]≔q\left[x,y,\mathrm{\theta }\left[1\right]\right]+q\left[x,y,\mathrm{\theta }\left[2\right]\right]-q\left[y,\mathrm{\theta }\left[1\right],\mathrm{\theta }\left[2\right]\right]=0$

${\mathrm{pde}}_1≔Q_{x,y,{\mathrm{\theta }}_1}+Q_{x,y,{\mathrm{\theta }}_2}-Q_{y,{\mathrm{\theta }}_1,{\mathrm{\theta }}_2}=0$

$\mathrm{pde}\left[2\right]≔q\left[\mathrm{\theta }\left[1\right]\right]=0$

${\mathrm{pde}}_2≔Q_{{\mathrm{\theta }}_1}=0$

$\mathrm{dsubs}\left(\mathrm{pde}\left[2\right]\,\mathrm{pde}\left[1\right]\right)$

$Q_{x,y,{\mathrm{\theta }}_2}=0$

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

$Q=\mathrm{f\_\_9}\left(x\,y\right)\mathrm{\_\λ1}+\mathrm{f\_\_10}\left(x\,y\right){\mathrm{\theta }}_1+\left(\mathrm{f\_\_6}\left(x\right)+\mathrm{f\_\_5}\left(y\right)\right){\mathrm{\theta }}_2+\left(\mathrm{f\_\_8}\left(x\right)+\mathrm{f\_\_7}\left(y\right)\right)\mathrm{\_\λ2}{\mathrm{\theta }}_1{\mathrm{\theta }}_2$