 Euler - Maple Help

PDEtools

 Euler
 Euler's operator - return the conditions that a DE system must satisfy in order to be the divergence of a current Calling Sequence Euler(PDESYS, DepVars, integrabilityconditions = ..., jetnotation = ..., simplifier = ...) Parameters

 PDESYS - a system consisting of a single equation or a set or list of them, involving partial or ordinary (possibly not) differential equations DepVars - (optional) a specification of the unknown(s) in PDESYS jetnotation = ... - (optional) can be false, jetvariables (default), jetvariableswithbrackets, jetnumbers or jetODE; to respectively return or not using the different jet notations available integrabilityconditions = ... - (optional) - can be true (default) or false; indicates whether or not to reduce the determining PDE output system taking into account its integrability conditions simplifier = ... - optional - indicates the simplifier to be used instead of the default simplify/size Description

 • The Euler command represents Euler's operator: it returns zero when applied to an expression that is the divergence of another expression also called "current". When this divergence is also equal to zero this current is called conserved current.
 • Denoting the unknowns of a DE system by U, their derivatives by dU and the independent variables by X, Euler's operator is constructed as (d/dU + (-1)^n d/dX d/ddU), where n is the differential order of dU, d/dX is the differentiation operator such that dU = d/dx U and in Euler's operator there is one term of the form (-1)^n d/dX d/ddU for each dU in the DE system.
 • If DepVars is not given, Euler will consider all the differentiated unknown functions in PDESYS as unknowns of the problems. Specifying DepVars however permits not only restricting the unknowns in different ways but also specifying unknowns of the problems which do not appear differentiated in PDESYS.
 • When applying Euler's operator to a DE system constituted by N equations in M unknowns, there will be N x M expressions; that is: each of the M unknowns generates an operator of the form just described and each of these operators generates one expression when applied to each of the N equations of the system.
 • This set of expressions resulting from applying Euler's operator to a DE system, representing the conditions for the DE system to be the divergence of a current, is by default simplified with respect to its integrability conditions, then each expression in the set is simplified with respect to its size. This default behavior can be changed by providing one or more of the arguments integrabilityconditions = false, simplifier = none, or simplifier = any_desired_simplification.
 • To avoid having to remember the optional keywords, if you type the keyword misspelled, or just a portion of it, a matching against the correct keywords is performed, and when there is only one match, the input is automatically corrected. Examples

Consider the following PDE "system" consisting of a single pde.

 > $\mathrm{with}\left(\mathrm{PDEtools}\right):$
 > $U≔\mathrm{diff_table}\left(u\left(x,t\right)\right):$
 > $\mathrm{declare}\left({U}_{[]}\right)$
 ${u}{}\left({x}{,}{t}\right){}{\mathrm{will now be displayed as}}{}{u}$ (1)
 > ${\mathrm{pde}}_{1}≔{U}_{t,t}+{U}_{x,x}+{U}_{x}{U}_{[]}=0$
 ${{\mathrm{pde}}}_{{1}}{≔}{u}{}{{u}}_{{x}}{+}{{u}}_{{t}{,}{t}}{+}{{u}}_{{x}{,}{x}}{=}{0}$ (2)

This system automatically satisfies the conditions for being a divergence mentioned in the Description:

 > $\mathrm{Euler}\left({\mathrm{pde}}_{1}\right)$
 $\left\{{0}\right\}$ (3)

Hence ${\mathrm{pde}}_{1}$ is the divergence of a current

 > $\mathrm{ConservedCurrents}\left({\mathrm{pde}}_{1}\right)$
 $\left[{{\mathrm{_J}}}_{{x}}{}\left({x}{,}{t}{,}{u}{,}{{u}}_{{x}}{,}{{u}}_{{t}}\right){=}{{\mathrm{D}}}_{{3}}{}\left(\mathrm{f__3}\right){}\left({x}{,}{t}{,}{u}\right){}{{u}}_{{t}}{+}{{\mathrm{D}}}_{{2}}{}\left(\mathrm{f__3}\right){}\left({x}{,}{t}{,}{u}\right){+}{\mathrm{f__5}}_{{t}}{+}\frac{{{u}}^{{2}}}{{2}}{+}{{u}}_{{x}}{,}{{\mathrm{_J}}}_{{t}}{}\left({x}{,}{t}{,}{u}{,}{{u}}_{{x}}{,}{{u}}_{{t}}\right){=}{-}{\mathrm{f__5}}_{{x}}{-}{{\mathrm{D}}}_{{3}}{}\left(\mathrm{f__3}\right){}\left({x}{,}{t}{,}{u}\right){}{{u}}_{{x}}{-}{{\mathrm{D}}}_{{1}}{}\left(\mathrm{f__3}\right){}\left({x}{,}{t}{,}{u}\right){+}\mathrm{f__6}{}\left({x}\right){+}{{u}}_{{t}}\right]{,}\left[{{\mathrm{_J}}}_{{x}}{}\left({x}{,}{t}{,}{u}{,}{{u}}_{{x}}{,}{{u}}_{{t}}\right){=}{{\mathrm{D}}}_{{3}}{}\left(\mathrm{f__3}\right){}\left({x}{,}{t}{,}{u}\right){}{{u}}_{{t}}{+}{{\mathrm{D}}}_{{2}}{}\left(\mathrm{f__3}\right){}\left({x}{,}{t}{,}{u}\right){+}{\mathrm{f__5}}_{{t}}{+}\frac{{{u}}^{{2}}{}{t}}{{2}}{+}{{u}}_{{x}}{}{t}{,}{{\mathrm{_J}}}_{{t}}{}\left({x}{,}{t}{,}{u}{,}{{u}}_{{x}}{,}{{u}}_{{t}}\right){=}{-}{\mathrm{f__5}}_{{x}}{-}{{\mathrm{D}}}_{{3}}{}\left(\mathrm{f__3}\right){}\left({x}{,}{t}{,}{u}\right){}{{u}}_{{x}}{-}{{\mathrm{D}}}_{{1}}{}\left(\mathrm{f__3}\right){}\left({x}{,}{t}{,}{u}\right){+}\mathrm{f__6}{}\left({x}\right){+}{t}{}{{u}}_{{t}}{-}{u}\right]$ (4)

and admits a constant integrating factor:

 > $\mathrm{IntegratingFactors}\left({\mathrm{pde}}_{1}\right)$
 $\left[{{\mathrm{_μ}}}_{{1}}{}\left({x}{,}{t}{,}{u}{,}{{u}}_{{x}}{,}{{u}}_{{t}}\right){=}{1}\right]{,}\left[{{\mathrm{_μ}}}_{{1}}{}\left({x}{,}{t}{,}{u}{,}{{u}}_{{x}}{,}{{u}}_{{t}}\right){=}{t}\right]$ (5)

When combined with the rest of the Maple library, the Euler operator can serve varied purposes. Consider for example deriving the most general form of the divergence of a current that is also a first order PDE in two variables. The starting point is a generic expression, $\mathrm{\Delta }\left(x,t,u,{u}_{x},{u}_{t}\right)$, so it depends only on the first order derivatives.

 > ${\mathrm{pde}}_{2}≔\mathrm{Δ}\left(x,t,{U}_{[]},{U}_{x},{U}_{t}\right)=0$
 ${{\mathrm{pde}}}_{{2}}{≔}{\mathrm{\Delta }}{}\left({x}{,}{t}{,}{u}{,}{{u}}_{{x}}{,}{{u}}_{{t}}\right){=}{0}$ (6)

The conditions that Delta must satisfy in order to be a divergence are:

 > $\mathrm{conditions_for_divergence}≔\mathrm{Euler}\left({\mathrm{pde}}_{2}\right)$
 ${\mathrm{conditions_for_divergence}}{≔}\left[{{\mathrm{\Delta }}}_{{u}{,}{{u}}_{{x}}}{=}\frac{{-}{{\mathrm{\Delta }}}_{{u}{,}{{u}}_{{t}}}{}{{u}}_{{t}}{+}{{\mathrm{\Delta }}}_{{u}}{-}{{\mathrm{\Delta }}}_{{t}{,}{{u}}_{{t}}}{-}{{\mathrm{\Delta }}}_{{x}{,}{{u}}_{{x}}}}{{{u}}_{{x}}}{,}{{\mathrm{\Delta }}}_{{{u}}_{{x}}{,}{{u}}_{{x}}}{=}{0}{,}{{\mathrm{\Delta }}}_{{{u}}_{{t}}{,}{{u}}_{{x}}}{=}{0}{,}{{\mathrm{\Delta }}}_{{{u}}_{{t}}{,}{{u}}_{{t}}}{=}{0}\right]$ (7)

These conditions can be integrated.

 > $\mathrm{Divergence_1PDE_general_form}≔\mathrm{pdsolve}\left(\mathrm{conditions_for_divergence}\right)$
 ${\mathrm{Divergence_1PDE_general_form}}{≔}\left\{{\mathrm{\Delta }}{}\left({x}{,}{t}{,}{u}{,}{{u}}_{{x}}{,}{{u}}_{{t}}\right){=}{\mathrm{f__3}}_{{u}}{}{{u}}_{{x}}{+}{\mathrm{f__4}}_{{u}}{}{{u}}_{{t}}{+}{\mathrm{f__3}}_{{x}}{+}{\mathrm{f__4}}_{{t}}{+}\mathrm{f__6}{}\left({x}{,}{t}\right)\right\}$ (8)

Verify that these conditions are sufficient by applying Euler's operator to this result. First convert the result from jet notation to function notation.

 > $\mathrm{answer}≔\mathrm{FromJet}\left(\mathrm{rhs}\left({\mathrm{Divergence_1PDE_general_form}}_{1}\right),u\left(x,t\right)\right)=0$
 ${\mathrm{answer}}{≔}{{\mathrm{D}}}_{{3}}{}\left(\mathrm{f__3}\right){}\left({x}{,}{t}{,}{u}\right){}{{u}}_{{x}}{+}{{\mathrm{D}}}_{{3}}{}\left(\mathrm{f__4}\right){}\left({x}{,}{t}{,}{u}\right){}{{u}}_{{t}}{+}{{\mathrm{D}}}_{{1}}{}\left(\mathrm{f__3}\right){}\left({x}{,}{t}{,}{u}\right){+}{{\mathrm{D}}}_{{2}}{}\left(\mathrm{f__4}\right){}\left({x}{,}{t}{,}{u}\right){+}\mathrm{f__6}{}\left({x}{,}{t}\right){=}{0}$ (9)

So the above is the most general form of a divergence that is also a first order PDE. The following verifies that this form is correct.

 > $\mathrm{Euler}\left(\mathrm{answer},u\left(x,t\right)\right)$
 $\left\{{0}\right\}$ (10)

The most general form of a second order linear PDE in two independent variables that is also a divergence of a current can be derived in a similar way, starting with the following definition.

 > $\mathrm{declare}\left(A\left(x,t\right)\right)$
 ${A}{}\left({x}{,}{t}\right){}{\mathrm{will now be displayed as}}{}{A}$ (11)
 > ${\mathrm{pde}}_{3}≔{A}_{0}\left(x,t\right){U}_{[]}+{A}_{1}\left(x,t\right){U}_{x}+{A}_{2}\left(x,t\right){U}_{t}+{A}_{3}\left(x,t\right){U}_{x,x}+{A}_{4}\left(x,t\right){U}_{x,t}+{A}_{5}\left(x,t\right){U}_{t,t}=0$
 ${{\mathrm{pde}}}_{{3}}{≔}{{A}}_{{0}}{}{u}{+}{{A}}_{{1}}{}{{u}}_{{x}}{+}{{A}}_{{2}}{}{{u}}_{{t}}{+}{{A}}_{{3}}{}{{u}}_{{x}{,}{x}}{+}{{A}}_{{4}}{}{{u}}_{{t}{,}{x}}{+}{{A}}_{{5}}{}{{u}}_{{t}{,}{t}}{=}{0}$ (12)

The conditions for divergence, in the form of equations satisfied by the A[j]( x, t ) are obtained by applying Euler's operator.

 > $\mathrm{Euler}\left({\mathrm{pde}}_{3},{U}_{[]}\right)$
 $\left\{{{A}}_{{0}}{-}{{A}}_{{2}{,}{t}}{-}{{A}}_{{1}{,}{x}}{+}{{A}}_{{5}{,}{t}{,}{t}}{+}{{A}}_{{4}{,}{t}{,}{x}}{+}{{A}}_{{3}{,}{x}{,}{x}}\right\}$ (13)

Note that in the above calculation, the dependent variable of the problem must be specified, otherwise A[j]( x, t ) for all j would be also picked up as dependent variables. The result above is a single expression from which A[j]( x, t ) for one j can be isolated; the simplest form is achieved by isolating A.

 > $\mathrm{condition}≔\mathrm{isolate}\left({1}_{},{A}_{0}\left(x,t\right)\right)$
 ${\mathrm{condition}}{≔}{{A}}_{{0}}{=}{{A}}_{{2}{,}{t}}{+}{{A}}_{{1}{,}{x}}{-}{{A}}_{{5}{,}{t}{,}{t}}{-}{{A}}_{{4}{,}{t}{,}{x}}{-}{{A}}_{{3}{,}{x}{,}{x}}$ (14)
 > $\mathrm{Divergence_linear_2PDE}≔\mathrm{subs}\left(\mathrm{condition},{\mathrm{pde}}_{3}\right)$
 ${\mathrm{Divergence_linear_2PDE}}{≔}\left({{A}}_{{2}{,}{t}}{+}{{A}}_{{1}{,}{x}}{-}{{A}}_{{5}{,}{t}{,}{t}}{-}{{A}}_{{4}{,}{t}{,}{x}}{-}{{A}}_{{3}{,}{x}{,}{x}}\right){}{u}{+}{{A}}_{{1}}{}{{u}}_{{x}}{+}{{A}}_{{2}}{}{{u}}_{{t}}{+}{{A}}_{{3}}{}{{u}}_{{x}{,}{x}}{+}{{A}}_{{4}}{}{{u}}_{{t}{,}{x}}{+}{{A}}_{{5}}{}{{u}}_{{t}{,}{t}}{=}{0}$ (15)

This result can be verified applying Euler's operator to it.

 > $\mathrm{Euler}\left(\mathrm{Divergence_linear_2PDE},{U}_{[]}\right)$
 $\left\{{0}\right\}$ (16)