Consider the following PDE "system" consisting of a single pde.
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This system automatically satisfies the conditions for being a divergence mentioned in the Description:
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Hence
is the divergence of a current
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 = u[x]+_F3(x, t, u)*u[t]+Int(_F3[t]+_a, _a = `` .. u)+_F5(x, t), _J[t](x, t, u, u[x], u[t]) = Int(-_F3[x], _a = `` .. u)+Int(-_F5[x], t)-_F3(x, t, u)*u[x]+_F6(x)+u[t]], [_J[x](x, t, u, u[x], u[t]) = t*u[x]+_F3(x, t, u)*u[t]+Int(_F3[t]+_a*t, _a = `` .. u)+_F5(x, t), _J[t](x, t, u, u[x], u[t]) = Int(-_F3[x], _a = `` .. u)+Int(-_F5[x], t)-_F3(x, t, u)*u[x]+_F6(x)+u[t]*t-u]](/support/helpjp/helpview.aspx?si=7537/file02497/math190.png)
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and admits a constant integrating factor:
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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,
, so it depends only on the first order derivatives.
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The conditions that Delta must satisfy in order to be a divergence are:
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These conditions can be integrated.
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![Divergence_1PDE_general_form := {Delta(x, t, u, u[x], u[t]) = _F3(x, t, u)*u[x]+_F4(x, t, u)*u[t]+Int(_F4[t]+_F3[x], u)+_F6(x, t)}](/support/helpjp/helpview.aspx?si=7537/file02497/math230.png)
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Verify that these conditions are sufficient by applying Euler's operator to this result. First convert the result from jet notation to function notation.
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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.
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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.
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The conditions for divergence, in the form of equations satisfied by the A[j]( x, t ) are obtained by applying Euler's operator.
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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[0].
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![Divergence_linear_2PDE := (A[2, t]+A[1, x]-A[5, t, t]-A[4, t, x]-A[3, x, x])*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](/support/helpjp/helpview.aspx?si=7537/file02497/math329.png)
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This result can be verified applying Euler's operator to it.
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