=PDEVAL(col1, exprn, stamp, opr, pt, [options])

• Use PDEVAL to compute a single value from a solution array obtained by any of the partial differential equations solvers. PDEVAL computes the value using a specified calculus operation including interpolation, differentiation or integration.

• In a dynamical optimization problem, PDEVAL can be used to define dynamic constraints on the solution array of the partial differential equations solver. A constraint formula is typically defined as the difference between the computed value by PDEVAL and a target value. The constraint formulas are input to NLSOLVE.

Required Inputs

col1 a reference to the first column in the solution array. This defines the time or spatial range for the calculus operation.

You may select a sub range of column 1 although this reduces the accuracy of the calculus operation.

exprn a reference to the operand for the calculus operation.

The reference may be to single variable or a formula implicating mulutiple variables. If your operand is a single variable, just pass its name in this argument.

stamp the time or spatial point value that identifies the target block within the solution array. The stamp must be selected from the available output points in the first row of the solution array.

See Output Format section for PDSOLVE OR PDASOLVE.

opr a string value specifying the calculus operation to perform. Supported values are INTERP, DERIV, and INTEG.

pt defines the point of interpolation or differentiation for the calculus operation.

Not required if the calculus operation is INTEG.

Optional Inputs

order controls the calculus operation as described in below:

• If opr is INTERP or INTEG
Enter 1 to use linear splines, or 3 to use cubic splines. Default is 3.

• If opr is DERIV
Enter the derivative order: 1, 2 or 3. Default is 1.

Consider the array result shown in Table 1 and obtained by PDSOLVE for heat transfer problem. The results are in the default format which gives you a snapshot of the system variables spatial distribution at selected temporal values. The first row specifies the time points (in this example 0, 0.5 and 1), and the first column specifies the spatial points.

 M N O P Q R S 8 t 0 0 0.5 0.5 1 1 9 x u ux u ux u ux 10 0 0 0 100 -58.2477 100 -16.9646 11 0.071429 0 0 95.8481 -57.8812 98.7907 -16.8578 12 0.142857 0 0 91.7486 -56.7862 97.5967 -16.5387 13 0.214286 0 0 87.7528 -54.9767 96.4330 -16.0114 14 0.285714 0 0 83.9112 -52.4755 95.3142 -15.2825 15 0.357143 0 0 80.2720 -49.3144 94.2544 -14.3614 16 0.428571 0 0 76.8811 -45.5333 93.2669 -13.2597 17 0.5 0 0 73.7809 -41.1799 92.3641 -11.9914 18 0.571429 0 0 71.0105 -36.3092 91.5574 -10.5726 19 0.642857 0 0 68.6047 -30.9823 90.8569 -9.0212 20 0.714286 0 0 66.5937 -25.2663 90.2713 -7.3566 21 0.785714 0 0 65.0028 -19.2329 89.8081 -5.5998 22 0.857143 0 0 63.8519 -12.9578 89.4730 -3.7727 23 0.928571 0 0 63.1556 -6.5198 89.2703 -1.8983 24 1 0 0 62.9225 0 89.2024 0

You can use PDEVAL to extract or compute values of interest from the simulation result:

 Objective Formula Result Find derivative of u(x, t=0.5) at x=0.5. Note that it matches the value of ux(x=0.5,t=0.5), cell Q17 of Table 1. =PDEVAL(A13:A27,u,0.5,"deriv",0.5) -41.18 Integrate u(x, t=0.5) from x=0 to 1. =PDEVAL(A13:A27,u,0.5,"integ") 76.39494 Interpolate u(x,t=0.5) at x = 0.75. =PDEVAL(A13:A27,u,0.5,"interp",0.75) 65.74446 Integrate the square of u(x, t=0.5) from x=0 to 1. =PDEVAL(A13:A27,M21,0.5,"integ") 5966.39 M21 =u^2

When solving a parametrized dynamical optimization problem, PDEVAL is used to define constraints on the dynamical system solution typically as the difference between a computed value by PDEVAL and a target value. You can then use the nonlinear solver NLSOLVE to find optimal values for the design parameters that will minimize the differences in your defined constraints.

Question or Comment? Email us:
support @ excel-works.com
ExceLab: Transforming Excel into a Calculus Power House
ExceLab functions and methods are protected by USA Patents 9286286, 9892108, 10114812 and pending.