boundary value problems and partial differential equations (pdes) 1daniel baur / numerical methods...
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1Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Boundary Value Problems andPartial Differential Equations (PDEs)
2
2
d ( , ) d ( , ) d ( , ), , , ,
d d d
y t z y t z y t zf y t z
t z z
Daniel Baur
ETH Zurich, Institut für Chemie- und Bioingenieurwissenschaften
ETH Hönggerberg / HCI F128 – Zürich
E-Mail: [email protected]
http://www.morbidelli-group.ethz.ch/education/index
2Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Boundary Value Problems (BVP) for ODEs
Problem definition: Find a solution for a system of ODEs
Subject to the boundary conditions (BCs):
The total number of BCs has to be equal to the number of equations!
d ( )( , )
d
y ty f t y
t
0 0( ( ), ) 0
( ( ), ) 0f f
g y t t
h y t t
3Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Shooting Method
A first approach is to transform the BVP into an initial value problem (IVP), by guessing the missing initial conditions and using the BC to refine the guess, until convergence is reached
This way, the same algorithms as for IVPs can be used, but the convergence can be very problematic
Target
Too high: reduce initial velocity!
Too low: increase initial velocity!
4Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Collocation Method
A more sophisticated approach is the collocation method;it is based on approximating the unknown function with a sum of polynomials multiplied with unknown coefficients
The coefficients are determined by forcing the approximated solution to satisfy the ODE at a number of points equal to the number of coefficients
Matlab has a built-in function bvp4c which implements this method; it can also solve singular value problems
1
1
( ) ( )N
i ii
y t a P t
d( , )
d
y yf y t
t t S
5Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Example of a BVP
Consider a tubular reactor
We can model it as a plug flow reactor (PFR) with back-mixing by using the following partial differential equation
Dax is the (effective) axial dispersion coefficient [m2/s],v is the linear flow velocity [m/s] and n is the reaction order
cin A Bnkn cout
2
2
( , ) nA A Aax n A
c t x c cD v k c
t x x
0 L
6Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Tubular Reactor: Dimensionless Form
Let us cast the model in dimensionless form by defining
Where Pe is the Peclet and Da is the Damköhler number
2
2
1 nA
in
t t v
L
c u u uu Da u
c Pe z z
xzL
1
,n
n in
ax
L k cL vPe Da
D v
The numerical solution of a problem is usually much simpler if it
is dimensionless (most variables will range from 0 to 1).
7Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Steady State Assumption, Boundary Conditions
By assuming steady state, the time variable vanishes and we get an ODE
This equation is subject to the Danckwerts BCs (mass balance over inlet, continuous profile at the outlet)
2
2
1 d d0
d dnu u
Da uPe z z
0
1
1 d( 0) 1
d
d0
d
z
z
uu z
Pe z
u
z
8Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Transformation into a first order ODE
bvp4c solves first order ODEs, so if we remember the «trick» and transform our ODE, we get
And the boundary conditions
1
12
d d
d d
y u
y uy
z z
12
22
2 12
d
d
d d d
d d dn n
yy
z
y u uPe Da u Pe y Da y
z z z
1 2
2
1( 0) 1 ( 0)
( 1) 0
y z y zPe
y z
9Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Partial Differential Equations
Problem definition:In a partial differential equation (PDE), the solution depends on more than one independent variable, e.g. space and time
The function is usually subject to both inital conditions and boundary conditions
In our example
plus the Danckwerts BCs which apply at all times
2
2
( , ) 1
( 0) 0
nu z u uDa u
Pe z zu z
10Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Characterization of Second Order PDEs
Second order PDEs take the general form
where A, B and C are coefficients that may depend on x and y
These PDEs fall in one of the following categories1. B2 – AC < 0: Elliptic PDE2. B2 – AC = 0: Parabolic PDE3. B2 – AC > 0: Hyperbolic PDE
There are specialized solvers for some types of PDEs, hence knowing its category can be useful for solving a PDE
2 0xx xy yyAu Bu Cu
11Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Numerical Solution of PDEs
In general, it can be very difficult to solve PDEs numerically One approach is to discretize all but one dimension of the
solution; this way a system of ODEs is obtained that can be solved more easily
Note that these ODE systems are usually very stiff There are different ways of discretizing a dimension, for
example the collocation method we saw earlier Sophisticated algorithms refine the discretization in places
where the solution is still inaccurate Matlab has a built-in solver for parabolic and elliptic PDEs
in two dimensions, pdepe
12Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Unsteady State Tubular Reactor
Let us consider the start-up of a tubular reactor, i.e.
We can easily see that this is always a parabolic PDE(B = C = 0), hence the Matlab solver is applicable
2
2
0
1
( , ) 1
( 0) 0
1( 0) 1
0
n
z
z
u z u uDa u
Pe z zu z
duu z
Pe dz
du
dz
13Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Method of Finite Differences
We can apply the so-called finite differences method, if we remember numerical differentiation
Also, we can easily derive a similar expression for the second order derivative
0 0( ) ( )d ( )
d 2
f x h f x hf x
x h
20 0 0
2 2
( ) 2 ( ) ( )d ( )
d
f x h f x f x hf x
x h
14Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Method of Finite Differences (Continued)
The PDE for the start-up of the tubular reactor reads
Applying the method of finite differences, we get
where Δz = 1/N is the discretization step, N being the number of grid points
If we number the grid points with i = 1...N, we get
2
d 21
d 2nz z z z z z z z z zz
u u u u u uDa u
Pe z z
1 1 1 12
d 21
d 1/ 2 /ni i i i i ii
u u u u u uDa u
Pe N N
2
2
( , ) 1 nu z u uDa u
Pe z z
15Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Method of Finite Differences (Continued)
What happens at the boundaries u1 and uN?
One possibility is to invent pseudo grid-points u0 and uN+1 that fulfill the boundary conditions
In our case
1 00
0
10
1
1 11 1
1 /
1 /
z
N N
u uuu
Pe z Pe z
N Pe uu
N Peu u
16Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Method of Finite Differences (Continued) Rearranging the equations gives us
With the initial conditions and «boundary conditions»
2 2 21
1 1
d2
d 2 2ni
i i i i
u N N N N Nu u Da u u
Pe Pe Pe
10
1
( 0) 0, 1
1 /
1 /
i
N N
u i N
N Pe uu
N Peu u
17Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Assignment 1
1. Solve the dimensionless tubular reactor using bvp4c for 50 different values of the Peclet number between 0.01 and 100 and for a reaction of first and second order
In both cases use Da = 1
2. Plot the conversion at the end of the reactor (1-cA/cin) vs. Peclet for both reaction orders; Also plot the ratio between the conversions of the first order and second order reaction
What is better for these reactions, a lot of back-mixing (Pe small, CSTR) or ideal plug flow (Pe large, PFR)?
What influence does the reaction order have overall and at low or high Peclet numbers?
18Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Usage of bvp4c
bvp4c uses a call of the form sol = bvp4c(@ode_fun, @bc_fun, solinit, options); ode_fun is a function taking as inputs a scalar t and a vector y,
returning as an output dy / dt bc_fun is a function taking as inputs vectors where the boundary
conditions are evaluated, returning as output the residual at the boundary
solinit initializes the solution by usingsolinit = bvpinit(range, @initfun)
options is an options structure resulting fromoptions = bvpset('FJacobian', @jac_fun)
sol is a struct containing the solution and other parameters
19Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Usage of bvp4c (Continued)
In our case use the following
1(1) 1 (2) ( 0)
residual(2) ( 0)
a a
b
y yPe
y
12
22 1
dd d
dd
dn
yy
y zyz
Pe y Da yz
function dy = ode_fun(t,y,...)
function res = bc_fun(ya,yb,...)
11
0 1nPe Da n y Pe
J function J = jac_fun(t,y,...)
20Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Assignment 21. Use pdepe to solve the startup of the tubular reactor.
Consider only the first order reaction with Pe = 100 and Da = 1. Plot the conversion at the end of the reactor vs. dimensionless time. At what time does the solution reach steady state, i.e. how many
reactor volumes of solvent will you need? Compare the solution to what you have found in assignment 1, if the difference is smaller than 0.1%, assume that steady state has been reached.
21Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Usage of pdepe
pdepe uses the following syntax sol = pdepe(m,@pde_fun,@ic_fun,@bc_fun,xmesh,tspan);
m is a parameter that describes the symmetry of the problem (slab = 0, cylindrical = 1, spherical = 2); in our case slab, so m = 0
@pde_fun is a function that describes the PDE in this form:
@ic_fun is a function that takes as input a vector x and returns the initial conditions at t = 0
@bc_fun is a function that describes the boundary conditions, taking as an input xl, ul, xr, ur and t, returning the BCs in a form:
that is, pl, ql, pr and qr
, , , , , , , , ,m mu u u uc x t u x x f x t u s x t u
x t x x x
, , , , , , 0u
p x t u q x t f x t ux
22Daniel Baur / Numerical Methods for Chemical Engineers / BVP and PDE
Usage of pdepe (Continued)
In our case, use the following
2
2
( , ) 1 nu z u uDa u
Pe z z
0
1
, , , , , , 0
11 ( 0) 0
0
z
z
up x t u q x t f x t u
x
uu z
Pe z
du
dz
, , , , , , , , ,m mu u u uc x t u x x f x t u s x t u
x t x x x
4 Variable Inputs
5 Variable Inputs