Introduction to Mathematical Modelling: Ordinary Differential ... - UiO

Introduction to Mathematical Modelling: Ordinary Differential Equations and Heat Equations Knut–Andreas Lie SINTEF ICT, Dept. Applied Mathematics

INF2340 / Spring 2005 – p.

Overview 1. Ordinary differential equations • Example of ODE models − radioactive decay, Newton’s second law, population models • Numerical solution of ODEs − forward and backward Euler, Runge-Kutta methods 2. Heat equations • from physical problem to simulator code • steady heat conduction − finite differences, linear algebra • unsteady heat conduction − finite differences, boundary conditions

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Ordinary Differential Equations (ODEs) y

(m)

= f t, y, y, ˙ y¨, . . . , y

(n−1)




Simple example: radioactive decay Given a quantity q of a radioactive matter, which decays at a certain constant rate k. The model reads dq(t) = r(t) ˙ = −k · q. dt Solution: q(t) = q(t0 )e−k(t−t0 ) .

In general: finding a solution is not so easy, although there are approaches in certain special cases −→ numerical approach!

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ODEs cont’d Very simple example from high school physics: Consider the motion of a body with mass m under constant force f , which is initially at rest at position x0 . Newton’s second law reads f = ma = m¨ x(t) where x(t) is the position of the body at time t. The corresponding ODE then reads f = m¨ x(t),

˙ =0 x(0) = x0 , x(0)

This equation is easily integrated x(t) = x0 +

f 2 2m t . INF2340 / Spring 2005 – p.

Population models Consider the dynamics of a single species (isolated or with no predators) • constant birth rate b per time and individual • constant death rate d per time and individual • hence, constant growth rate λ = b − d

The model (Maltus, 1798): p(t) ˙ = λ · p(t)

Solution p(t) = p0 eλt predicts exponential growth or decay.

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Population models cont’d Is there any realism in this? • between 1700 and 1960: growth rate of about 0.02,

population doubles in 34.67 years

• generally: limited resources on earth slows down growth

More realism (Verhulst et al., 19th century): • linear rates: b(t) = b0 − b1 p(t),

d(t) = d0 − d1 p(t)

−→ new model: p(t) ˙ = −k(p(t) − p∞ )

Solution: p(t) = p∞ + (po − p∞ )e−kt

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Population models cont’d What about realism now? • Populations tend to follow a S-shape (logistic model) • New model: • New solution:

p(t) ˙ = a · p(t) − b · p2 (t) p(t) =

a·p0 b·p0 +(a−b·p0 )e−at

And so on ... adding more than one species (predator–pray) ... Do we have equilibrium, is it attractive or repellent, ...?

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Population models cont’d An ODE model from modern research, describing the dynamics of HIV-1 infection in vivo (Perelson& Nelson, SIAM Review 41/1, 1999) : The rate of change of uninfected cells T , productively infected cells T ∗ , and virus V : ´ ` dT = s + pT 1 − T /Tmax − dT T − kV T dt dT ∗ = kV T − δT ∗ dt dV = N δT ∗ − cV. dt Here: dT – death rate of uninfected cells δ – death rate of infected cells p – rate of proliferation (continuous development of cells in tissue) N – virus production per infected cell c – clearance rate INF2340 / Spring 2005 – p.

Numerical solution of ODEs – Euler’s method We wish to solve the equation: y 0 = f (y, t),

y(a) = α,

a≤t≤b

Obvious solution – use finite differences: • generate a mesh: ti = a + ih, for each i = 0, . . . , N , where

h = (b − a)/N is called stepsize

• apply forward differences to the equation

y(ti+1 ) = y(ti ) + hf (y(ti ), ti ),

i = 0, . . . , N − 1

This gives an explicit formula for each y(ti+1 ) once y0 is known.

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Example: Euler’s method for u0 = 3u 20

Exact h=1/4 h=1/16 h=1/64

18 16 14 12 10 8 6 4 2 0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

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Another Euler method – backward Euler Once again we consider: y 0 = f (y, t),

y(a) = α,

a≤t≤b

and introduce a mesh: ti = a + ih, for each i = 0, . . . , N This time we apply backward differences y(ti+1 ) = y(ti ) + hf (y(ti+1 ), ti+1 ),

i = 0, . . . , N − 1

This gives an equation for each y(ti+1 ) once y0 is known.

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Two different methods • forward Euler: explicit method

y(ti+1 ) = y(ti ) + hf (y(ti ), ti )

the new value is given by a formula • backward Euler: implicit method

y(ti+1 ) = y(ti ) + hf (y(ti+1 ), ti+1 )

the new value is given by an algebraic equation

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Example: u0 = −4π sin(4πx) u(xi ) = u(xi−1 ) − 4π sin(4πxi−1 )

u(xi ) = u(xi−1 ) − 4π sin(4πxi )

Exact Forward Euler Backward Euler

1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

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Discretisation errors When using finite differences in forward Euler, we make a local discretisation error at each point y(ti+1 ) − y(ti ) = f (y(ti ), ti ) + error h Consider y˙ = f (y). Expanding y(ti+1 ) by a Taylor polynomial: ˙ i ) + y¨(τ ) 12 h2 − y(ti ) = hf (y(ti )) + h · error y(ti ) + hy(t ˙ i ) = f (y(ti )), for ti ≤ τ ≤ ti+1 . Now since y(t error = y¨(τ ) ·

1 2h

d = f (y(τ )) · 12 h dt

We say that forward Euler is a first-order method.

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Discretisation errors cont’d We say that the scheme is consistent if y(t + h) − y(t) − f (y(t), t) → 0 as h → 0 l(h) = max h t∈[a,b] Similarly, we define the global discretisation error as e(h) = max yi − y(ti ) , ti ∈[a,b]

where y(ti ) is the exact solution and yi is computed by our scheme. The scheme is convergent if e(h) → 0 for t → 0

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Higher order methods – Runge–Kutta Consider the forward Euler method and try to evaluate f (·) at some other point yi+1 = yi + hf (yi + β). Let us repeat the error analysis y(ti+1 ) − y(ti ) = f (yi )h + f˙(yi ) 12 h2 + f¨(τ ) 16 h3 = hf (yi + β) + h · error Assume now that the error equals 16 f¨(τ )h2 . Now f˙(yi ) = f 0 (yi )f (yi ) and we have f (yi ) + f 0 (yi )f (yi ) 12 h = f (yi + β) = f (yi ) + βf 0 (yi ) + β 2 + . . . This means that β = 12 h f (yi ).

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Runge-Kutta methods cont’d Thus, we have a new second-order method yi+1 = yi + hf yi +

1 2 h f (yi )




There are other alternatives also, e.g., yi+1 = yi +

1 2h



f (yi ) + f yi + hf (yi )




The Runge-Kutta methods are all on the form:

•

Approximate the solution at a point ti ≤ τ ≤ ti+1 by the intermediate step w = yi + (τ − ti )f (yi )

•

Combine yi , w, f (yi ), f (w) to get a more accurate solution yi+1 .

For higher-order methods we use more intermediate steps.

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From ODEs to PDEs So far, we have used ODEs: • involve derivatives with respect to only one variable • if the unknown function depends on more variables, these

have been treated as constants

Now, partial differential equations (PDEs): • involve (partial) derivatives with respect to more than one

variable

• the unknown function depends on more than one variable • stationary problems: no time-dependence • unsteady problems. time-dependence, but maybe steady

limit

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Mathematical model: ∂t u = ∆u + f • Models propagation of heat within a given object • Examples − a heated rod (1D) − a heated plate or smoothing of images (2D) − heat distribution in a furnace (3D) • Functions of interest − u(x,t), u(x,y), u(x,y,t), u(x,y,z), u(x,y,z,t), ... • More generally, heat propagation depends on the

conductivity of the material

ut = ∇(K(x, u)∇u) + f (x, t, u)

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Heat conduction in the continental crust x=0

Ts

x x=b -Q

• Knowing the temperature at the earth’s surface and the heat flow from the mantle, what is the temperature distribution through the continental crust? • Interesting question in geology and geophysics – and for those nations exploring oil resources...

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Physical and mathematical model • Our prototype differential equation (an ODE in 1D): ∂2u − 2 = f (x) ∂x Here u is the temperature. • Needs to be equipped with boundary conditions • Very simple equation, but it has applications to − fluid flow in channels − deflection of electric cables − strength analysis of beams − ... • In multidimensions −∆u = f is the elliptic Poission equation, which is fequently occuring in mathematical models

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Heat conduction in the continental crust cont’d x=0

Ts

x x=b -Q

Physical assumptions:

• • •

Crust of infinite area Steady state heat flow Heat generated by radioactive decay

Physical quantities:

• • •

u(x) : temperature q(x) : heat flux (velocity of heat) s(x) : heat release per unit time and mass

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Derivation of the model x=0

inflow

s(x)=R exp(-x/L)

x

outflow

x=b

Physical principles:

•

First law of thermodynamics: net outflow of heat = total generated heat

•

Fourier’s law: heat flows from hot to cold regions (i.e. heat velocity is poportional to changes in temperature) q(x) = −λu0 (x)

•

Heat generation due to radioactive decay: s(x) = R exp(−x/L)

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Derivation of the model cont’d x=0 q(x-h/2) h

s(x) q(x+h/2)

x x=b

From the first law of thermodynamics: q(x + h/2) − q(x − h/2) = s(x) h Using a Taylor expansion: q(x + h/2) − q(x − h/2) 1 000 = q 0 (x) + q (x)h2 + . . . h 24 Hence, as h → 0 we have q 0 (x) = s(x)

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Derivation of the model cont’d Ts

x=0

x x=b -Q

Combining the 1st law of thermodynamics (q 0 = s) with Fourier’s law (q = −λu), we get d − dx

„

du λ dx

«

= s(x)

Boundary conditions:

• •

u(0) = Ts (at the surface of the earth) q(b) = −Q (at the bottom of the crust)

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Mathematical model d  du  λ = Re−x/L , − dx dx

u(0) = Ts ,

λ(b)u0 (b) = −Q

Observe that u depends upon seven parameters: u = u(x; λ, R, L, b, Ts , Q)! Suppose that we want to investigate the influence of the different parameters. Assume (modestly) three values of each parameter: −→ Number of possible combinations: 36 = 729. Using scaling we can reduce the six physical parameters λ, R, L, b, Ts , Q to only two!

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Scaling We introduce dimensionless quantities (and assume that λ is constant): x=x ¯b,

u/λ, u = Ts + Qb¯

s(b¯ x) = R¯ s(¯ x)

This gives ¯ d2 u − 2 = γe−¯x/β , d¯ x

u ¯ = 0,

d¯ u (1) = 1 d¯ x

where we have two dimensionless quantities β = b/L,

γ = bR/Q

Dropping the bars, we get an equation on the form −u00 (x) = f (x), x ∈ (0, 1),

u(0) = 0,

u0 (1) = 1

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Discretisation • Introduce a grid xi = (i − 1)h and compute the unknown at grid points ui = u(xi )

u2 u1

x=0

u3 u4

u5

x=1

x

• Differential equation fulfilled at each node −u00 (xi ) = f (xi ) • Approximate by standard finite differences ui+1 − 2ui + ui−1 = −h2 fi ,

i = 1, . . . , n − 1

As opposed to the ODEs we have seen earlier, this is a linear system of unknowns

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Discretising boundary conditions • u(0) = 0 simply becomes u1 = 0 • u0 (1) = 1 can be approximated as un+1 − un−1 =1 2h • Problem: un+1 is not in the mesh! • Solution: Use the discrete differential equation for i = n: un−1 − 2un + ui+1 = −h2 fn and the discrete boundary condition to eliminate un+1 • The result is 2un−1 − 2un = −2h − h2 fn

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Linear system of equations u1 u1

=0 −2u2

+u3

u2

−2u3 .. .

= −h2 f2 +u4 .. . .. .

.. .. ..

= −h2 f3 .. . .. . .. . .. .

. .

..

.

.

..

.

..

.

..

.

..

.

un−2

..

.

−2un−1

+un

2un−1

−2un

= −h2 fn−1 = −2h − h2 fn

INF2340 / Spring 2005 – p. 3

Using linear algebra We write the system as Au = b: 2 1 6 6 61 6 6 6 60 6 6 6. 6 .. 6 6 6 .. 6. 6 6 6 .. 6. 6 6. 6. 6. 6 6. 6. 6. 4 0

0

0

0

−2

1

0

1 .. .

−2 .. .

..

.

..

.

..

.

..

.

..

.

···

···

···

··· .. .

··· .. .

··· .. .

..

..

.

..

.

..

.

1 .. .

..

.

..

.

..

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

.

···

···

···

.

1

−2

0

2

32

3

2

3

0 u1 7 6 7 76 7 6 7 76 6 7 6 7 2 0 7 6 u2 7 6 −h f2 7 7 7 6 7 76 7 6 7 76 6 7 6 7 2 0 7 6 u3 7 6 −h f3 7 7 7 6 7 76 7 6 7 .. 7 6 6 7 7 6 7 . . . 76 . 7 6 7 . . 7 76 . 7 6 . 76 6 7 7 .. 7 6 . 7 = 6 7 . 7 76 . 7 6 . . 7 76 . 7 6 . 76 6 7 7 . 76 . 7 6 7 . . 76 . 7 6 . 7 . 7 6 . 7 6 7 . 76 6 7 7 . 76 . 7 6 7 . . 76 . 7 6 . 7 . 7 76 . 7 6 7 6 7 76 2 6 7 6 7 1 5 4un−1 5 4 −h fn−1 7 5 −2 un −2h − h2 fn 0

This system can be solved by Gaussian elimination.

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Implementation Assembly of matrix in pseudocode: Matrix( real ) A(n,n) Vector(real ) b(n ), u(n) // Assemble matrix and left−hand side for i =1:n if i ==1 // Left boundary A (1,1) = 1; b(1) = 0 else if i ==n // Right boundary A(i , i −1) = 2; A(i , i ) = −2; b( i ) = −2∗h − h∗h∗f(x(i )); else // Interior A(i , i −1) = 1; A(i , i ) = −2; A(i , i +1) = 1; b( i ) = −h∗h∗f(x(i )); end

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In C++ int main(int argc, char ∗∗argv) { : double ∗b, ∗u; double ∗∗A; : b = new double[n]; u = new double[n]; A = new double∗[n]; A [0] = new double[n∗n]; for ( i =1; i