Introduction to Environment System Modeling

Transcription

1 Introduction to Environment System Modeling (3 rd week:modeling with differential equation) Department of Environment Systems, Graduate School of Frontier Sciences, the University of Tokyo Masaatsu AICHI

2 Contents Conservation law Flux/Storage/Source and Sink Balance Typical processes in environment modeling Attenuation Diffusion/Dispersion Advection Material derivative

3 Conservation law z Typical conservation laws dz Mass conservation Momentum conservation Energy conservation dx dy Charge conservation etc x Conservation law means the balance of flux, storage, source, and sink. + +,,,, +, +,,, +,, +,, + = 0 Dividing by and taking the infinitesimal limit, = + + = 0 y

4 Balance equation + + = 0 Change in storage Flux divergence Source/Sink ( > 0 denotes sink) This is a common form though the detailed laws for each term are different The characteristics of equation changes with the concrete formulae for storage, divergence, source, and sink terms.

5 Radial symmetry in a plane + = Dividing by 2 and taking a limit, + = = 1 Polar coordinate = Dividing by 4 and taking a limit, = 2 = 1

6 Typical storage terms Stored material in a solution Concentration: = Stored thermal energy in a material Product of specific heat density temperature: = Stored charge in a condenser Product of electric capacity and voltage: = Stored fluid mass of law compressibility by pressure change Pressure divided by bulk modulus: = Stored fluid mass due to the elastic deformation caused by the fluid pressure change Product of storage coefficient and pressure : = Typically the storage term is proportional to some potential. There are nonlinear models for advanced problems

7 Typical source/sink terms Artificial addition or removal = Disappearance with a constant probability(e.g., radioactive decay) Proportional to the amount: = Production with a constant probability(e.g., population increase) Proportional to the amount: = Production or loss through chemical reactions

8 Growth and attenuation growth: = Initial condition: 0 = The solution is = attenuation: = Initial condition: 0 = The solution is = Exponential increase or decrease Linear in semi-logarithmic chart

9 Typical diffusion flux Fick s first law Diffusion flux is proportional to the concentration gradient: = Fourier s law Thermal conduction is proportional to the temperature gradient: = Ohm s law Electric current is proportional to the electric potential gradient: = Darcy s law Fluid flux governed by viscos force is proportional to the fluid potential: = h Flux is proportional to some potential gradient

10 Typical diffusion equation Diffusion equation(fick s second law): = Thermal conduction equation: = Electric current equation: = Groundwater flow equation: = h

11 Example of the exact solution of a diffusion equation Boundary conc. 1.0 Decartes 1D = x Initial conc. 0.2 Initial condition, 0 = Boundary condition, = 0, = C = Diffusion coeff.: D= [m 2 /s] Time: t=86400 [s] = 0.2 = x [m]

12 Example of the exact solution of a diffusion equation Radial 1D h = 1 h Initial condition h, 0 = h Boundary condition h, = h h lim 2 = Pumping rate 500m 3 /day Initial head 0 Hydraulic conductivity 10-5 m/s Specific storage m -1 h [m] x m 1 day 1 week 1 month 3 months 1 year 10 years = 4 = r [m] + log + 2 2! 3 3! + +! + = (Euler const.)

13 Typical advection flux Solute transport by a flow verocity of the solution with a concentration Product of flow velocity and concentration: = Transport of momentum Product of velocity and mass: = Heat transport by a wind Product of the flow velocity of a wind and heat content: = Product of the field velocity and the content or mass

14 Advection equation Advection equation for a solute: = = If the flow field is in a steady-sate (or incompressible) and source/sink terms are zero, mass balance of flow becomes = 0, and hence, the equation is reduced to = The exact solution of advection equation is, =, It has no concrete functional form. The concentration directly comes from the upstream backward from to.

15 Relative velocity between observer and flow field The previous explanations are based on the view of an observer at the outside of the flow field If the observer moves on the flow, the advection term is not seen. That is, = 0. Based on 全微分 = + + +, = = + is obtained. If the observer (coordinate) moves on the flow of velocity, = 0 expresses the same process (principle of relativity) Because the conclusion is same, we can choose more convenient one depending on the case. = + is called as material derivative or Lagrange derivative, and it is used for the derivation of Navier-Stokes equation in the theory of fluid dynamics, for example.

16 Superposition of multiple processes Advection + diffusion flux: = Advection-diffusion equation: = Advection + diffusion + attenuation Advection-diffusion-attenuation equation: = + It is achieved by summing up the necessary processes to be considered in the model.

17 Continuum mechanics Force balance F = 0 x 3 σ 13 σ 31 σ 33 σ 32 σ dx 23 3 σ 21 σ F = 0 σ 11 σ 12 dx 1 dx 2 x F = 0 x 1 Let Σ =, and it becomes Σ + = 0 The formula is very similar to the conservation equation.

18 Static solid mechanics Force balance direction: Σ + = 0 Σ = Constitutive equation Stress ー Strain relation: Displacement-strain relation Strain tensor ε = + Putting them into together, we obtain = 0 2 2G K G 3 ij ij kk ij

19 Fluid dynamics Momentum conservation direction: Σ + = Constitutive equation Stress ー Strain relation: velocity-strain rate relation ij p Σ = ij 2 t ij Strain rate tensor: = + Putting them into together, we obtain v t i p v v v F 2 i i i xi The mass balance equation = 0

20 Special methods to solve differential Laplace transform equation example:for =, integrating both sides of equation as, it becomes 0 = where =. If the initial condition is 0 = 0, it is reduced to = Solving for is easy and = + If the boundary conditions are = at = 0, = 0 at =, they give = + =, = + = 0, and = is the final solution

21 Inverse Laplace transform To obtain the time-domain solution from, we need an inversion. Usually it can be achieved by referring a list of known Laplace transform If the solution is not listed in the table, it is necessary to calculate Bromwich integral =. This is a complex integral in gauss plane and calculated with the residue theorem. It is not easy and the analytical integration is not always possible. Numerical methods to calculate inverse Laplace transform Stehfest, Iseger, etc.

22 Stehfest s algorithm = 1,!!!!!! is the maximum integer less than Technique for numerical evaluation The factorial and exponent calculations appeared in increases explosively. Do not calculate them directly. Calculate with logarithmic scaling and invert it by exp function. Mathematically the accuracy becomes better as N becomes greater. Numerically, however, it is not true because the cancelation error becomes greater as becomes greater. Empirically, N~5 is a first choice and some trial-and-error might give the best accuracy around it.

23 Special methods to solve differential Fourier transform equation Example:for =, integrating the both sides of equation by, it becomes = where = can be easily obtained as = + If the boundary conditions are = 0 で =, = で = 0, they give 0 = + =, = + = 0, and = is the final solution

24 The meaning of the solution in Fourier transformed domain is a complex function The magnitude of,, means the amplitude of the signal of the angular frequency The argument of,, means the phase shift Fourier transform gives the amplitude and phase-shift of periodical signal For example, cos gives the component of angular frequency of the input signal Since all the periodic signal can be expressed with the series of trigonometric function(fourier series) and the solutions of linear differential equation can be superposed, this approach works all the periodic input signal.

25 Concept of Fourier series expansion Periodic signals can be expressed with a series of trigonometric function For example, a input signal of period T is described as = cos 2 + sin 2 = 2 cos 2 = 2 sin 2 In case the time series data is available only within the period T, the same method can be applied if we assume the unobserved data is periodic copies of the observed ones.

26 Brief proof for Fourier coefficients Only the integral of the product of cos or sin of same period are non-zeros and the others are zeros 2 2 cos sin 2 2 = cos 4 2 = 2 1 cos 4 cos sin = sin + sin = 0 cos 2 cos 2 1 = 2 cos cos 2 = 0 sin 2 sin 2 = cos cos 2 = 0 Infinite series of orthogonal base functions must be able to express all functions (Linear algebra) = =

27 Complex calculation for coefficients = exp and applications Let =, then this is the Fourier transform for the period 0, Though the integral can be defined for continuous data, usually we have discrete observation of time span Then, the approximation ~ exp is used The amplitude + and the phase tan of the signal components of angular frequency = are obtained. If the input signal is Fourier transformed, it can be used for the boundary condition of Fourier transformed differential equation. Then, the solution in Fourier domain gives the amplitude and phase shift of the solution.

28 Example of Discrete Fourier transform tide level [cm, TP] Tide (Kouzushima) 23-Jul 20-Aug 17-Sep 15-Oct 12-Nov 10-Dec amplitude [cm] S2 M2 O1 K1 Tide (Kouzushima) period [day]

29 Methods for Fourier transform MS-EXCEL Data Analysis? Fourier Transform The number of data should be 2 n R (x <- read.table("sample_data.txt")) ftx <- fft(x$v1) ftx <- 2*ftx/length(x$V1) ftx[1] <- ftx[1]*0.5 period <- 1:(length(x$V1)/2) period <- length(x$v1)/period absftx <- abs(ftx[1:(length(x$v1)/2)]) phaseftx <- atan2(im(ftx[1:(length(x$v1)/2)]),re(ftx[1:(length(x$v1)/2)])) write.table(cbind(period,absftx,phaseftx), "output.txt", quote=false, append=false) (The number of data is not limited to 2 n )

30 Method for Fourier transform fortran Basically, it is possible to make a simple code for Discrete Fourier Transform based on the previous handouts. Calculation of exp is the core part +α Complex number declaration complex(8) :: fx Pure imaginary number i (0,1) Magnitude of complex number abs(fx) Argument of complex number atan2(imag(fx),real(fx))

31 Summary Most processes of interest in environment studies are expressed with the differential equation of conservation law type. Though there are the exact solutions under several simple conditions, it is difficult to find the exact solution if the condition becomes a bit complex as it is observed in the actual situations. Laplace transform is effective to find the analytical solution. However the exact inverse Laplace transform is usually difficult and a numerical inversion is often necessary. Fourier transform is effective for analyzing and modeling periodic steady-state that is often observed in natural processes. However, it is not suitable for non-periodic processes. For general applications, numerical methods like FDM or FEM work. From the next lecture, numerical methods for conservation law type will be explained.