1. Ordinary differential equations

3. Finite element method Divide macroscopic object into small pieces
Can investigate real-size objects in real world time regimes
Disregarding of atomic nature of matter and long-range interactions

4. Case study
Reactor with reaction

5. More complicated problem Accumulation = inputs - outputs

6. More complicated problem Accumulation = inputs - outputs
𝑄 55 =2 𝑚 3 𝑚𝑖𝑛
inputs
outputs
𝑄 15 =3 𝑚 3 𝑚𝑖𝑛
outputs
𝑐 5
𝑄 54 =2 𝑚 3 𝑚𝑖𝑛
outputs
outputs
𝑄 25 =1 𝑚 3 𝑚𝑖𝑛
𝑄 44 =11 𝑚 3 𝑚𝑖𝑛
inputs
outputs
inputs
outputs
inputs
outputs
𝑄 01 =5 𝑚 3 𝑚𝑖𝑛
𝑐 01 =20 𝑚𝑔/ 𝑚 3
𝑄 12 =3 𝑚 3 𝑚𝑖𝑛
𝑐 1
𝑐 2
𝑐 4
outputs
𝑄 23 =1 𝑚 3 𝑚𝑖𝑛
outputs
𝑄 31 =1 𝑚 3 𝑚𝑖𝑛
inputs
outputs
inputs
𝑄 34 =8 𝑚 3 𝑚𝑖𝑛
𝑄 03 =8 𝑚 3 𝑚𝑖𝑛
𝑐 03 =20 𝑚𝑔/ 𝑚 3
𝑐 3

7. How to solve that problem? Accumulation = inputs - outputs So?
𝑄 15 =3 𝑚 3 𝑚𝑖𝑛
outputs
𝑉 1 𝑑 𝑐 1 𝑑𝑡 = 𝑐 01 𝑄 01 + 𝑐 3 𝑄 31 − 𝑐 1 𝑄 12 − 𝑐 1 𝑄 15
inputs
outputs
𝑄 01 =5 𝑚 3 𝑚𝑖𝑛
𝑐 01 =20 𝑚𝑔/ 𝑚 3
𝑄 12 =3 𝑚 3 𝑚𝑖𝑛
Q01c01 = 50 mg/min, Q03c03 = 160 mg/min,
V1 =50 m3, V2 = 20 m3, V3 = 40 m3, V4 = 80 m3, and V5 = 100 m3
𝑐 1
outputs
𝑄 31 =1 𝑚 3 𝑚𝑖𝑛

8. How to solve that problem? Accumulation = inputs - outputs So?
𝑄 15 =3 𝑚 3 𝑚𝑖𝑛
outputs
𝑉 1 𝑑 𝑐 1 𝑑𝑡 = 𝑐 01 𝑄 01 + 𝑐 3 𝑄 31 − 𝑐 1 𝑄 12 − 𝑐 1 𝑄 15
inputs
outputs
𝑄 01 =5 𝑚 3 𝑚𝑖𝑛
𝑐 01 =20 𝑚𝑔/ 𝑚 3
𝑄 12 =3 𝑚 3 𝑚𝑖𝑛
Q01c01 = 50 mg/min, Q03c03 = 160 mg/min,
V1 =50 m3, V2 = 20 m3, V3 = 40 m3, V4 = 80 m3, and V5 = 100 m3
𝑐 1
𝑉 1 𝑑 𝑐 1 𝑑𝑡 =−0.12 𝑐 𝑐 3 +1
outputs
𝑄 31 =1 𝑚 3 𝑚𝑖𝑛

9. How to solve that problem? Accumulation = inputs - outputs So?
𝑄 15 =3 𝑚 3 𝑚𝑖𝑛
outputs
𝑉 1 𝑑 𝑐 1 𝑑𝑡 = 𝑐 01 𝑄 01 + 𝑐 3 𝑄 31 − 𝑐 1 𝑄 12 − 𝑐 1 𝑄 15
inputs
outputs
𝑄 01 =5 𝑚 3 𝑚𝑖𝑛
𝑐 01 =20 𝑚𝑔/ 𝑚 3
𝑄 12 =3 𝑚 3 𝑚𝑖𝑛
𝑄 01 𝑐 01 =50 𝑚𝑔 𝑚𝑖𝑛 , 𝑄 03 𝑐 03 =160 𝑚𝑔 𝑚𝑖𝑛 ,
𝑉 1 =50 𝑚 3 , 𝑉 2 =20 𝑚 3 , 𝑉 3 =40 𝑚 3 , 𝑉 4 =80 𝑚 3 , 𝑉 5 =100 𝑚 3
𝑐 1
𝑑 𝑐 1 𝑑𝑡 =−0.12 𝑐 𝑐 3 +1
outputs
𝑄 31 =1 𝑚 3 𝑚𝑖𝑛

10. Accumulation = inputs - outputs
How to solve that problem?
Accumulation = inputs - outputs
For other reactors
𝑑 𝑐 1 𝑑𝑡 =−0.12 𝑐 𝑐 3 +1
Initially the concentration in all reactors
will be zero. What will be the evolution
of concentrations in time?
𝑑 𝑐 2 𝑑𝑡 =0.15 𝑐 1 −0.15 𝑐 2
𝑑 𝑐 3 𝑑𝑡 =0.025 𝑐 2 −0.225 𝑐 3 +4
𝑑 𝑐 4 𝑑𝑡 =0.1 𝑐 3 − 𝑐 𝑐 5
How to solve this problem?
𝑑 𝑐 5 𝑑𝑡 =0.03 𝑐 𝑐 2 −0.04 𝑐 5

11. A really simple example Converstion of cyclopropane to propane
𝑘= −4 1 𝑠 , 𝑇= 500 𝑜 𝐶
1st order
𝒅𝒄 𝒅𝒕 =−𝒌𝒄 𝒕
If the initial concentration c(0) of cyclopropane is 0.05 M, what is the concentration after 30 min?

12. How to calculate c(t)? Differential equation
𝒅𝒄 𝒅𝒕 =−𝒌𝒄 𝒕

13. How to calculate c(t)? Differential equation
𝒅𝒄 𝒅𝒕 =−𝒌𝒄 𝒕
𝟎 𝒕 𝟏 𝒄 𝒅𝒄 𝒅𝒕 𝒅𝒕 =−𝒌 𝟎 𝒕 𝒅𝒕
𝒄(𝟎) 𝒄 𝒕 𝒅𝒄 𝒄 =−𝒌 𝟎 𝒕 𝒅𝒕

14. How to calculate c(t)? Differential equation
𝒅𝒄 𝒅𝒕 =−𝒌𝒄 𝒕
𝒄(𝟎) 𝒄 𝒕 𝒅𝒄 𝒄 =−𝒌 𝟎 𝒕 𝒅𝒕
𝐥𝐧 𝒄 𝒕 − 𝐥𝐧 𝒄 𝟎 =−𝒌𝒕

15. How to calculate c(t)? Differential equation
𝒅𝒄 𝒅𝒕 =−𝒌𝒄 𝒕
𝒄(𝟎) 𝒄 𝒕 𝒅𝒄 𝒄 =−𝒌 𝟎 𝒕 𝒅𝒕
𝐥𝐧 𝒄 𝒕 − 𝐥𝐧 𝒄 𝟎 =−𝒌𝒕
𝒄 𝒕 𝒄 𝟎 = 𝒆 −𝒌𝒕

16. How to calculate c(t)? Differential equation
𝒅𝒄 𝒅𝒕 =−𝒌𝒄 𝒕
𝒄(𝟎) 𝒄 𝒕 𝒅𝒄 𝒄 =−𝒌 𝟎 𝒕 𝒅𝒕
𝐥𝐧 𝒄 𝒕 − 𝐥𝐧 𝒄 𝟎 =−𝒌𝒕
𝒄 𝒕 𝒄 𝟎 = 𝒆 −𝒌𝒕
𝒄 𝒕 =𝒄 𝟎 ∗ 𝒆 −𝒌𝒕

17. Different approach Euler’s scheme

18. Instead of solving function analytically  step by step procedure
𝒇 𝒕+𝚫𝒕 =𝒇 𝒕 + 𝒇 ′ 𝒕 ∗𝚫𝒕
Analytical solution f t

19. Instead of solving function analytically  step by step procedure
𝒇 𝒕+𝚫𝒕 =𝒇 𝒕 + 𝒇 ′ 𝒕 ∗𝚫𝒕
Analytical solution
Point-by-point solution f f t t

20. How does it works  we start from f(0)
1. Choose step 𝚫𝒕=𝟏
2. Compute next point using „chain rule”
𝒇 𝒕+𝚫𝒕 =𝒇 𝒕 + 𝒇 ′ 𝒕 ∗𝚫𝒕
Example
𝒇 𝒕 =𝟐∗𝒕  𝒇 ′ 𝒕 =𝟐

21. How does it works  we start from f(0)
𝒇 𝟎+𝚫𝒕=𝟏 =𝟎+𝟐∗𝟏 = 2 f
f(1) t 1

22. How does it works  we start from f(0)
𝒇 𝟏+𝟏 =𝟐+𝟐∗𝟏 = 4 f
f(2)
f(1) t 2 2

23. And so on… point by point we reproduce solution
f(t) = 2t f
f(2)
f(1) t 2 2

24. Problems? - Accuracy
Example – derivative varies (change of the function’s shape)
Wrong value! f t

25. Solution? – More points, shorter interval
Example – derivative varies (change of the function’s shape) f
Problem? Computing efficiency! t

26. Different approach – finite difference method
How to calculate c(t)?
Different approach – finite difference method
𝚫𝒄 𝚫𝒕 =−𝒄 𝒕

27. Different approach – finite difference method
How to calculate c(t)?
Different approach – finite difference method
𝚫𝒄 𝚫𝒕 =−𝒌𝒄 𝒕
𝒄 𝒕+𝚫𝐭 −𝐜 𝐭 𝚫𝒕 =−𝒌𝒄 𝒕

28. Different approach – finite difference method
How to calculate c(t)?
Different approach – finite difference method
𝚫𝒄 𝚫𝒕 =−𝒌𝒄 𝒕
𝒄 𝒕+𝚫𝐭 −𝐜 𝐭 𝚫𝒕 =−𝒌𝒄 𝒕
𝒄 𝒕+𝚫𝐭 −𝐜 𝐭 =−𝚫𝐭𝒌𝒄 𝒕

29. Different approach – finite difference method, explicit scheme
How to calculate c(t)?
Different approach – finite difference method, explicit scheme
𝚫𝒄 𝚫𝒕 =−𝒄𝒌 𝒕
𝒄 𝒕+𝚫𝐭 −𝐜 𝐭 𝚫𝒕 =−𝒌𝒄 𝒕
𝒄 𝒕+𝚫𝐭 −𝐜 𝐭 =−𝚫𝐭𝒌𝒄 𝒕
𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝐤𝒄 𝒕 =𝒄 𝒕 𝟏−𝐤𝚫𝐭

30. 𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝐤𝚫𝐭
𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝐤𝚫𝐭
𝐤=𝟏
𝒄 𝒏 = 𝒄 𝒏−𝟏 𝟏−𝚫𝐭 c t

31. 𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝚫𝐭 𝐤=𝟏 𝒄 𝒏 = 𝒄 𝒏−𝟏 𝟏−𝚫𝐭 𝚫𝐭=𝟎.𝟏 𝒄 𝟎 = 𝒄 𝟎 =𝟏
𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝚫𝐭
𝐤=𝟏
𝒄 𝒏 = 𝒄 𝒏−𝟏 𝟏−𝚫𝐭
𝚫𝐭=𝟎.𝟏
𝒄 𝟎 = 𝒄 𝟎 =𝟏 c 1 t

32. 𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝚫𝐭 𝐤=𝟏 𝒄 𝒏 = 𝒄 𝒏−𝟏 𝟏−𝚫𝐭 𝚫𝐭=𝟎.𝟏 𝒄 𝟎 = 𝒄 𝟎 =𝟏
𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝚫𝐭
𝐤=𝟏
𝒄 𝒏 = 𝒄 𝒏−𝟏 𝟏−𝚫𝐭
𝚫𝐭=𝟎.𝟏
𝒄 𝟎 = 𝒄 𝟎 =𝟏
𝒄 𝟏 = 𝒄 𝟎 𝟏−𝚫𝐭 =𝟏∗ 𝟏−𝟎.𝟏 =𝟎.𝟗 c 1 t
0.1

33. 𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝚫𝐭 𝐤=𝟏 𝒄 𝒏 = 𝒄 𝒏−𝟏 𝟏−𝚫𝐭 𝚫𝐭=𝟎.𝟏 𝒄 𝟎 = 𝒄 𝟎 =𝟏
𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝚫𝐭
𝐤=𝟏
𝒄 𝒏 = 𝒄 𝒏−𝟏 𝟏−𝚫𝐭
𝚫𝐭=𝟎.𝟏
𝒄 𝟎 = 𝒄 𝟎 =𝟏
𝒄 𝟏 = 𝒄 𝟎 𝟏−𝚫𝐭 =𝟏∗ 𝟏−𝟎.𝟏 =𝟎.𝟗
𝒄 𝟐 = 𝒄 𝟏 𝟏−𝚫𝐭 =𝟎.𝟗∗ 𝟏−𝟎.𝟏 =𝟎.𝟖𝟏 c 1
0.1
0.2 t

34. 𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝚫𝐭 𝐤=𝟏 𝒄 𝒏 = 𝒄 𝒏−𝟏 𝟏−𝚫𝐭 𝚫𝐭=𝟎.𝟏 𝒄 𝟎 = 𝒄 𝟎 =𝟏
𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝚫𝐭
𝐤=𝟏
𝒄 𝒏 = 𝒄 𝒏−𝟏 𝟏−𝚫𝐭
𝚫𝐭=𝟎.𝟏
𝒄 𝟎 = 𝒄 𝟎 =𝟏
𝒄 𝟏 = 𝒄 𝟎 𝟏−𝚫𝐭 =𝟏∗ 𝟏−𝟎.𝟏 =𝟎.𝟗
𝒄 𝟐 = 𝒄 𝟏 𝟏−𝚫𝐭 =𝟎.𝟗∗ 𝟏−𝟎.𝟏 =𝟎.𝟖𝟏 c 1
𝒄 𝟑 = 𝒄 𝟐 𝟏−𝚫𝐭 =𝟎.𝟖𝟏∗ 𝟏−𝟎.𝟏 =𝟎.𝟕𝟐𝟗
0.1
0.2
0.3 t

35. 𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝚫𝐭
𝐤=𝟏
𝒄 𝒏 = 𝒄 𝒏−𝟏 𝟏−𝚫𝐭
𝚫𝐭=𝟎.𝟏
𝒄 𝟎 = 𝒄 𝟎 =𝟏
𝒄 𝟏 = 𝒄 𝟎 𝟏−𝚫𝐭 =𝟏∗ 𝟏−𝟎.𝟏 =𝟎.𝟗
𝒄 𝟐 = 𝒄 𝟏 𝟏−𝚫𝐭 =𝟎.𝟗∗ 𝟏−𝟎.𝟏 =𝟎.𝟖𝟏
𝒄 𝟑 = 𝒄 𝟐 𝟏−𝚫𝐭 =𝟎.𝟖𝟏∗ 𝟏−𝟎.𝟏 =𝟎.𝟕𝟐𝟗

36. Problems?
𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 −𝚫𝐭𝒄 𝒕 =𝒄 𝒕 𝟏−𝐤𝚫𝐭
𝐤=𝟏
Say 𝚫𝐭=𝟏 𝐭𝐡𝐞𝐧
𝒄 𝟏 =𝒄 𝟎 𝟏−𝟏 =𝟎
𝒄 𝟐 =𝒄 𝟏 𝟏−𝟏 =𝟎
Say 𝚫𝐭>𝟏 𝚫𝐭=𝟐 𝐭𝐡𝐞𝐧
𝒄 𝟏 =𝒄 𝟎 𝟏−𝟐 =−𝒄 𝟎 <𝟎 negative concentration. Unphysical!!!
𝒄 𝒕+𝚫𝐭 =𝒄 𝒕 𝟏−𝐤𝚫𝐭 →𝟏−𝐤𝚫𝐭>𝟎→𝚫𝐭<𝟏/𝐤

37. Solution? Implicit scheme
Δ𝑡→0 f
𝑑𝑓 𝑑𝑡 𝑡 ∗ ≈ Δ𝑓 Δ𝑡 𝑡 ∗ 𝑜𝑟 Δ𝑓 Δ𝑡 𝑡 ∗ +Δ𝑡 or anywhere inbetween
𝑓 𝑡 ∗ Δ𝑓
Slope in point 𝑡 ∗
𝑓 𝑡 ∗ +Δ𝑡
𝑡𝑔𝛽= lim Δ𝑡→0 Δ𝑓 Δ𝑡 = 𝑑𝑓 𝑑𝑡 Δ𝑡 𝛼 𝛽
𝑡 ∗
𝑡 ∗ +Δ𝑡 t
𝑡𝑔𝛼= Δ𝑓 Δ𝑡
Average slope on Δ𝑡 interval

38. Solution? Implicit scheme
Before…
Now…
𝚫𝒄 𝚫𝒕 =−𝒌𝒄 𝒕
𝚫𝒄 𝚫𝒕 =−𝒌𝒄 𝒕+𝚫𝐭
𝒄 𝒕+𝚫𝐭 −𝐜 𝐭 𝚫𝒕 =−𝒌𝒄 𝒕+𝚫𝐭
𝒄 𝒕+𝚫𝐭 −𝐜 𝐭 𝚫𝒕 =−𝒌𝒄 𝒕
𝒄 𝒕+𝚫𝐭 =𝐜 𝐭 −𝚫𝐭𝒌𝒄 𝒕+𝚫𝐭
𝒄 𝒕+𝚫𝐭 =𝐜 𝐭 −𝚫𝐭𝒌𝒄 𝒕

39. Solution? Implicit scheme
𝒄 𝒕+𝚫𝐭 =𝐜 𝐭 −𝚫𝐭𝐤𝒄 𝒕+𝚫𝐭
𝒄 𝒕+𝚫𝐭 +𝚫𝐭𝐤𝒄 𝒕+𝚫𝐭 =𝐜 𝐭
𝒄 𝒕+𝚫𝐭 𝟏+𝐤𝚫𝐭 =𝐜 𝐭
𝒄 𝒕+𝚫𝐭 = 𝐜 𝐭 𝟏+𝐤𝚫𝐭
Always positive

40. Solutions, comparison 𝒅𝒄 𝒅𝒕 =−𝒄 𝒕 𝐤=𝟏 𝒄 𝒕+𝚫𝐭 = 𝐜 𝐭 𝟏+𝚫𝐭
𝒄 𝒕+𝚫𝐭 = 𝐜 𝐭 𝟏+𝚫𝐭
𝒄 𝒕+𝚫𝐭 =𝐜 𝐭 −𝚫𝐭𝒄 𝒕
𝒄 𝒕 =𝒄 𝟎 𝒆 −𝒕
𝚫𝐭=𝟎.𝟏
𝚫𝐭=𝟐

41. Problems? - Accuracy
Example – derivative varies (change of the function’s shape)
Wrong value! f t

42. But what about set of ODE?
Same approach combined with linear equations
𝑑 𝑐 1 𝑑𝑡 =−0.12 𝑐 𝑐 3 +1
𝑑 𝑐 1 𝑑𝑡 𝑑 𝑐 2 𝑑𝑡 𝑑 𝑐 3 𝑑𝑡 𝑑 𝑐 4 𝑑𝑡 𝑑 𝑐 5 𝑑𝑡 =
𝑑 𝑐 2 𝑑𝑡 =0.15 𝑐 1 −0.15 𝑐 2
-0.12
0.02
0.15
-0.15
0.025
-0.225
0.1
0.1375
0.03
0.01
-0.05
𝑐 1 𝑐 2 𝑐 3 𝑐 4 𝑐
𝑑 𝑐 3 𝑑𝑡 =0.025 𝑐 2 −0.225 𝑐 3 +4
𝑑 𝑐 4 𝑑𝑡 =0.1 𝑐 3 − 𝑐 𝑐 5
𝑑 𝑐 5 𝑑𝑡 =0.03 𝑐 𝑐 2 −0.04 𝑐 5

43. But what about set of ODE?
Same approach combined with linear equations
𝑑 𝑐 1 𝑑𝑡 𝑑 𝑐 2 𝑑𝑡 𝑑 𝑐 3 𝑑𝑡 𝑑 𝑐 4 𝑑𝑡 𝑑 𝑐 5 𝑑𝑡 =
-0.12
0.02
0.15
-0.15
0.025
-0.225
0.1
0.1375
0.03
0.01
-0.05
𝑐 1 𝑐 2 𝑐 3 𝑐 4 𝑐
-0.12
0.02
0.15
-0.15
0.025
-0.225
0.1
0.1375
0.03
0.01
-0.05
𝑑 𝑑𝑡 𝑐 1 𝑐 2 𝑐 3 𝑐 4 𝑐 5 =
𝑐 1 𝑐 2 𝑐 3 𝑐 4 𝑐

44. 𝒅 𝒄 𝒅𝒕 =𝑨 𝒄 + 𝒃 But what about set of ODE?
Same approach combined with linear equations
-0.12
0.02
0.15
-0.15
0.025
-0.225
0.1
0.1375
0.03
0.01
-0.05
𝑑 𝑑𝑡 𝑐 1 𝑐 2 𝑐 3 𝑐 4 𝑐 5 =
𝑐 1 𝑐 2 𝑐 3 𝑐 4 𝑐 𝒄 𝒄 𝒃 A
𝒅 𝒄 𝒅𝒕 =𝑨 𝒄 + 𝒃

45. What next? Finite difference approach but on vectors!
𝒅 𝒄 𝒅𝒕 =𝑨 𝒄
𝚫 𝒄 𝚫𝒕 =𝑨 𝒄

46. What next? Finite difference approach but with vectors!
𝚫 𝒄 𝚫𝒕 =𝑨 𝒄
𝒄 𝒕+𝚫𝐭 − 𝐜 (𝐭) 𝚫𝒕 =𝑨 𝒄
𝒄 𝒕+𝚫𝐭 = 𝒄 𝒕 +𝚫𝐭𝑨 𝒄

47. 𝒄 𝒕+𝚫𝐭 = 𝒄 𝒕 +𝚫𝐭𝑨 𝒄 (𝒕) 𝒄 𝒕+𝚫𝐭 = 𝑰+𝚫𝐭𝑨 𝐜 𝒕 𝒄 𝒕+𝚫𝐭 =𝑩 𝐜 𝒕
What next? Finite difference approach but with vectors!
Explicit scheme 𝒄 = 𝒄 (𝒕)
𝒄 𝒕+𝚫𝐭 = 𝒄 𝒕 +𝚫𝐭𝑨 𝒄 (𝒕)
𝒄 𝒕+𝚫𝐭 = 𝑰+𝚫𝐭𝑨 𝐜 𝒕
𝒄 𝒕+𝚫𝐭 =𝑩 𝐜 𝒕
𝑩=𝑰+𝚫𝐭𝑨

48. 𝒄 𝒕+𝚫𝐭 =𝑩 𝐜 𝒕 What next? Finite difference approach but with vectors!
Explicit scheme 𝒄 = 𝒄 (𝒕)
Define c(0)
𝒄 𝒕+𝚫𝐭 =𝑩 𝐜 𝒕
𝑩=𝑰+𝚫𝐭𝑨
Compute 𝑩=𝑰+𝚫𝐭𝐀
Problem?
Unstable!
Compute c in the next time frame
𝒄 𝒏𝚫𝐭 =𝐁𝐜( 𝐧−𝟏 𝚫𝐭))
𝒏+𝟏

49. 𝒄 𝒕+𝚫𝐭 = 𝒄 𝒕 +𝚫𝐭𝑨 𝒄 (𝒕+𝚫𝐭) 𝒄 𝒕+𝚫𝐭 −𝚫𝐭𝑨 𝒄 (𝒕+𝚫𝐭)= 𝒄 𝒕
What next? Finite difference approach but on vectors!
Implicit scheme 𝒄 = 𝒄 (𝒕+𝚫𝐭)
𝒄 𝒕+𝚫𝐭 = 𝒄 𝒕 +𝚫𝐭𝑨 𝒄 (𝒕+𝚫𝐭)
𝒄 𝒕+𝚫𝐭 −𝚫𝐭𝑨 𝒄 (𝒕+𝚫𝐭)= 𝒄 𝒕
𝑰−𝚫𝐭𝑨 𝒄 𝒕+𝚫𝐭 = 𝒄 𝒕

50. 𝑰−𝚫𝐭𝑨 𝒄 𝒕+𝚫𝐭 = 𝒄 𝒕 𝑩 𝒄 𝒕+𝚫𝐭 = 𝒄 𝒕 𝒄 𝒕+𝚫𝐭 = 𝑩 −𝟏 𝒄 𝒕 𝑩=𝑰−𝚫𝐭𝑨
What next? Finite difference approach but on vectors!
Implicit scheme 𝒄 = 𝒄 (𝒕+𝚫𝐭)
𝑰−𝚫𝐭𝑨 𝒄 𝒕+𝚫𝐭 = 𝒄 𝒕
𝑩 𝒄 𝒕+𝚫𝐭 = 𝒄 𝒕
𝒄 𝒕+𝚫𝐭 = 𝑩 −𝟏 𝒄 𝒕
𝑩=𝑰−𝚫𝐭𝑨

51. Implicit scheme 𝒄 = 𝒄 (𝒕) Compute c in the next time frame
Define c(0)
𝒄 𝒕+𝚫𝐭 = 𝑩 −𝟏 𝐜 𝒕
𝑩=𝑰−𝚫𝐭𝑨
Compute 𝑩=𝑰+𝚫𝐭𝐀
Problem?
Computing 𝑩 −𝟏
Compute 𝑩 −𝟏
Compute c in the next time frame
𝒄 𝒏𝚫𝐭 = 𝐁 −𝟏 𝐜( 𝐧−𝟏 𝚫𝐭))
𝒏+𝟏

52. Working example implicite scheme
𝑨 k 𝟏 =𝟏 𝑩 k 𝟐 =𝟏 𝑪
𝒅𝑨 𝒅𝒕 =−𝑨
Initially A = 1, B = 0, C = 0
𝒅𝑩 𝒅𝒕 =𝑨−𝑩
𝑑 𝑑𝑡 𝐴 𝐵 𝐶 = − − 𝐴 𝐵 𝐶
𝒅𝑪 𝒅𝒕 =𝑩
𝑐 0 =

53. Working example 1. Initiate, calculate B
𝑨 k 𝟏 =𝟏 𝑩 k 𝟐 =𝟏 𝑪
𝑑 𝑑𝑡 𝐴 𝐵 𝐶 = − − 𝐴 𝐵 𝐶
𝑐 0 =
𝑩=𝑰−𝚫𝐭𝑨
𝜟𝒕=1
𝑐 0 =
𝐵= −1 − − = − −1 1

54. Working example 2. Calculate 𝑩 −𝟏 𝐵= 2 0 0 −1 2 0 0 −1 1
𝐵= − −1 1
𝐵 −1 =

55. Working example 3. Iterate n=1
𝑐 𝑛Δ𝑡 = 𝐵 −1 𝑐 𝑛−1 Δ𝑡
𝑐 1 = 𝐵 −1 𝑐 0 = =

56. Working example 3. Iterate n=2
𝑐 𝑛Δ𝑡 = 𝐵 −1 𝑐 𝑛−1 Δ𝑡
𝑐 2 = 𝐵 −1 𝑐 1 = =

57. Working example 3. Iterate n=3
𝑐 𝑛Δ𝑡 = 𝐵 −1 𝑐 𝑛−1 Δ𝑡
𝑐 3 = 𝐵 −1 𝑐 2 = =

58. Working example 3. Iterate n=4
𝑐 𝑛Δ𝑡 = 𝐵 −1 𝑐 𝑛−1 Δ𝑡
𝑐 4 = 𝐵 −1 𝑐 3 = =

59. Partial ODE
Diffusion equation 𝜕𝑐 𝜕𝑡 =−𝐷 𝜕 2 𝑐 𝜕 𝑥 2

60. „Building blocks” for differentials
𝜕𝑐 𝜕𝑥 = 𝑐 𝑖+1 − 𝑐 𝑖−1 Δ𝑥 +𝑂 Δ 𝑥 2 𝜕𝑐 𝜕𝑥 = 𝑐 𝑖 − 𝑐 𝑖−1 Δ𝑥 +𝑂 Δ𝑥 𝜕 2 𝑐 𝜕 𝑥 2 = 𝑐 𝑖+1 −2 𝑐 𝑖 + 𝑐 𝑖−1 Δ 𝑥 2 +𝑂 Δ 𝑥 2

61. 𝜕𝑐 𝜕𝑡 =−𝐷 𝜕 2 𝑐 𝜕 𝑥 2
𝜕𝑐 𝜕𝑡 = 𝑐 𝑚,𝑖+1 − 𝑐 𝑚,𝑖 Δ𝑡
𝜕 2 𝑐 𝜕 𝑥 2 = 𝑐 𝑚+1,𝑖 −2 𝑐 𝑚,𝑖 + 𝑐 𝑚−1,𝑖 Δ 𝑥 2

62. Example – 1 D Diffusion equation (drug release)
Initial concentration profile
𝜙(𝑥)
Instantaneous
outtake
Instantaneous
outtake
𝑐 𝑡 =−𝐷 𝑐 𝑥𝑥 𝑐 0,𝑡 =0 𝑐 1,𝑡 =0 𝑐 𝑥,0 =𝜙 𝑥
Drug 1
Question: How the profile changes in time?

63. Example – 1 D Diffusion equation (drug release)
Initial concentration profile
𝜙(𝑥)
Instantaneous
outtake
Instantaneous
outtake
𝑐 𝑡 =−𝐷 𝑐 𝑥𝑥 𝑐 0,𝑡 =0 𝑐 1,𝑡 =0 𝑐 𝑥,0 =𝜙 𝑥
Drug 1
Question: How the profile changes in time?

64. Example – 1 D Diffusion equation (drug release)
Initial concentration profile
𝜙(𝑥)
Instantaneous
outtake
Instantaneous
outtake
𝑐 𝑡 =−𝐷 𝑐 𝑥𝑥 𝑐 0,𝑡 =0 𝑐 1,𝑡 =0 𝑐 𝑥,0 =𝜙 𝑥
Drug 1
Question: How the profile changes in time?

65. Example – 1 D Diffusion equation (drug release)
Initial concentration profile
𝜙(𝑥)
Instantaneous
outtake
Instantaneous
outtake
𝑐 𝑡 =−𝐷 𝑐 𝑥𝑥 𝑐 0,𝑡 =0 𝑐 1,𝑡 =0 𝑐 𝑥,0 =𝜙 𝑥
Drug 1
Question: How the profile changes in time?

66. Example – 1 D Diffusion equation (drug release)
Initial concentration profile
𝜙(𝑥)
Instantaneous
outtake
Instantaneous
outtake
𝑐 𝑡 =−𝐷 𝑐 𝑥𝑥 𝑐 0,𝑡 =0 𝑐 1,𝑡 =0 𝑐 𝑥,0 =𝜙 𝑥
Drug 1
Question: How the profile changes in time?

67. Example – 1 D Diffusion equation (drug release)
Initial concentration profile
𝜙(𝑥)
𝑐 𝑡 =−𝐷 𝑐 𝑥𝑥 𝑐 0,𝑡 =0 𝑐 1,𝑡 =0 𝑐 𝑥,0 =𝜙 𝑥
𝑐 𝑛,𝑖
concentration in position 𝑛Δ𝑥
and time 𝑖Δ𝑡 𝑇 Δ𝑡 1 Δ𝑥 𝐿
𝑥 𝑖 =𝑛Δ𝑥, Δ𝑥= 𝐿 𝑀
𝑡 𝑛 =𝑖Δ𝑡, Δ𝑡= 𝑇 𝑁

68. 𝑐 𝑡 =−𝐷 𝑐 𝑥𝑥 𝑐 0,𝑡 =0 → 𝑐 0,𝑗 =0 𝑐 1,𝑡 =0→ 𝑐 𝑀,𝑗 =0 𝑐 𝑥,0 =𝜙 𝑥 → 𝑢 𝑖,0 =𝜙 𝑥

69. 𝑥 𝑖 =𝑛Δ𝑥, Δ𝑥= 𝐿 𝑀 𝑡 𝑛 =𝑖Δ𝑡, Δ𝑡= 𝑇 𝑁

70. 𝑐 𝑚,𝑖+1 − 𝑐 𝑚,𝑖 Δ𝑡 =−𝐷 𝑐 𝑚+1,𝑖 −2 𝑐 𝑚,𝑖 + 𝑐 𝑚−1,𝑖 Δ 𝑥 2 𝑐 𝑚,𝑖+1 = 𝑐 𝑚,𝑖 − 𝐷Δ𝑡 Δ 𝑥 2 𝑐 𝑚+1,𝑖 −2 𝑐 𝑚,𝑖 + 𝑐 𝑚−1,𝑖 𝑐 𝑚,𝑖+1 = 𝑐 𝑚,𝑖 −𝑟 𝑐 𝑚+1,𝑖 −2 𝑐 𝑚,𝑖 + 𝑐 𝑚−1,𝑖

71. 𝑐 𝑚,𝑖+1 = 𝑐 𝑚,𝑖 −𝑟 𝑐 𝑚+1,𝑖 −2 𝑐 𝑚,𝑖 + 𝑐 𝑚−1,𝑖 𝑐 𝑚,𝑖+1 =−𝑟 𝑐 𝑚−1,𝑖 + 1+2𝑟 𝑐 𝑚,𝑖 −𝑟 𝑐 𝑚+1,𝑖

72. 𝑐 𝑡 =−𝐷 𝑐 𝑥𝑥 𝑐 0,𝑡 =0 → 𝑐 0,𝑗 =0 𝑐 1,𝑡 =0→ 𝑐 𝑀,𝑗 =0 𝑐 𝑥,0 =𝜙 𝑥 → 𝑢 𝑖,0 =𝜙 𝑥 𝑐 1,𝑖+1 =−𝑟 𝑐 0,𝑖 + 1+2𝑟 𝑐 1,𝑖 −𝑟 𝑐 2,𝑖 𝑐 1,𝑖+1 = 1+2𝑟 𝑐 1,𝑖 −𝑟 𝑐 2,𝑖 𝑐 𝑀−1,𝑖+1 =−𝑟 𝑐 𝑀−2,𝑖 + 1+2𝑟 𝑐 𝑀−1,𝑖 −𝑟 𝑐 𝑀,𝑖 𝑐 𝑀−1,𝑖+1 =−𝑟 𝑐 𝑀−2,𝑖 + 1+2𝑟 𝑐 𝑀−1,𝑖

73. 𝑐 1,𝑖+1 = 1+2𝑟 𝑐 1,𝑖 −𝑟 𝑐 2,𝑖 𝑐 𝑚,𝑖+1 =−𝑟 𝑐 𝑚−1,𝑖 + 1+2𝑟 𝑐 𝑚,𝑖 −𝑟 𝑐 𝑚+1,𝑖 𝑐 𝑀−1,𝑖+1 =−𝑟 𝑐 𝑀−2,𝑖 + 1+2𝑟 𝑐 𝑀−1,𝑖 𝑐 1,𝑖+1 𝑐 2,𝑖+1 ⋮ 𝑐 𝑀−1,𝑖+1 = 1+2𝑟 −𝑟 0 ⋯ 0 −𝑟 1+2𝑟 −𝑟 ⋯ 0 0 ⋮ 0 ⋱ 0 ⋱ ⋯ ⋱ −𝑟 0 ⋮ 1+2𝑟 𝑐 1,𝑖 𝑐 2,𝑖 ⋮ 𝑐 𝑀−1,𝑖

74. 𝑐 1,𝑖+1 𝑐 2,𝑖+1 ⋮ 𝑐 𝑀−1,𝑖+1 = 1+2𝑟 −𝑟 0 ⋯ 0 −𝑟 1+2𝑟 −𝑟 ⋯ 0 0 ⋮ 0 ⋱ 0 ⋱ ⋯ ⋱ −𝑟 0 ⋮ 1+2𝑟 𝑐 1,𝑖 𝑐 2,𝑖 ⋮ 𝑐 𝑀−1,𝑖 𝑐 𝑖+1 =𝐴 𝑐 𝑖