6. Introduction Descriptions Statements

7. A string can be defined as a rigid body whose dimensions are small when compared with its length.

8. The string in our model will be stretched between two fixed pegs that are separated by a distance of length L. L Peg 1Peg 2

9. Tension (T 0 ) will be the force of the two pegs pulling on the string. For our model, we will assume near constant tension.

10. Density can be defined as the ratio of the mass of an object to its volume. For a string, density is mass per unit length. In our model, we will also assume near constant density for the string.

11. Derivation of the Wave Equation Basic modeling assumptions Review of Newton’s Law Calculus prereqs Equational Derivation

12. Transverse : Vibration perpendicular to the X-axis Model Assumptions L

13. Density is assumed constant  = 1 Initial Deformation is small

14. Model Assumptions Tension T is constant and tangent to the curve of the string T = 1 L

15. Newton’s Second Law F = ma

16. Calculus Prerequisites T = [] 1  1 + dy dx () ² () i + ( dy dx  1 + dy dx () ² ) j |T| y = f(x) Angle of Inclination  T y x

17. Equational Derivation x x+  x ss u x

18. Equational Derivation x x+  x ss u x Vertical Forces Horizontal Forces ss

19. T [ uu xx (x +  x, t)  1 + uu xx (x, t) () 2 ] - uu xx  1 + uu xx (x +  x, t) () 2 Vertical Forces: Get smaller and go to zero

20. Vertical Forces: T [ uu xx (x +  x, t)  ] - uu xx (x, t) 1  1 =  (  s)  ²u t²t² (x,t)

21. Vertical Forces: uu xx (x +  x, t) - uu xx (x, t) =  s  ²u t²t² (x,t) Mass Acceleration Net Force

22. uu xx (x +  x, t) - uu xx (x, t) xx Vertical Forces =  s  ²u t²t² (x,t) xx

23. ²u²u x²x² =  ²u t²t² (x,t) Vertical Forces One dimensional wave equation

24. Solution to the Wave Equation Partial Differential equations Multivariate Chain rule D’Alemberts Solution Infinite String Case Finite String Case Connections with Fourier Analysis

25. 2nd Order Homogeneous Partial Differential Equation ABCFED 0  ²y  x² xxxx +++++  t² tt yy yy xx y =

26. Classification of P.D.E. types  = B² - AC Hyperbolic  > 0 Parabolic  = 0 Elliptic  < 0

27. Boundary Value Problem Finite String Problem Fixed Ends with 0 < x < l [u] = 0 and [u] = 0 X = 0X = l

28. Cauchy Problem Infinite String Problem Initial Conditions [u] =  (x) and [du/dt]  (x) t=0 0 l =

29. Multi- Variable Chain Rule example f(x,y) = xy² + x² g(x,y) = y sin(x) h(x) = e F(x,y) = f(g(x,y),h(x)) x

30. Let u = g(x,y) v = h(x) So F = f(u,v) = uv² + u²

31. F u v x y ff uu ff vv gg gg xx xx hh yy Variable Dependency Diagram

32. ff uu FF xx = gg xx + ff vv uu xx =((v² + 2u)(y cos(x)) + (2uv)e ) x = (e )² + 2y sin(x) (y cos(x)) + 2(y sin(x) e e ) x xx

33. Multi-Variable Chain Rule for Second Derivative Very Similar to that of the first derivative

34. Our Partial Differential Equation ξ = x – t η = x + t So ξ + η = 2x x = (ξ + η)/2 And - ξ + η = 2t t = (η – ξ)/2

35. Using Multi-Variable Chain Rule uu ξξ  u tt ηη =+ ²u²u t²t²  tt = ξξ ηη -  u + [ ]

36. Using Algebra to reduce the equation ²u²u t²t²  ξ² 2  ²u ηξηξ ²u ²u - =  ²u  η² +

37.  ηη uu uu ξξ xx =+  ²u ηη uu uu ξξ  x² =+ uu xx [ ]

38. Using Algebra to simplify  ²u  η²  ²u  ξ²  x² =+ ηξηξ 2  ²u +

39. Substitute What we just found  ²u  t²  ²u  x² =

40.  η²  ²u  ξ² + ηξηξ 2  ²u  η²  ²u  ξ² + ηξηξ 2  ²u + = ηξηξ = ηξηξ ηξηξ 4  ²u = 0

41. We finally come up with ηξηξ  ²u = 0 When u = u(ξ, η) η = x + t ξ = x - t

42. Intermission

43. Can I get a Beer? Sorry, we don’t sell to strings here A String walks into a bar

44. Can I get a beer? Again we can’t serve you because you are a string I’m afraid not!

45. And Now Back to the Models Presentation

46. D’Alemberts Solution

49. Then unsubstituting Relabeling in more conventional notation gives Integrating with respect to Ada Next integrating with respect to Xi

50. Infinite String Solution Which is a cauchy problem Reasonable initial conditions

51. So we have And we have to solve for f and g

52. When we solved for f and g, we found

53. Then when we plug those into U

54. Finite Solution Boundary Value Problem Boundary conditions

56. This is a periodic function with period 2L. If the boundary conditions hold this above is true. This equation relates to the sin and cos functions.

57. NEED A CONCLUSION

58. A Special thanks To Dr. Steve Deckelman for all your help and support S.L. Sobolev “Partial Differential Equations of Mathematical Physics Scott A. Banaszynski for use of his wonderful guitar

59. ?

60. Thank you for coming, enjoy the rest of the presentations.