Mathematical Physics P-334 - PowerPoint PPT Presentation

mathematical physics p 334 n.
Skip this Video
Loading SlideShow in 5 Seconds..
Mathematical Physics P-334 PowerPoint Presentation
Download Presentation
Mathematical Physics P-334

play fullscreen
1 / 39
Mathematical Physics P-334
Download Presentation
Download Presentation

Mathematical Physics P-334

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Mathematical PhysicsP-334 Dr. HatimDirar Department of Physics, College of Science Imam MohamadIbn Saud Islamic University

  2. Scopes • Differential equations lead to: • Definition • Interpretation • Forecasting of any physical system of any dimension

  3. Scopes in Physics • Mechanics • Free Fall • let us considering free fall system defined by the following equation of motion ÿ=-g • The above equation is second order DE

  4. Classification of Differential Equations • DE is defined as an equation contains derivative of an unknown function, which express the relationship we seek • the equation includes total derivative like is called ordinary differential equation (ODE) • If the function is of multivariables , the differential equation is known as partial differential equation (PDE) • Order of DE indicates the order of the highest derivative appear in the equation • Degree of the DE equations indicates the power of the derivative of highest order after the equation has been rationalized e.g. is of second order and first degree

  5. Classification of DE • Example 2 The above equation has said to be of a first order and second degree since it will contains after it is rationalized • DE has said to be linear if each term of it is such that the dependant variable or its derivative occur only once and only to the first power e.g. whereas, the following equation is nonlinear

  6. Classification of DE • If the right hand side of the DE is equal to zero it has said to be homogenous, otherwise it is inhomogeneous • If you know two solutions of any homogenous equation, other solution can be constructed as the linear combination of these solutions • Trivial change of unfamiliar DE might convert it to a recognizable DE • Many DEs are too difficult to be solved and few can be solved in closed form

  7. Solution of DE Solution of first-order DE • DE of general form or is clearly first-order DE Such equation can be solved by separation of variable method If are reducible to then we have

  8. Solution of DE • The above equation can be solved easily by integrating directly Example: Solve the following DE Solution: Let us rearrange the equation or to separate the variables hence it will be Now let us integrate as;

  9. Solution of DE the solution will be Where, C is constant • Some times, the variable may not be separable. In such case we need to change these variable so as to be separable, the following is the general form; Where f is an arbitrary function and a and b are constants. If we let

  10. Solution of DE Then, then the DE will be From which we obtained, such DE became separable

  11. Solution of DE Example: Solve the following DE Solution: Let, then and the DE will be

  12. Solution of DE or, hence the equation is separable can easy be solved by integrating the variables

  13. Solution of DE • Homogenous DE, which has the general form may also be reduced to an equation with separable variables Example: Solve the equation Solution: The right hand side can be written as

  14. Solution of DE hence we get function of single variable, let and our equation become from which we have Integration gives

  15. Solution of DE • Ex: Solve the following DE: 1- 2- 3-

  16. Exact equation • Example: Solve the equation Solution: The above equation has said to be homogenous if we will be able to change the variable, such that to eliminate -5 and -1 from numerator and denominator, respectively. Let,

  17. Exact equation • The above equation has said to be homogenous if we will be able to change the variable, such that to eliminate -5 and -1 from numerator and denominator, respectively. • Where, 𝛂 and 𝛃 are constants, chosen in such away our equation will be homogenous. Substitute for both x and y we get.

  18. Now let, then. • Hence our equation will be,

  19. herefore, the above equation becomes separable and hence solvable. • Example 2.5 Fall of a skydiver. Solution: Assuming the parachute opens at the beginning of the fall, there are two forces acting on the parachute: the downward force of gravity mg, and the

  20. upward force of air resistance. If we choose a coordinate system that has at the earth's surface and increases upward, then the equation of motion of the falling diver, according to Newton's second law, is where m is the mass, g the gravitational acceleration, and k a positive constant. In general the air resistance is very complicated, but the power-law approximation is useful in many instances in which the velocity does not vary appreciably.

  21. Experiments show that for subsonic velocity up to 300 m/s, the air resistance is approximately proportional to . The equation is separable To make the equation more simple let Assume that,

  22. Therefore, Let us apply the above reduction we get, Upon integration Where C is the constant of integration

  23. let, Fro1m the above solution it is obviously seen that, as; ,

  24. This implies that if the fall is from sufficient high, the fallen particle will fall with constant velocity known as terminal velocity. To determine C we need to know the value of k, which is being 30Kg/m

  25. Exact DE The following equation has said to integrable because it is homogenous The solution of the above equation is of the form The above equation has said to be exact if To check this let us let us consider the above equation, By performing differential we get, which the general properties of partial derivative of any well behave f unction

  26. If our exact DE is of the form, Then it is easy to identify that, Then it follows from the above equation that, Hence, it is the same relation we had obtained previously

  27. Examples: Show that the equation, is exact and find its general solution Solution: Let us rearrange the above equation so as to look like the following standard form Since,

  28. Therefore, the given equation is exact and its general solution will be, This implies that, Where, are the integration constants, comparing the above solution it is clearly seen that,

  29. Thus the general solution is, It is very interesting to consider a DE of this type The left hand side is an exact differential equation and on the right hand side is the function of x only. Then the general solution can be written as, Alternatively we can write the above solution as;

  30. Since the LHS of the above equation is exact, then Let us test the exactness of the above equation; Hence, the above equation satisfies the requirement for being exact; we can write the solution as: Where,

  31. Integrating Factor (IF) Inexact DE can be converted to exact by multiplying it by an IF. Different DE have different Ifs, which are rather difficult to find. Let us consider the following DE, Let, be the IF of the above DE. Now multiply the DE by IF, we get

  32. Now the RHS of the above equation has said to be integrable; the condition is that the LHS is exactHence the equation becomes, Which yield, Or,

  33. Hence it is possible to write the general solution as; Where, Therefore, the solution will be;

  34. Example: Show that the given equation is not exact; then find a suitable IF that could makes the DE exact and look for its general solution Solution: First of all let us write the above equation in the standard form .i.e. It is obviously seen that, This implies that, the given equation is not exact

  35. Let us divide the equation by x, we get; From the above equation we found that, Hence, the required IF is; Now apply this IF to the given equation we get, Integrating the above equation we get, and

  36. Application in Physics: For the given LCR circuit, shown below, find the current in the circuit as the function of time t

  37. Solution: First of all we need to setup the differential equation for the current flowing in the circuit. R and L represents the constant resistance and inductance, respectively. and where, is the voltage dropped across R is the voltage dropped across L

  38. Let us apply Kirchhoff’s second law here t represents the independent variable and stand for the dependent. The general solution will be, E and k are constants, and the evaluation of the above integral gives,

  39. Regardless the value of k, we see that, At , the solution will be,