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- 1. 160280102051 160280102052 160280102053 160280102055 160280102057 160280102059 160280102060 160280102061
- 2. • Definition • One of the classical partial differential equation of mathematical physics is the equation describing the conduction of heat in a solid body (Originated in the 18th century). And a modern one is the space vehicle reentry problem: Analysis of transfer and dissipation of heat generated by the friction with earth’s atmosphere.
- 3. • A differential equation is an equation for an unknown function of one or several variables that relates the values of the function itself and of its derivatives of various orders. • Ordinary Differential Equation: Function has 1 independent variable. • Partial Differential Equation: At least 2 independent variables.
- 4. • General (implicit) form for one function u(x,y) : • Highest derivative defines order of PDE • Explicit PDE => We can resolve the equation to the highest derivative of u. • Linear PDE => PDE is linear in u(x,y) and for all derivatives of u(x,y) • Semi-linear PDEs are nonlinear PDEs, which are linear in the highest order derivative. 6
- 5. Partial Differential Equations are those which involve partial derivatives with respect to two or more independent variables. Partial derivatives 𝜕𝑧 𝜕𝑥 , 𝑑𝑧 𝑑𝑦 , 𝜕2 𝑧 𝜕𝑥2, 𝜕2 𝑧 𝜕𝑦2, 𝜕2 𝑧 𝜕𝑥𝜕𝑦 are denoted by p,q,r,t,s respectively. The ORDER of a PDE is the order of the highest derivatives in the equation. The DEGREE of a PDE is the power of the highest order partial derivative present in the equation.
- 6. Partial differential equations can be formed using the following methods. 1) By elimination of arbitrary constants in the equation of the type f(x, y, z, a, b)=0 where a and b are arbitrary constants. 2) By elimination of arbitrary functions in the equation of the type z = f(u) where u is a function of x, y and z
- 7. Consider the function f(x, y, z, a, b) = 0 ..(i) where a and b are two independent arbitrary constants. To eliminate two constants, we require two more equations. Differentiating eq. (i) partially with respect to x and y in turn, we obtain 𝜕𝑓 𝜕𝑥 . 𝜕𝑥 𝜕𝑥 + 𝜕𝑓 𝜕𝑧 . 𝜕𝑧 𝜕𝑥 = 0 ⇒ 𝜕𝑓 𝜕𝑥 + 𝜕𝑓 𝜕𝑧 . 𝑝 = 0 ……(ii) 𝜕𝑓 𝜕𝑦 . 𝜕𝑦 𝜕𝑦 + 𝜕𝑓 𝜕𝑧 . 𝜕𝑧 𝜕𝑦 = 0 ⇒ 𝜕𝑓 𝜕𝑦 + 𝜕𝑓 𝜕𝑧 . 𝑞 = 0 ……(iii) Eliminating a and b from the set of equations (i), (ii) and (iii) , we get a PDE of the first order of the form F(x, y, z, p, q) = 0. .. (iv)
- 8. Consider a relation between x, y and z of the type (u; v) = 0 ..(v) where u and v are known functions of x, y and z and is an arbitrary function of u and v. Differentiating equ. (v) partially with respect to x and y, respectively, we get 𝜕𝜑 𝜕𝑢 𝜕𝑢 𝜕𝑥 + 𝜕𝑢 𝜕𝑧 𝑝 + 𝜕𝜑 𝜕𝑣 𝜕𝑣 𝜕𝑥 + 𝜕𝑣 𝜕𝑧 𝑝 = 0 …….(vi) 𝜕𝜑 𝜕𝑢 𝜕𝑢 𝜕𝑦 + 𝜕𝑢 𝜕𝑧 𝑞 + 𝜕𝜑 𝜕𝑣 𝜕𝑣 𝜕𝑦 + 𝜕𝑣 𝜕𝑧 𝑞 = 0 .…..(vii)
- 9. Eliminating 𝜕𝜑 𝜕𝑢 and 𝜕𝜑 𝜕𝑣 from the eq. (vi) and (vii), we get the equations 𝜕 𝑢,𝑣 𝜕(𝑦,𝑧) 𝑝 + 𝜕 𝑢,𝑣 𝜕(𝑧,𝑥) 𝑞 = 𝜕 𝑢,𝑣 𝜕(𝑥,𝑦) …..(viii) which is a PDE of the type (iv). since the power of p and q are both unity it is also linear equation, whereas eq. (iv) need not be linear.
- 10. partial differential equation of first order is said to be linear if the dependent variable and its derivatives are degree of one and the products of the dependent variable and its derivatives do not appear in the equation . The equation is said to be quasi-linear if the degree of the highest order derivative is one and the products of the highest order partial derivatives are not present. A quasi-linear partial differential equation is represented as P(x,y,z)*P+Q(X,Y,Z)*q=R(x,y,z) This equation is known as lagrange’s linear equation.
- 11. Working rule 1. write the given differential equation in the standard form Pp+Qq=R 2. Form the lagrange’s auxiliary equation d𝑥/P=dy/Q=R (1.1) 3.Solve the simultaneous equation in Eq. to obtain its two independent solutions u=c1 , v=c2 4.Write the general solution of the given equation as, f(u,v) = 0.
- 12. A homogeneous linear partial differential equation of the nth order is of the form homogeneous because all its terms contain derivatives of the same order.
- 13. Equation (1) can be expressed as As in the case of ordinary linear equations with constant coefficients the complete solution of (1) consists of two parts, namely, the complementary function and the particular integral. The complementary function is the complete solution of f (D,D') z = 0------- (3), which must contain n arbitrary functions as the degree of the polynomial f(D,D'). The particular integral is the particular solution of equation (2).
- 14. Finding the complementary function Let us now consider the equation f(D,D') z = F (x,y) The auxiliary equation of (3) is obtained by replacing D by m and D' by 1. Solving equation (4) for „m‟, we get „n‟ roots. Depending upon the nature of the roots, the Complementary function is written as given below:
- 15. Finding the particular Integral
- 16. Expand [f (D,D')]-1 in ascending powers of D or D' and operate on xm yn term by term. Case (iv) : When F(x,y) is any function of x and y. into partial fractions considering f (D,D') as a function of D alone. Then operate each partial fraction on F(x,y) in such a way that where c is replaced by y+mx after integration
- 17. Non Homogeneous Linear PDES If in the equation )1...()........,(),( yxFzDDf the polynomial expression is not hom ogeneous, then (1) is a non- homogeneous linear partial differential equation ),( DDf Complete Solution = Complementary Function + Particular Integral To find C.F., factorize into factors of the form ),( DDf )( cDmD yx ezDDDD 3222 )43( Ex
- 18. If the non homogeneous equation is of the form )()(. ),())(( 21 2211 21 xmyexmyeFC yxFzcDmDcDmD xcxc 1.Solve 22 )( xzDDDD Solution )1(),( 2 DDDDDDDDDf )()(. 21 yxyeFCx
- 19. 6.5.125.4.34.3123 1 ...... )1()1(1 )1( 1 1 . 65443 2 2 2 2 22 2 2 1 22 2 xxxxx x D x D D x D D x D x D D DDDDD x IP
- 20. 2.Solve 4)32)(1( zDDDD Solution 3 4 )2()( 1 3 1 xyexyez xx
- 21. Linear Second Order P.D.E : Examples (Classification) Hyperbolicis ACBCBcA t txu x txu c Parabolicis ACBCBA t txu x txu EquationWave 041,0,0 0 ),(),( EquationWave ______________________________________ EquationHeat 040,0, 0 ),(),( EquationHeat 22 2 2 2 2 2 2 2 2
- 22. • Consider an elastic string stretched to a length l along the x-axis with its two fixed ends at x=0 and x=l. • Assumption made: i. The string is homogeneous with constant density ρ. ii. The string is perfectly elastic and offers no resistance to bending. iii. The tension in the string is so large that the force due to weight of the string can be neglected. • Consider the motion of the small portion PQ of length δx of the string. Since the string produces no resistance to bending, the tension T₁ and T₂ at points P and Q will act tangentially at P and Q respectively. • Assuming that the point on the string move only in the vertical direction, there is no motion in the horizontal direction. • Hence, the sum of the forces in the horizontal direction must be zero.
- 23. • −𝑇₁cos 𝛼 + 𝑇₂ sin 𝛽 = 0 • 𝑇₁cos 𝛼 = 𝑇₂ sin 𝛽 = 𝑇(constant ),say ………..(1) • The forces acting vertically on the string are the vertical components of tension at points P and Q. Thus, the resultant vertical force acting on PQ is 𝑇₂ sin 𝛽- 𝑇₁ sin 𝛼. • By newtons second law of motion, • Resultant Force=Mass×Accelaration • −𝑇₁sin 𝛼 + 𝑇₂ sin 𝛽 = 𝜌𝛿𝑥 𝜕2 𝑦 𝜕𝑡2 ………(2) • Dividing by T, • 𝑇₂ sin 𝛽 T − 𝑇₁sin 𝛼 T = 𝑃𝛿𝑥 𝜕2 𝑦 𝜕𝑡2 • 𝑇₂ sin 𝛽 𝑇₂ cos 𝛽 − 𝑇₁sin 𝛼 𝑇₁cos 𝛼 = 𝑃𝛿𝑥 𝜕2 𝑦 𝜕𝑡2 • tan 𝛽 − tan 𝛼 = 𝑃𝛿𝑥 𝑇 𝜕2 𝑦 𝜕𝑡2 ……….(3) • Since tan 𝛼 and tan 𝛽 are the slopes of the curve at point P and Q respectively. • tan 𝛼 = 𝜕𝑦 𝜕𝑥 𝑃 = 𝜕𝑦 𝜕𝑥 𝑥 • tan 𝛽 = 𝜕𝑦 𝜕𝑥 𝑄 = 𝜕𝑦 𝜕𝑥 𝑥+𝛿𝑥
- 24. • Substitute in Eq.(3) • 𝜕𝑦 𝜕𝑥 𝑥+𝛿𝑥 − 𝜕𝑦 𝜕𝑥 𝑥 = 𝑃𝛿𝑥 𝑇 𝜕2 𝑦 𝜕𝑡2 • Dividing by 𝛿𝑥 and taking limit 𝛿𝑥 → 0, • lim 𝛿𝑥→0 𝜕𝑦 𝜕𝑥 𝑥+𝛿𝑥 − 𝜕𝑦 𝜕𝑥 𝑥 𝛿𝑥 = 𝑃𝛿𝑥 𝑇 𝜕2 𝑦 𝜕𝑡2 • 𝜕2 𝑦 𝜕𝑥2 = 1 𝑐2 𝜕2 𝑦 𝜕𝑡2 where 𝑐2 = 𝑇 𝑃 • 𝜕2 𝑦 𝜕𝑡2 = 𝑐2 𝜕2 𝑦 𝜕𝑥2 • This equation is known as one dimensional wave equation.
- 25. Solution of One –Dimensional Wave Equation • The one dimensional wave equation is • 𝜕2 𝑦 𝜕𝑡2 = 𝑐2 𝜕2 𝑦 𝜕𝑥2 ………………(4) • Let 𝑦 = 𝑋 𝑥 𝑇 𝑡 be a solution of Eq.(4) • 𝜕2 𝑦 𝜕𝑡2 = 𝑋𝑇′′ , 𝜕2 𝑦 𝜕𝑥2 = 𝑋′′ 𝑇 • Substituting in Eq.(4) • 𝑋𝑇′′ = 𝑐2 𝑋′′ 𝑇 • 𝑋`` 𝑋 = 1 𝑐2 𝑇`` 𝑇 • Since 𝑋 and 𝑇 are only the functions of x and t respectively, this equation holds good if each term is constant. • Let 𝑋`` 𝑋 = 1 𝑐2 𝑇`` 𝑇 = 𝑘,say • Considering 𝑋`` 𝑋 = 𝑘, 𝑑2 𝑋 𝑑𝑥2 − 𝑘𝑋 = 0 …….(5)
- 26. • Considering 1 𝑐2 𝑇`` 𝑇 = 𝑘, 𝑑2 𝑇 𝑑𝑡2 − 𝑘𝑐2 𝑇 = 0 ………(6) • Solving Equations (5) & (6), the following cases arise: I. When k is positive • Let 𝑘 = 𝑚2 • 𝑑2 𝑋 𝑑𝑥2 − 𝑚2 𝑋 = 0 and 𝑑2 𝑇 𝑑𝑡2 − 𝑚2 𝑐2 𝑇 = 0 • 𝑋 = 𝑐1 𝑒 𝑚𝑥 + 𝑐2 𝑒−𝑚𝑥 and T = 𝑐3 𝑒 𝑚𝑐𝑡 + 𝑐4 𝑒−𝑚𝑐𝑡 • Hence , the solution of Eq.(4) is • 𝑦 = (𝑐1 𝑒 𝑚𝑥 + 𝑐2 𝑒−𝑚𝑥)(𝑐3 𝑒 𝑚𝑐𝑡 + 𝑐4 𝑒−𝑚𝑐𝑡) ……(7) • II. When k is negative • Let 𝑘 = −𝑚2 • 𝑋 = 𝑐1cos 𝑚𝑥 + 𝑐2 sin 𝑚𝑥 and T = 𝑐3cos 𝑚𝑐𝑡 + 𝑐4 sin 𝑚𝑐𝑡 Hence the solution of Eq.(4) is 𝑦 = 𝑐1cos 𝑚𝑥 + 𝑐2 sin 𝑚𝑥 𝑐3cos 𝑚𝑐𝑡 + 𝑐4 sin 𝑚𝑐𝑡 ……(8)
- 27. III. When k=0 𝑋 = 𝑐1 𝑥 + 𝑐2 and T = 𝑐3 𝑡 + 𝑐4 Hence, the solution of Eq.(4) is • 𝑦 = (𝑐1 𝑥 + 𝑐2)(𝑐3 𝑡 + 𝑐4) ……(9) • Out of these three solution , we need to choose that solution which is consistent with the physical nature of the problem. Since we are dealing with problems on vibrations, y must be a periodic function of x and t. Thus , the solution must involve trigonometric terms. • Hence, the solution of the form given by Eq.(8) • 𝑦 = 𝑐1cos 𝑚𝑥 + 𝑐2 sin 𝑚𝑥 𝑐3cos 𝑚𝑐𝑡 + 𝑐4 sin 𝑚𝑐𝑡
- 28. ( 1 ) ( 2 ) ( 3 ) ( 4 ) ( 5 ) d'Alembert's Solution The method of d'Alembert provides a solution to the one-dimensional wave equation that models vibrations of a string. The general solution can be obtained by introducing new variables and , and applying the chain rule to Using (4) and (5) to compute the left and right sides of (3) then gives