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B Tech Mathematics III Lecture Note PARTIAL DIFFERENTIAL EQUATION A differential equation containing terms as partial derivatives is called a partial differential equation (PDE). The order of a PDE is the order of highest partial derivative. The dependent variable z depends on independent variables x and y. p = z , q= z , r= 2z , s= 2z , t= 2z x y x2 x y y2 For example: q + px = x + y is a PDE of order 1 s + t = x2 is a PDE of order 2 Formation of PDE by eliminating arbitrary constant: For f(x,y,z,a,b) = 0 differentiating w.r.to x,y partially and eliminating constants a,b we get a PDE 2 2 2 Example 1: From the equation x + y + z = 1 form a PDE by eliminating arbitrary constant. 2 2 2 Solution: z = 1 - x - y Differentiating w.r.to x,y partially respectively we get 2z z 2x and 2z z 2y x y p = z = - x/z and q= z = - y/z x y z = - x/p = -y/ q qx = py is required PDE Example 2 From the equation x/2 + y/3 + z/4 = 1 form a PDE by eliminating arbitrary constant. Solution: Differentiating w.r.to x,y partially respectively we get 1 1 z 0 and 1 1 z 0 2 4 x 2 4 y 1z z 0 4x y B Tech Mathematics III Lecture Note p = z = q = z x y p = q is required PDE Formation of PDE by eliminating arbitrary function Let u= f(x,y,z), v= g(x,y,z) and ϕ(u,v) = 0 We shall eliminate ϕ and form a differential equation Example 3 From the equation z = f(3x-y)+ g(3x+y) form a PDE by eliminating arbitrary function. Solution: Differentiating w.r.to x,y partially respectively we get p z 3f '(3x y)3g'(3x y) and q z f '(3x y) g'(3x y) x y 2z 2z r x2 9f ''(3x y)9g''(3x y) and t y2 f ''(3x y) g''(3x y) From above equations we get r = 9t which is the required PDE. 11.1 An equation involving atleast one partial derivatives of a function of 2 or more independent variable is called PDE. A PDE is linear if it is of first degree in the dependent variable and its partial derivatives. If each term of such an equation contains either dependent variable or one of its derivatives the equation is called homogeneous. Important Linear PDE of second order 2 U = c U (One dimensional Wave equation) tt xx 2 U = c U (One dimensional Heat equation) t xx U + U = 0 (Two dimensional Laplace equation) xx yy U + U + U = 0 (Three dimensional Laplace equation) xx yy zz U + U = f(x,y) (Two dimensional Poisson equation) xx yy PROBLEMS -t 1. Verify that U = e Sin 3x is a solution of heat equation. -t -t Solution: U = -e Sin 3x and U = -9e Sin 3x t xx 2 U = c U (One dimensional heat equation) .......... (1) t xx B Tech Mathematics III Lecture Note Putting the partial deivativers in equation (1) we get -t 2 -t -e Sin 3x = -9c e Sin 3x 2 Hence it is satisfied for c = 1/9 2 One dimensional heat equation is satisfied for c = 1/9. Hence U is a solution of heat equation. 2. Solve Uxy = -Uy Solution: Put U = p then p p y x p x p Integrating we get ln p = - x + ln c(y) –x U/ y = p = e c(y) –x U = e c(y) y –x Integrating we get U = e (y) ϕ(y) + D(x) where ϕ(y) = ∫c(y) y 11.2 Modeling: One dimensional Wave equation We shall derive equation of small transverse vibration of an elastic string stretch to length L and then fixed at both ends. Assumptions. 1. The string is elastic and does not have resistance to bending. 2. The mass of the string per unit length is constant. 3. Tension caused by stretching the string before fixing it is too large. So we can neglect action of gravitational force on the string. 4. The string performs a small transverse motion in vertical plane. So every particle of the string moves vertically. Consider the forces acting on a small portion of the string. Tension is tangential to the curve of string at each point.Let T and T be tensions at end points. Since there is no motion in horizontal direction, horizontal 1 2 components of tension are T Cos α= T Cos β = T = Constant ..... ...... (1) 1 2 The vertical components of tension are - T Sin α and T Sin β of T and T 1 2 1 2 By Newton’s second law of motion, resultant force = mass x acceleration 2u T Sin β - T Sin α = x 2 1 t2 T Sin T Sin x 2u 2 - 1 = T T T t2 B Tech Mathematics III Lecture Note T Sin T Sin x 2u 2 - 1 = T Cos T Cos T t2 2 1 x 2u tan - tan = T t2 ...... .................(2) As tan β = ( u/ x) = Slope of the curve of string at x x tan α = ( u/ x) = Slope of the curve of string at x+x x+Δx x 2u Hence from equation (2) ( u/ x) - ( u/ x) = x+Δx x T t2 2u [ ( u/ x) - ( u/ x) ]/x = Dividing both sides by x x+Δx x T t2 Taking limit as x →0 we get 2u Lim x→0 [ ( u/ x) - ( u/ x) ]/x = x+Δx x T t2 u 2u = 2 x x T t 2u 2u x2 = T t2 2u T 2u t2 x2 2u 2u T OR C2 whereC2 t2 x2 which is One dimensional Wave equation 11.3 Solution of One dimensional Wave equation (separation of variable method) 2 One dimensional wave equation is u = c u .............................................(1) tt xx Boundary Condition u( 0, t) = 0, u(L,t) = 0 .............................................(2) Initial Condition u( x,0) = f(x) = initial deflection ..................(3) u (x, 0) = g(x) = initial velocity ..................(4) t Step I Let u(x,t) = F(x) A(t) Then u = F(x) Ӓ(t) and u = F″ (x) A(t) tt xx 2 Equation (1) becomes F(x) Ӓ(t) = C F″ (x) A(t)
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