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
           1z  z 0 
                         
           4x        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)