Partial Differential Equations: Method of Separation of Variables
Explore the powerful method of separation of variables for solving partial differential equations like wave, heat, and Laplace equations. Learn how to break down the problems, solve ordinary differential equations, and combine solutions to tackle complex physical phenomena efficiently.
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Partial Differential Equations (MAT-203) UNIT IV Method of Separation of variables
THE METHOD OF SEPERATION OF VARIABLE The method of separation of variables is a very powerful method for obtaining solution for certain problem involving partial deferential equation problems those are of great physical interest can be solved by this method . For example, wave equation ,heat equation , and Laplace equation can be solved by this method. this method involves a solution which breaks up into a product of functions, each of which contains only one of the variable, if the partial differential equation involves n independent variables x1, x2, , xn , we first assume that the equation possesses solution of the form X1, X2, , Xn, where Xiis a function of only Xi(i = 1,2,, ,n). This basic assumption will produce ordinary differential equation , one in the each of the unknown function Xi (i=1,2, ,n). We solve these n problems will produce particular solution of the form X1,X2, , Xnsatisfying some supplementary condition of the original problem. Then these particular solution are combined by superposition rule to produce a solution of the problem. during the application of this method we shall come across the second order linear partial differential equation
A?2? ??2+ ? ?2? ????+ ??2? ??2+ ??? ??+ ??? (1) ??+ ? ? = 0 Where A,B,C,D,E and F are real constants. The equation (1) is said to be (a) hyperbola if B2- 4AC > 0 (b) parabolic if B2 4AC = 0 (c) elliptic if B2 4AC < 0. For example the equation ?2? ??2-?2? A=1,B=0,C=-1 and B2- 4AC > 0. this equation is satisfied by small transverse displacement of the points of a vibrating string. on the other hand the equation ??2= 0 (special case of wave equation ) is hyperbolic , since ?2? ??2+ ?2? ??2= 0 (two-dimensional Laplace equation ) is elliptic since A=1,B=0,C=1 and B2- 4AC < 0. this equation is satisfied by the steady state temperature of a thin rectangular plate . similarly, the heat equation ?? ??= c?2? ??2is parabolic
Example . Solve ?2? Solution :- Let z = X(x) Y(y) (1) Be a trial solution of the given partial differential equation . Then the given equation reduces to X (x)Y(y) 2 X (x)Y(y) + X(x)Y (y) = 0 separating the variable, we get = ? (?) ?(?) The left hand side is a function of x only, where as the right hand side is a function y only since x and y are different variables, equality in (2) can occur only if left-hand side and right-hand side both equal to a constant ,say a. thus we get two ordinary differential equation X (x) 2 X (x) - a X(x)= 0 (3) And Y (y) + a Y(y)= 0 The auxiliary equation for (3) is m 2 - 2m a = 0. ??2 2?? ??+ ?? ??= 0. ? ? 2? ? ?(?) (2) (4)
which yields m = 1 X(x) = C1?1+ 1+? ?+ C2?1 1+? ? The auxiliary equation for (4) is m + a = 0 And so m = -a . Therefore, the solution of (4) is Y(y) = C3? ??. substitution the value of X(x) and Y(y) into (4), We get z = X(x)Y(y) = [C1?1+ 1+? ?+ C2?1 1+? ?]C3? ?? = [C4?1+ 1+? ?+ C5?1 1+? ?]? ??, Which is the required solution of the given differential equation . 1 + ? . Therefore, the solution of (3)
?? ??= 3? subject to the condition that u (0 , y) = 3 ? ? 5 ? 5?. Solution :- The given equation involves two variables x and y. so, let u(x , y) = X(x)Y(y). Then the given equation reduces to 4 ? (x)Y(y) + X(x) ? (y) = 3X(x)Y(y) Or equivalently , 4? ? ?(?) . Since x and y are the independent variables, the left hand side and right hand side are both equal to some constant , say a . Thus ,we get two differential equations 4? ? ?(?)= ?, (1) And 3?(?) ? (?) ?(?) The solution of (1) is log(x) = ? 4? + ????1 Example :- 4 ?? ??+ ?(?)=3?(?) ? (?) (2)
? 4? ?(?) ?1= ? ? 4? or X(x) = C1 ? The equation (2) can be written as (3) ? (?) ?(?)= 3 ? So its solution is log Y(y) = (3-a)y + logC2 or Y(y) = C2?3 ? ? Using (3) and (4) , the solution is (4) ? 4? . ?3 ? ? u(x , y) = C1C2? . therefore , using the initial condition u(0, y) = 3? ?- 3? 5? 3? ?- 3? 5? gives one value of set as C1C2 = 3 and a-3 = 1, that is C1C2 = 3 and a = 4. , we get = C1C2? ? 3 ?+ 0 ? 5 ?
The other set of values is given by 3? ?- 3? 5?= 0 ? ?+ C1C2 ? (? 3)? And that set is C1C2 = -1 and a-3 = 5, That is , C1C2 = -1 and a = 8, Thus we get two solution 4 4? . ?3 4 ?= 3??? ?= 3 ?? ? 8 4? . ?3 8 ?= ?2?? 5 ?= ?2? 5?. u1 (x, y)=C1C2? And u2 (x, y) = -1 ? Hence the required solution is u(x ,y) = u1 (x, y) + u2 (x, y) = 3 ?? ? ?2? 5?.
