Demonstrate the method of Solution by Separation of Variable

Demonstrate the method of \'Solution by Separation of Variables\' by solving the partial differential equation defined below: partial differential^2 u/partial differential t^2 = 1/c^2 partial differential^2 u/partial differential x^2 Hence write the genetic solution for the case when mu>0, and two Eigen values k1 and k2.

Solution

Solution: Given equation is

\\frac{\\partial^{2}u }{ \\partial t^{2}} = (1/c²) \\frac{\\partial^{2}u }{ \\partial x^{2}}

or \\frac{\\partial^{2}u }{ \\partial x^{2}} = c² \\frac{\\partial^{2}u }{ \\partial t^{2}} ..................(1)

Suppose that (1) has the solution of the form

u(x,t)= X(x)T(t)          .............................(2)

where X and T are respectively functions of x and t alone.

Substituting this value of u in (1), we have

X\'\'T = c²XT\'\'

or X\'\'/X = c²T\'\'/T

So X\'\' - \\mu X = 0 .......................(3)

and T\'\' - (1/c²)\\mu T = 0 ............................(4)

Solving (3) and (4), we obtain

(i) when \\mu = 0, then X = a₁x + a₂, T = a₃t + a₄

(ii) when \\mu is positive and = (\\lambda)² (say), then

X = b₁e^{\\lambda x} + b₂e^{-\\lambda x} , T = b₃e^{(1/c)\\lambda t} + b₄e^{-(1/c)\\lambda t}

(iii) when \\mu is negative and = -(\\lambda)² (say), then

X= c₁\\cos \\lambda x + c₂ \\sin \\lambda x, T = c₃\\cos (\\lambda t)/c + c₄\\sin (\\lambda t)/c

Thus the various possible solutions are:

u(x,t) = (a₁x + a₂)(a₃t + a₄)..................(5)

u(x,t) = (b₁e^{\\lambda x} + b₂e^{-\\lambda x})(b₃e^{(1/c)\\lambda t} + b₄e^{-(1/c)\\lambda t}).......................(6)

u(x,t) = (c₁\\cos \\lambda x + c₂ \\sin \\lambda x)(c₃\\cos (\\lambda t)/c + c₄\\sin (\\lambda t)/c )....................(7)

Now we have to choose the solution which is consistent with the physical nature of the problem.

Second Part

when \\mu is positive and = (\\lambda)² (say), then

X = b₁e^{\\lambda x} + b₂e^{-\\lambda x} , T = b₃e^{(1/c)\\lambda t} + b₄e^{-(1/c)\\lambda t}

and u(x,t) = (b₁e^{\\lambda x} + b₂e^{-\\lambda x})(b₃e^{(1/c)\\lambda t} + b₄e^{-(1/c)\\lambda t}) is solution for the given problem.

Avalue of \\lambda for which the given problem has a non-trivial solution is called an eigen-value and the non-trivial solution is called an eigen-function.

Here \\lambda can be considered as eigenvalue k.