Question

Let G(x,t) be the Green's function satisfying

          G"(x,t) + G(x,t) = δ(x-t)

with G(0,t) = 0 and G(pi/2) = 0, and the neccessary continuity conditions.

Here t is fixed, 0 < t < pi/2; the derivative means differentiation with respect to x, 0 < x < pi/2

Fill in the blanck:

G(x,t)=                                                   when 0<x<t

                                                             when t<x<pi/2

Answer 1

Given greens function is satisfying

G^"(x,t)+G(x,t) =d(x-t). I am taking delta symbol d.this is impulse function.

First of all,we know about impulse function d(x-t)=1 If and only if x>=t other wise it is 0.

First we have to find out,G(x,t) when 0<x<t.In this condition d(x-t)=0 and it forms homogeneous differential equation

G^"(x,t)+G(x,t)=0 and this become X^"t+Xt=0 and (x^"+x)t=0 because derivative with respect to x and t is fixed.

We can write this equation as

(D²+1)x=0. And t=0

D²=-1

D₁=√-1=+-j. And D₂=0.

Now,solution become

G(x,t)=(c₁cosx+c₂sinx)(c₃e^0t).

Given initial conditions are

G(0,t)=(c₁)(c₃)=0 and c₁=0.

G(π/2,t)=c₂(c₃)=0 and c₂=0.

So,It is giving trivial solution that is G(x,t)=0 when 0<x<t.

d(x-t)=1 for t<x<pi/2.

Now,Equation become

G^"(x,t)+G(x,t)=1 when t<x<pi/2.and it forms non homogenous equation

solution contains two parts those are homogeneous solution and particular solution

G(x,t)=G_h(x,t)+G_p(x,t).

Already we know that G_h(x,t)=(c₁cosx+c₂sinx)(c₃).

Given G_p(x,t) on right side equation is G_p(x,t)=1=k.assume some constant k.With out assumption also G_p(x,t)=1 only.

Gp'(x,t)=G_p^"(x,t)=0.

Substitute in the original differential equation

We will get k=1.

So,particular solution G_p(x,t)=1.

Finally

G(x,t)=(c₁cosx+c₂sinx)(c₃)+1.

Now,apply given initial conditions, G(0,t)=0 and G(pi/2,t)=0

G(0,t)=c₁c₃+1=0 and c₁=(-1/c₃).

G(pi/2,t)=c₂c₃+1=0 and c₂=(-1/c₃)=c₁.

Now,substitute above values,then equation become

G(x,t)=((-1/c₃)cosx+(-1/c₃)sinx)c₃+1.and it can be simplified as

G(x,t)=(-1/c₃)(c₃)(cosx+sinx)+1.c₃ will get cancel

and G(x,t)=1-(cosx+sinx).

So,Finally Non trivial solution G(x,t)=1-cosx-sinx. When t<x<pi/2.