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Boundary conditions of ODE

  • Thread starter Mr Boom
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Homework Statement



Given w'' - w = f(x)

w'(0) = 1
w'(1) = 0


Homework Equations



Find the Green's Function

The Attempt at a Solution



The solution to the homogeneous equation is known as:

w(x) = A*exp(-x) + B*exp(x)

For G's function we have:

u(x) = A1*exp(-x) + B*exp(x), u'(0) = 1
v(x) = A2*exp(-x) + B*exp(x), v'(1) = 0

The homogeneous equation is easily solved by plugging in two B.C.s. However, with Green's function we have two equations with two unknowns and one B.C. each. This leaves:

u'(0) = 1 -> B1 = 1 + A1 -> u(x) = A1*exp(-x) + (1+A1)*exp(x)
v'(1) = 0 -> B2 = A2*exp(-2) -> v(x) = A2*exp(-x) + A2*exp(-2)*exp(x)

Since G1 = A(c)*u(x) and G2 = B(c)*v(x), and we have two jump conditions at point c, I don't see how this can be solved. I've done other problems where A(c) and B(c) can be merged with A1 and A2, but the "1+A1" term keeps me from reducing this two two unknowns and two equations. Am I missing something? Is there an implied BC that I'm not seeing?

Thanks
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



Given w'' - w = f(x)

w'(0) = 1
w'(1) = 0


Homework Equations



Find the Green's Function

The Attempt at a Solution



The solution to the homogeneous equation is known as:

w(x) = A*exp(-x) + B*exp(x)

For G's function we have:

u(x) = A1*exp(-x) + B*exp(x), u'(0) = 1
v(x) = A2*exp(-x) + B*exp(x), v'(1) = 0

The homogeneous equation is easily solved by plugging in two B.C.s. However, with Green's function we have two equations with two unknowns and one B.C. each. This leaves:

u'(0) = 1 -> B1 = 1 + A1 -> u(x) = A1*exp(-x) + (1+A1)*exp(x)
v'(1) = 0 -> B2 = A2*exp(-2) -> v(x) = A2*exp(-x) + A2*exp(-2)*exp(x)

Since G1 = A(c)*u(x) and G2 = B(c)*v(x), and we have two jump conditions at point c, I don't see how this can be solved. I've done other problems where A(c) and B(c) can be merged with A1 and A2, but the "1+A1" term keeps me from reducing this two two unknowns and two equations. Am I missing something? Is there an implied BC that I'm not seeing?

Thanks
The Green's function is the solution of G_a''(x) - G_a(x) = Delta(x-a), where Delta is the Dirac delta fcn. We can regard G_a as G_a(x) = H(x-a), where H''(x) - H(x) = Delta(x). Don't worry about boundary conditions at x = 0 and x = 1 for now. Since Delta(x) = 0 if x < 0 and if x > 0 we have H''-H=0 on x<0 and on x>0. What happens at x = 0? If we integrate the DE from 0 - h to 0 + h, for small h > 0 we get H'(0+h) - H'(0-h) + O(h) = 1. So, we want: (i) H(x) continuous at x = 0; and (ii) H'(0+)-H'(0-) = 1.

We have H = A1*exp(x)+ B1*exp(-x) for x < 0 and H = A2*exp(x)+B2*exp(-x) for x > 0 we need A1+B1 = A2+B2 [continuity] and A2* - B2* - A1 + B1 = 1 [jump condition]. Now, usually we want a Green's fcn on all of the real line, so we usually impose conditions at +-infinity. Let's assume boundedness. Thus, we have B1 = 0 and A2 = 0. Therefore, we now have A1 = B2 and B2-A1=1, so now we can get A1 and B2. The solution of w''-w = f is w(x) = w_0(x) + integral(H(x-t) f(t) dt, t=-infinity..infinity), and where w_0 is any solution of the homogeneous equation. (Here, you could take f(t) = 0 for t < 0 and t > 1, but if your original f(x) is defined for all x, you could use f on the whole line.) We can choose the constants in w_0 to give the desired boundary conditions.

RGV
 
Last edited:
  • #3
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Ray,

Thank you for the reply. If it's not too much trouble, could you explain where the w(x)=w_0 + int(...) equation comes from?
 
  • #4
Ray Vickson
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Ray,

Thank you for the reply. If it's not too much trouble, could you explain where the w(x)=w_0 + int(...) equation comes from?
Sure: the general solution of y'' - y = f is any _particular_ solution plus the solution of the homogeneous DE; this is an absolutely standard result that you should have burned into your soul. The integral involving H is just one particular solution.

RGV
 

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