Non-Homogeneous Boundary Conditions: How to Solve PDEs with Green's Function?

[Col.Buendia]

Hey Guys;

I'm solving PDE's with the use of Green's function where all the boundary conditions are homogeneous.
However, how do you solve ones in which we have non-homogeneous b.c's.

In case it helps, the particular PDE I'm looking at is:

y'' = -x^2

y(0) + y'(0) = 4, y'(1)= 2

Thanks for any help.

 

[isabelle]

That is an ODE, the general solution is

y = c_1 + c_2x - \frac{1}{12}x^4

The equations are:

y'(1) = c_2 - \frac{1}{3}(1)^3 = 1

Therefore we have c_2 = 4/3. Therefore the second equation reads:

c_1 + c_2 = 4

Therefore we have c _1 = 8/3.

Doing this problem with the method of Green functions, we begin with the piecewise solution c_1 + c_2x, x < x' , and b_1 + b_2 x, x > x'. The first boundary condition says that c_1 + c_2 = 4, and the second one says b_2 = 1. Continuity at x' yields c_1 + (4-c_1)x' = b_1 + x'. Finally, the jump condition on the first derivative yields y'(x')_right - y'(x')_left = 1 implies 1 - c_2 = 1, and so c_2 is zero, at which point I'm stuck too.

 

[matematikawan]

Hey Guys;

I'm solving PDE's with the use of Green's function where all the boundary conditions are homogeneous.
However, how do you solve ones in which we have non-homogeneous b.c's.

In case it helps, the particular PDE I'm looking at is:

y&#039;&#039; = -x^2

y(0) + y&#039;(0) = 4, y&#039;(1)= 2

Thanks for any help.

Try making the substitution:
Y(x) = y(x) - 2x - 2.

Then
Y(0) = y(0) - 2 , Y'(0) = y'(0) - 2 and Y'(1) = y'(1) - 2.
These will make your bc homogeneous.

 

[Col.Buendia]

Thanks, that seems to work perfectly.