Green's Functions for Boundary Value Problems
For an invertible linear boundary value problem with homogeneous boundary conditions, a Green's function gives the response to a unit point source. Integrating it against the forcing gives the solution. For separated second-order problems, its two pieces come from homogeneous solutions satisfying their respective endpoint conditions.
What a learner can do afterwards
- Constructs the Green's function for -c u'' on [0,1], with constant c>0 and stated homogeneous Dirichlet or mixed endpoint conditions
- Recovers the solution for a given forcing term as an integral against the Green's function
- Explains the jump condition on the derivative and where it comes from; checks the zero-total-forcing compatibility condition for -u'' on [0,1] with u'(0)=u'(1)=0
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Sturm-Liouville problems used homogeneous equations with conditions at both endpoints. Now solve -u''=f on [0,1], with u(0)=u(1)=0. G(x,s) is the response at x to a unit point source at s, where 0<s<1. Away from s, solve -G_xx=0. Choose pieces meeting their respective endpoint conditions. Match their values at s and impose the source's slope jump.
The homogeneous pieces are straight lines: Ax for x<s and B(1-x) for x>s. Continuity gives As=B(1-s). Their x slopes are A and -B. A unit source for -u'' requires right slope minus left slope to equal -1, so A+B=1. Solve the two equations: A=1-s and B=s. Continuity alone would leave their scale undetermined.
Thus G(x,s)=x(1-s) for 0<=x<=s, and G(x,s)=s(1-x) for s<=x<=1. Both expressions give s(1-s) at x=s. The endpoint values are G(0,s)=G(1,s)=0. To evaluate G, compare x with s before substituting. To differentiate in x, hold s fixed: the left slope is 1-s and the right slope is -s.
For continuous or piecewise continuous f, use u(x)=∫ from 0 to 1 of G(x,s)f(s) ds. Fix x and split at s=x: u(x)=(1-x)∫ from 0 to x of sf(s) ds + x∫ from x to 1 of (1-s)f(s) ds. This solves -u''=f where f is continuous. Use f(s), not f(x). At either endpoint every kernel value is zero, so the solution integral is zero.
For f(s)=1, the first integral contributes (1-x)x^2/2. The second contributes x(1-x)^2/2. Adding gives u(x)=x(1-x)/2. At x=1/3, the contributions are 1/27 and 2/27, giving u(1/3)=1/9. Check: u''=-1 and both endpoint values are zero. Multiplying the forcing by a constant multiplies the solution by that constant.
A unit point source δ(x-s) has integral 1 across s. Integrating -G_xx=δ(x-s) gives -[G_x(s+,s)-G_x(s-,s)]=1. Here s+ and s- mean right and left limits. G stays continuous, while its x derivative jumps by -1. For -c u''=f with constant c>0 and the same homogeneous conditions, divide the kernel by c. Its derivative jump is -1/c.
Replace the right condition by u'(1)=0, keeping u(0)=0. The right homogeneous piece must now be constant. Write G=Ax for x<s and G=B for x>s. Continuity gives As=B. The jump gives 0-A=-1, so A=1 and B=s. Thus G=x for x<=s and G=s for x>=s. Changing this endpoint condition changes the kernel.
The two-zero-endpoint kernel is symmetric: G(x,s)=G(s,x). For x<=s, both sides equal x(1-s); exchange x and s for the other case. For piecewise forcing, integrate only where it is nonzero. If f(s)=1 on [0,a] and zero elsewhere, then for x>a every contributing source satisfies s<x. Hence u(x)=(1-x)∫ from 0 to a of s ds=(1-x)a^2/2.
An ordinary inverse Green's function requires a unique solution. With u'(0)=u'(1)=0, integrating -u''=f requires ∫ from 0 to 1 of f(x) dx=0. Even when this holds, adding a constant gives another solution. The problem has no ordinary inverse. The kernels above use homogeneous boundary data. For nonzero data, first subtract a function meeting those data and adjust the forcing.
Choose pieces for the endpoint conditions, match their values, set the operator's slope jump, then integrate against the forcing.
This lesson, question bank and flashcards were written for this dataset.
These readings, additional questions and flashcards were written for this dataset.
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Find the coefficients of the kernel
Take -u''=f on [0,1], with zero values at both endpoints. Fix a source location s strictly inside the interval. Away from s, the kernel satisfies G_xx=0, so its pieces are straight lines.
The left endpoint selects the form Ax. The right endpoint selects B(1-x). At the source their values must match: As=B(1-s). Their x derivatives are A and -B. A unit source requires a derivative jump of -1, giving A+B=1.
Use B=1-A in the continuity equation. Then As=(1-A)(1-s), so A=1-s and B=s. The resulting kernel is x(1-s) for x<=s and s(1-x) for x>=s.
These checks have different jobs. Endpoint conditions select the forms of the pieces. Continuity joins their values. The derivative jump fixes their scale. Checking only the endpoint values or only continuity leaves part of the construction unfinished.
Keep the observation point fixed
In u(x)=∫ from 0 to 1 of G(x,s)f(s) ds, x is the observation point and s labels the source locations. Hold x fixed while integrating in s.
For sources with s<x, use s(1-x). For sources with s>x, use x(1-s). The solution is therefore (1-x)∫ from 0 to x of sf(s) ds + x∫ from x to 1 of (1-s)f(s) ds. The forcing must be evaluated at s because it weights the source being integrated.
If f=1, evaluating both integrals gives u=x(1-x)/2. If the forcing is zero on part of the interval, that part contributes nothing. Split at any change in the forcing formula and at s=x when needed. Piecewise continuous forcing gives a solution of the differential equation wherever the forcing is continuous.
This construction is a weighted integral, not generally an average. The weights need not sum or integrate to one.
Check the jump and the endpoint conditions
A unit point source has total integral 1 across its location. Integrating -c G_xx=δ(x-s), with constant c>0, shows that -c times the difference between the right and left x derivatives equals 1. The derivative jump is therefore -1/c. G itself remains continuous.
Changing an endpoint condition changes the construction. For -u'' with u(0)=0 and u'(1)=0, use Ax on the left and a constant B on the right. Continuity gives As=B, and the jump gives 0-A=-1. Thus the kernel is x for x<=s and s for x>=s.
Some conditions prevent an ordinary inverse. If both endpoint derivatives are zero, integrating -u''=f requires the total forcing to be zero. When this condition holds, adding any constant still gives another solution. Such a problem needs an additional convention and a modified representation. The ordinary inverse kernel used for the Dirichlet and mixed problems does not apply.
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