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Green's functions and the fundamental solution

The fundamental solution of Laplace's equation, Newtonian potentials, Green's functions, images and the 1D case.

The fundamental solution of the Laplacian is the potential of a unit point source: \[ \Phi(x) = \frac{1}{4\pi|x|}\ (n = 3), \qquad \Phi(x) = -\frac{1}{2\pi}\ln|x|\ (n = 2). \] Away from the origin it is harmonic, which the example checks for \( 1/r \). At the origin all of its "Laplacian" is concentrated: \( \nabla\Phi = -x/(4\pi|x|^3) \), and its flux through the sphere of radius \( r \) is \( -\frac{1}{4\pi r^2}\cdot 4\pi r^2 = -1 \) whatever \( r \) is. In the language of distributions, \( -\Delta\Phi = \delta \).

Superposing point sources gives the Newtonian potential \( u(x) = \int\Phi(x - y)f(y)\,dy \), which solves \( -\Delta u = f \) in all of space for nice \( f \). Gravity, electrostatics and the Biot-Savart law (velocity recovered from vorticity) are all this convolution.

On a bounded domain \( \Omega \) we need a Green's function \( G(x, y) = \Phi(x - y) - h(x, y) \), where for each \( y \) the corrector \( h \) is harmonic in \( x \) and equals \( \Phi(x - y) \) on \( \partial\Omega \), so \( G = 0 \) on the boundary. Then the solution of \( -\Delta u = f \) in \( \Omega \), \( u = g \) on \( \partial\Omega \), is \[ u(x) = \int_\Omega G(x,y)f(y)\,dy - \int_{\partial\Omega}g(y)\,\partial_{\nu_y}G(x,y)\,dS_y. \] For a half-space the corrector is an image charge: subtract the potential of a source reflected across the boundary. In one dimension, \( -u'' = f \) on \( (0,1) \) with \( u(0) = u(1) = 0 \) has \( G(x,y) = x(1 - y) \) for \( x \le y \) and \( y(1 - x) \) for \( y \le x \); with \( f = 1 \) this gives \( u = \int_0^1 G(x,y)\,dy = x(1 - x)/2 \).

Picture it: \( G(x, y) \) is the shape of a taut string pushed up by a unit point force at \( y \): a tent with straight sides pinned at both ends. Any load is a sum of point forces, so any deflection is a sum of tents.

Think it: a Green's function is the kernel of the inverse operator, \( (-\Delta)^{-1} \), just as a matrix inverse has entries. Its symmetry \( G(x,y) = G(y,x) \) reflects the symmetry of \( -\Delta \). In incompressible flow the pressure is \( p = (-\Delta)^{-1}\partial_i\partial_j(u_iu_j) \), a nonlocal function of the velocity, and it is this operator that couples every point of the fluid to every other.

Вработен пример · laplacian of 1/sqrt(x^2+y^2+z^2)

Laplacian of 1/sqrt(x^2+y^2+z^2)

\nabla^2\left(\frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}}\right)

Чекор по чекор

  1. f(x, y, z) = \frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}},\quad \nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}

    The Laplacian adds the second partial derivative in each direction.

  2. \frac{\partial f}{\partial x} = - \frac{x}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}},\quad \frac{\partial^2 f}{\partial x^2} = \frac{3 x^{2}}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{5}{2}}} - \frac{1}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}}

    Differentiate twice with respect to x, holding the other variables constant.

  3. \frac{\partial f}{\partial y} = - \frac{y}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}},\quad \frac{\partial^2 f}{\partial y^2} = \frac{3 y^{2}}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{5}{2}}} - \frac{1}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}}

    Differentiate twice with respect to y, holding the other variables constant.

  4. \frac{\partial f}{\partial z} = - \frac{z}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}},\quad \frac{\partial^2 f}{\partial z^2} = \frac{3 z^{2}}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{5}{2}}} - \frac{1}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}}

    Differentiate twice with respect to z, holding the other variables constant.

  5. \nabla^2 f = \left(\frac{3 x^{2}}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{5}{2}}} - \frac{1}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}}\right) + \left(\frac{3 y^{2}}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{5}{2}}} - \frac{1}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}}\right) + \left(\frac{3 z^{2}}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{5}{2}}} - \frac{1}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}}\right) = \frac{3 x^{2}}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{5}{2}}} + \frac{3 y^{2}}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{5}{2}}} + \frac{3 z^{2}}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{5}{2}}} - \frac{3}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}}

    Add the second partials.

  6. \nabla^2 f = 0

    Simplify.

  7. \nabla^2 f = 0

    The Laplacian is zero, so f is harmonic: it solves Laplace's equation.

Одкриј го одговорот
\nabla^2 f = 0

Сега ти. Секој чекор, слика, одговорот е скриен додека не прашаш.

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Како да го направиш тоа?: Green's functions and the fundamental solution

  1. Start from the fundamental solution for the dimension: 1/(4π|x|) or -(1/2π) ln|x|.
  2. Find a harmonic corrector that cancels it on the boundary (images for half-spaces and balls).
  3. Check G vanishes on the boundary and is symmetric.
  4. Write u as the integral of G against the source, plus the boundary term for non-zero boundary data.
  5. In one dimension, build G from two straight pieces that meet continuously with a unit jump in slope.

Прашања што луѓето ги поставуваат

Why the minus sign in -ΔΦ = δ?

Because -Δ is the positive operator: ∫(-Δu)u = ∫|∇u|² ≥ 0. With the sign chosen this way, the fundamental solution in three dimensions is positive.

Do Green's functions exist for every domain?

For bounded domains with reasonably smooth boundaries, yes, but explicit formulas exist only for special shapes: half-spaces, balls, rectangles (as series) and domains conformally equivalent to a disc in the plane.

What should I know before starting?

Partial derivatives, the divergence theorem and multiple integrals from multivariable calculus; linear second-order ODEs; eigenvalues from linear algebra; and enough analysis to be comfortable with uniform convergence and integrals over infinite intervals.

Why are there so few formulas for solutions?

Explicit solutions exist for linear equations with constant coefficients on simple domains. Almost everything else (curved domains, variable coefficients, nonlinear terms) has none, so the modern subject proves that a solution exists and estimates its size and smoothness without ever writing it down.

How does this course lead to the Navier-Stokes problem?

The Navier-Stokes equations are a heat equation for the velocity with a transport term and a pressure that is found from a Poisson equation. Reading the Millennium problem needs energy estimates, weak solutions, Sobolev spaces, local existence with a blow-up criterion, and the scaling argument that explains why three dimensions is hard. The last six lessons build those.

What do elliptic, parabolic and hyperbolic mean?

Elliptic equations (Laplace) describe equilibrium and smooth everything; parabolic equations (heat) describe diffusion forward in time; hyperbolic equations (waves) carry signals at finite speed and keep their sharp edges. The type decides which data make a sensible problem.

Повеќе во Partial Differential Equations