如何用分离变量法求解半无限域上热方程的Dirichlet问题？

Absolutely, you can definitely use separation of variables to solve the Dirichlet problem for the heat equation on a semi-infinite domain—though it looks a bit different from the finite-domain case because we're dealing with a continuous spectrum of eigenvalues instead of discrete ones. Let's walk through this step by step.

First, let's define the standard Dirichlet problem for the heat equation on ( x \geq 0 ), ( t > 0 ):

• Heat equation: ( u_t = k u_{xx} ) (where ( k > 0 ) is thermal diffusivity)
• Boundary condition: ( u(0, t) = g(t) ) (Dirichlet condition at the domain's edge)
• Asymptotic constraint: ( \lim_{x \to \infty} u(x, t) = 0 ) (or at least bounded, to avoid physically unrealistic unbounded solutions)
• Initial condition: ( u(x, 0) = f(x) ) for ( x > 0 )

Assume a solution of the form ( u(x, t) = X(x)T(t) ). Substitute this into the heat equation:
[
X(x)T'(t) = k X''(x)T(t)
]
Divide both sides by ( k X(x)T(t) ) to split the equation into spatial and temporal parts:
[
\frac{T'(t)}{k T(t)} = \frac{X''(x)}{X(x)} = -\lambda
]
Here, ( -\lambda ) is our separation constant (we use the negative sign to prioritize physically meaningful, time-decaying solutions).

We now have two independent ordinary differential equations to solve:

1. Time ODE: ( T'(t) + k\lambda T(t) = 0 )
2. Spatial ODE: ( X''(x) + \lambda X(x) = 0 )

We analyze possible values of ( \lambda ) while enforcing the asymptotic boundedness constraint:

• Case 1: ( \lambda < 0 )
  Let ( \lambda = -\omega^2 ) (where ( \omega > 0 )). The spatial ODE becomes ( X'' - \omega^2 X = 0 ), with solutions ( X(x) = A e^{\omega x} + B e^{-\omega x} ). Since ( e^{\omega x} ) blows up as ( x \to \infty ), we set ( A = 0 ), leaving ( X(x) = B e^{-\omega x} ).
• Case 2: ( \lambda = 0 )
  The spatial ODE simplifies to ( X'' = 0 ), with solutions ( X(x) = Ax + B ). Again, ( Ax ) grows without bound as ( x \to \infty ), so ( A = 0 ), leaving a constant solution ( X(x) = B ). This only works if ( g(t) ) is constant, so it's a special case, not the general solution.
• Case 3: ( \lambda > 0 )
  Let ( \lambda = \omega^2 ) (where ( \omega > 0 )). The spatial ODE becomes ( X'' + \omega^2 X = 0 ), with oscillating solutions ( X(x) = A \cos(\omega x) + B \sin(\omega x) ). These are bounded as ( x \to \infty ), so we keep this case—especially useful for homogeneous boundary conditions (( g(t) = 0 )).

For the time ODE, the solution for ( \lambda = \omega^2 > 0 ) is ( T(t) = C e^{-k \omega^2 t} ). Combining with spatial solutions gives a family of valid solutions:

• For non-homogeneous boundaries: ( u_\omega(x, t) = D(\omega) e^{-\omega x} e^{-k \omega^2 t} )
• For homogeneous boundaries (( u(0,t)=0 )): ( u_\omega(x, t) = D(\omega) \sin(\omega x) e^{-k \omega^2 t} ) (since ( X(0)=0 ) requires ( A=0 ))

Since ( \omega ) can take any positive real value (continuous spectrum), we combine these solutions with an integral instead of a discrete series:

• Non-homogeneous boundary (( g(t) \neq 0 )):
  [
  u(x, t) = \int_0^\infty D(\omega) e^{-\omega x} e^{-k \omega^2 t} d\omega
  ]
• Homogeneous boundary (( g(t) = 0 )):
  [
  u(x, t) = \int_0^\infty D(\omega) \sin(\omega x) e^{-k \omega^2 t} d\omega
  ]

We determine ( D(\omega) ) using our problem's constraints:

• Homogeneous boundary, non-homogeneous initial condition:
  Apply ( u(x, 0) = f(x) ), which gives ( f(x) = \int_0^\infty D(\omega) \sin(\omega x) d\omega ). This is the Fourier sine transform of ( f(x) ), so we solve for ( D(\omega) ):
  [
  D(\omega) = \frac{2}{\pi} \int_0^\infty f(\xi) \sin(\omega \xi) d\xi
  ]
  Substituting back gives a solution equivalent to the odd-domain extension method, but derived directly via separation of variables.

• Non-homogeneous boundary, homogeneous initial condition:
  Apply ( u(0, t) = g(t) ), which gives ( g(t) = \int_0^\infty D(\omega) e^{-k \omega^2 t} d\omega ). This is a Laplace transform relation; solving for ( D(\omega) ) and substituting back leads to the solution:
  [
  u(x,t) = \frac{x}{2\sqrt{\pi k^3 t^3}} \int_0^t g(\tau) e^{-x2/(4k(t-\tau))} d\tau
  ]

To answer your core question: yes, separation of variables is completely feasible for this problem. The key distinction from finite domains is that we use an integral (continuous sum) instead of a discrete series, since eigenvalues form a continuous spectrum rather than a discrete set. You don't need to extend the domain to the full real line—though the final solution often aligns with extension method results (due to links between integral transforms and domain extensions), you're directly applying separation of variables on the semi-infinite domain itself.

内容的提问来源于stack exchange，提问作者will_cheuk