证明热方程柯西问题的解在t>0时关于x为偶函数

Alright, let's work through this problem step by step. We're given the heat equation Cauchy problem:

$$\left{ \begin{array}{l l} u_{t} - \kappa u_{xx}=0 & \quad \mbox{$x \in \mathbb{R}, t>0$,}\ \quad u(x,0) = \psi(x), \end{array} \right. $$

where $\kappa > 0$ is a constant, and $\psi(x)$ is an even function (meaning $\psi(-x) = \psi(x)$ for all $x \in \mathbb{R}$). We need to prove that for every $t>0$, the solution $u(x,t)$ is also even in $x$: $u(-x,t) = u(x,t)$.

Method 1: Using Solution Uniqueness (Construct an Auxiliary Function) #

This is a standard symmetry argument for PDEs:

1. Define the auxiliary function: Let $v(x,t) = u(-x, t)$. Our goal is to show $v(x,t) = u(x,t)$.
2. Verify $v$ satisfies the heat equation:
   • Compute the time derivative: $v_t = \frac{\partial}{\partial t}u(-x,t) = u_t(-x,t)$ (since differentiating with respect to $t$ doesn't affect the $-x$ term).
   • Compute the second spatial derivative: First, $v_x = \frac{\partial}{\partial x}u(-x,t) = -u_x(-x,t)$. Then, $v_{xx} = \frac{\partial}{\partial x}\left(-u_x(-x,t)\right) = u_{xx}(-x,t)$.
   • Substitute into the heat equation:
     $$v_t - \kappa v_{xx} = u_t(-x,t) - \kappa u_{xx}(-x,t)$$
     Since $u$ solves the original heat equation, $u_t(y,t) - \kappa u_{xx}(y,t) = 0$ for all $y \in \mathbb{R}$. Letting $y = -x$, this gives us $v_t - \kappa v_{xx} = 0$. So $v$ satisfies the same heat equation as $u$.
3. Check the initial condition:
   When $t=0$, $v(x,0) = u(-x,0) = \psi(-x)$. Since $\psi$ is even, $\psi(-x) = \psi(x) = u(x,0)$. So $v$ has the same initial data as $u$.
4. Apply uniqueness:
   The Cauchy problem for the heat equation has a unique solution (under mild conditions on $\psi$, like bounded continuity or integrability, which are implied here since a solution exists). Therefore, $v(x,t) = u(x,t)$, which means $u(-x,t) = u(x,t)$.

Method 2: Direct Calculation Using the Fundamental Solution #

If you prefer working with the explicit solution formula, here's another way:
The solution to the heat equation Cauchy problem can be written using the fundamental solution (heat kernel):
$$u(x,t) = \frac{1}{\sqrt{4\pi\kappa t}} \int_{\mathbb{R}} \psi(y) e^{-(x-y)2/(4\kappa t)} dy$$
Now compute $u(-x,t)$:
$$u(-x,t) = \frac{1}{\sqrt{4\pi\kappa t}} \int_{\mathbb{R}} \psi(y) e^{-(-x-y)2/(4\kappa t)} dy = \frac{1}{\sqrt{4\pi\kappa t}} \int_{\mathbb{R}} \psi(y) e^{-(x+y)2/(4\kappa t)} dy$$
Make a substitution: let $z = -y$, so $dy = -dz$. When $y \to -\infty$, $z \to \infty$, and vice versa—this flips the integral bounds, canceling out the negative sign:
$$u(-x,t) = \frac{1}{\sqrt{4\pi\kappa t}} \int_{\mathbb{R}} \psi(-z) e^{-(x-z)2/(4\kappa t)} dz$$
Since $\psi$ is even, $\psi(-z) = \psi(z)$. Substitute this in, and we get:
$$u(-x,t) = \frac{1}{\sqrt{4\pi\kappa t}} \int_{\mathbb{R}} \psi(z) e^{-(x-z)2/(4\kappa t)} dz = u(x,t)$$
This directly shows $u(-x,t) = u(x,t)$.

内容的提问来源于stack exchange，提问作者KBG