含空间一阶、二阶导数的PDE分离变量法求解技术咨询

Got it, let's work through this PDE step by step — that first-order spatial derivative looks tricky at first glance, but we can use a simple exponential substitution to turn it into a standard heat equation you probably already know how to solve.

Your PDE is of the form:
$$u_t - u_{xx} - 2u_x = 0$$
We can use an exponential transformation to eliminate the $u_x$ term. Let's define a new function $v(x,t)$ such that:
$$u(x,t) = e^{ax + bt}v(x,t)$$
Our goal is to choose $a$ and $b$ to cancel out the first-order derivative (and any constant terms if possible).

First, compute the necessary derivatives of $u$:

• $u_x = e^{ax+bt}(av + v_x)$
• $u_{xx} = e^{ax+bt}(a2v + 2av_x + v_{xx})$
• $u_t = e^{ax+bt}(bv + v_t)$

Substitute these into the original PDE, then divide both sides by $e^{ax+bt}$ (since it's never zero):
$$bv + v_t - (a^2v + 2av_x + v_{xx}) - 2(av + v_x) = 0$$

Now group terms by the order of derivatives of $v$:
$$v_t - v_{xx} + (-2a - 2)v_x + (b - a^2 - 2a)v = 0$$

To eliminate the $v_x$ term, set its coefficient to zero:
$$-2a - 2 = 0 \implies a = -1$$

Next, substitute $a=-1$ into the coefficient of the $v$ term and set it to zero to simplify further:
$$b - (-1)^2 - 2(-1) = 0 \implies b - 1 + 2 = 0 \implies b = -1$$

So our substitution simplifies to:
$$u(x,t) = e^{-x - t}v(x,t)$$
And substituting back into the original PDE gives us the standard heat equation:
$$v_t - v_{xx} = 0$$

Now let's convert the original conditions to apply to $v$:

• Boundary conditions: $u(0,t)=0$ and $u(\ell,t)=0$ translate directly to $v(0,t)=0$ and $v(\ell,t)=0$ (since $e^{-x-t}$ is never zero)
• Initial condition: Substitute $t=0$ into the substitution:
  $$u(x,0) = e^{-x}v(x,0) = 2e^{-x}\sin\left(\frac{3\pi x}{\ell}\right)$$
  Divide both sides by $e^{-x}$ to get:
  $$v(x,0) = 2\sin\left(\frac{3\pi x}{\ell}\right)$$

Now the problem for $v$ is the familiar homogeneous Dirichlet heat equation:
$$\begin{cases} v_{t}(x,t)-v_{xx}(x,t)=0\qquad \mathrm {for};(x,t)\in(0,\ell)\times\mathbb R^{+} \ v(x,0)=2\sin(\frac{3\pi x}{\ell})\qquad\qquad;;; \mathrm{for};x \in (0,\ell) \ v(0,t)=v(\ell,t)=0\qquad\qquad\qquad;\mathrm{for} ;t\in\mathbb{R^{+}} \end{cases} $$

For the heat equation with homogeneous Dirichlet boundary conditions, we know the solution is a sum of separated solutions:
$$v(x,t) = \sum_{n=1}^\infty C_n e^{{-\left(\frac{n\pi}{\ell}\right)}2 t} \sin\left(\frac{n\pi x}{\ell}\right)$$

Using the initial condition to find coefficients $C_n$:
$$v(x,0) = \sum_{n=1}^\infty C_n \sin\left(\frac{n\pi x}{\ell}\right) = 2\sin\left(\frac{3\pi x}{\ell}\right)$$

This is a Fourier sine series, so we can directly read off the coefficients:

• $C_3 = 2$
• All other $C_n = 0$ for $n \neq 3$

So the solution for $v$ is:
$$v(x,t) = 2e^{{-\left(\frac{3\pi}{\ell}\right)}2 t} \sin\left(\frac{3\pi x}{\ell}\right)$$

Now substitute $v(x,t)$ back into our original substitution:
$$u(x,t) = e^{-x - t} \cdot 2e^{-\frac{9\pi2}{\ell^2}t} \sin\left(\frac{3\pi x}{\ell}\right)$$

Combine the exponential terms:
$$u(x,t) = 2e^{-x - t\left(1 + \frac{9\pi^2}{\ell2}\right)} \sin\left(\frac{3\pi x}{\ell}\right)$$

To make sure this works, you can:

• Plug $u(x,t)$ back into the original PDE to confirm it satisfies the equation
• Check that $u(x,0)$ matches the initial condition
• Confirm $u(0,t)=u(\ell,t)=0$ for all $t>0$

All of these should hold true!

内容的提问来源于stack exchange，提问作者muserock92