使用拉普拉斯变换求解PDE时存疑，请求检查推导过程

Hey there! Let's break down your work and solve this PDE step by step.

First: Your Laplace Transform Steps Are Correct! #

Let's verify each step you took:

• You applied the Laplace transform (with respect to $t$) to both sides of the PDE $u_x -4xu_t=0$, defining $W(x,s) = \mathcal{L}[u(x,t)]$. This is valid since we're treating $x$ as a parameter during the transform.
• For $\mathcal{L}[u_x]$: Since differentiation with respect to $x$ commutes with the Laplace transform (assuming sufficient smoothness), this becomes $W_x(x,s)$ — correct.
• For $\mathcal{L}[4xu_t]$: Using the derivative property $\mathcal{L}[u_t] = sW(x,s) - u(x,0)$, multiplying by $4x$ gives $4x(sW(x,s)-u(x,0))$. Since $u(x,0)=0$, this simplifies to $4xsW(x,s)$ — also correct.
• The boundary condition transform $\mathcal{L}[u(0,t)] = \mathcal{L}[t+3] = \frac{1}{s^2} + \frac{3}{s}$ is perfect.

So your reduced ODE system:
$$
\begin{cases}
W_x(x,s)-4xsW(x,s)=0\
W(0,s)=\frac{1}{s^2}+\frac{3}{s}
\end{cases}
$$
is 100% accurate.

Next: Solve the ODE for $W(x,s)$ #

This is a first-order linear separable ODE. Let's solve it properly:

1. Separate variables: $\frac{dW}{W} = 4xs,dx$
2. Integrate both sides: $\ln|W(x,s)| = 2sx^2 + C(s)$ (the constant depends on $s$, since $x$ is our variable here)
3. Exponentiate: $W(x,s) = C(s)e^{2sx2}$
4. Apply the boundary condition $W(0,s)=C(s)=\frac{1}{s^2}+\frac{3}{s}$, so:
   $$
   W(x,s) = \left(\frac{1}{s^2} + \frac{3}{s}\right)e^{2sx2}
   $$

Critical Note: Incompatible Initial & Boundary Conditions #

Before computing the inverse Laplace transform, we need to address a problem: your initial condition $u(x,0)=0$ and boundary condition $u(0,t)=t+3$ conflict at $(x,t)=(0,0)$:

• From $u(x,0)=0$, we get $u(0,0)=0$
• From $u(0,t)=t+3$, we get $u(0,0)=3$

This means the problem as stated is ill-posed (no classical smooth solution exists). However, we can still find a generalized solution (including distributional terms like delta functions) or assume there was a typo in the initial condition.

Case 1: Corrected Initial Condition (Compatible with Boundary) #

If we assume the initial condition was meant to be $u(x,0)=2x^2+3$ (which matches $u(0,0)=3$), then we can use the method of characteristics to find a classical solution quickly:

• The PDE $u_x=4xu_t$ has characteristic curves defined by $\frac{dx}{4x} = dt$, integrating gives $t = 2x^2 + C$ (or $C = t - 2x^2$).
• $u$ is constant along characteristic curves, so $u(x,t) = u(0, t+2x^2) = (t+2x^2)+3 = t+2x^2+3$.
• Verify: $u_x=4x$, $4xu_t=4x\cdot1=4x$, so $u_x-4xu_t=0$; $u(x,0)=2x^2+3$; $u(0,t)=t+3$. Perfect.

Case 2: Original (Conflicting) Conditions — Generalized Solution #

If we stick with $u(x,0)=0$, we need to compute the inverse Laplace transform of $W(x,s)$:
$$
u(x,t) = \mathcal{L}^{-1}\left[\left(\frac{3}{s} + \frac{1}{s^2}\right)e{2sx^2}\right]
$$
Using Laplace transform properties for exponential terms (and recognizing that $e^{as}$ corresponds to "forward shifts" which introduce distributional terms for $t < -a$):

• For $t + 2x^2 > 0$ (which is always true since $t\geq0$ and $2x^2\geq0$), we can use the property $\mathcal{L}^{-1}[e{as}F(s)] = f(t+a)u(t+a)$ where $u(\cdot)$ is the unit step function. But since $t+a = t+2x^2 >0$, $u(t+a)=1$, so:
  $$
  u(x,t) = 3u(t+2x^2) + (t+2x^2)u(t+2x2) = t+2x^2+3
  $$
  However, this violates the initial condition $u(x,0)=0$. To resolve the conflict, we need to subtract the part that doesn't satisfy the initial condition, which introduces a delta function term:
  $$
  u(x,t) = (t+2x^2+3) - (2x^2+3)\delta(t)
  $$
  This generalized solution satisfies both the PDE, the boundary condition, and the initial condition (since $\delta(t)=0$ for $t>0$, and at $t=0$, the delta term cancels the $2x^2+3$ part to give $u(x,0)=0$).

内容的提问来源于stack exchange，提问作者Clyde A. Jansen