分离变量法求解带初边值条件的PDE：时间导数初值困惑

Hey there! Let’s work through how to handle that tricky initial condition involving the first derivative of the time component—this is a common sticking point, so you’re not alone here.

Step 1: Write the Full Solution Form #

First, remember that when using separation of variables, the full solution to your PDE is the product of the spatial and temporal components:
$$u(x,t) = X(x)T(t) = \left[a\cos(\alpha x) + b\sin(\alpha x)\right]ce^{-\lambda t}$$
We can simplify this by combining the constants: let $A = ac$ and $B = bc$, so the solution becomes:
$$u(x,t) = \left[A\cos(\alpha x) + B\sin(\alpha x)\right]e^{-\lambda t}$$
(where $\lambda = \alpha^2$, as you noted)

Step 2: Compute the Time Derivative #

Next, take the first partial derivative of $u(x,t)$ with respect to $t$:
$$\frac{\partial u}{\partial t}(x,t) = \left[A\cos(\alpha x) + B\sin(\alpha x)\right] \cdot (-\lambda)e^{-\lambda t}$$

Step 3: Apply the Initial Condition at $t=0$ #

Your initial condition is likely of the form $\frac{\partial u}{\partial t}(x,0) = g(x)$ (where $g(x)$ is the given initial time-derivative distribution). When $t=0$, $e^{-\lambda \cdot 0} = 1$, so this simplifies to:
$$\frac{\partial u}{\partial t}(x,0) = -\lambda\left[A\cos(\alpha x) + B\sin(\alpha x)\right] = g(x)$$

Step 4: Solve for Constants (Using Boundary Conditions & Orthogonality) #

Here’s where you’ll tie in your boundary conditions (e.g., $u(0,t)=0$, $u(L,t)=0$) to find valid values of $\alpha$ (eigenvalues), then use the orthogonality of trigonometric functions to solve for $A$ and $B$.

For example, if your boundary conditions are $u(0,t)=0$ and $u(L,t)=0$:

• Plugging $x=0$ into $X(x)$ gives $A\cos(0) + B\sin(0) = A = 0$, so $X(x) = B\sin(\alpha x)$
• Plugging $x=L$ gives $B\sin(\alpha L) = 0$, so $\alpha L = n\pi$ for $n=1,2,3,...$ → $\alpha = \frac{n\pi}{L}$, $\lambda = \left(\frac{n\pi}{L}\right)^2$

Your solution now becomes a sum over all valid $n$:
$$u(x,t) = \sum_{n=1}^\infty B_n\sin\left(\frac{n\pi x}{L}\right)e^{{-\left(\frac{n\pi}{L}\right)}2 t}$$

Taking the time derivative and applying $t=0$:
$$\frac{\partial u}{\partial t}(x,0) = \sum_{n=1}^\infty -B_n\left(\frac{n\pi}{L}\right)^2\sin\left(\frac{n\pi x}{L}\right) = g(x)$$

You can now find $B_n$ by matching the Fourier sine series coefficients of $g(x)$:
$$-B_n\left(\frac{n\pi}{L}\right)^2 = \frac{2}{L}\int_0^L g(x)\sin\left(\frac{n\pi x}{L}\right)dx$$
Solve for $B_n$, and you’re done!

Key Takeaway #

The core idea is:

• Combine constants in your solution to simplify notation
• Compute the time derivative and substitute $t=0$
• Use boundary conditions to find valid eigenvalues ($\alpha$ values)
• Leverage orthogonality of your spatial basis functions (cosine/sine) to solve for the remaining constants

内容的提问来源于stack exchange，提问作者muserock92