经典二阶偏微分方程的解法 (Solutions to Classical Second-Order PDEs)
拉普拉斯方程 (Laplace’s Equation)
定义 (Definition)
拉普拉斯方程的形式为：
The form of Laplace’s Equation is:
其中 是拉普拉斯算子。
where is the Laplacian operator.
解法 (Solution Method)
分离变量法 (Method of Separation of Variables)
-
假设解 (Assume the Solution)
假设解为
Assume the solution is
-
分离变量 (Separate Variables)
将拉普拉斯方程代入假设解并分离变量：
Substitute the assumed solution into Laplace’s equation and separate variables:
\frac{X”(x)}{X(x)} = -\frac{Y”(y)}{Y(y)} = \lambda
这里 $\lambda$ 是分离常数。 Here $\lambda$ is the separation constant. 3. **求解 ODE (Solve the ODEs)** 得到两个常微分方程： Obtain two ordinary differential equations:X”(x) - \lambda X(x) = 0
Y”(y) + \lambda Y(y) = 0
根据 $\lambda$ 的值，解这些方程得到 $X(x)$ 和 $Y(y)$ 的通解。 Solve these equations for $X(x)$ and $Y(y)$ depending on the value of $\lambda$ to get the general solutions. 4. **构造通解 (Construct the General Solution)** 将 $X(x)$ 和 $Y(y)$ 的解结合起来，构造 $u(x, y)$ 的通解。 Combine the solutions of $X(x)$ and $Y(y)$ to construct the general solution of $u(x, y)$. 5. **应用边界条件 (Apply Boundary Conditions)** 使用边界条件确定特解。 Use boundary conditions to determine the particular solution. ## 热传导方程 (Heat Equation) ### 定义 (Definition) 热传导方程的形式为： The form of the Heat Equation is:u_t = \alpha \Delta u
其中 $u_t = \frac{\partial u}{\partial t}$，$\alpha$ 是热扩散系数。 where $u_t = \frac{\partial u}{\partial t}$, $\alpha$ is the thermal diffusivity. ### 解法 (Solution Method) **分离变量法 (Method of Separation of Variables)** 1. **假设解 (Assume the Solution)** 假设解为 $u(x, t) = X(x)T(t)$ Assume the solution is $u(x, t) = X(x)T(t)$ 2. **分离变量 (Separate Variables)** 将热传导方程代入假设解并分离变量： Substitute the assumed solution into the heat equation and separate variables:X(x)T’(t) = \alpha X”(x)T(t)
除以 $\alpha X(x)T(t)$ 并重排： Divide by $\alpha X(x)T(t)$ and rearrange:\frac{T’(t)}{\alpha T(t)} = \frac{X”(x)}{X(x)} = -\lambda
这里 $\lambda$ 是分离常数。 Here $\lambda$ is the separation constant. 3. **求解 ODE (Solve the ODEs)** 得到两个常微分方程： Obtain two ordinary differential equations:T’(t) + \alpha \lambda T(t) = 0
X”(x) + \lambda X(x) = 0
解这些方程得到 $X(x)$ 和 $T(t)$ 的通解。 Solve these equations for $X(x)$ and $T(t)$ to get the general solutions. 4. **构造通解 (Construct the General Solution)** 将 $X(x)$ 和 $T(t)$ 的解结合起来，构造 $u(x, t)$ 的通解。 Combine the solutions of $X(x)$ and $T(t)$ to construct the general solution of $u(x, t)$. 5. **应用初始条件和边界条件 (Apply Initial and Boundary Conditions)** 使用初始条件和边界条件确定特解。 Use initial and boundary conditions to determine the particular solution. ## 波动方程 (Wave Equation) ### 定义 (Definition) 波动方程的形式为： The form of the Wave Equation is:u_{tt} = c^2 \Delta u
其中 $u_{tt} = \frac{\partial^2 u}{\partial t^2}$，$c$ 是波速。 where $u_{tt} = \frac{\partial^2 u}{\partial t^2}$, $c$ is the wave speed. ### 解法 (Solution Method) **分离变量法 (Method of Separation of Variables)** 1. **假设解 (Assume the Solution)** 假设解为 $u(x, t) = X(x)T(t)$ Assume the solution is $u(x, t) = X(x)T(t)$ 2. **分离变量 (Separate Variables)** 将波动方程代入假设解并分离变量： Substitute the assumed solution into the wave equation and separate variables:X(x)T”(t) = c^2 X”(x)T(t)
除以 $c^2 X(x)T(t)$ 并重排： Divide by $c^2 X(x)T(t)$ and rearrange:\frac{T”(t)}{c^2 T(t)} = \frac{X”(x)}{X(x)} = -\lambda
这里 $\lambda$ 是分离常数。 Here $\lambda$ is the separation constant. 3. **求解 ODE (Solve the ODEs)** 得到两个常微分方程： Obtain two ordinary differential equations:T”(t) + c^2 \lambda T(t) = 0
X”(x) + \lambda X(x) = 0
解这些方程得到 $X(x)$ 和 $T(t)$ 的通解。 Solve these equations for $X(x)$ and $T(t)$ to get the general solutions. 4. **构造通解 (Construct the General Solution)** 将 $X(x)$ 和 $T(t)$ 的解结合起来，构造 $u(x, t)$ 的通解。 Combine the solutions of $X(x)$ and $T(t)$ to construct the general solution of $u(x, t)$. 5. **应用初始条件和边界条件 (Apply Initial and Boundary Conditions)** 使用初始条件和边界条件确定特解。 Use initial and boundary conditions to determine the particular solution. ## 解法示例 (Example Solutions) ### 拉普拉斯方程示例 (Laplace's Equation Example) #### 方程 (Equation) 求解以下拉普拉斯方程： Solve the following Laplace's equation:\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0
在矩形区域 $0 \leq x \leq a$ 和 $0 \leq y \leq b$，边界条件为： In a rectangular region $0 \leq x \leq a$ and $0 \leq y \leq b$, with boundary conditions:u(0, y) = u(a, y) = 0, \quad u(x, 0) = 0, \quad u(x, b) = f(x)
#### 解法 (Solution Method) 1. **假设解 (Assume the Solution)** 假设解为 $u(x, y) = X(x)Y(y)$ Assume the solution is $u(x, y) = X(x)Y(y)$ 2. **分离变量 (Separate Variables)** 将假设解代入拉普拉斯方程： Substitute the assumed solution into Laplace's equation:X”(x)Y(y) + X(x)Y”(y) = 0
除以 $X(x)Y(y)$ 并重排： Divide by $X(x)Y(y)$ and rearrange:\frac{X”(x)}{X(x)} = -\frac{Y”(y)}{Y(y)} = \lambda
X”(x) + \lambda X(x) = 0
Y”(y) - \lambda Y(y) = 0
设 $\lambda = \left(\frac{n\pi}{a}\right)^2$，得到解： Assume $\lambda = \left(\frac{n\pi}{a}\right)^2$, obtain solutions:X_n(x) = \sin\left(\frac{n\pi x}{a}\right)
Y_n(y) = A_n \sinh\left(\frac{n\pi y}{a}\right)
4. **构造通解 (Construct the General Solution)** 结合边界条件 $u(x, 0) = 0$ 和 $u(x, b) = f(x)$，构造解： Combine with boundary conditions $u(x, 0) = 0$ and $u(x, b) = f(x)$, construct the solution:u(x, y) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{a}\right) \sinh\left(\frac{n\pi y}{a}\right)
利用傅里叶级数展开 $f(x)$ 来确定 $B_n$： Determine $B_n$ using Fourier series expansion of $f(x)$:f(x) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{a}\right)
### 热传导方程示例 (Heat Equation Example) #### 方程 (Equation) 求解以下热传导方程： Solve the following heat equation:u_t = \alpha \frac{\partial^2 u}{\partial x^2}
在区间 $0 \leq x \leq L$，初始条件和边界条件为： In the interval $0 \leq x \leq L$, with initial and boundary conditions:u(x, 0) = f(x), \quad u(0, t) = u(L, t) = 0
#### 解法 (Solution Method) 1. **假设解 (Assume the Solution)** 假设解为 $u(x, t) = X(x)T(t)$ Assume the solution is $u(x, t) = X(x)T(t)$ 2. **分离变量 (Separate Variables)** 将假设解代入热传导方程： Substitute the assumed solution into the heat equation:X(x)T’(t) = \alpha X”(x)T(t)
除以 $X(x)T(t)$ 并重排： Divide by $X(x)T(t)$ and rearrange:\frac{T’(t)}{\alpha T(t)} = \frac{X”(x)}{X(x)} = -\lambda
T’(t) + \alpha \lambda T(t) = 0
X”(x) + \lambda X(x) = 0
设 $\lambda = \left(\frac{n\pi}{L}\right)^2$，得到解： Assume $\lambda = \left(\frac{n\pi}{L}\right)^2$, obtain solutions:X_n(x) = \sin\left(\frac{n\pi x}{L}\right)
T_n(t) = e^{-\alpha \left(\frac{n\pi}{L}\right)^2 t}
4. **构造通解 (Construct the General Solution)** 结合初始条件 $u(x, 0) = f(x)$，构造解： Combine with initial condition $u(x, 0) = f(x)$, construct the solution:u(x, t) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha \left(\frac{n\pi}{L}\right)^2 t}
利用傅里叶级数展开 $f(x)$ 来确定 $B_n$： Determine $B_n$ using Fourier series expansion of $f(x)$:f(x) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{L}\right)
### 波动方程示例 (Wave Equation Example) #### 方程 (Equation) 求解以下波动方程： Solve the following wave equation:u_{tt} = c^2 \frac{\partial^2 u}{\partial x^2}
在区间 $0 \leq x \leq L$，初始条件和边界条件为： In the interval $0 \leq x \leq L$, with initial and boundary conditions:u(x, 0) = f(x), \quad u_t(x, 0) = g(x), \quad u(0, t) = u(L, t) = 0
#### 解法 (Solution Method) 1. **假设解 (Assume the Solution)** 假设解为 $u(x, t) = X(x)T(t)$ Assume the solution is $u(x, t) = X(x)T(t)$ 2. **分离变量 (Separate Variables)** 将假设解代入波动方程： Substitute the assumed solution into the wave equation:X(x)T”(t) = c^2 X”(x)T(t)
除以 $c^2 X(x)T(t)$ 并重排： Divide by $c^2 X(x)T(t)$ and rearrange:\frac{T”(t)}{c^2 T(t)} = \frac{X”(x)}{X(x)} = -\lambda
T”(t) + c^2 \lambda T(t) = 0
X”(x) + \lambda X(x) = 0
设 $\lambda = \left(\frac{n\pi}{L}\right)^2$，得到解： Assume $\lambda = \left(\frac{n\pi}{L}\right)^2$, obtain solutions:X_n(x) = \sin\left(\frac{n\pi x}{L}\right)
T_n(t) = A_n \cos\left(\frac{n\pi c t}{L}\right) + B_n \sin\left(\frac{n\pi c t}{L}\right)
4. **构造通解 (Construct the General Solution)** 结合初始条件 $u(x, 0) = f(x)$ 和 $u_t(x, 0) = g(x)$，构造解： Combine with initial conditions $u(x, 0) = f(x)$ and $u_t(x, 0) = g(x)$, construct the solution:u(x, t) = \sum_{n=1}^{\infty} \left[ A_n \cos\left(\frac{n\pi c t}{L}\right) + B_n \sin\left(\frac{n\pi c t}{L}\right) \right] \sin\left(\frac{n\pi x}{L}\right)
利用傅里叶级数展开 $f(x)$ 和 $g(x)$ 来确定 $A_n$ 和 $B_n$： Determine $A_n$ and $B_n$ using Fourier series expansion of $f(x)$ and $g(x)$:f(x) = \sum_{n=1}^{\infty} A_n \sin\left(\frac{n\pi x}{L}\right)
g(x) = \sum_{n=1}^{\infty} \frac{n\pi c}{L} B_n \sin\left(\frac{n\pi x}{L}\right)