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# Partial Differential Equations ## 1. Basic Concepts ### 1.1 Definition * A partial differential equation (PDE) is an equation involving an unknown function of two or more variables and certain of its partial derivatives. * The **order** of a PDE is the order of the highest derivative appea...

# Partial Differential Equations ## 1. Basic Concepts ### 1.1 Definition * A partial differential equation (PDE) is an equation involving an unknown function of two or more variables and certain of its partial derivatives. * The **order** of a PDE is the order of the highest derivative appearing in the equation. ### 1.2 Examples 1. $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 0$ (1st order) 2. $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ (2nd order, Laplace equation) 3. $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ (2nd order, Heat equation) 4. $\frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2}$ (2nd order, Wave equation) * **Linear PDE**: A PDE is linear if the unknown function and its derivatives appear linearly. * Examples 1-4 above are linear. * **Nonlinear PDE**: A PDE is nonlinear if it is not linear. * Example: $\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} = 0$ (Burger's equation, 2nd order) ### 1.3 Solutions * A **solution** of a PDE is a function that satisfies the equation in some region of the independent variables. * In general, the solution of a PDE is not unique. We need additional conditions to determine a unique solution. * **Initial conditions**: Conditions on the unknown function and its derivatives at a particular value of time. * Example: $u(x, 0) = f(x), \frac{\partial u}{\partial t}(x, 0) = g(x)$ * **Boundary conditions**: Conditions on the unknown function and its derivatives at a particular location in space. * Example: $u(0, t) = a(t), u(L, t) = b(t)$ ## 2. First-Order Linear PDEs ### 2.1 General Form $a(x, y) \frac{\partial u}{\partial x} + b(x, y) \frac{\partial u}{\partial y} + c(x, y) u = f(x, y)$ ### 2.2 Method of Characteristics * Consider the PDE: $a(x, y) \frac{\partial u}{\partial x} + b(x, y) \frac{\partial u}{\partial y} = 0$ * Let $x = x(s), y = y(s)$, and $u(x, y) = u(x(s), y(s)) = v(s)$. * Then, $\frac{dv}{ds} = \frac{\partial u}{\partial x} \frac{dx}{ds} + \frac{\partial u}{\partial y} \frac{dy}{ds}$ * If we choose $x(s)$ and $y(s)$ such that $\frac{dx}{ds} = a(x, y), \frac{dy}{ds} = b(x, y)$ * Then, $\frac{dv}{ds} = 0 \implies v(s) = const.$ * The curves defined by $\frac{dx}{a} = \frac{dy}{b}$ are called **characteristic curves**. * The general solution is of the form $u(x, y) = F(\phi(x, y))$ where $\phi(x, y) = C$ defines the characteristic curves and $F$ is an arbitrary function. ### 2.3 Example * Solve $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 0$ with $u(x, 0) = f(x)$. * Here, $a = 1, b = 1$. The characteristic equations are $\frac{dx}{ds} = 1, \frac{dy}{ds} = 1$ * Integrating, we get $x = s + c_1, y = s + c_2 \implies x - y = c_1 - c_2 = C$ * Thus, $\phi(x, y) = x - y$. * The general solution is $u(x, y) = F(x - y)$ for some arbitrary function $F$. * Using the initial condition, $u(x, 0) = f(x) = F(x - 0) = F(x)$. Thus, $F(x) = f(x)$. * Therefore, the solution is $u(x, y) = f(x - y)$.