Method of Characteristics Examples

PDEs · Axiom Academy

EXAMPLE Method of Characteristics Solving first-order PDEs by following characteristic curves Problem 1: Linear Transport Equation Solve the first-order PDE: u x + 2u y = 0 with initial condition u(x, 0) = sin(x) Characteristic Curves in the xy-plane The characteristic lines y = 2x + C show the directions along which the solution propagates Solve the first-order PDE: xu x + yu y = u with initial condition u(x, 0) = x 2 Method of Characteristics Summary Excellent work! You've successfully solved first-order PDEs using the method of characteristics. Here's what we learned: General Form: For a first-order PDE of the form a(x,y,u)u x + b(x,y,u)u y = c(x,y,u), we transform it into a system of ODEs along characteristic curves. Characteristic Equations: The system dx/ds = a, dy/ds = b, du/ds = c describes how x, y, and u vary along characteristic curves parametrized by s. Linear Case: When a and b are constant, the characteristics are straight lines. The solution is constant along these lines for homogeneous PDEs (c = 0). With Source Term: When c is not zero, u changes along characteristics according to du/ds = c. This leads to exponential growth or decay along characteristic curves. Initial Conditions: The initial condition determines the value of u at s = 0 for each characteristic. These values propagate along their respective characteristic curves.

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