Math Problem Statement

Solve the 1st order quasilinear PDE with the equation u ∂u/∂x + y ∂u/∂y = x and the initial condition u(x, 0) = h(x) for some C¹ function h(x).

Solution

This is a first-order quasilinear partial differential equation (PDE). To solve it, we will use the method of characteristics. The problem can be summarized as:

uux+yuy=x,with initial condition u(x,0)=h(x), for some C1 function h(x).u \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x, \quad \text{with initial condition } u(x, 0) = h(x), \text{ for some } C^1 \text{ function } h(x).


Step-by-step solution:

1. Characteristic equations:

The PDE can be rewritten in terms of characteristic equations:

dxu=dyy=dux.\frac{dx}{u} = \frac{dy}{y} = \frac{du}{x}.

This gives us three equations to solve.


2. Solve dxu=dyy\frac{dx}{u} = \frac{dy}{y}:

Cross-multiply: ydx=udy    dyy=dxu.y \, dx = u \, dy \quad \implies \quad \frac{dy}{y} = \frac{dx}{u}.

Integrate both sides: lny=lnu+C1,where C1 is a constant.\ln y = \ln u + C_1, \quad \text{where } C_1 \text{ is a constant.}

Exponentiate: y=ueC1.y = u \cdot e^{C_1}.

Let eC1=k1e^{C_1} = k_1, so: y=k1u.y = k_1 u.


3. Solve dxu=dux\frac{dx}{u} = \frac{du}{x}:

Cross-multiply: xdx=udu.x \, dx = u \, du.

Integrate both sides: x22=u22+C2.\frac{x^2}{2} = \frac{u^2}{2} + C_2.

Simplify: x2u2=2C2.x^2 - u^2 = 2C_2.

Let 2C2=k22C_2 = k_2, so: x2u2=k2.x^2 - u^2 = k_2.


4. Combine the solutions:

From Step 2, we know y=k1uy = k_1 u. Substituting this into Step 3, we have: x2y2k12=k2.x^2 - \frac{y^2}{k_1^2} = k_2.

Rewriting: x2y2k12=k2    x2y2u2=k2.x^2 - \frac{y^2}{k_1^2} = k_2 \quad \implies \quad x^2 - \frac{y^2}{u^2} = k_2.


5. Apply the initial condition:

The initial condition is u(x,0)=h(x)u(x, 0) = h(x). At y=0y = 0, the characteristic equations reduce to: x2u2=k2,and u=h(x).x^2 - u^2 = k_2, \quad \text{and } u = h(x).

Substitute u=h(x)u = h(x) into the first equation: x2h(x)2=k2.x^2 - h(x)^2 = k_2.

Thus, k2=x2h(x)2k_2 = x^2 - h(x)^2.


6. Final solution:

The general solution is obtained by substituting k2=x2h(x)2k_2 = x^2 - h(x)^2 into the characteristic equation: x2u2=x2h(x)2.x^2 - u^2 = x^2 - h(x)^2.

Simplify: u=h(x).u = h(x).

This determines u(x,y)u(x, y) as a function of h(x)h(x).


Would you like detailed clarification on any part of this derivation? Here are five related questions to explore further:

  1. What are the general steps in the method of characteristics for solving quasilinear PDEs?
  2. How do we verify if a solution satisfies the original PDE and initial condition?
  3. Why do we solve characteristic equations in pairs during this method?
  4. What is the significance of the initial curve in solving quasilinear PDEs?
  5. How does the form of the initial condition affect the solution to the PDE?

Tip: The method of characteristics works best for first-order quasilinear PDEs where characteristics can be explicitly parameterized. Always check if the problem is hyperbolic before applying this technique!

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Math Problem Analysis

Mathematical Concepts

Partial Differential Equations
Quasilinear PDEs
Method of Characteristics

Formulas

Characteristic equations: dx/u = dy/y = du/x
Solution of characteristic equations: y = k₁u, x² - u² = k₂

Theorems

Method of Characteristics: A technique for solving first-order quasilinear PDEs by reducing them to a system of ordinary differential equations.

Suitable Grade Level

Undergraduate - Advanced Calculus or Introductory PDEs