Math Problem Statement

B Find the solution of the Cauchy problem ux + uy = u2, with u(x, 0) = x2. [30 marks]

Solution

The given Cauchy problem is:

ux+uy=u2,u(x,0)=x2.u_x + u_y = u^2, \quad u(x, 0) = x^2.

Step 1: Setting Up the Method of Characteristics

To solve this first-order partial differential equation using the method of characteristics, we start by rewriting the equation in the following form:

ux+uy=u2.\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = u^2.

We now use the method of characteristics, which involves finding curves along which the solution is constant. The characteristics are defined by the system of equations:

dxds=1,dyds=1,duds=u2.\frac{dx}{ds} = 1, \quad \frac{dy}{ds} = 1, \quad \frac{du}{ds} = u^2.

Here, ss is the parameter along the characteristic curves.

Step 2: Solving the Characteristic Equations

  1. Solving for x(s)x(s) and y(s)y(s):

From the system dxds=1\frac{dx}{ds} = 1 and dyds=1\frac{dy}{ds} = 1, we integrate with respect to ss:

x(s)=s+x0andy(s)=s+y0,x(s) = s + x_0 \quad \text{and} \quad y(s) = s + y_0, where x0x_0 and y0y_0 are the initial conditions.

  1. Solving for u(s)u(s):

Now, we solve the equation duds=u2\frac{du}{ds} = u^2. This is a separable differential equation, so we can write:

duu2=ds.\frac{du}{u^2} = ds.

Integrating both sides:

1u=s+C,-\frac{1}{u} = s + C, where CC is a constant of integration. Solving for u(s)u(s), we get:

u(s)=1Cs.u(s) = \frac{1}{C - s}.

Step 3: Using the Initial Condition

The initial condition is given as u(x,0)=x2u(x, 0) = x^2. At y=0y = 0, we have s=0s = 0, so:

u(x0,0)=1C=x02.u(x_0, 0) = \frac{1}{C} = x_0^2.

Thus, C=1x02C = \frac{1}{x_0^2}.

Step 4: Expressing the Solution

Now that we have the value of CC, we can write the solution for u(s)u(s):

u(s)=11x02s.u(s) = \frac{1}{\frac{1}{x_0^2} - s}.

Using the relationship x(s)=s+x0x(s) = s + x_0 and y(s)=s+y0y(s) = s + y_0, we express ss in terms of xx and yy. Specifically, from x=s+x0x = s + x_0, we have s=xx0s = x - x_0. Therefore, the solution becomes:

u(x,y)=11x02(xx0).u(x, y) = \frac{1}{\frac{1}{x_0^2} - (x - x_0)}.

This gives the solution of the PDE in terms of the initial condition.

Final Solution:

The final form of the solution is:

u(x,y)=11x2(xx0).u(x, y) = \frac{1}{\frac{1}{x^2} - (x - x_0)}.

Conclusion

This method provides the general solution to the Cauchy problem. Would you like more details on any of the steps or need help with another part of the problem?

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Math Problem Analysis

Mathematical Concepts

Partial Differential Equations
Method of Characteristics
First-Order PDEs

Formulas

∂u/∂x + ∂u/∂y = u^2
dx/ds = 1, dy/ds = 1, du/ds = u^2
∫ du/u^2 = ds

Theorems

Method of Characteristics

Suitable Grade Level

University/Advanced Undergraduate