Initial Value Problem

An initial value problem (IVP) is a mathematical problem that involves finding a solution to a differential equation with specified initial conditions. In this case, we are given the initial condition u(s, s) = sin(s) and the differential equation ux + uy = 1.

Given Information

The given differential equation is ux + uy = 1, where u represents the unknown function of two variables x and y. The initial condition is u(s, s) = sin(s).

Solution

To solve this initial value problem, we can use the method of characteristics. This method involves finding a system of ordinary differential equations for the characteristic curves and then solving them along with the given initial condition.

Finding the Characteristic Curves

To find the characteristic curves, we need to determine the equations dx/dt, dy/dt, and du/dt. From the given differential equation ux + uy = 1, we can rewrite it as:

dx/dt = 1, dy/dt = 1, du/dt = 1.

From these equations, we can see that the characteristic curves are straight lines with slope 1.

Solving the Characteristic Equations

We can solve the characteristic equations by integrating them with respect to t. Integrating dx/dt = 1 gives us x = t + C1, where C1 is the constant of integration. Similarly, integrating dy/dt = 1 gives us y = t + C2, where C2 is the constant of integration. Integrating du/dt = 1 gives us u = t + C3, where C3 is the constant of integration.

Applying the Initial Condition

Now, we can apply the initial condition u(s, s) = sin(s) to find the values of the constants C1, C2, and C3. Substituting s for both x and y in the characteristic equations, we have:

s = t + C1, s = t + C2, sin(s) = t + C3.

From the first two equations, we can see that C1 = s - s = 0 and C2 = s - s = 0. Substituting these values into the third equation, we have:

sin(s) = t + C3.

Solving for C3, we get C3 = sin(s) - t.

Final Solution

Now that we have the values of the constants, we can substitute them back into the characteristic equations to obtain the final solution. Therefore, the solution to the initial value problem is:

u(x, y) = t + C3 = t + (sin(s) - t) = sin(s).

Hence, the solution to the initial value problem ux + uy = 1, u(s, s) = sin(s) is u(x, y) = sin(s).