Solve equation using the Method of Characteristics

[sandrayuki]

1. Solve the following equation using the Method of Characteristics

xu_x +yu_y=-u
subject to u(cos(s), sin(s)) = 1, 0 ≤ s ≤ pi

Homework Equations

3. Characteristics for this equation are
dx/dt=x...(1)
dy/dt=y...(2)
du/dt=-u...(3)
From(3) get ln(u)=-t+c, u=k*exp(-t)

But i get stuck on the u(cos(s), sin(s)) = 1, i don't know how to use the initial contidion (u(cos(s), sin(s)) = 1 )to do the (1) and (2).

Can anyone help me to finish the rest of the question?

Many thanks

 

[mighty2000]

!



Hi there,

The Method of Characteristics is a powerful technique for solving partial differential equations like the one given in this problem. In order to use this method, we need to first find the characteristics of the equation, which are defined by the equations (1), (2), and (3) provided in the problem.

Now, in order to use the initial condition u(cos(s), sin(s)) = 1, we need to substitute this into the equations (1) and (2) to get:

dx/dt = cos(s) ...(4)
dy/dt = sin(s) ...(5)

Next, we can solve these equations by integrating with respect to t, which gives us:

x = sin(s)t + c1 ...(6)
y = -cos(s)t + c2 ...(7)

Now, we can use the initial condition again to solve for c1 and c2. Plugging in s = 0 (since 0 ≤ s ≤ pi) into equations (6) and (7), we get:

x = 0 + c1 = c1
y = -0 + c2 = c2

Therefore, our final solution for the characteristics is:

x = sin(s)t + x0 ...(8)
y = -cos(s)t + y0 ...(9)

Where x0 and y0 are constants determined by the initial condition.

Now, we can use these characteristics to solve the original equation. We can rewrite the equation as:

du/dt = -u(x,y) ...(10)

Using the chain rule, we can rewrite this as:

du/dt = (du/dx)*(dx/dt) + (du/dy)*(dy/dt)

Substituting in our expressions for dx/dt and dy/dt from equations (8) and (9), we get:

du/dt = u*sin(s) - u*cos(s)

This is a separable differential equation, which can be solved by separating the variables and integrating. This gives us:

ln(u) = -cos(s) + c3

Solving for u, we get:

u = k*exp(-cos(s))

Finally, we can use the initial condition u(cos(s), sin(s)) = 1 to solve for k. Substituting in the values for cos(s) and sin(s), we get:

1 = k*exp(-1)

Solving for k, we get:

k = exp(1)

Therefore