Solving an IVP for 1st Order Quasilinear PDE with Method of Characteristics

[sapiental]

Homework Statement

Solve the following IVP for 1st order quasilinear PDE
s using the method of characteristics.

u*u_x + y*u_y = x

u = 2s, y = s, x = s

Homework Equations

a(x,y,z)*u_x + b(x,y,z)*u_y = c(x,y,z)

z = u(x_o,y_o) = 2s

The Attempt at a Solution

The problem confuses me in the sense that this is the first one where x and y aren't coupled with their respective partial derivatives. for example I can solve

x*u_x + y*u_y = x*u

I'm just looking for more insight into the method of characteristics. does it imply that I can replace u with x with the parameters?

back to the original problem


u*u_x + y*u_y = x

u = 2s, y = s, x = s


is this an acceptable approach

dx/dk = u

and using the characteristic to write

dx/dk = 2x (where s is the first parameter, and k the second one because the equation has 2 independent vars)

dy/dk = y

dz/dk = x


this is the only part that confuses me and makes me believe I am missing the bigger picture for this method.

Thanks!

 

[mighty2000]

Thank you for your question. The method of characteristics is a useful technique for solving first order quasilinear PDEs. It involves finding a set of curves, known as characteristics, along which the PDE can be reduced to an ODE. These characteristics are determined by the coefficients a, b, and c in the PDE.

In your case, the PDE is u*u_x + y*u_y = x, and according to the method of characteristics, we can set up the following system of equations:

dx/dt = u
dy/dt = y
du/dt = x

where t is the parameter along the characteristic curves. We also have the initial conditions u(0) = 2s, y(0) = s, and x(0) = s. Solving this system of equations, we get the following characteristic curves:

x = s + t*u
y = s*e^t
u = 2s + t*x

Now, we can rewrite the PDE in terms of t and the characteristic curves:

u*u_x + y*u_y = x becomes (2s + t*x)*(1 + t*x) + s*e^t*(2s + t*x) = x

Simplifying, we get:

t^2*s*x + t*x^2 + 4s^2 + 2s*t*x = x

Rearranging, we get:

x*(t^2*s + t) + 2s*(2s + t) = 0

This is a separable ODE, and solving it, we get the solution:

x = c*e^(-t/(2s)) - 2s

where c is an arbitrary constant. Now, we can use this solution to eliminate t from the characteristic curves, and we get:

x = s + (c - 2s)*u
y = s*e^(c/(2s))
u = 2s + (c - 2s)*x

Finally, using the initial conditions, we can solve for c and get the solution for u:

x = s + (2s - 2s)*u = s
y = s*e^(2s/(2s)) = s*e
u = 2s + (2s - 2s)*s = 2s

Therefore, the solution to the given IVP is u =