1st order quasi-linear pde

[sapiental]

Homework Statement

u*u_x + y*u_y = xInitial condition:

u = 2*s on the parametric curve given by x = s, y = s, s is any real number.

Homework Equations

Given the equation:

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

Here, u(x,y) is an unknown function which we're trying to find.

a(x,y,z),b(x,y,z),c(x,y,z) are given functions of 3 variables.

The initial condition is a requirement that the unknown function u has a prescribed value when restricted to the parametric curve:

u(x_o(s),y_o(s)) = u_o(s), u_o(s) is a given function.

The Attempt at a Solution

My attempt crumbles early on.

From looking at the equation I get:a(x,y,z) = u

b(x,y,z) = y

c(x,y,z) = x

We alos have x(0) = s, y(0) = s, and u(x(0),y(0)) = u(s,s) = 2s

My professor mentioned we need to introduce a 2nd variable, t, for the following relations

dx(t)/dt = a(x(t),y(t),z(t))
dy(t)/dt = b(x(t),y(t),z(t))
dz(t)/dt = c(x(t),y(t),z(t))
1.dx/dt = u , x(0) = s

2.dy/dt = y , y(0) = s

3.dz/dt = x , u(x(0),y(0)) = 2s
I have a feeling that this might be wrong, and my other hypothesis is the following set up:dz(t)/dt = a(x(t),y(t),z(t))
dy(t)/dt = b(x(t),y(t),z(t))
dx(t)/dt = c(x(t),y(t),z(t))

*note the swap between z and x.

my main problem is relating a,b,c to the equation. For example, are

a(x(t),y(t),z(t))
b(x(t),y(t),z(t))
c(x(t),y(t),z(t))

fixed with respect to the partial derivattives u_x and u_y, or can I switch them around like I did in my second hypothesis?

This clarification will allow me to finish the problem. Any input is appreciated. Thank you!

 

[mighty2000]

Thank you for your post. Your attempt at finding a solution to the given equation and initial condition is on the right track, but there are a few things that can be clarified.

Firstly, in your initial attempt, you correctly identified the functions a(x,y,z), b(x,y,z), and c(x,y,z) as u, y, and x respectively. However, the given equation is in the form of a first-order partial differential equation, so it is not necessary to introduce a second variable t. Instead, we can use the chain rule to express the partial derivatives of u with respect to x and y in terms of the variable s, which is already given in the initial condition.

Secondly, in your second attempt, it is not necessary to swap the variables z and x. The given equation and initial condition are in terms of x and y, so it is more straightforward to use these variables in our solution.

To clarify, the functions a(x,y,z), b(x,y,z), and c(x,y,z) are not fixed with respect to the partial derivatives u_x and u_y. They are simply given functions of three variables that we can use to express u_x and u_y in terms of the variable s. This will enable us to find the solution u(x,y) that satisfies the given equation and initial condition.

I hope this helps in your solution. If you have any further questions, please do not hesitate to ask.