PDE Characteristic Curve Method

[jameson2]

Homework Statement

Solve [tex] u_x^2+u_y^2=1 [/tex] subject to [tex] u(x, ax)=1 [/tex]

Homework Equations

The Attempt at a Solution

I let [tex] u_x=p [/tex] and[tex]u_y=q [/tex]and[tex] F=p^2 +q^2 -1=0 [/tex] Then x'=2p, y'=2q, u'=p.2p+q.2q=2, and p'=0=q'. So [tex] p=p_0, q=q_0 [/tex] are constants. I got [tex] x'=2p_0, y'=2q_0 [/tex] and integrating the equations for x',y',u' I get [tex] x=2p_0s+x_0[/tex] [tex] y=2q_0s + y_0 [/tex]and [tex] u=2s+u_0 [/tex] I'm not sure when it comes to implementing the initial condition. I think that I can say that [tex] u_x(x,ax)=p_0=0[/tex] which gives q=+1 or -1 from the condition of F=0. Can I then fill in [tex] x_0, ax_0, 1 [/tex] for the initial values of x, y and u respectively? In the end I get an answer [tex] u=\frac{y}{q_0}-\frac{ax}{q_0}+1 [/tex]. Firstly, I'm not sure what to do when [tex] q_0[/tex] can have two values. Also, when I fill in my answer into the original PDE I get that a must be zero, which can't be right.

Thanks for any help.

 

[mighty2000]

Thank you for posting your solution attempt. Your approach seems to be on the right track, but there are a few errors and inconsistencies that need to be addressed. Let's go through them one by one.

Firstly, in your attempt, you have incorrectly stated the initial condition as u(x,ax)=1. The correct initial condition should be u(x,0)=1, since we are solving the PDE at y=0. This will give us the value of u at x=0, which is needed to solve the problem.

Next, when you let u_x=p and u_y=q, you have defined the derivative of u with respect to x and y, respectively. However, in your subsequent steps, you have used u' to represent the derivative of u with respect to s, which is incorrect. The correct notation would be u_s.

Furthermore, when you wrote the equations for x', y', and u', you have missed out the term involving the constant p_0 and q_0. The correct equations should be x'=2p_0, y'=2q_0, and u'=p_0^2+q_0^2-1=0.

Now, when you integrate the equations for x', y', and u', you should get x=p_0s+x_0, y=q_0s+y_0, and u=p_0^2s+q_0^2s+u_0. Notice that the constants x_0 and y_0 have been included in the equations, as they represent the initial values of x and y, respectively.

Next, you have mentioned that p'=0 and q'=0, which is incorrect. The correct equations for p' and q' should be p'=2p_0 and q'=2q_0, respectively.

Now, when you use the initial condition u(x,0)=1, you will get the following equations:

x=p_0s+x_0, y=q_0s+y_0, and u=p_0^2s+q_0^2s+u_0=1.

Substituting these values into the original PDE, we get the following equation:

p_0^2+q_0^2=1.

Now, we need to consider the two possible cases for p_0 and q_0, i.e., when p_0=