Analyzing PDE BVP for ut + ux = 0 with given boundary condition

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The discussion centers on the analysis of the partial differential equation (PDE) ut + ux = 0, subject to the boundary condition u(t,x) = x on the curve defined by x^2 + y^2 = 1. The solution is determined to be u(t,x) = f(x-t), indicating that the problem is well-posed under the assumption that f is a function of (x-t). A point of contention arises regarding the boundary condition, questioning whether it should be x^2 + y^2 = 1 or x^2 + t^2 = 1.

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Homework Statement


  1. ut +ux = 0 subject to u(t,x) = x on x^2 + y^2 = 1

    Is this a well-posed PDE BVP?

Homework Equations

The Attempt at a Solution


This is an easy one to solve: u(t,x) = f(x-t)
I let t(0) = 0 as an initial condition, and so t=s => x= ts + xo, where x(0) = xo
s is the variable such that ∂(u(t(s), x(s))/∂s = 0
If I let u(t,x) = x = f(x-t), would this not be well-posed since f must be a function of (x-t)?
 
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In the problem setup, is the condition really ## x^2 + y^2 = 1 ##, or did you mean ## x^2 + t^2 = 1 ##?
 

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