I need to prove that the Partial Differential Equation $u_x + u_t = 0 $ where $u(x,t) = x$ on $x^2 + t^2 = 1$ has no solution.
I solved this equation with method of characteristics and got one value of $u(x,t)$ as
$u(x,t) = x - t + \sqrt{1-(u(x,t))^2}$ ,
how do I show that this is not the solution of given equation.
Any hints?
I agree with you for this particular solution : $$u = x - t + \sqrt{1-u^2} \tag 1$$ There is another particular solution : $$u = x - t - \sqrt{1-u^2} \tag 2$$ In fact the solution is made of the union of the two branches which is described by the implicit equation : $$\boxed{2u^2-2(x-t)u+(x-t)^2-1=0}\tag 3$$ CHECKING : $$4uu_x-2u-2(x-t)u_x+2(x-t)=0 \quad\implies\quad u_x=\frac{u-x+t}{2u-x+t}$$ $$4uu_t+2u-2(x-t)u_t-2(x-t)=0 \quad\implies\quad u_t=\frac{-u+x-t}{2u-x+t}$$ $$u_x+u_t=\frac{u-x+t}{2u-x+t}+\frac{-u+x-t}{2u-x+t}=0\quad\text{thus eq.}(3)\text{ satisfies the PDE}.$$ And putting $u=x$ into Eq.$(3)$ : $$2x^2-2(x-t)x+(x-t)^2-1=0 \quad \to\quad x^2+t^2=1 \quad\text{thus the condition is satisfied.}$$ This seems contradictory with the assumption that there is no solution. Why ?
Comment:
According to Eq.(3) the solution $u(x,t)$ is not a one-to-one function but is a multivalued function. As a consequence one have to discuss about the wording of the problem.
If the question is "Has the PDE and condition a solution on the form of a one-to-one function? ", the answer is NO because the condition is not satisfied on the WHOLE circle $x^2+t^2=1.$
If the question is "Has the PDE and condition a solution on the form of a multivalued function ? ", the answer is YES.