# Uniqueness of solution to 1st order pdes

I am given a 1st order partial differential equation $y{\partial \psi\over\partial x}+x{\partial \psi\over\partial y}=0$ subjected to boundary condition $\psi(x,0)=\exp(-x^2)$. I have found that a solution is $\psi(x,y)=\exp(y^2-x^2)$. But I am asked when the solution is unique. Could someone please explain how to answer this? Thanks.

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It would be useful if you could tell us how you came up with your solution. Perhaps by the method of characteristics? – Jeff Dec 18 '11 at 17:21
Also, please include the domain of your PDE. Is it all of $\mathbb{R}^2$? The answer to your question will depend on the domain. – Jeff Dec 18 '11 at 18:13

Consider the parametric curves $x = A e^t + B e^{-t}$, $y = A e^t - B e^{-t}$, which satisfy $x' = y$, $y' = x$. Along such a curve any solution $\psi$ must be constant, according to the chain rule: $$\frac{d}{dt} \psi(x(t),y(t)) = \psi_x \frac{dx}{dt} + \psi_y \frac{dy}{dt} = 0$$ Now the curve intersects $y=0$ if and only if $A$ and $B$ are either both positive (i.e. $x > |y|$), both negative ($x < -|y|$), or both $0$ ($x=y=0$). So a boundary condition on $y=0$ produces uniqueness only in the regions $|x| \ge |y|$. In the region $|y| > |x|$ the solution is not unique. For example, you could add $f(y^2 - x^2)$ to $\psi(x,y)$ where $f$ is differentiable with $f(s) = 0$ for $s \le 0$.

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how do we know that if the characteristic curve intersects the initial curve then there is a unique solution?. i.e., how do we know that there is a unique solution in this case iff the curve intersects $y = 0$? – user27182 Mar 26 '13 at 18:26
Since the solution must be constant on the curve, the value on $y=0$ determines the solution on the curve if the curve intersects $y=0$. Note by the way that (except for the trivial case $A=B=0$) the curve will intersect $y=0$ at only one point $t = -\ln(A/B)/2$. – Robert Israel Mar 29 '13 at 23:03

Uniqueness can be addressed in the following way. Let us suppose that exist another solution $\phi(x,y)$ such that $\phi(x,0)=e^{-x^2}$ then, being your equation linear then also $\phi'(x,y)=a\psi(x,y)+b\phi(x,y)$, with two arbitrary coefficients $a$ and $b$ is a solution. The boundary condition will give $a+b=1$. So, unless you do not give another condition on $y$ your solution cannot be unique.

Of course, you have also the other condition in the given problem. From the fact that $\psi(x,0)=e^{-x^2}$ and from the other fundamental result that your equation has the general solution (characteristic method cited in the comments) $\psi(x,y)=\psi(x^2-y^2)$ for you is enough to set $\psi(0,0)=1$ and your solution is unique.

Finally, I would like to point out the simple way the solution OP proposed can be found. One have to search for a solution in the form

$$\psi(x,y)=\phi(y)e^{-x^2}$$

with $\phi(0)=1$ and the solution is immediately obtained, consistent with the characteristic method.

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You are assuming the existence of another solution $\phi \neq \psi$. – Jeff Dec 18 '11 at 17:06
If I want to prove uniqueness I have to guess that another solution does exist and then, to prove that this is not independent from the other. This is standard matter and I do not see the reason to downvote unless my argument is wrong. – Jon Dec 18 '11 at 17:32
Maybe I am mistaken, but your argument seems to be: assume another solution exists, try to prove it is linearly dependent with the first solution, and then if you can't succeed, it must imply non-uniqueness? This is wrong. – Jeff Dec 18 '11 at 18:09
I think you have some difficulties with foundations en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem. Here the idea is taken from the fixed point uniqueness. I just repeat: Before to downvote, think! – Jon Dec 18 '11 at 18:14
Your solution is still wrong, and has nothing to do with the Picard Lindelof Theorem. Try to think about why your solution would imply non-uniqueness (and remember this is not an ODE). – Jeff Dec 18 '11 at 18:19