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I have a problem where I am supposed to solve a PDE and then tell where in the plane that it has a unique solution, I found an almost identical question but didn't understand the answer that was given. I will post both below and I ask for an explanation of the answer.

It seems like the basic premise of the answer is that if the characteristic curves do not intersect the intial curve then we cannot guarantee a unique solution. Why is this? Why would there necessarily be a unique solution if our characteristic curves did intersect the initial curve?


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


Consider the parametric curves $x=Aet+Be−t$, $y=Aet−Be−t$, which satisfy $x′=y$, $y′=x$. Along such a curve any solution $ψ$ must be constant, according to the chain rule: $\frac{d}{dt}ψ(x(t),y(t))=ψ_x\frac{dx}{dt}+ψ_y\frac{dy}{dt}=0$ 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 $|x|≥|y|$. In the region $|y|>|x|$ the solution is not unique. For example, you could add $f(y^2−x^2)$ to $ψ(x,y)$ where f is differentiable with $f(s)=0$ for $s≤0$.


Maybe it is worth mentioning that I commented on the answer in the original question, but I did not get a reply. So asking again is necessary....

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marked as duplicate by 900 sit-ups a day, Cookie, Adam Hughes, M Turgeon, hardmath Aug 5 '14 at 2:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I asked this a year before the duplicate, so it wasn't already answered... – user27182 Aug 8 '14 at 23:04
up vote 1 down vote accepted

There are perhaps 3 things to understand about characteristics here. Firstly the idea is to propagate information from the boundary or initial data into the main region, so those which don't connect the data to the region don't do what you want. Secondly think about what "constant on characteristics" means: it is the simplest way in which information can be propagated, to wit, it doesn't change at all. That is where the uniqueness comes from. Thirdly, characteristics give some information about what kind of data is appropriate for a given pde. In this case a reasonable problem would be to specify that you want a solution in, say, the quadrant where $x$ and $y$ are positive, and specify the value on some curve which intersects each characteristic uniquely, such as along $y=1/x>0$. Or, specify the value on $x>1, y=0$ and look for a solution in $x>1+|y|$. The problem was likely designed to bring out this idea.

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So is it that: in the region where our characteristics intersect the data curve there is necessarily a unique solution, because we know how the boundary data is propagated through our region and it is done so uniquely. However, in the regions where the characteristics do not intersect the data curve, there may or may not be a unique solution (or maybe not even a solution at all) and we cannot tell merely from the characteristics; or, in the latter region, can we be sure that there is not a unique solution merely from the characteristics? – user27182 Mar 29 '13 at 14:56

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