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I am trying to solve the partial differential equation $x\ u_ x - u\ u_y = y$ with the initial condition $u(1,y) = y$ , using the mathod of characteristics. My problem is with y and z , I mean

$$\frac{dy}{dt} = -z$$ $$\frac{dz}{dt} = y$$ How can I solve these two? Is the solution unique? What is the maximal domain where it is defined?

Math12
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2 Answers2

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You have a linear system for $x,y,z$ That splits into two subsystems. You can solve the system for $(y,z)$ using matrix exponential methods or just taking the t-derivative in one of the equations and combining with the other. Either way your solution is $$ y=C_1\cos t+C_2\sin t,\quad z=C_1\sin t-C_2\cos t $$ You also have $x=C_0e^t$ so you can find the constants from your initial data on the initial curve.

The solution of a linear system is defined everywhere and unique through any given point.

GReyes
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  • @GReyesThanks GReyes. The question about the uniqueness and the maximal domain was about the PDE. What is your idea about it? – Math12 Feb 15 '19 at 19:02
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    The projected characteristics clearly intersect so you do not have a unique smooth global solution. You do have a unique solution in the neighborhood of any point of your initial curve. – GReyes Feb 15 '19 at 20:34
  • @GReyesWe have $$x(s,t)=e^t$$ $$y(s,t)=s(\ cost - \ sint)$$ $$z(s,t)=s( \ sint + \ cost)$$. How can we identify the characteristics and their projections to $xy-plane$. – Math12 Feb 15 '19 at 20:35
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    The projections are just $x(s,t)$ and $y(s,t)$. You can have $x(s_1,t_1)=x(s_2,t_2)$ only if $t_1=t_2=t$ and then $y(s,t)=0$ with, say, $t=\pi/4$ and any $s$. What that means is that the projection of the characteristics issued at $(1,s,s)$ intersect at $(e^{\pi/4},0)$. At that point the value of the solution $u$ must be $\sqrt{2}s$, thus different along each characteristic. Whenever that happens you have to issue a shock and resort to weak solutions. You can also see the collision of the characteristics from Cesareo's answer, choosing $C_1$ and $C_3$ appropriately. – GReyes Feb 16 '19 at 04:40
  • Thank you. It was really helpful. – Math12 Feb 16 '19 at 07:56
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Hint.

Regarding the PDE (it remembers a Hopf type PDE)

$$ x u_x-u u_y = y $$

we have

$$ \dot x = x\\ \dot y = -u\\ \dot u = y $$

from then we can extract

$$ y\dot y + u\dot u = 0\\ u\dot x+x\dot y = 0 $$

representing the characteristic families

$$ y^2+u^2 = C_1\\ $$

and after substitution

$$ \pm\frac{\dot y}{\sqrt{C_1-y^2}} =-\frac{\dot x}{x} $$

or

$$ \pm\arctan\left(\frac{y}{\sqrt{C_1-y^2}}\right) = -\ln x + C_2 $$

as characteristic curves.

Cesareo
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