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I need to solve the following problem

$$xu_x + u_y = \sqrt{u}, \, u>0 $$

$$ u(x,0)=2+\sin(x).$$

Solving the characteristic equations implies

$x(t,s)=c_1(s)e^t, \, y(t,s)=t+c_2(s), \, u(t,s)=(\frac{t}{2} + c_3 (s) ) ^2 .$

Using $(x(0,s),y(0,s),u(0,s)) = (s,0,2+\sin(s))$ implies $c_1(s)=s, c_2(s)=0, c_3^2(s)=2+\sin(x)$, so that $u(x,y)=[\frac{y}{2}\pm \sqrt{2+\sin(xe^{-y})}]^2$.

My question is -

Which of the two options $c_3(s)=\sqrt{2+\sin(s)}$ or $c_3(s)=-\sqrt{2+\sin(s)}$ is the unique solution? (Transversality condition in this case implies that we have a unique solution )

Is it true that if we add the constraint $y>0$, then from $\sqrt{u}=\frac{t}{2}+c_3(s)$ we can deduce that $t>-2c_3(s)$, so that we must have $c_3(s)=\sqrt{2+\sin(s)}$ for a continuous solution?

Thank you!

Dr. John
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  • Did you check your solution $u(x,y)=[\frac{y}{2}\pm \sqrt{2+\sin(x)}]^2$ in puting it into the PDE ? If I made no mistake the checking fails. So this function is not solution of the PDE. The correct solution is : $$u(x,y)=\left(\frac{y}{2}\pm \sqrt{2+\sin(x:e^{-y})}\right)^2$$ – JJacquelin Aug 08 '21 at 13:50
  • @JJacquelin : You are right. It is a type. I edited the post. Thanks for noticing! – Dr. John Aug 08 '21 at 13:52
  • If I’m not mistaken, one of the solutions goes negative for $y$ small but nonzero. – Deane Aug 08 '21 at 14:54
  • @Deane : after thinking about it , I see that the solution can't go negative, because of the power... – Dr. John Aug 08 '21 at 16:23
  • Yes, you’re right. – Deane Aug 08 '21 at 16:39

1 Answers1

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When you solve for $z(t)=u(x(t),y(t))$ along the characteristic curve through $(s,0)$ for some fixed $s$, you have $$ dz/dt = \sqrt{z} ,\qquad z(0) = 2+\sin(s) . $$ This gives $$ \frac{d}{dt} \sqrt{z} = \frac{dz/dt}{2 \sqrt{z}} = \frac12 , $$ which after integration from $0$ to $t$ becomes $$ \sqrt{z(t)} - \sqrt{2+\sin(s)} = \frac{t}{2} , $$ and the spurious solution with the wrong sign never even appears; you get $$ z(t) = \left( \frac{t}{2} + \sqrt{2+\sin(s)} \right)^2 $$ right away.

Hans Lundmark
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  • Thank you! Before accepting the answer, I was wondering - do you see any way by which we can get rid of the solution with the negative sign using another way? (which is not just substituting back in the equation) – Dr. John Aug 08 '21 at 17:35
  • I suppose in your computations you must have had $\sqrt{u} = \tfrac{t}{2} + c_3(s)$ before reaching the final expression for $u$? Then you know that $\tfrac{t}{2} + c_3(s)$ can't be negative, and in particular at time $t=0$ it can't be negative, and therefore $c_3(s) \ge 0$ gives you the correct choice for the sign. – Hans Lundmark Aug 08 '21 at 20:11
  • Wow, you are right! I completely missed it! Thank you very much! – Dr. John Aug 09 '21 at 06:05