Suppose the problem ${yu_x-xu_y=0}$, find all the solutions for the given conditions:
- ${u(x,0)=x^{2}}$
- ${u(x,1)=x^2+1}$
- ${u(x,0)=x}$
where ${x}$ is a real number.
I solved the problem and reached the general solution given by ${u(x,y)=F(x^{2}+y^{2})}$
Applying the conditions gave me: ${u(x,y)=x^{2}+y^{2}}$ for the three conditions! But the question is, is it unique?
So checking the Jacobian for the three gives that: ${J = s}$ where it can be zero at one point only, I am stuck at finding the difference for the three cases.
Thanks in advance!
Answer to your question about finding the difference for the three cases.
For the cases 1 and 2 you correctly find the solution $$u(x,y)=x^2+y^2$$ Obviously this solution satisfies the PDE and the respective conditions. This would be the same for any condition such as $u(x,a)=x^2+a^2$ .
But this is different for case 3 :
Condition $u(x,0)=x=F(x^2+0^2)=F(x^2)$
Let $\quad x^2=X \quad;\quad x= \pm\sqrt{X}$
Putting this function $F(X)$ into the above general solution where $X=x^2+y^2$ gives : $u(x,y)=\pm\sqrt{x^2+y^2} $ and in order to satisfy the condition $u(x,0)=x$, the solution is : $$u(x,y)=\begin{cases} \sqrt{x^2+y^2} & \text{if } x>0 \\ -\sqrt{x^2+y^2} & \text{if } x<0 \end{cases}$$