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I have the linear PDE

$$yu_x - xu_y = 0 \qquad x^2 + y^2 < a^2 \\ u(0,y) = (a^2-y^2)^{\frac{1}{2}} \qquad y \in(-a,a)$$

where $a > 0$ is a constant.

So what I have done is to say that

$$a_1 = y \quad a_2=-x \quad b=0 \\ \gamma(x_0(s), y_0(s)) = \gamma(0,s) = (a^2-s^2)^{\frac{1}{2}}$$

with $\gamma(0,s)$ being the Cauchy curve.

From this I say that

$$\widetilde{x}_\tau = \widetilde{y} \qquad \widetilde{x}(0,s) = 0 \\ \widetilde{y}_\tau = -\widetilde{x} \qquad \widetilde{y}(0,s) = s \\ \widetilde{z}_\tau = 0 \qquad \widetilde{z}(0,s) = (a^2-s^2)^{\frac{1}{2}}$$

Now I see that

$$ \widetilde{x}_{\tau\tau} = \widetilde{y}_\tau = -\widetilde{x}$$

this gives

$$\widetilde{x} = A\,cos(\tau) + B\,sin(\tau)$$

and from the initial boundary condition it can be shown that $A=0$. Similarly,

$$ \widetilde{y}_{\tau\tau} = \widetilde{x}_\tau = -\widetilde{y}$$

giving

$$\widetilde{y} = C\,e^{\tau} + B\,e^{\tau}$$

which using the boundary condition gives

$$s = C + D$$

now I know I have made a mistake somewhere because I get stuck here. I don't know how to find the values of $B,\, C,\,D$. But I don't know what I ahve done wrong.

I know that there is a second method where you multiply $x_\tau$ and $y_\tau$ together but I don't want to do it that way.

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$$yu_x-xu_y=0$$ $$\frac{\partial u}{x\partial x}-\frac{\partial u}{y\partial y}=0$$ With $X=x^2$ and $Y=y^2$ : $$\frac{\partial u}{\partial X}-\frac{\partial u}{\partial Y}=0$$ The general solution is well known : $$u=F(X+Y)$$ $$u(x,y)=F(x^2+y^2) \tag 1$$ $F$ is an arbitrary differentiable function, to be determined according to the boundary condition.

Condition : $\quad u(0,y)=\sqrt{a^2-y^2}$ $$F(0^2+y^2)=\sqrt{a^2-y^2}$$ Let $X=y^2$ $$F(X)=\sqrt{a^2-X}$$ Now, the function $F(X)$ is determined. We put it into the above general solution Eq.$(1)$, where $X=x^2+y^2$. Thus $F(x^2+y^2)=\sqrt{a^2-(x^2+y^2)}$ $$u(x,y)=\sqrt{a^2-(x^2+y^2)}$$

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  • $\begingroup$ Wow. This is a completely different way to what I've ever seen before $\endgroup$ – Gragbow May 10 '18 at 16:03
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Even with your method it should work

$$ \begin{cases} yu_x - xu_y = 0 \qquad x^2 + y^2 < a^2 \\ u(0,y) = (a^2-y^2)^{\frac{1}{2}} \qquad y \in(-a,a) \end{cases} $$ $$dz=0 \implies z=c_2$$ For the second integral curve $$\frac {dx}{y}=\frac {dy}{-x} \implies -xdx-ydy=0$$ $$ \implies -\frac {x^2}2-\frac {y^2}2=c_1 \implies {x^2}+ {y^2}=c_1 $$ $$f(c_1)=c_2 \implies f(x^2+y^2)=z$$ Therefore $$u(x,y) =f(x^2+y^2)$$ with IC we have that $$u(0,y) =f(y^2)=(a^2-y^2)^{1/2}$$ So that $$u(x,y) =(a^2-y^2-x^2)^{1/2}$$

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