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Solve the following Goursat problem

$xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0,$

$u(x,y)=f(x) \; \; on \; \; y^2+x^2=16 \; \; for \; \; 0 \leq x \leq 4 > $-- eq 1

$u(x,y)=g(y) \; \; on \; \; x=0 \; \; for \; \; 0 \leq y \leq4$ -- eq 2

$f(0) = g(4)$

My attempt :

Using $\alpha = y^2 - x^2 \; \; and \; \; \beta = y^2 +x^2 $ I've reduced the given problem to

$u_{\alpha \beta} = 0 \Rightarrow u= \phi(\alpha) + \gamma(\beta)$

$ \Rightarrow u(x,y) = \phi(y^2-x^2) + \gamma(y^2+x^2)$ - eq 3

Using 1 $\Rightarrow f(x) = \phi(16 - 2x^2) + \gamma(16) \Rightarrow \phi(x)= f(\sqrt{\frac{16-x}{2}}) - \gamma(16)$-- eq 4

Using 2 $\Rightarrow g(y) = \phi(y^2) + \gamma(y^2) \Rightarrow \gamma(y) = g(\sqrt{y}) - \phi(y) = g(\sqrt(y)) - f(\sqrt{\frac{16-x}{2}}) + \gamma(16)$ -- eq 5

$f(0)= \phi(16) + \gamma(16) = g(4)$ -- eq 6

using 4 and 5

$u(x,y)= \phi(y^2-x^2) + \gamma(y^2-x^2) = f(\sqrt{\frac{16-y^2+x^2}{2}}) - \gamma(16) + g(\sqrt{y^2+x^2}) - f(\sqrt{\frac{16-x^2-y^2}{2}}) + \gamma(16)$

$\Rightarrow u(x,y)= f(\sqrt{\frac{16-y^2+x^2}{2}}) - f(\sqrt{\frac{16-x^2-y^2}{2}}) + g(\sqrt{y^2+x^2})$

Is this correct ?

I didn't get to use the given condition f(0)=g(4) anywhere.

vishu
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1 Answers1

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$$xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0$$ HINT :

We observe that the change of $x$ into $-x$ doesn't change the equation, as well as the change of $y$ into $-y$. This draw us to change of variables :

$\begin{cases}X=x^2\\Y=y^2\end{cases} \quad;\quad u_x=2xu_X\quad;\quad u_y=2yu_Y$

$u_{xx}=2u_X+4x^2u_{XX}\quad;\quad u_{yy}=2u_Y+4y^2u_{YY}$

$$xy^3(2u_X+4x^2u_{XX}) - x^3y(2u_Y+4y^2u_{YY}) -y^3(2xu_X) + x^3(2yu_Y)=0$$ $$Y(2u_X+4Xu_{XX}) - X(2u_Y+4Yu_{YY}) -2Yu_X + 2Xu_Y=0$$ $$u_{XX}-u_{YY}=0$$ The well-known general solution is : $$u(X,Y)=F(X+Y)+G(X-Y)$$ with arbitrary functions $F$ and $G$. $$u(x,y)=F(x^2+y^2)+G(x^2-y^2)$$ Then, determine the functions $F$ and $G$ according to the boundary conditions.

JJacquelin
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