I'm having hard time solving this pde.
$$U_{xx} – 2\cos(x) U_{xy} – (15+\sin^2(x)) U_{yy} + \sin(x) U_{y} = 0$$
With following initial conditions
$$U(x,y)|_{y = -\sin(x)} = 4\sin(x)$$
$$\frac{\partial U(x,y)}{\partial y}|_{y = -\sin(x)} = 2 + 2\sin(x)$$
Solving the characteristic equation we get.
$$\frac{dy}{dx} = – \cos(x) \pm 4x$$
So we need to use this kind of substitution
$\xi = y + \sin(x) – 4x $
$\eta = y + \sin(x) + 4x $
and then computing partial derivitives we get
$U_{x} = (\cos(x) – 4)U_{\xi} + (\cos(x) + 4)U_{\eta}$
$U_{y} = (U_{\xi} + U_{\eta})$
$U_{xx} = (\cos(x) – 4)^2 U_{\xi\xi} +2(\cos^2(x) – 16)U_{\xi\eta} + (\cos(x) + 4)^2U_{\eta\eta} – \sin(x)U_{\xi} – \sin(x)U_{\eta}$
$U_{xy} = (\cos(x) – 4)U_{\xi\xi} + 2\cos(x)U_{\xi\eta} + (\cos(x) + 4)U_{\eta\eta}$
$U_{yy} = U_{\xi\xi} + 2U_{\xi\eta} + U_{\eta\eta}$
After substituting them into original equation we get.
$$U_{\xi\eta} = 0$$
After integrating wrt $\xi$ and wrt $\eta$
$$U(\xi,\eta) = f(\xi) + g(\eta)$$
After substituting $x$ and $y$
$$U(x,y) = f(y + \sin(x) – 4x) + g(y + \sin(x) + 4x )$$
$$U_{y}(x,y) = f'_{y}(y + \sin(x) – 4x) + g'_{y}(y + \sin(x) + 4x )$$
Now it's time to solve the Cauchy problem. I've got this system, and this is where I'm stuck
$\left\{\begin{matrix}
f(-4x) + g(+ 4x ) = 4\sin(x)
\\
f'_{y}(- 4x) + g'_{y}(+ 4x ) = 2 + 2 \sin(x)
\end{matrix}\right.$
Any hints how to proceed? Thank you 🙂
Best Answer
Note first that $f$ and $g$ are single variable functions. Then, the partial derivative of $U$ wrt $y$ is:
$$U_y(x,y)=f'(\xi)\frac{\partial\xi}{\partial y}+g'(\eta)\frac{\partial\eta}{\partial y}=f'( y + \sin(x) - 4x)·1+g'( y + \sin(x) + 4x)·1$$
The boundary conditions are:
$$\begin{cases} f(-4x) + g(4x ) = 4\sin(x)\\ f'(- 4x) + g'(4x ) = 2 + 2 \sin(x) \end{cases}$$
Deriving the first wrt $x$:
$$\begin{cases} -4f'(-4x)+4g'(4x)=4\cos(x)\\ f'(- 4x) + g'(4x ) = 2 + 2 \sin(x) \end{cases}$$
$$8g'(4x)=8+8\sin(x)+4\cos(x)\;;g'(4x)=1+\sin(x)+(1/2)\cos(x)$$
$$8f'(-4x)=8+8\sin(x)-4\cos(x)\;;f'(4x)=1+\sin(x)-(1/2)\cos(x)$$
We have now isolated $f'$ and $g'$, so they can be found by integration.