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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 :)

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  • $\begingroup$ Try write the boundary conditions for the variables $\xi$ and $\eta$ from the inverse change of variables. I did, but not sure about the result because it is messy. $\endgroup$ Mar 18, 2017 at 11:41
  • $\begingroup$ I couldn't get anything helpful. Also, the answer given here math.stackexchange.com/questions/1190242/… explains how to proceed this step. It's confusing me though. $\endgroup$
    – shcolf
    Mar 18, 2017 at 12:55
  • $\begingroup$ You almost get it. $\endgroup$ Mar 18, 2017 at 13:19

1 Answer 1

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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.

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  • $\begingroup$ Thank you. Your first formula cleared up everything. After substituation I was thinking that $f$ and $g$ as only functions of $x$ $\endgroup$
    – shcolf
    Mar 18, 2017 at 14:00
  • $\begingroup$ Actually my $f'_{y}$ notation hides significant details inside which I've missed :) $\endgroup$
    – shcolf
    Mar 18, 2017 at 14:08
  • $\begingroup$ Yes, that's the point that makes things go. Anyway, the solution is a complicated function. $\endgroup$ Mar 18, 2017 at 14:11

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