Issues for solving a linear partial differential equation of second order

Question

I'm currently solving this partial differential equation:

$$u_{xx}+3u_{xy}-4u_{yy}=xy,$$ with $u(x,x)=\sin(x)$ and $\dfrac{\partial }{\partial x} u(x,y) \mid_{y=x} = 0. $

I am just learning how to solve this kind of equations and I found that linear PDEs of second order have a classification and the method of solving depends of this classification (the equation above is hyperbolic). So I calculate this canonical form for the equation:

$$-625 w _{\xi \eta} = (\eta - \xi)(4 \eta + \xi),$$

and get $$w= -\frac{1}{625} \left( \frac{4}{3} \xi \eta^3 - \frac{3}{4}\xi^2\eta^2 - \frac{1}{3} \xi^3 \eta \right) + f_1(\xi) + f_2 (\eta).$$

where $\xi (x,y)= 4y -x$ and $\eta (x,y) = x+y.$

I'm having issues by the computation of $f_1 (\xi)$ and $f_2 (\eta).$ I can't find that functions satisfying the initial conditions and the PDE. Could someone please help me to find this functions?

And... what is the solution $u(x,y)$ ?

Thanks in advance!

Answer 1

This wouldn't happen to be in reference to this meme, would it? https://pics.me.me/an-i-get-you-guy-give-me-one-such-that-66095644.png

I recently saw it myself and was able to find a solution after struggling with it for a bit. Here's how I did it:

First, from the method of characteristics with $a=1, 2b=3,$ and $c=-4$, I got that $$\xi(x,y) = -4x + y$$ $$\eta(x,y) = x + y$$ $$x = \frac{1}{5}\left(-\xi +\eta \right)$$ $$y = \frac{1}{5}\left( \xi + 4\eta \right)$$

Since both boundary conditions are at $y = x$, we also note that

$$\xi(x,x) = -3x$$

$$\eta(x,x) = 2x$$

For the first boundary condition, $f(x,x) = \sin(x) = \frac{41x^4}{625} + g(-3x) + h(2x)$, and isolating the arbitrary functions gives us \eqref{eq1}:

$$g(-3x) + h(2x) = \sin(x) - \frac{41x^4}{625} \tag{1}\label{eq1}$$

Now, for the second boundary condition $\frac{\partial}{\partial x}f(x,y)\mid_{y=x} = 0$, we make use of the fact that

$$ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial \xi}\frac{\partial \xi}{\partial x} + \frac{\partial f}{\partial \eta}\frac{\partial \eta}{\partial x}$$ $$= \left[\frac{1}{625}\left(\xi^2\eta + \frac{3\xi\eta^2}{2} - \frac{4\eta^3}{3} \right) + g^{\prime}(\xi)\right](-4) \space + \left[\frac{1}{625}\left(\frac{\xi^3}{3} + \frac{3\xi^2\eta}{2} - 4\xi\eta^2 \right) + h^{\prime}(\eta)\right](1) = 0$$

Factoring everything and plugging in the definitions for $\xi(x,x)$ and $\eta(x,x)$ gives

$$4g^{\prime}(-3x) - h^{\prime}(2x) = \frac{326x^3}{1875}$$

Now we can integrate both sides with respect to $x$ (note that we'll have to use integration by substitution for the arbitrary functions) and multiply the result by 6 to get \eqref{eq2}:

$$-8g(-3x) - 3h(2x) = \frac{163x^4}{625} + 6k \tag{2}\label{eq2}$$

where $k$ is some arbitrary constant.

If we solve for each arbitrary function by solving the system of equations that $\eqref{eq1}$ and $\eqref{eq2}$ make, we find that

$$g(-3x) = -\frac{3}{5}\sin(x) - \frac{8x^4}{625} - \frac{6k}{5}$$

and

$$h(2x) = \frac{8}{5}\sin(x) - \frac{33x^4}{625} + \frac{6k}{5}$$

Since $\xi(x,x) = -3x$ and $\eta(x,x) = 2x$, we can plug in $-\frac{\xi}{3}$ for all $x$s in $g$ and $\frac{\eta}{2}$ for all $x$s in $h$.

Now we can finally build the solution for $f(x,y)$. Combining terms and substituting $\xi$ and $\eta$ with their original definitions gives

$$ f(x,y) = \frac{8}{5}\sin\left(\frac{x+y}{2}\right) - \frac{3}{5}\sin\left(\frac{4x-y}{3}\right) + \frac{-65x^4 + 92x^3y + 6x^2y^2 - 28xy^3 - 5y^4}{1296} $$

Hopefully this helps!