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I'm stuck when im trying to solve this equation: $\frac{1-n}{2}xu_x+\frac{n-1}{2}u-2x^2u_{xx}+b n\cdot x\cdot y \cdot u_{xy}=0$, where $u=u(x,y)$ and $b$ and $n>0$ are constant parameters.

Comparing this with general 2-variable PDE: $A(x, y)u_{xx} + B(x, y)u_{xy} + C(x, y)u_{yy} = Φ(x, y,u,u_x,u_y)$

I get: $A=-2x^2$ ; $B=bnxy\\$ and $C=0$.

This is hyperbolic equation because $B^2-AC=(bnxy)^2>0$

From $\frac{dy}{dx} = \frac{B \pm\sqrt{B^2-4AC}}{2A} = \frac{+bnxy-bnxy}{4x^2}=0$ or $\frac{dy}{dx}=\frac{2bnxy}{-4x^2}=-\frac{1}{2}\frac{bny}{x}$, characteristic curves are: $y=const$ and $y=x^{\frac{-bnx}{2}}\cdot constant$

Using integration constants as coordinates: $\xi=y$ and $\eta=x^{\frac{bnx}{2}}y$

Then:

$ξ_x=0$, $ξ_y=1$ and $\eta_x=\frac{bn}{2}x^{\frac{bn}{2}-1}y$, $ \eta_y=x^{\frac{bnx}{2}}$

Using chain rule:

$u_x = u_ξξ_x + u_ηη_x=0+u_\eta\frac{bn}{2}x^{\frac{bn}{2}-1}y=\frac{bn}{2}x^{\frac{bn}{2}-1}y\cdot u_\eta$ $u_{xx} = u_{ξξ}ξ^2_x + 2u_{ξη}ξ_xη_x + u_{ηη}η^2_x+ u_ξξ_{xx} + u_ηη_{xx}=0+0+\frac{(bn)^2}{4}x^{bn-2} y^2\cdot u_{\eta\eta}+0+\frac{bn}{2}(\frac{bn}{2}-1)x^{\frac{bn}{2}-2}y\cdot u_\eta=\frac{(bn)^2}{4}x^{bn-2} y^2\cdot u_{\eta\eta}+\frac{bn}{2}(\frac{bn}{2}-1)x^{\frac{bn}{2}-2}y\cdot u_\eta$ $u_{xy} = u_{ξξ}ξ_xξ_y + u_{ξη} (ξ_xη_y + ξ_yη_x) + u_{ηη}η_xη_y + u_ξξ_{xy} + u_ηη_{xy}=\frac{bn}{2}x^{\frac{bn}{2}-1}u_\eta+\frac{bn}{2}x^{bn-1}yu_{\eta\eta}+\frac{bn}{2}x^{\frac{bn}{2}-1}y u_{\eta\xi}$

Plugging this to equation, and simplifying this with expressions for $\eta$ and $\xi$: $\Big[\frac{1-n}{2}(\frac{bn}{2}\eta)+bn\cdot\eta \Big]u_\eta+\frac{(bn)^2}{2}\xi\eta \cdot u_{\xi\eta}+\frac{n-1}{2}u=0$

After further rearrangements equation takes form:

$a_1u_\eta+a_2\xi u_{\xi\eta}+a_3\frac{1}{\eta}u=0$ with $a_1=(\frac{1-n}{4}bn+bn)$, $ a_2=bn^2/2$, $a_3=(n-1)/2$

Now i don't know what to do - i tried use Mathematica software to solve this but Mathematica doesn't support 2nd order PDE's with lower order terms. Can i use separation of variables to solve it? If so - how to separate this equation? Im trying to follow simillar aproach as Canonical form of PDE

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  • $\begingroup$ Edit: there was additional symbol in $u_x$ - variable $x$ which simplified equation in canonical form (so i started with wrong equation). Now i corrected this but still i don't know how to solve it. $\endgroup$ – StarPlatinumZaWardo Apr 16 '19 at 12:39
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After hours of intensified thinking - i found satisfying me solution. Assuming: $u=f(\xi)g(\eta)$, I recasted equation in form:

$a_1+a_2\xi\frac{f'(\xi)}{f(\xi)}+a_3\frac{1}{\eta}\frac{g(\eta)}{g'(\eta)}=0$

then i split it into 2 equations:

$a_1+a_2\xi\frac{f'(\xi)}{f(\xi)}=-k$ and $a_3\frac{1}{\eta}\frac{g(\eta)}{g'(\eta)}=k$.

From this ODE's functions $f$ and $g$ are given by:

$f(\xi)=\xi^{-(k+a_1)/a_2}C_1$ and $g(\eta)=\eta^{a_3/k}C_2$

Where $C_i$ are integration constants and $k$ is arbitrary constant different from $0$ (it comes from separation of variables). Pluging back $\eta=x^{\frac{bnx}{2}}y$ together with $\xi=y$ should solve this PDE.

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