how reduce the following elliptic equation to form of $v_{xx}+v_{yy}+cv=0$

Question

how reduce the elliptic equation $$u_{xx}+3u_{yy}-2u_{x}+24u_{y}+5u=0$$ to the form $$v_{xx}+v_{yy}+cv=0$$ by change of dependent ?

Thanks in advance

Answer 1

If we scale a function $u$'s value by $e^{ax}$, then: $$ \frac{\partial^2 u}{\partial x^2} = a^2e^{ax}u + 2ae^{ax}u_x + e^{ax}u_{xx}. $$ Hence multiplying the equation by $e^{-x+4y}$, you could check the equation can be written as: $$ \frac{\partial}{\partial x^2}(e^{-x+4y}u) + 3\frac{\partial}{\partial y^2}(e^{-x+4y}u) -44 e^{-x+4y}u= 0 $$ Let $$w = e^{-x+4y}u$$ we then have: $$ w_{xx} + 3w_{yy} - 44 w = 0 $$

Now if we scale $w$'s variable $y$ by $b$, i.e., letting $v(x,y) = w(x,by)$, then: $$ v_y = \frac{\partial w(x,by)}{\partial (by)} \frac{d(by)}{dy} = bw_y $$ and $$ v_{yy} = b^2 w_{yy} $$ Notice $w_y$ is taking derivative with respect to $y$, while $w_y$ is rather taking derivative with respect to the second variable, if to avoid any confusion.

Therefore, we could try letting $$ v = w(x,\sqrt{3}y) = e^{-x+4\sqrt{3}y}u(x,\sqrt{3}y) $$ and the rest is left for you to check.