Let $$u_{xx}-4u_{xy}+3u_{yy}=0.$$ Find the general solution given the solution $u(x,y)=f(\lambda x+y).$

My attempt was as follows: let $u(x,y)=e^{\lambda x+y}$. Then by computing $u_{xx},u_{xy}, \text{ and } u_{yy}$ we get $e^{\lambda x+y}(\lambda^2-4\lambda+3).$ This shows us that $\lambda =1$ or $\lambda =3$.

Is this the right track?

  • 4
    $\begingroup$ You should not take a specific function $e^{\lambda x+y}$: in other words, just take $u(x,y)=f(\lambda x+y)$ as given. You should find that $u_{xx}$ etc are equally easy as what you did already, and you get the same values for $\lambda$. $\endgroup$ – David Sep 8 '14 at 1:47
  • 1
    $\begingroup$ Then take $t=\lambda x+y$ and you get an ODE after rewriting the x,y derivatives as t derivatives. $\endgroup$ – mathematician Sep 8 '14 at 1:50
  • $\begingroup$ So you mean, for example, $u_{xx}=f_{xx}(\lambda x+y)\lambda^2$ and so forth? Doesn't the $f_{xx}$ and sort forth complicate factoring? $\endgroup$ – emka Sep 8 '14 at 1:51
  • $\begingroup$ Do you know what they mean by the general solution in the question? $\endgroup$ – Mhenni Benghorbal Sep 8 '14 at 2:13
  • 1
    $\begingroup$ Instead of arbitrary constants, general solutions of PDE's may involve arbitrary functions. $\endgroup$ – Robert Israel Sep 8 '14 at 2:36

The idea is that the differential operator $\partial_{xx} - 4 \partial_{xy} + 3 \partial_{yy}$ decomposes as a product of two commuting operators of order $1$: $$\partial_{xx} - 4 \partial_{xy} + 3 \partial_{yy} = ( \partial_x - \partial_y)(\partial_x - 3 \partial_y)= (\partial_x - 3 \partial_y)( \partial_x - \partial_y)$$ Now for any functions $f$, $g$ in $1$ variable we have $$( \partial_x - \partial_y)( f(x+y) )= 0$$ and $$(\partial_x - 3 \partial_y) ( g (3x + y) ) =0$$ Therefore, any function $u$ of form $$u(x,y) = f(x+y) + g(3x + y)$$ is a solution of the equation. It is not hard to show that in this way we get all the solutions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.