I need to solve this Partial Differential Equation for $\lambda(x,y)$,

$$\frac{\partial \lambda}{\partial x} + h(x,y)\frac{\partial \lambda}{\partial y} - \lambda \frac{\partial h}{\partial y} = 0$$ where $$\frac{dy}{dx} = h(x,y).$$

The additional information given is $\lambda$ is a bivariate polynomial in $x$ and $y$. My initial approach was to try using the method of characteristics, but I know I can't since $y$ is dependent on $x$.

So I guess I should use some sort of degree bound, and find the coefficients by equating powers on both sides, However I just wanted to know if there is actually a better method to do this before I proceed? And even if it has to be done by powers, how do I get the degree?

Additional information: And this PDE is part of a Symmetry Solver to find the infinitesimals, $\xi$ and $\eta$ of the transformed canonical co-ordinates of a first order differential equation.

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    $\begingroup$ It is hard to answer this question without knowing $h(x,y)$. Note, though, that since $y$ depends on $x$, you could solve the second equation explicitly and substitute the solution into the first one, which then becomes an ODE. However, having a polynomial solution is a special property for any differential equation. So, unless you start with that assumption and just solve the first PDE "by powers", you're not guaranteed to end up with a polynomial using some other solution method. $\endgroup$ – Igor Khavkine Jun 18 '13 at 9:18
  • $\begingroup$ @Igor Khavkine Thanks for the reply. Actually though it might sound strange, the first equation is actually used as a step to solve the second one. So making the assumption that h(x,y) is also a polynomial, would the only way to do this be, trial and error, that is choose λ to be a polynomial of a certain degree, substitute and then find the coefficients? $\endgroup$ – Manoj Jun 18 '13 at 11:17

Because $y$ depends on $x$, this should be an ODE. In fact, $$ \frac{\partial \lambda }{\partial x}=\frac{\partial \lambda }{\partial y} \frac{\partial y}{\partial x}=h\left( x,y\right) \frac{\partial \lambda }{ \partial y} $$ which transforms the "pde" into the ode $$ 2h\frac{\partial \lambda }{\partial y}-\lambda \frac{\partial h}{\partial y} =0 $$ This, in turn, leads to $$ 2\frac{\lambda ^{\prime }}{\lambda }=\frac{h^{\prime }}{h} $$ where $^{\prime }=\frac{\partial }{\partial y}.$ Finally, $$ \lambda ^{2}=Ph $$ where $P$ is a constant.

For example, if $h\left( x,y\right) =\left( x+y\right) ^{2},$ then $\lambda \left( x,y\right) =x+y$ and $$ \frac{\partial \lambda }{\partial x}=1,\ \ \frac{\partial \lambda }{\partial y}=1,\ \ \frac{\partial h}{\partial y}=2\left( x+y\right) $$ which transforms the original "pde" into $$ 2h\left( x,y\right) -\lambda \frac{\partial h}{\partial y}=2\left( x+y\right) ^{2}-\left( x+y\right) 2\left( x+y\right) =0 $$

  • $\begingroup$ Thanks. This answer seems really simple, I tried out a few examples, I cannot really find any fault with it except that $\lambda^{2} = h(x,y) + f(y)$. However I don't know why the research paper I use, makes it necessary that $\lambda$ has to be a bivariate in $x$ and $y$, it seems completely unnecessary. Even I don't think that it matters if $f(y)$ is a polynomial or not. P.S : I cannot upvote, I don't have enough reputation. $\endgroup$ – Manoj Jun 18 '13 at 14:53
  • $\begingroup$ No, this doesn't work. For the example that you have given. that is $\lambda(x,y) = x + y$, substituting in the PDE, I get $\frac{\partial \lambda}{\partial x}$ = 1 $\frac{\partial \lambda}{\partial y}$ = 1 $\frac{\partial h}{\partial y} = 2*(x + y)$ which gives me $1 + (x+y){^2} - 2(x+y){^2}$ which is not equal to zero. $\endgroup$ – Manoj Jun 18 '13 at 15:15

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