1
$\begingroup$

I am trying to gain some information on a non-linear partial differential equation by issuing the following Mathematica command:

DSolve[ f[x, y] - x D[f[x, y], x, y] == 0 , f[x, y] , {x, y} ]

As an answer I just get the same expression. Does that mean that Mathematica cannot come up with any useful information regarding the solution of such equations?

Are there other things I can try? Is anything known about such differential equations?

Thanks in advance,

Uri

P.S. The function Exp[ 2 Sqrt[ y Log[x] ] ] is "almost" a solution of this differential equation.

$\endgroup$
5
  • $\begingroup$ this pde is linear. Why do you call it non-linear? $\endgroup$
    – Nasser
    Aug 10, 2023 at 6:49
  • $\begingroup$ The second derivative is multiplied by x, which makes it non-linear. $\endgroup$
    – Uri Z
    Aug 10, 2023 at 6:51
  • $\begingroup$ No. that is wrong. Nonlinearity is on the dependent variable, not the independent variables. The dependent variable enters the pde as linear. $\endgroup$
    – Nasser
    Aug 10, 2023 at 6:53
  • $\begingroup$ Thanks for correcting me. Is there anything you can say regarding the queation itself? $\endgroup$
    – Uri Z
    Aug 10, 2023 at 7:40
  • $\begingroup$ By $ x \partial_x \to \partial_z $ it is the wave equation $\partial_{z,y}f =f$. $\endgroup$
    – Roland F
    Aug 10, 2023 at 11:37

1 Answer 1

2
$\begingroup$

Mathematica can't solve this directly. One way around is to help it a little by using separation of variables. By assuming $f(x,y)= X(x) Y(y)$

f[x_,y_]:=X[x]*Y[y];
pde=f[x,y]-x D[f[x,y],x,y]==0
Expand[pde[[1]]/f[x,y]]==0

Mathematica graphics

The above shows that we have

$$ x \frac{X'}{X} \frac{ Y'}{Y} = 1 $$ Which is the same as $$ x \frac{X'}{X} = \frac{ Y}{Y'} $$ Hence these both are same and equal to some constant, say $\lambda$ which is the eigenvalue. $$ x \frac{X'}{X} = \frac{ Y}{Y'} = \lambda $$

So we have two ode's to solve

ode1=x*X'[x]/X[x]==λ
ode2=Y[y]/Y'[y]==λ
solX  =  DSolveValue[ode1,X[x],x]
solY  =  DSolveValue[ode2,Y[y],y]/.C[1]->C[2]

Hence the final solution is

solX*solY

Mathematica graphics

The eigenvalue $\lambda$ can be determined from boundary conditions.

Verification

To verify that the above solution is correct

ClearAll["Global`*"]
sol=f->Function[{x,y},E^(y/λ) x^λ C[1] C[2]]
pde=f[x,y]-x D[f[x,y],x,y]==0
pde/.sol

Mathematica graphics

$\endgroup$
2
  • $\begingroup$ Thanks. That's interesting! As I said in the post, the function $e^{\sqrt{y \ln x}}$ is "close" to being a solution of the equation. I wonder whether there is an exact solution of a similar form. Such a solution, if it exists, is not separable. Is anything known about non-separable solutions? $\endgroup$
    – Uri Z
    Aug 10, 2023 at 7:51
  • $\begingroup$ The more general solution is Integrate[c[λ] E^(y/λ) x^λ, λ], where c is an arbitrary function of λ. $\endgroup$
    – bbgodfrey
    Aug 11, 2023 at 1:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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