Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

hi there, i have the following pde: $$(\partial_x x^4 \partial_x - \partial_t^2)y(x,t)=0$$ and found the solution $$y=a+t^2-1/x^2$$, with a a constant. Is this solution unique? Does anyone know of any other solution, tricks to generate other solutions? Thankful for help. Regina

share|improve this question
    
Your solution can be slightly generalised to $y=A+Bt^2-B/x^2$ (both $A$ and $B$ are arbitrary constants). This equation can also be solved by using the standard method of separation of variables. –  Aleksey Pichugin Mar 22 '11 at 8:35

3 Answers 3

Since your equation is linear and homogeneous, linear combinations of solutions are solutions. Basic solutions include $1$ and $t^2 - 1/x^2$ from your solution, as well as $1/x^3$, $t$, and $(1-a/x) \exp(a/x) \exp(a t)$ and $(1-b/x) \exp(b/x) \exp(-b t)$ for any constants a and b.

share|improve this answer

PDEs look like ODEs, but only look like. The solution set of an ODE of order $n$ is usually parametrized by $n$ scalar (integration constant). On the contrary, the solution set of a PDE of order $n$ in $d$ independent variables ($d=2$ in your case) is usually parametrized by $n$ functions of $d-1$ variables. This is clear in the hyperbolic case because you just solve a Cauchy problem with initial data on a non-characteristic hypersurface. More generally, if the equation has analytic coefficients, you can apply the Cauchy-Kowalevskaia Theorem.

In conclusion, your explicit solutions are far from unique.

share|improve this answer

More generally, $F(t + 1/x) - x F'(t + 1/x)$ and $F(t - 1/x) + x F'(t - 1/x)$ for any differentiable function F.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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