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
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$\begingroup$ 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. $\endgroup$– Aleksey PichuginCommented Mar 22, 2011 at 8:35
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3 Answers
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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.
answered Mar 22, 2011 at 7:27
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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.
answered Mar 22, 2011 at 9:38
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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.
answered Mar 22, 2011 at 7:55