most recent 30 from http://mathoverflow.net 2013-05-24T10:15:04Z http://mathoverflow.net/feeds/question/59142 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59142/another-solution-to-pde-possible Regina 2011-03-22T05:40:11Z 2011-03-22T09:38:12Z

<p>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</p>

http://mathoverflow.net/questions/59142/another-solution-to-pde-possible/59146#59146 Robert Israel 2011-03-22T07:27:46Z 2011-03-22T07:27:46Z

<p>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.</p>

http://mathoverflow.net/questions/59142/another-solution-to-pde-possible/59148#59148 Robert Israel 2011-03-22T07:55:07Z 2011-03-22T07:55:07Z

<p>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. </p>

http://mathoverflow.net/questions/59142/another-solution-to-pde-possible/59158#59158 Denis Serre 2011-03-22T09:38:12Z 2011-03-22T09:38:12Z

<p>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.</p> <p>In conclusion, your explicit solutions are far from unique.</p>