Talk:Burgers' equation

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[edit] Factor before the integral

It is \ln(a \cdot b) = \ln(a) + \ln(b) and thus the factor (4πνt) − 1 / 2 in the solution of the initial value problem

u(x,t)=-2\nu\frac{\partial}{\partial x}\ln\Bigl\{(4\pi\nu t)^{-1/2}\int_{-\infty}^\infty\exp\Bigl[-\frac{(x-x')^2}{4\nu t} -\frac{1}{2\nu}\int_0^{x'}u(x'',0)dx''\Bigr]dx'\Bigr\}.

could be splitted from the integral. The differentiation \frac{\partial}{\partial x} then removes that summand. Therefore the solution is given by

u(x,t)=-2\nu\frac{\partial}{\partial x}\ln\Bigl\{\int_{-\infty}^\infty\exp\Bigl[-\frac{(x-x')^2}{4\nu t} -\frac{1}{2\nu}\int_0^{x'}u(x'',0)dx''\Bigr]dx'\Bigr\}

which is simpler (but hides the origin of the solution, which is the heat equation). --Hero Wanders (talk) 22:37, 14 March 2008 (UTC)