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Given the Fokker-Planck equation $$D\frac{\partial^2}{\partial x^2}\rho(x;t)=\frac{\partial}{\partial t}\rho(x;t)$$ the paper I'm reading said to have taken the Laplace transform, resulting $$D\frac{\partial^2}{\partial x^2}\rho(x;s)=s\rho(x;s)-\rho(x;0)$$ I just can't get this result.

Maybe I don't understand what the author meant by the notation "$;t$", because it seems that $t$ is a given constant, but if I assume so the transform results: $$\frac{1}{s}D\frac{\partial^2}{\partial x^2}\rho(x;s)=sL[\rho(x;t)]-\rho(x;0)=s\frac{\rho(x;s)}{s}-\rho(x;0)$$ If this is right the result should be $$D\frac{\partial^2}{\partial x^2}\rho(x;s)=s(\rho(x;s)-\rho(x;0))$$ Is it a typing error or I missed something?

Capivara Cometa
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  • You can get to the below answer just by directly using the definition of Laplace transform and integration by parts. Though the identity used is common in lower levels courses​ on the subject of ODEs – Triatticus Jun 06 '17 at 06:56

1 Answers1

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$$\mathcal{L}(s)\{\frac{\partial}{\partial t}\rho(x,t)\}=s\mathcal{L}(s)\{\rho(x,t)\}-\rho(x,0)=s\rho(x,s)-\rho(x,0)\\ \mathcal{L}(s)\{D{\partial^2 \over \partial x^2} \rho(x,t)\}=D{\partial^2 \over \partial x^2}\mathcal{L}(s)\{ \rho(x,t)\}=D{\partial^2 \over \partial x^2}\rho(x,s)$$

Please note how the notation is overloaded into $\rho$.

Brethlosze
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