I am stuck on a problem for a while. How do I solve $f'(x)=-kf(x-1)$? Unlike normal questions which has $f(x)$ on the RHS, this has $f(x-1)$ which has me stumped.
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1Apply Fourier transform as in https://mathoverflow.net/questions/114875/on-equation-fz1-fz-fz/114878#114878 – Alexandre Eremenko Sep 28 '19 at 13:15
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2Since, according to the question, the term $f(x-1)$ on the right hand side seems to be a source of confusion here, it might be worthwhile to mention that this is a special case of so-called delay differential equations. – Jochen Glueck Sep 28 '19 at 13:51
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1Even simpler that Fourier Transform, use Laplace transform. – Jean Marie Becker Sep 29 '19 at 21:35
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Let $\lambda_k$ be all roots of the equation $\lambda+ke^{-\lambda}=0$, real or complex. The general solution is a linear combination $$f(x)=\sum_k c_ke^{\lambda_kx}$$ When $k=1/e$, there is a multiple root, $\lambda_0=-1$, add $cxe^{-x}$.
Alexandre Eremenko
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