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Is it possible to transform the inhomogenous heat equation:

$ u_t = u_{xx} + h(x,t)$ for $ - \infty < x< \infty , t > 0$ and $u(x,0) = 0$

to the integral equation:

$$\int_0^t \int_{-\infty}^{\infty} {\frac{1}{2\sqrt{\pi(t-s)}} \exp({\frac{-(x-y)^2}{4(t-s)}} ) \times h(y,s)\, dy\, ds}$$

using the Fourier transform?

So far I have:

$$U(y,t) = e^{-y^2 t}\int_0^t e^{y^2 s} H(y,s) ds $$

where capitalization is notation for Fourier transform (eg $ F(u) = U $)

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If $\hat u(\xi,t)$ is the Fourier transform of $u$, then the IVP becomes $$ \hat u_t(\xi,t)=-\xi^2 \hat u(\xi,t) +\hat h(\xi,t), \quad \hat u(\xi,0)=0. $$ Which implies that $$ \hat u(\xi,t)=\int_0^t \mathrm{e}^{-\xi^2(t-s)}\hat h(\xi,s)\,ds. $$ The only thing that remains is to use the inverse Fourier transform, \begin{align} u(x,t)&=\frac{1}{2\pi} \int_{-\infty}^\infty\int_0^t \mathrm{e}^{2\pi x\xi i-\xi^2(t-s)}\hat h(\xi,s)\,ds\, d\xi=\frac{1}{2\pi}\int_0^t\int_{-\infty}^\infty\mathrm{e}^{2\pi x\xi i-\xi^2(t-s)}\hat h(\xi,s)\,d\xi\,ds \\ &=\frac{1}{2\pi}\int_0^t\int_{-\infty}^\infty\mathrm{e}^{2\pi x\xi i-\xi^2(t-s)}\int_{-\infty}^\infty \mathrm{e}^{-2\pi i\xi yi}h(y,s)\,dy\,d\xi\,ds\\ &=\frac{1}{2\pi}\int_0^t\int_{-\infty}^\infty h(y,s)\left(\int_{-\infty}^\infty \mathrm{e}^{2\pi x(\xi-y) i-\xi^2(t-s)}\,d\xi\right)\,dy\,ds. \end{align} It remains to show that \begin{align} \frac{1}{2\pi}\int_{-\infty}^\infty\mathrm{e}^{2\pi x(\xi-y) i-\xi^2(t-s)}\,d\xi= \frac{1}{2\sqrt{\pi(t-s)}}\mathrm{e}^{-\frac{(x-y)^2}{4(t-s)}}. \end{align}

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  • $\begingroup$ sorry but would you be able to elaborate on the last bit? i tried but it wouldn't come out $\endgroup$
    – KJC
    Dec 28, 2013 at 1:05
  • $\begingroup$ THANKS! the bit i was missing was where we sub back in for F(h) . i have it now. $\endgroup$
    – KJC
    Dec 28, 2013 at 1:43
  • $\begingroup$ Hello @YiorgosS.Smyrlis !!! Could I ask you something about distribution theory? $\endgroup$
    – Evinda
    Mar 28, 2016 at 21:50
  • $\begingroup$ @Evinda: [email protected] $\endgroup$ Mar 29, 2016 at 9:24

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