siden ingen har prøvd seg så slenger jeg ut løsning
setter: [tex]\hat{u}(\omega,t)=\mathcal{F}\left{u(x,t)\right}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u(x,t)e^{-i\omega x}\rm{d}x[/tex]
siden: [tex]\lim_{|x|\to\infty}u(x,t)=0[/tex] og [tex]\lim_{|x|\to\infty}u_{x}(x,t)=0[/tex]
har en at: [tex]\mathcal{F}\left{u_{xx}(x,t)\right}=-\omega^{2}\hat{u}(x,t)[/tex]
deretter har en: [tex]\mathcal{F}\left{u_{t}(x,t)\right}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\partial d}{\partial t}\left(u(x,t)e^{-i\omega x}\right)\rm{d}x\Rightarrow \frac{\partial}{\partial t}\left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u(x,t)e^{-i\omega x}\rm{d}x\right]=\frac{\partial}{\partial t}\hat{u}(\omega,t)[/tex]
dette gir: [tex]tu_{xx}(x,t)=u_{t}(x,t)\Rightarrow -t\omega^{2}\hat{u}(\omega,t)=\frac{\partial}{\partial t}\hat{u}(\omega,t)\Rightarrow \frac{d \hat{u}(\omega,t)}{\hat{u}(\omega,t)}=-t\omega^{2}dt\Rightarrow \hat{u}(\omega,t) = C(\omega)e^{-t^{2}\omega^{2}/2}[/tex]
fra dette kan en se at: [tex]\hat{u}(\omega,0) = C(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u(x,0)e^{-i\omega x}\rm{d}x = \hat{f}(\omega)\Rightarrow C(\omega)=\hat{f}(\omega)\Rightarrow \hat{u}(\omega,t)=\hat{f}(\omega)e^{-t^{2}\omega^{2}/2[/tex]
bruker: [tex]\mathcal{F}\left{f\ast g\right}=\sqrt{2\pi}\hat{f}\cdot\hat{g}\Leftrightarrow \mathcal{F}^{-1}\left{\hat{f}\cdot\hat{g}\right}=\frac{1}{\sqrt{2\pi}}\hat{f}\ast\hat{g}[/tex]
slik at en kan finne en funksjon [tex]h(x)[/tex] som er slik at [tex]\hat{h}(\omega) = e^{-t^{2}\omega^{2}/2}[/tex]
har så:
[tex]\int_{-\infty}^{\infty}e^{i\omega x}e^{-\lambda x^{2}}\rm{d}x = \sqrt{\frac{\pi}{\lambda}}e^{-\omega^{2}/4\lambda} \Rightarrow \mathcal{F}\left{e^{-\lambda x^{2}}\right}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\omega x}e^{-\lambda x^{2}}\rm{d}x=\frac{1}{\sqrt{2\pi}}\sqrt{\frac{\pi}{\lambda}}e^{-(-\omega)^{2}/4\lambda}=\frac{1}{\sqrt{2\lambda}}e^{-\omega^{2}/4\lambda}[/tex]
for t>0: [tex]\frac{1}{4\lambda}=\frac{t^2}{2}\Rightarrow \lambda=\frac{1}{2t^{2}}\Rightarrow e^{-t^{2}\omega^{2}/2}=e^{-\omega^{2}/4\lambda}=\sqrt{2\lambda}\cdot\frac{1}{\sqrt{2\lambda}}e^{-\omega^{2}/4\lambda}[/tex]
[tex]\hat{h}(\omega)=e^{-t^{2}\omega^{2}/2}\Rightarrow h(x)=\sqrt{2\lambda}e^{-\lambda x^{2}}=\frac{1}{t}e^{-x^{2}/2t^{2}}\Rightarrow u(x,t)=\mathcal{F}\left{\hat{f}\cdot\hat{h}\right}=\frac{1}{\sqrt{2\pi}}\hat{f}\ast\hat{h}=\frac{1}{t\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x-p)e^{-p^{2}/2t^{2}}\rm{d}p[/tex]
for t>0: [tex]p = ts\,, \rm{d}p=t\rm{d}s\Rightarrow u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x-st)e^{-s^{2}/2}\rm{d}s\Leftrightarrow u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x-st)g(s)\rm{d}s[/tex]
som gir: [tex]\underline{\underline{g(s) = e^{-s^{2}/2}}}[/tex]
for t=0: [tex]u(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-s^{2}/2}\rm{d}s=f(x)\cdot\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-s^{2}/2}\rm{d}s=f(x)[/tex]
