# LT1

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 $u_{t}=u_{{xx}}\,$ $u(x,0)=\partial (x)\,$$-\infty $t>0\,$

$L(u)={\bar {u}}\,$

$L(u_{t})=s{\bar {u}}-u(x,0)=s{\bar {u}}-\partial (x)\,$

$L(u_{{xx}})={\bar {u}}_{{xx}}\,$

Rewrite the DE after the transform:

$s{\bar {u}}-\partial (x)={\bar {u}}_{{xx}}\,$

${\bar {u}}_{{xx}}-s{\bar {u}}=-\partial (x)\,$

Find the homogeneous solution to the transformed DE.

${\bar {u}}_{{xx}}-s{\bar {u}}=0\,$

${\bar {u}}=c_{1}e^{{{\sqrt {s}}x}}+c_{2}e^{{-{\sqrt {s}}x}}\,$

${\bar {u}}(x,s)={\begin{cases}c_{1}e^{{{\sqrt {s}}x}}&x<0\\c_{2}e^{{-{\sqrt {s}}x}}&x>0\end{cases}}\,$

$u\,$ is continuous at $x=0\,$ and so ${\bar {u}}(0^{+},s)={\bar {u}}(0^{-},s)\implies c_{1}=c_{2}\,$

$\int _{{x=0^{-}}}^{{0^{+}}}{\bar {u}}_{{xx}}-s{\bar {u}}=-\int _{{0^{-}}}^{{0^{+}}}\partial (x)\,dx\,$

${\bar {u}}_{x}{\Big |}_{{x=0^{-}}}^{{0^{+}}}=-1\,$

${\bar {u}}_{x}(0^{+},s)-{\bar {u}}_{x}(0^{-},s)=-1\,$

${\bar {u}}(x,s)={\begin{cases}{\frac {1}{2{\sqrt {s}}}}e^{{{\sqrt {s}}x}}&x<0\\{\frac {1}{2{\sqrt {s}}}}e^{{-{\sqrt {s}}x}}&x>0\end{cases}}\,$

${\bar {u}}(x,s)={\frac {1}{2{\sqrt {s}}}}e^{{-{\sqrt {s}}|x|}}\,$

$u(x,t)=L^{{-1}}({\bar {u}}(s,t))=L^{{-1}}\left({\frac {1}{2{\sqrt {s}}}}e^{{-{\sqrt {s}}|x|}}\right)={\frac {1}{2{\sqrt {\pi t}}}}e^{{\frac {-x^{2}}{4t}}}\,$