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

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}}}\,

Partial Differential Equations

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