LT1

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

Partial Differential Equations

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