LT3

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u_{t}=u_{{xx}}\,u_{x}(0,t)=\alpha (t)\,
u(x,0)=0\,
x>0\,
t>0\,


The transformed BVP is

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

{\bar  {u}}_{x}(0,s)={\bar  {\alpha }}(s)\,

{\bar  {u}}(x,0)=0\,

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

To have a solution that's bounded at infinity, set c_{1}=0\,

{\bar  {u}}_{x}(0,s)=-c_{2}{\sqrt  {s}}={\bar  {\alpha }}(s)\,

c_{2}={\frac  {-{\bar  {\alpha }}(s)}{{\sqrt  {s}}}}\,

{\bar  {u}}(s,t)={\frac  {-{\bar  {\alpha }}(s)}{{\sqrt  {s}}}}e^{{-{\sqrt  {s}}x}}\,

Use convolution to find the inverse of this function.

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

L^{{-1}}\left(F(s)G(s)\right)=\int _{0}^{t}f(\tau )g(t-\tau )\,d\tau \,

The solution is

u(x,t)=\int _{0}^{t}-\alpha (\tau ){\frac  {e^{{{\frac  {-x^{2}}{4(t-\tau )}}}}}{{\sqrt  {\pi (t-\tau )}}}}\,d\tau \,

Partial Differential Equations

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