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at\leq x<\infty \,

Create a new moving coordinate system {\bar  {x}}=x-at,\,\,\,{\bar  {t}}=t\,

Translate the derivatives to the new system.

u_{x}={\frac  {\partial u}{\partial x}}={\frac  {\partial u}{\partial {\bar  {x}}}}{\frac  {\partial {\bar  {x}}}{\partial x}}+{\frac  {\partial u}{\partial {\bar  {t}}}}{\frac  {\partial {\bar  {t}}}{\partial x}}\,

=u_{{\bar  {x}}}(1)+u_{{\bar  {t}}}(0)=u_{{\bar  {x}}}\,

u_{{xx}}=u_{{{\bar  {x}}{\bar  {x}}}}\,

u_{t}=-au_{{\bar  {x}}}+u_{{\bar  {t}}}\,

The DE written in the new variables is

-au_{{\bar  {x}}}+u_{{\bar  {t}}}-u_{{{\bar  {x}}{\bar  {x}}}}=0\,

Now transform {\bar  {t}}\rightarrow s\,. The new BVP is

{\bar  {u}}_{{{\bar  {x}}{\bar  {x}}}}+a{\bar  {u}}_{{\bar  {x}}}-s{\bar  {u}}=0\,

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

{\bar  {u}}(0,{\bar  {t}})={\bar  {g}}(s)\,

Solve this ODE


r={\frac  {-a\pm {\sqrt  {a^{2}+4s}}}{2}}\,

{\bar  {u}}({\bar  {x}},s)=c_{1}e^{{{\frac  {-a+{\sqrt  {a^{2}+4s}}}{2}}}}+c_{2}e^{{{\frac  {-a-{\sqrt  {a^{2}+4s}}}{2}}}}\,

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

Plugging in the BC,

{\bar  {u}}(0,t)=c_{2}={\bar  {g}}(s)\,

{\bar  {u}}({\bar  {x}},s)={\bar  {g}}(s)e^{{{\frac  {-a-{\sqrt  {a^{2}+4s}}}{2}}{\bar  {x}}}}\,

The solution is the inverse Laplace transform of this function. It can be done with convolution.

L^{{-1}}\left(e^{{-{\bar  {x}}{\sqrt  {{\frac  {a^{2}}{4}}+s}}}}\right)={\frac  {e^{{-{\frac  {a^{2}t}{4}}}}{\bar  {x}}e^{{{\frac  {-{\bar  {x}}^{2}}{4t}}}}}{{\sqrt  {4\pi t^{3}}}}}\,

L^{{-1}}\left(e^{{{\frac  {a{\bar  {x}}}{2}}}}e^{{-{\bar  {x}}{\sqrt  {{\frac  {a^{2}}{4}}+s}}}}\right)={\frac  {e^{{-{\frac  {a{\bar  {x}}}{2}}-{\frac  {a^{2}t}{4}}-{\frac  {{\bar  {x}}^{2}}{4t}}}}}{{\sqrt  {4\pi t^{3}}}}}{\bar  {x}}\,

u({\bar  {x}},t)=\int _{0}^{t}g(\tau ){\frac  {e^{{-{\frac  {a{\bar  {x}}}{2}}-{\frac  {a^{2}(t-\tau )}{4}}-{\frac  {{\bar  {x}}^{2}}{4(t-\tau )}}}}}{{\sqrt  {4\pi (t-\tau )^{3}}}}}{\bar  {x}}\,d\tau \,

u(x,t)=\int _{0}^{t}g(\tau ){\frac  {e^{{-{\frac  {a(x-at)}{2}}-{\frac  {a^{2}(t-\tau )}{4}}-{\frac  {(x-at)^{2}}{4(t-\tau )}}}}}{{\sqrt  {4\pi (t-\tau )^{3}}}}}(x-a(t-\tau ))\,d\tau \,

Everything in the integrand except g(\tau )\, is called the Green's function for this problem.

Partial Differential Equations

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