LT2

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$u_t-u_{xx}=0\,$

$u(x,0^+) = 0\,$
$u(at,t) = g(t)\,$
$at\le x<\infty\,$
$t>0\,$

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

Translate the derivatives to the new system.

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

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

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

$u_t = -a u_\bar{x}+u_\bar{t}\,$

The DE written in the new variables is

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

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

$r^2 + a r - s = 0\,$

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

$\bar{u}(\bar{x},s) = c_1 e^{\frac{-a+\sqrt{a^2+4s}}{2}} + c_2 e^{\frac{-a-\sqrt{a^2+4s}}{2}}\,$

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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LT2

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