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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