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Can anyone help me with this IVP heat equation problem? I have


where $x \in \mathbb{R}$, $t>0$, $u(x,0)=0$

So i've found by taking a Fourier transformation that


I understand this is a standard technique. The homogenous solution of this PDE is then


Can anyone help me solve the non-homogenous form so I can show it must be the result of a convolution and hence find an integral equation for $u$ by inversion.

If I didn't explain well enough part d on page 81 of this set of notes explains the homogeneous case

EDIT: So I have found

$$\hat{u}=e^{-w^2t} \Big[ \int_0^t e^{w^2s}\hat{g}(w,s) ds +A(w)\Big]$$

And as $u(x,0)=0 \Rightarrow \hat{u}(w,0)=0$ then $A(w)=0$ and

$$\hat{u}=e^{-w^2t}\int_0^t e^{w^2s}\hat{g}(w,s) ds$$

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up vote 2 down vote accepted

Your procedure cannot find all of the solutions that satisfy $u(x,0)=0$ .

Because the PDE is inhomogeneous generally and with only one condition $u(x,0)=0$ , So it is better to take Laplace transform on $t$ :

$\mathcal{L}_{t\to s^2}\{u_t\}-\mathcal{L}_{t\to s^2}\{u_{xx}\}=\mathcal{L}_{t\to s^2}\{g(x,t)\}$




$u(x,t)=\mathcal{L}^{-1}_{s^2\to t}\{C_1(s)e^{xs}\}+\mathcal{L}^{-1}_{s^2\to t}\{C_2(s)e^{-xs}\}-\mathcal{L}^{-1}_{s^2\to t}\left\{\dfrac{e^{xs}}{2s}\int_0^xG(x,s)e^{-xs}~dx\right\}+\mathcal{L}^{-1}_{s^2\to t}\left\{\dfrac{e^{-xs}}{2s}\int_0^xG(x,s)e^{xs}~dx\right\}$

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Can you do the same when $u_t-u_{xx}=g(u)$? – ilciavo Jul 10 '15 at 14:45

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