# Solving the heat equation with Fourier Transformations

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

$$u_t-u_{xx}=g(x,t)$$

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

So i've found by taking a Fourier transformation that

$$\hat{u_t}(w,t)=-w^2\hat{u}(w,t)+\hat{g}(w,t)$$

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

$$\hat{u}=A(w)e^{-w^2t}$$

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 http://www.maths.ox.ac.uk/system/files/coursematerial/2011/979/36/DEnotes-fin.pdf

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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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 $t$ :

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

$s^2U(x,s)-u(x,0)-U_{xx}(x,s)=G(x,s)$

$U_{xx}(x,s)-s^2U(x,s)=-G(x,s)$

$U(x,s)=C_1(s)e^{xs}+C_2(s)e^{-xs}-\dfrac{e^{xs}}{2s}\int_0^xG(x,s)e^{-xs}~dx+\dfrac{e^{-xs}}{2s}\int_0^xG(x,s)e^{xs}~dx$

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

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