Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|cite|improve this question

1 Answer 1

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 $t$ :

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




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

share|cite|improve this answer
Can you do the same when $u_t-u_{xx}=g(u)$? – ilciavo Jul 10 at 14:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.