
# 6.2: Inhomogeneous Heat Equation

Here we consider the initial value problem
for $$u=u(x,t)$$, $$u\in C^\infty(\mathbb{R}^n\times R_+)$$,
\begin{eqnarray*}
u_t-\triangle u&=&f(x,t)\ \ \mbox{in}\ x\in\mathbb{R}^n,\ t\ge0,\\
u(x,0)&=&\phi(x),
\end{eqnarray*}
where $$\phi$$ and $$f$$ are given. From
$$\widehat{u_t-\triangle u}=\widehat{f(x,t)}$$
we obtain an initial value problem for an ordinary differential equation:
\begin{eqnarray*}
\frac{d\widehat{u}}{dt}+|\xi|^2\widehat{u}&=&\widehat{f}(\xi,t)\\
\widehat{u}(\xi,0)&=&\widehat{\phi}(\xi).
\end{eqnarray*}
The solution is given by
$$\widehat{u}(\xi,t)=e^{-|\xi|^2 t}\widehat{\phi}(\xi)+\int_0^t\ e^{-|\xi|^2(t-\tau)}\widehat{f}(\xi,\tau)\ d\tau.$$
Applying the inverse Fourier transform and a calculation as in the proof of Theorem 5.1, step (vi), we get}
\begin{eqnarray*}
u(x,t)&=&(2\pi)^{-n/2}\int_{\mathbb{R}^n}\ e^{ix\cdot\xi}\Big(e^{-|\xi|^2t}\widehat{\phi}(\xi)\\
&&\ \ +\int_0^t\ e^{-|\xi|^2(t-\tau)}\widehat{f}(\xi,\tau)\ d\tau\Big)\ d\xi.
\end{eqnarray*}
From the above calculation for the homogeneous problem and calculation as in the proof of Theorem 5.1, step (vi), we obtain the formula
\begin{eqnarray*}
u(x,t)&=&\frac{1}{(2\sqrt{\pi t})^n}\int_{\mathbb{R}^n}\ \phi(y)e^{-|y-x|^2/(4t)}\ dy\\
& &+\int_0^t \int_{\mathbb{R}^n}\ f(y,\tau)\frac{1}{\left(2\sqrt{\pi(t-\tau)}\right)^n}\ e^{-|y-x|^2/(4(t-\tau))}\ dy\ d\tau.
\end{eqnarray*}

This function $$u(x,t)$$ is a solution of the above inhomogeneous initial value problem provided
$$\phi\in C(\mathbb{R}^n),\ \ \sup_{\mathbb{R}^n}|\phi(x)|<\infty$$
and if
$$f\in C(\mathbb{R}^n\times[0,\infty)),\ \ M(\tau):=\sup_{\mathbb{R}^n}|f(y,\tau)|<\infty,\ 0\le\tau<\infty.$$

### Contributors

• Integrated by Justin Marshall.