Consider the following PDE, $$\frac{\partial u}{\partial t} − κ \frac{\partial^2 u}{\partial x^2} = S_0δ(x)δ(t),$$ subject to the initial condition, u(x, 0) = δ(x), with κ > 0, and $S_0$ > 0. Find the solution of this PDE by taking both a Fourier and a Laplace transformation. You may use that fact that the Laplace transformation of the Dirac delta function is one, i.e. L{δ(t)} = 1

Can someone explain to me what happens when taking a fourier transform after a laplace transform.

e.g. How to find $ \mathcal{F}( \mathcal{L}(u))$? Can you show in detail what this gives using the definitions of the fourier and laplace transform. The question above is just the context of why i need to use this but I'm not sure how to find it. Thanks.


What the question means, I think, is to take the Fourier Transform in the spatial variable and the Laplace Transform in the time variable.

First, taking the Laplace Transform $\mathcal{L}[u(x,t](s) = w(x, s)$ and letting $u_0(x) := u(x, 0)$ be a known initial condition, we obtain:

$$sw(x,s) - u_0(x) -\kappa\frac{\partial^2 w(x,t)}{\partial x^2} = S_0 \delta(x)$$

which is equivalent to

$$\left(s - \kappa\frac{\partial^2}{\partial x^2}\right)w(x,s) = S_0\delta(x) + u_0(x)$$

Taking the Fourier Transform of this equation, letting $\mathcal{F}[v(x, s)](k) = v(k, s)$, we obtain

$$\left(s+\kappa k^2\right)v(k, s) = S_0 + \int_{-\infty}^{\infty}u_0(x)e^{-ikx}dx$$

or, equivalently

$$v(k,s) = \frac{1}{s + \kappa k^2}\left(S_0 + \int_{-\infty}^{\infty}u_0(x)e^{-ikx}dx\right)$$

(depending on your Fourier Transform convention, various factors of ${2\pi}^{-1}$ may appear).

Now, we only have to invert the two transforms, and then we are done.

The spatial stuff looks pretty messy, but the Laplace Transform can be done easily, as we only have one $s$ in there.

In fact, we have a function that looks something like this:

$$v(k, s) = \frac{F(k)}{s + a(k)}$$

which has an Inverse Laplace Transform of $p(k, t) = \mathcal{L}^{-1}[v(k,s)](t) = F(k)\exp(-a(k)t)$, so we are left with

$$p(k, t) = \exp\left(-\kappa k^2 t\right)\left(S_0 + \int_{-\infty}^{\infty}u_0(x)e^{-ikx}dx\right)$$

The Inverse Fourier Transform of the first term can be reduced to calculating a Gaussian integral, giving us the typical Gaussian solution one would expect from the Heat Equation as a homogeneous solution. This calculation is outlined below:

\begin{align} u_H(x, t) &= \mathcal{F}[S_0 \exp\left(-\kappa k^2 t\right)](x, t)\\ &= \frac{S_0}{2\pi}\int_{-\infty}^\infty \exp(-\kappa t k^2 + i x k)dk\\ &= \frac{S_0}{2\pi}\int_{-\infty}^\infty \exp\left(-\kappa t\left(k^2 - \frac{ixk}{\kappa t}\right)\right)dk\\ &= \frac{S_0}{2\pi}\int_{-\infty}^\infty\exp\left(-\kappa t\left(k-\frac{ix}{2\kappa t}\right)^2 - \frac{x^2}{4\kappa t}\right)dk\\ &= \frac{S_0}{2\pi}\exp\left(-\frac{x^2}{4\kappa t}\right)\int_{-\infty}^\infty \exp\left(-\kappa t\left(k-\frac{ix}{2\kappa t}\right)^2\right)dk \end{align}

Letting $z := k - \frac{ix}{2\kappa t}$, we get that $dk = dz$. The contour on integration shifts to a line parallel to the real axis, but it is not too difficult to show that if one were to consider a a rectangular closed contour the runs along both this line and the real axis, the contribution of the short edges of the rectangle would go to zero as the length of the line approaches infinity. Thus, we can shift the integral back onto the real axis, as the integral along the real axis and along the line parallel to it must be equal in value.

\begin{align} u_H(x, t) &= \frac{S_0}{2\pi}\exp\left(-\frac{x^2}{4\kappa t}\right)\int_{-\infty}^\infty \exp\left(-\kappa tz^2\right)dz\\ &= \frac{S_0}{2\pi}\exp\left(-\frac{x^2}{4\kappa t}\right)\sqrt{\frac{\pi}{\kappa t}}\\ &= \frac{S_0}{\sqrt{4\pi\kappa t}}\exp\left(-\frac{x^2}{4\kappa t}\right) \end{align}

This is the known solution to the Heat Equation as a homogeneous solution.

The inhomogeneous solution can only be written down in integral form, as we do not know anything about the initial distribution $u_0(x)$. It is given by:

\begin{align} u_I(x, t) &= \frac{1}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^{\infty}\exp\left(-\kappa k^2 t\right)e^{ikx} e^{-iky} u_0(y) dy dk\\ &= \frac{1}{2\pi}\int_{-\infty}^\infty u_0(y)\int_{-\infty}^\infty \exp\left(-\kappa t k^2 + ik(x-y)\right)dkdy\\ &= \frac{1}{\sqrt{4\pi \kappa t}}\int_{-\infty}^\infty u_0(y)\exp\left(-\frac{(x-y)^2}{4\kappa t}\right) dy \end{align}

which is preciously the convolution of the initial condition $u_0(x)$ with the fundamental homogeneous solution $u_H(x, t)$.

The final solution is then given as the sum of the homogeneous and the inhomogeneous solution.

EDIT: The above was done for a general $u_0(x)$, because I missed that $u_0(x)$ had been specified in the questions. Given that we have $u_0(x) = \delta(x)$, we can calculate the final integral easily:

\begin{align} u_I(x,t) &= \frac{1}{\sqrt{4\pi \kappa t}}\int_{-\infty}^\infty \delta(y)\exp\left(-\frac{(x-y)^2}{4\kappa t}\right) dy\\ &= \frac{1}{\sqrt{4\pi \kappa t}}\exp\left(-\frac{x^2}{4\kappa t}\right) \end{align}

This gives us the full solution

$$u(x,t) = \frac{S_0 + 1}{\sqrt{4\pi \kappa t}}\exp\left(-\frac{x^2}{4\kappa t}\right)$$

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  • $\begingroup$ Great answer thanks. Except the initial condition was given in the question. $u_0(x) = \delta(x)$ @Tom $\endgroup$ – stutsy Oct 6 '18 at 11:48
  • $\begingroup$ Whoops, completely missed that. In that case, you can actually evaluate the convolution integral easily and arrive at a completely explicit solution. Anyway, glad I could help. $\endgroup$ – Tom Oct 6 '18 at 14:38
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    $\begingroup$ I've added an edit dealing with the given initial condition, just for completeness. $\endgroup$ – Tom Oct 8 '18 at 10:42
  • $\begingroup$ @Tom Would you expect the same result for $u(x,t)$ if you took a Fourier transform of the PDE and solved the resulting ODE using a Laplace transform (basically the opposite order as in your response)? This relates to my post math.stackexchange.com/questions/3183552/… $\endgroup$ – user557493 Apr 11 '19 at 11:39

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