Distribution theory - Find a solution to the linear partial differential equation $\partial u+au=\phi$

Question

This was a problem on my final last semester. I am trying to learn how this is done for the future. I thought that I figured it out and I emailed the instructor and he said there was a simpler solution without using Fourier transforms. I was wondering if anyone knows what to do as I am very stuck...

What I did:

We want a solution of the form $$LE=\delta$$ where $\delta$ is the delta function. Then, $$u=E * \phi$$ So, we have $$\partial E +aE = \delta$$ This is where I took the Fourier transform to get, $$-i\xi\hat{E}+a\hat{E}=1$$ $$\Rightarrow (a-i\xi)\hat{E}=1$$ $$\Rightarrow \frac1{(a-i\xi)}+c\delta=\hat{E}$$ is a family of solutions. Then we can take $c=0$.

Then I have $$E=\int_{-\infty}^\infty \frac{e^{-i\xi x}}{a-i \xi} \frac{d\xi}{2\pi}$$

Again, I was told there was a much simpler solution. I am hoping someone can help me out.

Thanks.

Answer 1

Using the method of integrating factor: $$E'+aE = \delta$$ $$(Ee^{ax})' = \delta e^{ax} = \delta$$ $$Ee^{ax} = H$$ $$E = He^{-ax},$$ where $H$ is the Heavyside function.

Answer 2

The equation $\partial u + au = \phi$ is functionally equivalent to the ODE $u'(x)+au(x)=\phi(x)$. Since we are only required to find a solution, we will take the additional condition $u(x_0) = 0$ (this can be relaxed but it makes the calculations messier). We look for a fundamental solution $E(x;\xi)$ which satisfies $E'(x;\xi)+aE(x;\xi) = \delta(x-\xi)$ and $E(x_0;\xi) = 0$ for all $\xi>x_0$. When $x\neq\xi$, $\delta(x-\xi) = 0$, so $E(x;\xi) = \begin{cases} c_<\exp(-ax) & x_0\leq x<\xi \\ c_> \exp(-ax) & \xi<x \end{cases}$

Integrating the ODE on the interval $(\xi-\epsilon,\xi+\epsilon)$ gives us

$$ \int_{\xi-\epsilon}^{\xi+\epsilon} E'(x;\xi)\text dx + a\int_{\xi-\epsilon}^{\xi+\epsilon}E(x;\xi)\text dx = \int_{\xi-\epsilon}^{\xi+\epsilon}\delta(x-\xi)\text dx $$

The second term vanishes and the RHS $\to 1$ as $\epsilon\to 0$. The first term is $E(\xi+\epsilon;\xi)-E(\xi-\epsilon;\xi)$, which gives us the size of the jump discontinuity of $E$ at $x=\xi$, we'll call it $J_E(\xi)$. Then, the above integral expression becomes

$$ J_E(\xi) + 0 = 1 $$

So, this jump condition gives us the constraint $c_> \exp(-a\xi) - c_< \exp(-a\xi) = 1$. This, combined with the initial condition, gives us two constraints with which we can solve for the two constants.

By the initial condition, $c_< \exp(-ax_0) = 0$, so $c_< = 0$. This reduces the second equation to $c_> \exp(-a\xi) = 1$, so $c_> = \exp(a\xi)$. Plugging these constant back in, we get an expression for the fundamental solution

$$ E(x;\xi) = \begin{cases} 0 & x_0\leq x<\xi \\ \exp(-a(x-\xi)) & \xi<x \end{cases} = H(x-\xi)\exp(-a(x-\xi)) $$

where $H$ is the Heaviside function. We notice that $E(x;\xi) = F(x-\xi)$ with $F(t) = H(t)\exp(-at)$.

Thus, the solution to the original equation can be written as

$$ u(x) = \int_{x_0}^\infty \phi(\xi)E(x;\xi)\text d\xi = (\phi * F)(x)$$

(Note: your definition of $*$ may require $x_0 = 0$, and as such that particular value can be chosen)