Poisson's equation in 1D with point source

[lampCable]

Homework Statement

Solve $\Delta\phi = -q\delta(x)$ on $\mathbb{R}$.

Correct answer: $\phi = -\frac{q}{2}|x| + Ax + B$

Homework Equations

The Attempt at a Solution

In one dimension the equation becomes $\frac{d^2 \phi}{d x^2} = -q\delta(x)$. We integrate from $-\infty$ to $x$ to obtain

$$\int_{-\infty}^x \frac{d^2 \phi}{d x^2} dx = \frac{d \phi}{dx}(x) - \frac{d \phi}{dx}(-\infty) = \int_{-\infty}^x -q\delta(x) dx.$$

If $x<0$, the last integral becomes $0$, from which it follows that $\frac{d \phi}{dx}(x) = \frac{d \phi}{dx}(-\infty)$. Since this holds for all $x<0$ we find that $\frac{d \phi}{dx}(-\infty) = A = \text{constant}$.

If $x>0$, the last integral becomes $-q$, so we find that $\frac{d \phi}{dx}(x) = -q + \frac{d \phi}{dx}(-\infty) = -q + A$.

The same procedure with an integral from $x$ to $\infty$ we find that for $x<0$ the potential is $\frac{d \phi}{dx}(x) = q + B$, where $C$ is some other constant.

(1) Where do I go from here? My idea is that I need to connect the two parts and identify that $\frac{d \phi}{dx}(x) = -\frac{q}{2}\text{sign}(x) + C$, where $C$ some constant.

(2) Are there any other approaches to this problem?

 

[Orodruin]

You are complicating things by having two different approaches. Each approach works perfectly well by itself. Just integrate one of the approaches again.

 

[lampCable]

Okay, so if I use the first approach and integrate again I get for $x<0$ that $\phi = -qx + Ax + B$ and for $x>0$ that $\phi = Ax + C$. But can I really use a continuity argument at $x=0$ to conclude that $B = C$?

 

[Orodruin]

Okay, so if I use the first approach and integrate again I get for $x<0$ that $\phi = -qx + Ax + B$ and for $x>0$ that $\phi = Ax + C$. But can I really use a continuity argument at $x=0$ to conclude that $B = C$?

You do not need a continuity argument, it follows from the integration. Also, note that you have mixed up when the $-qx$ term enters here.

A primitive function for $\delta(x)$ is the Heaviside function $\theta(x)$ and the Heaviside function has $x\theta(x)$ as a primitive function. Hence, the general primitive function for $\delta(x)$ is $\theta(x) + A$ and the general primitive function of that is therefore $x\theta(x) + Ax + B$. Note that $\theta'(x) = \delta(x)$ and therefore $d(x\theta(x))/dx = \theta(x) + x \delta(x) = \theta(x)$. Really, your problem only involves the knowledge of these primitive functions and the addition of the integration constants.

 

[Ray Vickson]

Homework Statement

Solve $\Delta\phi = -q\delta(x)$ on $\mathbb{R}$.

Correct answer: $\phi = -\frac{q}{2}|x| + Ax + B$

Homework Equations

The Attempt at a Solution

In one dimension the equation becomes $\frac{d^2 \phi}{d x^2} = -q\delta(x)$. We integrate from $-\infty$ to $x$ to obtain

$$\int_{-\infty}^x \frac{d^2 \phi}{d x^2} dx = \frac{d \phi}{dx}(x) - \frac{d \phi}{dx}(-\infty) = \int_{-\infty}^x -q\delta(x) dx.$$

If $x<0$, the last integral becomes $0$, from which it follows that $\frac{d \phi}{dx}(x) = \frac{d \phi}{dx}(-\infty)$. Since this holds for all $x<0$ we find that $\frac{d \phi}{dx}(-\infty) = A = \text{constant}$.

If $x>0$, the last integral becomes $-q$, so we find that $\frac{d \phi}{dx}(x) = -q + \frac{d \phi}{dx}(-\infty) = -q + A$.

The same procedure with an integral from $x$ to $\infty$ we find that for $x<0$ the potential is $\frac{d \phi}{dx}(x) = q + B$, where $C$ is some other constant.

(1) Where do I go from here? My idea is that I need to connect the two parts and identify that $\frac{d \phi}{dx}(x) = -\frac{q}{2}\text{sign}(x) + C$, where $C$ some constant.

(2) Are there any other approaches to this problem?

A more-or-less standard approach is to note that for $x < 0$ or $x > 0$ we have $d^2 \phi(x)dx^2 = 0$, so
$$\phi(x) = \begin{cases} ax+b & \text{for} \; x < 0\\
cx+d & \text{for} \; x > 0
\end{cases}
$$
If we integrate the DE from $x = -\epsilon$ to $x = +\epsilon$ we get
$$\phi^{'}(\epsilon) - \phi^{'}(-\epsilon) = -q, $$
so $\phi^{'}$ has a jump discontinuity at $x=0$. Continuity of $\phi(x)$ plus this jump condition gives:
$$\begin{array}{ccc}
\phi(-0) = \phi(+0) &\Rightarrow& b = d\\
\phi^{'}(-0) = \phi^{'}(+0)-q &\Rightarrow& -a = -c - q
\end{array}
$$
You need two more conditions, such as conditions at $\pm \infty$, and different ways of choosing these will yield different solutions to the problem. For example, we could take $a = b = 0$ and then get $d=0$ and $c = -q$. That would give
$$\phi_1(x) = \begin{cases} 0 & \text{for} \; x < 0 \\
-qx & \text{for} \; x > 0
\end{cases}
$$
Alternatively, we could take $b = d = 0$, $a=q/2, c=-q/2$, giving $\phi_2(x) = -(q/2) |x|$.
Of course, the general form of solution would be $\phi(x) = A + B x + \phi_1(x)$, where $\phi_i(x)$ is any special solution. Using $\phi_2$ gives you the form you seek.

 

[lampCable]

Continuity of ϕ(x)ϕ(x)\phi(x) plus this jump condition gives:

ϕ(−0)=ϕ(+0)⇒b=dϕ′(−0)=ϕ′(+0)−q⇒−a=−c−​

How did you know that $\phi$ was continuous at $x=0$?

 

[Orodruin]

I always found that argument ad hoc and a bit wanting. Not in the least because it pre-assumes that the function is continuous at $x = 0$, usually without showing that this must be the case (which actually follows from the differential equation). However, you can make the argument more rigorous by assuming that $\phi(x) = g(x)\theta(x) + h(x) \theta(-x)$, with $g(x)$ and $h(x)$ being the solutions for $x>0$ and $x < 0$, respectively. Inserting this into the differential equation will give you both the discontinuity of the derivative and the continuity of the function itself.

 

[Orodruin]

How did you know that $\phi$ was continuous at $x=0$?

Well spotted. I suggest that you try inserting the ansatz I just gave in #7 into the differential equation. Remember the following identities:
$$
\theta'(x) = \delta(x), \quad \delta(x) f(x) = \delta(x) f(0),
$$
and that $\delta'(x)$ is a distribution that is different from $\delta(x)$ but also is zero except for in $x = 0$.

 

[Ray Vickson]

How did you know that $\phi$ was continuous at $x=0$?

(1) I am looking for A solution, so I am entitled to put other conditions on it, provided that they do not create contradictions, etc.
(2) We can actually prove continuity, by integrating again around 0 to get that $\phi(+0) -\phi(-0) = 0$ (essentially as pointed out by Orodruin in #7).

 

[lampCable]

Letting $g=Ax+B$ and $h=Cx+D$, and inserting the ansatz into the equation we find $\phi '' = (A-C)\delta(x) + (B-D)\delta'(x) = -q\delta \Rightarrow A = -q+C,\ B = C$. So it does indeed follow that $g = (-q+C)x + D$ and $h = Cx + D$.

This shows that $\phi\to D$ in the limit as $x\to0$. But in my understanding the Heaviside function is not well defined at $x=0$ so I am not convinced that this is a proof of continuity at $x=0$. Or am I missing something?

 

[Orodruin]

Nor is the delta function well defined in $x = 0$. When you are dealing with differential equations that involve deltas and Heaviside functions, you really should consider them as distributions rather than functions. A distribution does not necessarily have a well defined value in every point, but that really does not matter. In the distributional sense, you can give the Heaviside function any value at $x = 0$ and it will still be the same.

 

[Ray Vickson]

Letting $g=Ax+B$ and $h=Cx+D$, and inserting the ansatz into the equation we find $\phi '' = (A-C)\delta(x) + (B-D)\delta'(x) = -q\delta \Rightarrow A = -q+C,\ B = C$. So it does indeed follow that $g = (-q+C)x + D$ and $h = Cx + D$.

This shows that $\phi\to D$ in the limit as $x\to0$. But in my understanding the Heaviside function is not well defined at $x=0$ so I am not convinced that this is a proof of continuity at $x=0$. Or am I missing something?

The integral of the Heaviside function is continuous; that is, for $a>0$ the function $F(x) = \int_{-a}^x H(t) \, dt$ is just a "ramp" function: $F(x) = 0$ for $-a \leq x \leq 0$ and $F(x) = x$ for $x > 0$. Note that we can define $\int_{-a}^0 H(t)\,dt$ even though $H(0)$ may not be defined (or be defined in different ways by different people), because when you look at the integral for what it really IS (a limit of Riemann sums) the single point $x=0$ does not matter. (Note, however, that this fails if we replace $H(t)$ by $\delta(t)$ because $\delta$ is not really an ordinary function at all.)