# An intuitive explanation of the Poisson equation?

The Poisson equation $$\nabla^2 u = \Delta u = f$$.

When $$f = 0$$ we obtain the Laplacian equation which to me has an intuitive interpretation with the mean-value property. However, how can we form an intuitive understanding of the Poisson equation from the function $$f$$? How does $$f$$ impact $$u$$?

• Read read read. Maybe any introductory book on physics will be a good start. Commented Dec 8, 2023 at 4:57
• Physics viewpoint: in electrostatics, for instance, if you have a charge density $\rho$, the electric potential $\phi$ satisfies $\Delta \phi = - \rho/\varepsilon$. Similar equations appear in fluids (potential flow) and Newtonian gravitation. The idea behind the Poisson equation is that we have an irrational field $E = -\mathrm{grad}(\phi)$ with non-zero divergent. The divergent $\mathrm{div}(E) = - \Delta \phi$ computes the flux of $E$ on the space via the Gauss theorem.
– Hugo
Commented Dec 8, 2023 at 14:10
• I mean, from the mean value property we get a similar understand of Poisson equation. If $f$ is large and positive, then $u$ is much smaller at $x$ than its average value around $x$. Vice versa for negative $f$. Commented Feb 23 at 16:29
• The Poisson equation has similar averaging properties to Laplace's equation, however the equalities turn into inequalities. In Laplace's equation, the function is always equal to its spherical averages. In the Poisson equation, it will either be strictly greater or strictly lesser, depending on the sign of $f$. Points in which $f>0$ will be "sources" whereas points where $f<0$ will be "sinks". Commented Feb 25 at 21:23

1. Poisson Equation: The Poisson equation,$$\nabla^2 u = f$$, describes how a scalar function (u) changes in response to a given "influence" represented by (f). It's like describing how a landscape changes based on certain influences scattered across it.

2. Source Term (f): Think of (f) as a distribution of "influence" or "impact" spread across space. It could represent anything from the concentration of a substance to the intensity of a force.

Now, let's consider how (f) impacts (u):

• Local Effect: At any point in space, the value of (f) determines the local impact or contribution to the function (u). If (f) is high at a point, it suggests a strong influence, causing (u) to be higher in that vicinity. On the other hand, if (f) is low or zero, there's little to no influence, so (u) remains relatively unaffected.

• Global Effect: The overall distribution of (f) across space influences the overall behavior of (u). High values of (f) in certain regions will tend to push (u) higher in those areas, while low values of (f) will have less impact.

• Boundary Conditions: Similar to the edges of a canvas influencing a painting, the behavior of (u) at the boundaries of the domain, along with any imposed conditions, can significantly shape the final solution.

In essence, the Poisson equation describes how a function (u) responds to influences represented by (f), with the solution reflecting a balance between these influences across space. Think of (f) as painting different regions of a canvas with varying intensities, where the final picture (u) emerges as a response to these influences.

Again, not an expert, but just intuitively explained as I got it.

Here is my understanding from a pure math point of view. But I am not sure if this is what you want for the ''intuition''.

In analysis and PDE, that $$\Delta u=f$$ says that $$u$$ ''gain 2 derivatives than'' $$f$$.

Assuming $$u$$ and $$f$$ are both defined on $$\mathbb R^n$$ and has at most polynomial growth, if we take Fourier transform on both sides then we have $$-4\pi^2|\xi|^2\hat u(\xi)=\hat f(\xi)$$. Basically it says that if $$\hat f(\xi)=O(|\xi|^{-m}$$ then $$\hat u(\xi)=O(|\xi|^{-m-2})$$ decay two more degree faster than $$\hat f$$ as $$\xi\to\infty$$.

The decay of Fourier coefficients is equivalent as saying the smoothness of the original functions. By taking Fourier transform back we see that this is saying $$u$$ is smoother than $$f$$ with order 2.

To get a better idea, we can consider $$L^2$$ Sobolev spaces. By the so-called Plancherel identity if $$f\in W^{k,2}(\mathbb R^n)$$ if and only if $$\int (1+|\xi|^2)^{k/2}|\hat f(\xi)|^2d\xi<\infty$$. If we igonore the "+1" in the coefficient, i.e. the low frequency, this is saying that $$u\in W^{k,2}(\mathbb R^n)$$.

In practice we can also make the same description locally, namely if $$f$$ ''looks nice near $$x_0$$'' then $$u$$ ''looks 2-order nicer near $$x_0$$''. This is what the Schauder's estimate comes into play.

Certainly thinking that, if $$f$$ is order $$k$$ differentiable (i.e. $$C^k$$) near $$x_0$$ then $$u$$ is order $$k+2$$ differentiable (i.e. $$C^{k+2}$$) near $$x_0$$, is a good starting point. Although that is not true in general. If you replace $$C^k$$ by Sobolev regularity $$W^{k,p}$$ for $$1 or H"older regularity $$C^{k,\alpha}$$ for $$0<\alpha<1$$, it will be correct.

I think checking the word elliptic differential operator can be a further path to get the intuition.

• Here is the trouble for $f\in \mathrm{L}^2(\mathbb{R}^n)$, one CANNOT have in general $u\in \mathrm{W}^{2,2}(\mathbb{R}^n)$ with continuous dependency with respect to $f$. If one assume such a corresponding estimates, arguing by contradiction, then a dilation argument implies that necessarily $u=0$. However, everything still holds locally. Commented Mar 1 at 10:55
• @ToGle that's why I say 'if you ignore +1'. If we replace Delta by 1-Delta, we can always get a such u. Commented Mar 1 at 16:17

It's everywhere in field theories, like in electrostatics or gravity. This means of course that you also have it for often-used the gravity analogue where force is applied on a rubber sheet, with pictures like this: .

So the function $$u({\bf x})$$ is the amount of downward displacement of the sheet at each point, if $$f({\bf x})$$ gives the force at each point. In the picture $${\bf x}$$ is in a two-dimensional space, which is the best choice if you want to visualize it. But the concept can of course be generalized.