Why can't the solution to Laplace's equation be derived from points?

[meldraft]

As the title suggests, the general solution of Laplace's equation has 4 arbitrary constants. One would imagine that if you e.g. have the potential at 4 points in a domain, you could get the specific solution by replacing:

V(x1,y1)=V1, V(x2,y2)=V2, V(x3,y3)=V3, V(x4,y4)=V4,

and solving the 4x4 system to find the 4 arbitrary constants. I tried it but it doesn't yield a solution.

Intuitively I understand that since I have not provided equations for the boundaries I should not be able to get the solution, but does anyone know the actual mathematical cause (I suspect that the equations are linearly dependent, but no sure why)?

 

[espen180]

Laplace's equation is elliptic. Therefore you need either Dirichlet or Neumann boundary conditions on a closed boundary surrounding you are interested in. If you try to use other boundary conditions, you will find that they are either too lenient, meaning that the solution is not unique, or they are too restrictive and no nice solution exists.

If your case, you have fixed the potential at 4 points. However, this is not enough to propagate a solution because you don't know the value of the potential at neighbouring points.

 

[meldraft]

Thank you for your answer!

My book on PDEs also suggests that for uniqueness to be guaranteed, we need to proceed as you described. It doesn't, however, state why. Do you know of any source where I can actually read the proof of that statement?

 

[chrisbaird]

Perhaps one of my previous posts on this subject will help.

 

[Meir Achuz]

Most advanced EM textbooks have proofs of the uniqueness theorem.

 

[Studiot]

and solving the 4x4 system to find the 4 arbitrary constants.

As you have observed Laplace equation is a partial dif =ion.

You add arbitrary functions not arbitrary constants when you integrate (solve) partial diffs.

You add constants of integration to ODEs.

 

[meldraft]

Thank you all for your answers!

I understand your points, but there is one thing I can't really explain. It is related to Studiot's remark that we have arbitrary functions:

Let's take the general solution:

V(x,y)=[A1cosh(ax) + A2sinh(ax)][B1cos(ay) + B2sin(ay)]

The above result would suggest that if I have 5 solutions of V(x,y) at 5 different points (let's consider them on different contours of the function), I should be able to calculate the constants A1,A2,B1,B2, and a.

Of course, I can't find a solution to that neither through algorithms, nor by hand, and from the theory and your remarks I know it shouldn't be possible because I do not have enough information.

The question is: Why? What is the mathematical explanation that, given a 5x5 (forgive my original 4x4 statement) system I cannot find the 5 unknowns A1,A2,B1,B2, and a?

The only answer I can come up with is that the equations are dependent, but I'm not sure how I can prove non-linear dependence.

 

[Meir Achuz]

The general solution is a sum over an infinite Fourier series with parameters A_n, etc., not just one term with a single A. You need an infinite number of points (or a continuous function on the surface.

 

[meldraft]

First of all, happy new year to everyone!

The general solution is a sum over an infinite Fourier series with parameters A_n, etc., not just one term with a single A. You need an infinite number of points (or a continuous function on the surface.

I'm sorry if I'm asking something obvious, but how is this a Fourier series? That solution comes from separating to potential to V(x,y) = X(x)Y(y).

By solving the two 2nd order differential equations that result from this, we get:

X(x) = c_1e^{ax}+c_2e^{-ax}
Y(y) = c_3e^{jay}+c_4e^{-jay}

Therefore, since V(x,y) = X(x)Y(y):

V(x,y)=X(x)Y(y)=(c_1e^{ax}+c_2e^{-ax})(c_3e^{jay}+c_4e^{-jay})

and this transforms into:

V(x,y)=[A_1cosh(ax) + A_2sinh(ax)][B_1cos(ay) + B_2sin(ay)]

You could of course solve it using Fourier series, but this solution is derived without it. I can't imagine how this involves Fourier series, so I would be grateful if you could give me a hint! :)

 

[Studiot]

I understand your points, but there is one thing I can't really explain. It is related to Studiot's remark that we have arbitrary functions:

OK look at this, I aim to convince you that there is no clear cut relationship between the order of a partial differential equation and the number of unknown constants or functions.

This is unlike ordinary differential equations where you say that one unknown constant is eliminated in the formation of a first order, two for a second and so on.

Consider the first order PDE

z = x\frac{{\partial z}}{{\partial x}} + y\frac{{\partial z}}{{\partial y}}

Now let both partials be constants

\begin{array}{l}<br /> \frac{{\partial z}}{{\partial x}} = a \\ <br /> \frac{{\partial z}}{{\partial y}} = b \\ <br /> \end{array}

By substitution we can see that this corresponds to the equation (solution)

z = ax + by

and that two arbitrary constants have been eliminated in the formation of a first order PDE

Now consider the second order PDE

\frac{{{\partial ^2}z}}{{\partial {t^2}}} = {c^2}\frac{{{\partial ^2}z}}{{\partial {x^2}}}

with a solution

z = f(x - ct) + g(x + ct)

where f and g are arbitrary functions of (x-ct) and (x+ct)

Then

\begin{array}{l}<br /> \frac{{\partial z}}{{\partial x}} = f&#039;(x - ct) + g&#039;(x + ct) \\ <br /> \frac{{{\partial ^2}z}}{{\partial {x^2}}} = f&#039;&#039;(x - ct) + g&#039;&#039;(x + ct) \\ <br /> \frac{{{\partial ^2}z}}{{\partial {t^2}}} = {c^2}f&#039;&#039;(x - ct) + {c^2}g&#039;&#039;(x + ct) \\ <br /> \end{array}

Comparison lead to the original equation, but note that two arbitrary functions have been eliminated in the formation of this second order PDE.

You could, of course, say that in the first example a and b are constant functions, preserving my original statement that you need to find arbitrary functions when you solve PDEs.

 

[Studiot]

I guess you internet access is intermittent since there is no further post but I am going to assume you are with my last post.

This means that we are ready to address your original question, which I propose to do by solving Laplaces equation to reach your solution (which is not a general solution) and thence to a real general solution which can be used to explain the remark about an infinite number of constants.

Firstly the Laplace equation in vector form : I will use rectangular coordinates as you have done.

{\nabla ^2}\varphi = 0

Assuming that the a solution can be found that is an independant function {say X(x), Y(y), Z(z)} of each of the coordinate axes the solution will be the product of these functions.

Thus

\varphi = X(x)Y(y)Z(z)

forming the second differentials and substituting into Laplace

YZ\frac{{{d^2}X}}{{d{x^2}}} + ZX\frac{{{d^2}Y}}{{d{y^2}}} + XY\frac{{{d^2}Z}}{{d{z^2}}} = 0

divide through by 1/XYZ

\frac{1}{X}\frac{{{d^2}X}}{{d{x^2}}} + \frac{1}{Y}\frac{{{d^2}Y}}{{d{y^2}}} + \frac{1}{Z}\frac{{{d^2}Z}}{{d{z^2}}} = 0

The only possible way this sum can always add up to zero is if the three parts are real constants, independant of x,yor z. Since we want real constants use squares.

\begin{array}{l}<br /> \frac{1}{X}\frac{{{d^2}X}}{{d{x^2}}} = {a^2} \\ <br /> \frac{1}{Y}\frac{{{d^2}Y}}{{d{y^2}}} = {b^2} \\ <br /> \frac{1}{Z}\frac{{{d^2}Z}}{{d{z^2}}} = {c^2} \\ <br /> \end{array}

with the condition that

{a^2} + {b^2} + {c^2} = 0

These can be seen to be three ordinary differential equations that have solutions

\begin{array}{l}<br /> X = {A_1}{e^{ax}} + {A_2}{e^{ - ax}} \\ <br /> Y = {B_1}{e^{bx}} + {B_2}{e^{ - bx}} \\ <br /> Z = {C_1}{e^{cx}} + {C_2}{e^{ - cx}} \\ <br /> \end{array}

This means we have found a particular solution that has 9 constants.
Your solution is only one of these equations.

a, b, c, A₁, A₂, B₁, B₂, C₁, C₂,

But any linear combination of these functions will also satisfy the original equation - there are an infinite number of these so if we add them all (infinite) up we obtain the general solution

\varphi = \sum\limits_1^\infty {{{\left( {{A_1}{e^{ax}} + {A_2}{e^{ - ax}}} \right)}_n}{{\left( {{B_1}{e^{bx}} + {B_2}{e^{ - bx}}} \right)}_n}{{\left( {{C_1}{e^{cx}} + {C_2}{e^{ - cx}}} \right)}_n}}

Hence the remark that you will have to find an infinite number of constants.

 

[meldraft]

You guessed right, I had a problem with my ISP

Your explanation is excellent, I understand the mathematics perfectly now :!)

This generates one physical question however. Please correct my reasoning if I am mistaken:

I have seen in the simulations that no matter the magnitude of the boundary conditions, the field generated looks exactly the same, albeit with different scaling (so if I have one plate with 0 Volts on one side and 10 Volts on another, if I change the BCs to 0 and 1000000 the field looks exactly the same, but the scale is different).

From what I understand, the infinite number of combinations can create an infinite number of fields. This goes for both "shape" and "magnitude".

If we consider that any linear combination of X(x), Y(y), Z(z) is a valid solution, this means that a function L(x)M(y)N(z), where:

L=t_1X(x), M=t_2Y(y), N=t_3Z(z) (t1,t2,t3 are arbitrary constants)

is t_1t_2t_3XYZ=t_1t_2t_3φ(x,y,z)=tφ(x,y,z)

Therefore, all combinations of X Y Z satisfy the original equation. This explains the "magnitude" part perfectly.

If we pick a single solution out of this family of infinite solutions, e.g. for t=1, then is the "shape" of the field only dependent on the 9 constants? (This would translate into the statement that a proper combination of these 9 constants can give a field of just about any shape (provided of course it's still smooth etc.)).

 

[Studiot]

That's really good that you are back with us and even better that you followed my short form derivation - it required the reader to work some things out for themselves between the lines.

You need to be aware that you will get different forms of solution depending upon your coordinate system.

In particular if you use spherical coordinates there is a singularity at the origin so that any region that includes the origin will 'blow up' and fail.

go well

 

[meldraft]

Thank you so much for your help! Time to hit the notebooks (again!)