# Generalized solution of Laplace equation with unknown boundary conditions

Background: For example, the 2D Laplace's equation: $$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} =0$$ , when solved by separation of variable $V=X(x)Y(y)$, has the solution of the form: $$V_k(x,y)=(Ae^{kx}+Be^{-kx}) + (C\sin ky+D\cos ky) \tag{1}$$ Usually, a set of (periodic) boundary conditions are given, which might eliminate some terms in $V_k$, and the general solution is the discrete sum of all the eigenvectors: $$V(x,y)=\sum_{k=-\infty}^{\infty}V_k(x,y) \tag{2}$$ , where $k$'s have a specific form depending on the boudary conditions given.

Situation: When the boundary conditions are not given, $V_k$ in (1) is of the most general form and $k$ can take any possible values in $(-\infty, \infty)$, which means the sum in (1) must be changed into an integral (correct me if I am wrong!).

Question: When the boundary conditions are not yet given, $k$ can be continuous. Thus, what is the generalized version of the sum in (2)?

Attempt: All I have tried so far is coming up with the (obvious?) following integral: $$V(x,y)=\int_{-\infty}^{\infty}[(A(k)e^{kx}+B(k)e^{-kx}) + (C(k)\sin ky+D(k)\cos ky)]dk$$ However, I have a feeling that this integral is not the correct answer.

$$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} =0 \tag 1$$ You say that you solved by separation of variable $V=X(x)Y(y)$ and found the solution on the form: $$V_k(x,y)=(Ae^{kx}+Be^{-kx}) + (C\sin ky+D\cos ky)$$ This is not correct. Put it into Eq.$(1)$ and observe that the equality is not satisfied.

Unfortunately the detail of the calculus is not shown. So one cannot check it and see where is the mistake or deficiency.

For particular solutions on the form $\quad V_k=X(x)Y(y)\quad$ we have $\quad X''Y+Y''X=0\quad$ which implies : $$\frac{X''}{X}=-\frac{Y''}{Y}=\lambda^2 \qquad \begin{cases} X=e^{\pm\lambda x}\\ Y=e^{\pm i\lambda y} \end{cases}$$ where $\lambda$ can be complex. So the functions $X(x)$ and $Y(y)$ can be exponential or sinusoidal or combination of both. In order to simplify the writing, let the solution be expressed on exponential form without forgetting that the exponential can be replaced by sinusoidal or product of exponential and sinusoidal functions. $$V_\lambda(x,y)=e^{\lambda(x\pm iy)}$$ Any linear combinations of $V_\lambda(x,y)$ with different $\lambda$ are solutions of Eq.$(1)$. For example : $$V(x,y)=\sum_\lambda A_\lambda e^{\lambda(x+ iy)}+\sum_\lambda B_\lambda e^{\lambda(x- iy)}$$ where the coefficients $A_\lambda$ and $B_\lambda$ are arbitrary constants.

A more general solution than a discret sum is expressed on integral form : $$V(x,y)=\int f(\lambda) e^{\lambda(x+ iy)}d\lambda+\int g(\lambda) e^{\lambda(x- iy)}d\lambda$$ $f(\lambda)$ and $g(\lambda)$ are arbitrary function insofar the integral be convergent.

We observe that the first integral is a function of $(x+iy)$ and the second is a function of $(x-iy)$. Thus the above solution can be written on the form : $$V(x,y)=F(x+iy)+G(x-iy)$$ $F$ and $G$ are arbitrary functions to be determined according to some boundary conditions.

• Thank you for your answer. Regardless of the validity of my solution, let's focus on the integral in your answer. You provided the sum: $$V(x,y)=\sum_\lambda A_\lambda e^{\lambda(x+ iy)}+\sum_\lambda B_\lambda e^{\lambda(x- iy)} \tag{1}$$ , and you said the follwing intergral is the more general form of (1): $$V(x,y)=\int f(\lambda) e^{\lambda(x+ iy)}d\lambda+\int g(\lambda) e^{\lambda(x- iy)}d\lambda \tag{2}$$ Could you provide the derivation from (1) to (2)? If you read the Attempt section in my OP, you will see that this is the source of confusion in my question. – A Slow Learner Apr 8 '18 at 7:43
• I don't think that going from the discret form to the integral form is a source of confusion in your "Attempt" section. On the contrary, your attempt is on the right way. The source of confusion is that you start from false particular solutions $V_k(x,y)=(Ae^{kx}+Be^{-kx}) + (C\sin ky+D\cos ky)$. If you had start from correct particular solutions, your final result would be the same as mine. So, don't worry about the derivation from (1) to (2). You have well done. But check the separation of variables part of your work : this the only cause of trouble. – JJacquelin Apr 8 '18 at 8:09
• Thank you. Actually, it is one of the sources of my confusion. For example, going from (1) to (2) is similar to going from $\sum f(x)$ to $\int f(x) dx$. However, I don't understand why it is the case. – A Slow Learner Apr 8 '18 at 8:39
• $\sum_n f(x_n)=\sum f(x_n)\delta x$ with $\delta x=1$. Compare to $\int f(x)dx$. This is similar to the computation of the area under a curve in considering the sum of an infinite number off small areas of high $f(x)$ and width $dx$. In the discret case, the function considered is a step function. In the integral case it is any continuous function, which could also be eventually a step function. – JJacquelin Apr 8 '18 at 9:11
• I am not sure about your explanation. Considering this: $$\begin{split} \int_a^b f(x) dx &= \lim_{{k \to \infty}} \sum_{n=1}^k f(x_n)\Delta x\\ &\neq \lim_{k \to \infty} \sum_{n=1}^k f(x_n)\delta x \text{ with \delta x =1} \end{split}$$ because in the first limit, $\Delta x \to 0$ as $k \to \infty$; while in the second limit, $\delta x$ must be kept to be constant $=1$ in order for $\sum_{n=1}^k f(x_n)=\sum_{n=1}^k f(x_n)\delta x$. In other words, it is reasonable to say that the second limit diverges: $$\lim_{k \to \infty} \sum_{n=1}^k f(x_n)\delta x \to \infty$$ – A Slow Learner Apr 8 '18 at 21:37