# Uniqueness of solution of Laplace's equation

Consider steady-state temperature in a rectangular plate; we get solution when we solve Laplace's equation.

In the midway of solving this equation, we take linear combination of basic solutions and then find the coefficients in this combination by Fourier series method.

e.g. In present case basic solution is

$\mathtt{T}(x,y) = e^{-ny/10}\sin(\dfrac{n\pi x}{10})$

where $\mathtt{T}(x,y)$ is temprature at $(x,y)$ location of plate.

We know that Laplace's equation has only one solution in the given region with specified boundary conditions. For each $n$ there is a different solution. Then why do we say that there is only one solution?

Why do we take linear combination?

• The solution is linear combination because Laplace operator is linear. – Alexander Jul 8 '15 at 4:08
• What boundary conditions are we trying to satisfy? This also looks like it doesn't even solve Laplace's equation, $\nabla^2 T = n^2 (1 - \pi^2) T/100$ – ZachMcDargh Jul 8 '15 at 4:09
• @ZachMcDargh : A long reactangular metal plate has its two long sides and far end is at 0 deg cel and the base is at 100 deg cel. Width of plate is 10 cm. One corner is at origin (0,0) and shorter side is along x-axis. – atom Jul 8 '15 at 5:25

So, it looks like you're trying to answer a question like this: you've got a plate which is held at $T=0$ at $x = 0, x = L$ for all $y$, with $T= T_0(x)$ at $y=0$. What is $T(x, y)$ in this space $y \ge 0, 0\le x\le L?$

Then the answer is that on this strip we can solve Laplace's equation $\nabla^2T = 0$ with the functions $T(x, y) = A \exp(-\alpha y) \sin(\alpha x)$. Here $A$ is arbitrary, the choice of $\sin$ and $\exp$ gives us the $x=0$ and $y\rightarrow\infty$ conditions directly, and we must restrict ourselves to $\alpha = n~\pi/L$ for integer $n$ if we want to enforce the boundary condition at $x=L$.

We use a sum of these, carefully weighted by choosing appropriate $A$, to guarantee $$T_0(x) = T(x, 0) = \sum_{n=1}^\infty A_n \sin(n~\pi~x/L).$$Because $\nabla^2(f + g) = \nabla^2 f + \nabla^2 g,$ if we simply extend this choice of $A_n$ into a solution:$$f(x, y) = \sum_n A_n \exp(-n~\pi~y/L)\sin(n~\pi~x/L)$$ we find that actually, $\nabla^2 f = 0.$

The uniqueness of the solution given the boundary conditions therefore tells us: this is not merely a solution, but the solution, so $T(x, y) = f(x, y)$. So it's a clever way to answer the question for all $T_0(x)$.

That is why we perform this summation.

In a comment you mention that $T_0(x) = T_0$, some constant, in your particular case. So the essential problem here is that you have temperature discontinuities at $(x, y) = (0, 0)$ and $(L, 0)$. If you use "image space" methods (aka "periodic boundary conditions") for characterizing this $T_0$, it is a "square wave" boundary condition. We're therefore using the $A_n$ that are characteristic of a square wave to obtain the solution.

We know that $\sin^2$ has average value $1/2$, so we can work out that $$\int_0^L dx~ T_0(x) ~\sin(m~\pi~x/L) = \frac{1}{2}~A_m L$$ which is in your case: $$A_m = \frac{2 T_0}{L} \int_0^L dx~ \sin(m~\pi~x/L) = \frac{2 T_0}{m \pi} \left[-\cos\left(\frac{m~\pi}{L}~x\right)\right]_0^L = \frac{4 T_0}{m \pi} ~ \operatorname{odd}(m)$$where $\operatorname{odd}(m) = (1 - (-1)^m)/2$ is 1 if $m$ is odd, else it is 0. As promised, this matches exactly the Fourier Series of a Square Wave and has well-known approximation artifacts having to do with that.

The Laplacian is a linear operator - it does't include any non linear functions of the derivatives, thus includes only linear combination of (second order) derivatives.
For the laplacian in cartesian coordinates - $\nabla^2=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})$ i.e. $\nabla^2 \phi=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}) \phi =\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}+\frac{\partial^2 \phi}{\partial z^2}$

if $\nabla^2 \phi=f(x)$ and $\nabla^2 \psi=g(x)$ then $\zeta=\phi+\psi$ sutisfies

$\nabla^2 \zeta=\nabla^2(\phi+\psi)=\nabla^2\phi+\nabla^2\psi=f(x)+g(x)$

by linearity of the derivative operator ( $\frac{\partial}{\partial x}(f(x)+g(x))=\frac{\partial}{\partial x}f(x)+\frac{\partial}{\partial x}g(x)$ )