solving the simplest PDE (non trivial): $\Delta u(x,t) = 0$

Question

(Also the PDE can be written like $\partial_{xx}u+\partial_{tt}u = 0$)

Of course for unambiguity this needs some condition:

$\bullet\quad u(0,t) = 0, \quad u(1,t) = 0$

$\bullet \quad u(x,0) = 0 \quad u(x,1) = 1$

Finding the function should be fairly easy due to the linearity:

it alludes the function $u$ is separable: $u(x,t) = X(x)\cdot T(t)$

Respectively for the PDE: $$\begin{align} &\partial_{xx} X(x)\,T(t) + X(x)\,\partial_{tt} T(t) \quad \Leftrightarrow \quad\dfrac{\partial_{xx}X(x)}{X(x)} = \dfrac{\partial_{tt}T(t)}{T(t)} \end{align}$$

Both sides should equal a constant:

$$\begin{array}{cc} &\dfrac{\partial_{xx}X(x)}{X(x)} = \lambda\\\\ &\dfrac{\partial_{tt}T(t)}{T(t)} = \lambda \end{array}$$

It turns out, only $\lambda = -\lambda$ fulfils the starting conditions. For elegancy one choses $\lambda = -\lambda^2$

After all 2 ODE are to be solved:

$$\begin{array}{cc} &\partial_{xx}X(x)+\lambda^2\,X(x) = 0 &\quad \Rightarrow X(x) = A\,\cos(\lambda\,x)+B\,\sin(\lambda\,x)\\\\ &\partial_{tt}T(t)+\lambda^2\,T(t) = 0 &\quad \Rightarrow T(t) = C_1\,\cos(\lambda\,t)+C_2\,\sin(\lambda\,t) \end{array}$$

This yields the total solution: $$\displaystyle{u(x,t) = \left(A\,\cos(\lambda\,x)+B\,\sin(\lambda\,x)\right)\, \left(C_1\,\cos(\lambda\,t)+C_2\,\sin(\lambda\,t)\right)}$$

Now using starting Condition:

How do I actually use these?

$\begin{array}{ccc} &\bullet &X(0) = A\,\cos(\lambda\,0)+B\,\sin(\lambda\,0) = 0 &\Rightarrow A = 0\\\\ & & X(1) = A\,\cos(\lambda\,1)+B\,\sin(\lambda\,1) = 0 &\Rightarrow \lambda = n\,\pi \end{array}$

so?

It seems like there are endless discrete solution, since $\lambda = n\,\pi \quad n \in \mathbb{N}$

The solution has to be the sum of all of them:

$u(x,t) = \displaystyle{\sum_{n = 1}^{\infty} B\,\sin(\lambda\,x)\,\left(C_1\,\cos(\lambda\,t)+C_2\,\sin(\lambda\,t)\right)}$

at this point let's call $B\,C_1 = a_n$ and $B\,C_2 = b_n$:

$u(x,t) = \displaystyle{\sum_{n = 1}^{\infty} \sin(\lambda\,x)\,\left(a_n\,\cos(\lambda\,t)+b_n\,\sin(\lambda\,t)\right)}$

Now using the other starting conditions:

$\begin{array}{ccc} &\bullet \quad &u(x,0) = \displaystyle{\sum_{n = 1}^{\infty} \sin(\lambda\,x)\,\left(a_n\,\cos(\lambda\,0)+b_n\,\sin(\lambda\,0)\right) = 0} &\Rightarrow a_n = 0 ?\\\\ & &u(x,1) = \displaystyle{\sum_{n = 1}^{\infty} \sin(\lambda\,x)\,\left(a_n\,\cos(\lambda\,1)+b_n\,\sin(\lambda\,1)\right) = 1} &\Rightarrow b_n = 0 ? \end{array}$

For the solution process I kept myself inspired by the solution of the wave equation, but since the starting conditions are different I can't really transfer it all. I don't even know it is relatable at all

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Addendum

Looked like I was caught by a disastrous sign error: going to be:

$$\begin{align} \dfrac{\partial_{xx}X(x)}{X(x)} = \color{red}{{-}}\dfrac{\partial_{tt}Y(y)}{Y(y)} \end{align}$$

hence the solutions:

$$\begin{array}{cc} &\partial_{xx}X(x)+\lambda^2\,X(x) = 0 &\quad \Rightarrow X(x) = A\,\cos(\lambda\,x)+B\,\sin(\lambda\,x)\\\\ &\partial_{tt}Y(y)\color{red}{{-}}\lambda^2\,Y(y) = 0 &\quad \Rightarrow Y(y) = C_1\,\cosh(\lambda\,t)+C_2\,\sinh(\lambda\,t) \end{array}$$

luckily the constant $\lambda$ is still the same so the solution can be straight written as sum:

$$u(x,t) = \displaystyle{\sum_{n = 1}^{\infty} \sin(\lambda_n\,x)\,\left(a_n\,\cosh\lambda_n\,t)+b_n\,\sinh(\lambda_n\,t)\right)}$$

Now, again trying to use starting condition:

$\begin{array}{ccc} &\bullet \quad &u(x,0) = \displaystyle{\sum_{n = 1}^{\infty} \sin(\lambda_n\,x)\,\left(a_n\,\cosh(\lambda_n\,0)+b_n\,\sinh(\lambda_n\,0)\right) = 0} &\Rightarrow a_n = 0 \\\\ & &u(x,1) = \displaystyle{\sum_{n = 1}^{\infty} \sin(\lambda\,x)\,\left(a_n\,\cosh(\lambda_n\,1)+b_n\,\sinh(\lambda_n\,1)\right) = 1} &\Rightarrow b_n = ? \end{array}$

Here I'm tripping once more.

Normally for determing coefficients one multiplies by $\sin(m\,\pi\,x)$ and integrates $[0, 1]$.

Even if $a_n = 0$ it's still heavy:

$$\displaystyle{\sum_{n= 1}^{\infty}b_n\,\int_{0}^{1}\sin(n\,\pi\,x)\,\sinh(\lambda_n\,1)\cdot \sin(m\,\pi\,x)\,\mathrm{dx} = \int_{0}^{1}1 \cdot\sin(m\,\pi\,x)}\,\mathrm{dx}$$

The difficulty consists of evaluation that one big integral.

If you had assumed $\sinh(\lambda_n\,1)$ with $\lambda_n = \pi\,n$ depends on x, like I did this would be pretty awkward. But since it's a constant just draw it out:

$$\begin{array}{cc}\displaystyle{\sum_{n=1}^{\infty}b_n\,\sinh(n\,\pi\,1)\,\int_{0}^{1}\sin(n\,\pi\,x)\,\cdot \sin(m\,\pi\,x)\,\mathrm{dx} = \frac{1}{2}\,b_n\,\sinh(n\,\pi\, 1)\,\delta_{n\,m}}\\\\ \text{Now bringing this to the other side (and assuming $n = m$ hence $\delta_{n\,m} = 1$)}:\\\\ \displaystyle{b_n =} \displaystyle{\dfrac{2}{\sinh(n\,\pi\,1)}\,}\int_{0}^{1}1 \cdot\sin(n\,\pi\,x)\,\mathrm{dx} = \dfrac{2}{\sinh(n\,\pi\,1)}\,\left(\dfrac{1-\cos(n\,\pi)}{n\,\pi}\right) \end{array}$$

With the predetermined solution:

$$u(x,y) = \displaystyle{\sum_{n = 1}^{\infty} \sin(n\,\pi\,x)\,\left(b_n\,\sinh(n\,\pi\,y)\right)} \quad \mathrm{\vert right?}$$

Answer 1

First of all, you shouldn't use the wave equation $u_{tt}=u_{xx}$ as a source of intuition for the Laplace equation $u_{xx}+u_{yy}=0$, since they behave completely differently. So let me use the letter $y$ instead of $t$ (and $Y$ instead of $T$).

The main problem with your attempt is that you have a sign error when you're separating the variables; it should be $X_{xx}/X = - Y_{yy}/Y$, so the ODEs for $X$ and for $Y$ are not the same.

A set of separated solutions which satisfy the three boundary conditions where $u=0$ is $$ u_n(x,y) = \sin(n \pi x) \sinh(n \pi y) , $$ so you should be able to finish by determining the constants $c_n$ such that the Fourier sine series $$ u(x,y) = \sum_{n=1}^{\infty} c_n u_n(x,t) = \sum_{n=1}^{\infty} c_n \sinh(n \pi y) \sin(n \pi x) $$ satisfies the remaining boundary condition $u(x,1)=1$ for $0<x<1$.