I am solving this PDE using separation of variables but stuck

$$u_{tt} − α^2 \ u_{xx} = 0, \\ 0 ≤ x ≤ 1, 0 ≤ t < ∞$$

$u(0, t) = 0$
$u(1, t) = sin(t)$
$u(x, 0) = 0$
$u_{t}(x, 0) = 0$\




what is the way forward

  • $\begingroup$ that didn't work. That is why I showed how far I went $\endgroup$
    – user245746
    Jun 7, 2023 at 8:39

1 Answer 1


The general solution of the wave equation is

$$u(x,t)=f_1(x-\alpha t)+f_2(x+\alpha t)$$

where $f_1(x-\alpha t)$ is a wave moving to the right and $f_2(x+\alpha t)$ moving to the left with the velocity $\alpha$.

My solution approach for this particular problem is

$$u(x,t)=x \sin (t)+\frac{1}{\pi}\sum _{k=1}^{\infty} c_{k} (d_{k} (\cos (\pi k (x-\alpha t))-\cos (\pi k (x+\alpha t)))+(\cos (\pi k x+t)-\cos (\pi k x-t)))$$

already satisfying 3 conditions (2 Dirichlet BC, 1 IC):

$$u(0,t)=0$$ $$u(x,0)=0$$ $$u(1,t)=\sin(t)$$

Now we have a look at the Neumann boundary condition $u_{t}(x,0)=0$:

$$u_{t}(x,0)=x+\underbrace{\frac{1}{\pi} \sum_{k=1}^{\infty} 2 c_{k}\cdot (-1+2\pi\alpha\cdot d_{k})\cdot \sin(k \pi x)}_{-x}=0$$

In order to force this equation to $0$, we use the Taylor series approximation:

$$-x=\frac{1}{\pi}\sum_{k=1}^{\infty} a_{k} \sin(\pi k x)=\frac{1}{\pi}\sum_{k=1}^{\infty} \frac{2 (-1)^k}{k} \sin(\pi k x)$$

In order to fit the coefficients we have to solve the equation $2 c_{k}(-1+2\pi\alpha d_{k})=a_{k}$. With $d_{k}=\pi k \alpha$ we get

$$c_{k}=\frac{(-1)^k}{k \left(\pi ^2 \alpha ^2 k^2-1\right)}$$

No we are going to visualize our solution for velocity $\alpha=1$:

enter image description here


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