Differential equations – Partial differential equations – Wave equation

The wave equation is: \(a^2 u_{xx}=u_{tt}\) for \(0 < x < L\) and \(t>0\).

Here \(u=u(x,t)\) is a function of two variables (the position variable \(x\) and the time variable \(t\)). This describes for instance the position of a vibrating string with length \(L > 0\), which thickness can be neglected, relative to the equilibrium position. The positive constant \(a^2\) is called the spring constant and depends on the properties of the string. At time \(t=0\) the string might have a certain initial position (shape) and might have a certain initial velocity. These can be described by a function \(f(x)\) and a function \(g(x)\) respectively for \(0\leq x\leq L\). Since both ends of the string are fixed at the equilibrium position, this leads to homogeneous boundary values. The position of the string might be both positive and negative; the string might be both above and below the equilibrium position (the motion is seen as in the plane).

This leads to the following initial-boundary-value problem with homogeneous boundary conditions:

\[\left\{\begin{array}{l}a^2u_{xx}=u_{tt},\quad 0 < x < L,\quad t>0\\[2.5mm] u(0,t)=0,\quad u(L,t)=0,\quad t > 0\\[2.5mm] u(x,0)=f(x),\quad u_t(x,0)=g(x),\quad 0\leq x\leq L.\end{array}\right.\]

We use the method of separation of variables: let \(u(x,t)=X(x)T(t)\), then we have:

\[a^2u_{xx}=u_{tt}\quad\Longleftrightarrow\quad a^2X''(x)T(t)=X(x)T''(t).\]

For \(u(x,t)=X(x)T(t)\neq 0\) we now have (divide by \(X(x)T(t)\)): \(\displaystyle\frac{X''(x)}{X(x)}=\frac{1}{a^2}\cdot\frac{T''(t)}{T(t)}\). The left-hand side only depends on \(x\) (and not on \(t\)), while the right-hand side only depends on \(t\) (and not on \(x\)). Therefore these can only be equal for all \(x\in(0,L)\) and \(t>0\) if they are constant. Hence:

\[\frac{X''(x)}{X(x)}=\frac{1}{a^2}\cdot\frac{T''(t)}{T(t)}=\sigma\quad(\text{separation constant}).\]

This implies:

\[X''(x)-\sigma X(x)=0\quad\text{en}\quad T''(t)-\sigma a^2 T(t)=0.\]

These are ordinary differential equations for \(X(x)\) and for \(T(t)\). Using the initial conditions we have (with \(T(t)\neq 0\)):

\[u(0,t)=0:\quad X(0)T(t)=0\quad\Longrightarrow\quad X(0)=0\quad\quad\text{and}\quad\quad u(L,t)=0:\quad X(L)T(t)=0\quad\Longrightarrow\quad X(L)=0.\]

This leads to the homogeneous boundary-value problem

\[\left\{\begin{array}{l}X''(x)-\sigma X(x)=0,\quad 0 < x < L\\[2.5mm]X(0)=0,\quad X(L)=0.\end{array}\right.\]

We distinguish three possibilities: \(\sigma=0\), \(\sigma > 0\) and \(\sigma < 0\):

• \(\sigma=0\): \(X''(x)=0\;\Longrightarrow\;X(x)=a_1x+a_2\). Using \(X(0)=0\) and \(X(L)=0\) we then have: \(a_2=0\) and \(a_1L+a_2=0\). Hence: \(a_1=a_2=0\) (the trivial solution).


• \(\sigma=\mu^2 > 0\): \(X''(x)-\mu^2X(x)=0\;\Longrightarrow\;X(x)=b_1\cosh(\mu x)+b_2\sinh(\mu x)\). Using \(X(0)=0\) and \(X(L)=0\) we then have: \(b_1=0\) and \(b_1\cosh(\mu L)+b_2\sinh(\mu L)=0\). Hence: \(b_1=b_2=0\) (the trivial solution), since \(\mu L\neq 0\) (and therefore: \(\sinh(\mu L)\neq 0\)).


• \(\sigma=-\mu^2 < 0\): \(X''(x)+\mu^2X(x)=0\;\Longrightarrow\;X(x)=c_1\cos(\mu x)+c_2\sin(\mu x)\). Using \(X(0)=0\) and \(X(L)=0\) we then have: \(c_1=0\) and \(c_1\cos(\mu L)+c_2\sin(\mu L)=0\). So there are nontrivial solutions if \(\sin(\mu L)=0\), so if \(\mu L=\pm n\pi\) met \(n\in\{1,2,3,\ldots\}\).

So there are only nontrivial solutions for \(\sigma_n=-\mu_n^2 < 0\) with \(\mu_n=\displaystyle\frac{n\pi}{L},\;n=1,2,3,\ldots\).

This leads to the eigenvalues \(\displaystyle\sigma_n=-\frac{n^2\pi^2}{L^2}\;n=1,2,3,\ldots\) and the eigenfunctions \(X_n(x)=\displaystyle\sin\left(\frac{n\pi x}{L}\right),\;n=1,2,3,\ldots\).

For \(T(t)\) we then have: \(T_n''(t)-\sigma_n a^2 T_n(t)=0\) with eigenfunctions \(T_n(t)=\displaystyle c_n\cos\left(\frac{n\pi at}{L}\right)+k_n\sin\left(\frac{n\pi at}{L}\right)\).

Then we have: \(u_n(x,t)=X_n(x)T_n(t)=\displaystyle\sin\left(\frac{n\pi x}{L}\right)\left(c_n\cos\left(\frac{n\pi at}{L}\right)+k_n\sin\left(\frac{n\pi at}{L}\right)\right),\;n=1,2,3,\ldots\). Now we use the super position principle:

\[u(x,t)=\sum_{n=1}^{\infty}\sin\left(\frac{n\pi x}{L}\right)\left(c_n\cos\left(\frac{n\pi at}{L}\right)+k_n\sin\left(\frac{n\pi at}{L}\right)\right).\]

Then we have:

\[u_t(x,t)=\sum_{n=1}^{\infty}\sin\left(\frac{n\pi x}{L}\right)\left(-\frac{n\pi a}{L}c_n\sin\left(\frac{n\pi at}{L}\right)+\frac{n\pi a}{L}k_n\sin\left(\frac{n\pi at}{L}\right)\right).\]

Using the initial position we then have for \(0\leq x\leq L\):

\[u(x,0)=f(x)\quad\Longleftrightarrow\quad\sum_{n=1}^{\infty}c_n\sin\left(\frac{n\pi x}{L}\right)=f(x).\]

This is a Fourier sine series for \(f(x)\). Using the Euler-Fourier formulas we then have:

\[c_n=\frac{2}{L}\int_0^Lf(x)\sin\left(\frac{n\pi x}{L}\right)\,dx,\quad n=1,2,3,\ldots.\]

Using the initial velocity we then have for \(0\leq x\leq L\):

\[u_t(x,0)=g(x)\quad\Longleftrightarrow\quad\sum_{n=1}^{\infty}\frac{n\pi a}{L}k_n\sin\left(\frac{n\pi x}{L}\right)=g(x).\]

This is a Fourier sine series for \(g(x)\). Using the Euler-Fourier formulas we then have:

\[\frac{n\pi a}{L}k_n=\frac{2}{L}\int_0^Lg(x)\sin\left(\frac{n\pi x}{L}\right)\,dx,\quad n=1,2,3,\ldots.\]

In the case of an initial velocity \(g(x)=0\) for all \(x\in[0,L]\) we might extend the function \(f(x)\) with \(0\leq x\leq L\) to an odd function on the interval \((-L,0)\) and outside that interval periodically with period \(2L\), hence

\[h(x)=\left\{\begin{array}{rl}-f(-x),&-L < x <0\\[2.5mm]f(x),&o\leq x\leq L\end{array}\right.\quad\text{en}\quad h(x+2L)=h(x)\quad\text{voor alle}\;x\in\mathbb{R},\]

then we have:

\[h(x)=\sum_{n=1}^{\infty}c_n\sin\left(\frac{n\pi x}{L}\right)\quad\text{met}\quad c_n=\frac{2}{L}\int_0^Lf(x)\sin\left(\frac{n\pi x}{L}\right)\,dx,\quad n=1,2,3,\ldots.\]

This implies that

\[h(x-at)=\sum_{n=1}^{\infty}c_n\left(\sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{n\pi at}{L}\right)-\cos\left(\frac{n\pi x}{L}\right)\sin\left(\frac{n\pi at}{L}\right)\right)\]

and

\[h(x+at)=\sum_{n=1}^{\infty}c_n\left(\sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{n\pi at}{L}\right)+\cos\left(\frac{n\pi x}{L}\right)\sin\left(\frac{n\pi at}{L}\right)\right).\]

This implies that

\[\frac{1}{2}\left[h(x-at)+h(x+at)\right]=\sum_{n=1}^{\infty}c_n\sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{n\pi at}{L}\right)=u(x,t).\]

Hence, the solution can be written as \(u(x,t)=\frac{1}{2}\left[h(x-at)+h(x+at)\right]\), where \(h(x)\) is the function, that is obtained from \(f(x)\) by extending it as an odd function which is periodic with period \(2L\).

More general: suppose that \(u(x,t)=\varphi(x-at)+\psi(x+at)\), then we have that

\[u_{xx}=\varphi''(x-at)+\psi''(x+at)\quad\text{and}\quad u_{tt}=(-a)^2\varphi''(x-at)+a^2\psi''(x+at).\]

Hence: \(u(x,t)=\varphi(x-at)+\psi(x+at)\) isa solution of \(a^2u_{xx}=u_{tt}\) for every function \(\varphi\) and every function \(\psi\).

However, this is not of practical use for solving an initial-boundary-problem based on such a wave equation. It is much beter to use the method of separation of variables instead.

The two-dimensional wave equation is: \(a^2\left(u_{xx}+u_{yy}\right)=u_{tt}\) with \(u=u(x,y,t)\).

For this we can also use the method of separation of variables: let \(u(x,y,t)=X(x)Y(y)T(t)\), then we have:

\[\alpha^2\left(u_{xx}+u_{yy}\right)=u_{tt}\quad\Longleftrightarrow\quad \alpha^2\left(X''(x)Y(y)T(t)+X(x)Y''(y)T(t)\right)=X(x)Y(y)T''(t)\]

For \(u(x,y,t)=X(x)Y(y)T(t)\neq 0\) have (divide by \(X(x)Y(y)T(t)\)): \(\displaystyle\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}=\frac{1}{\alpha^2}\cdot\frac{T''(t)}{T(t)}\). The left-hand side only depends on \(x\) and \(y\) (and not on \(t\)), while the right-hand side only depends on \(t\) (and not not \(x\) and \(y\)). Therefore these can only equal for all \(x\) and \(y\) in the domain and \(t>0\) if they are constant. Hence:

\[\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}=\frac{1}{\alpha^2}\cdot\frac{T''(t)}{T(t)}=\sigma\quad(\text{separation constant}).\]

This implies: \(T''(t)-\sigma\alpha^2 T(t)=0\) and \(\displaystyle\frac{X''(x)}{X(x)}=\sigma-\frac{Y''(y)}{Y(y)}\). The left-hand side only depends on \(x\) (and not on \(y\)), while the right-hand side only depends on \(y\) (and not on \(x\)). Therefore these can only be equal for all \(x\) and \(y\) in the domain if they are constant, so:

\[\frac{X''(x)}{X(x)}=\sigma-\frac{Y''(y)}{Y(y)}=\tau\quad(\text{separation constant}).\]

Hence:

\[X''(x)-\tau X(x)=0,\quad Y''(y)-(\sigma-\tau)Y(y)=0\quad\text{and}\quad T''(t)-\sigma\alpha^2 T(t)=0.\]

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Last modified on August 23, 2021