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## 1D Wave equation reloaded: discussion and examples

It will be a short lecture as two previous ones were a bit too long.

#### Domains of dependence and influence

Recall formula (4.4): \begin{multline} u(x,t)= \frac{1}{2}\bigl[ g(x+ct)+g(x-ct)\bigr]+\frac{1}{2c}\int_{x-ct}^{x+ct} h(x')\,dx'+\\[3pt] \frac{1}{2c} \iint_{\Delta (x,t)} f(x',t' )\,dx’d t' \label{equ-6.1} \end{multline} where $\Delta (x,t)$ is a characteristic triangle:

Therefore

Proposition. Solution $u(x,t)$ depends only on the right hand expression $f$ in $\Delta(x,t)$ and on the initial data $g,h$ on the base of $\Delta(x,t)$.

Definition. $\Delta(x,t)$ is a triangle of dependence for point $(x,t)$.

Conversely, if we change functions $g,h$ only near some point $(\bar{x},0)$ then solution can change only at points $(x,t)$ such that $(\bar{x},0)\in \Delta(x,t)$; let $\Delta^+(\bar{x},0)$ be the set of such points $(x,t)$:

Definition. $\Delta^+(\bar{x},0)$ is a triangle of influence for point $(\bar{x},0)$.

Remark. 1. We can introduce triangle of influence for each point (not necessary at $t=0$). 2. These notions work in much more general settings and we get domain of dependence and domain of influence which would not be triangles. 3. F.e. for $3D$ wave equation $u_{tt}- c^2 (u_{xx}+u_{yy}+u_{zz})=f$ those would be the backward light cone and forward light cone respectively and if $c$ is variable instead of cones we get conoids (imagine tin coin and then deform it but the vertex should remain).

Proposition. Solution propagates with the speed not exceeding $c$.
Example. \begin{align*} &u_{tt}-4u_{xx}= \sin(x)\cos( t),\\ &u|_{t=0}= 0,\\ &u_t|_{t=0}= 0. \end{align*} Then $c=2$ and according to (\ref{equ-6.2}) \begin{multline*} u(x,t)= \frac{1}{4} \int_0^t \int_{x-2(t-t')}^{x+2(t-t')} \sin (x')\cos( t')\,dx' d t'=\\ \frac{1}{4} \int_0^t \Bigl[\cos \bigl(x-2(t-t')\bigr)-\cos \bigl( x+2(t-t')\bigr) \Bigr] \cos( t')\, d t'= \\ \frac{1}{2} \int_0^t \sin(x) \sin (2(t-t'))\cos( t')\, d t'=\\ \frac{1}{4}\sin(x) \int_0^t [\sin (2(t-t')+t')+\sin( 2(t-t')-t')]\, d t'=\\ \frac{1}{4}\sin(x) \int_0^t [\sin (2t-t')+\sin( 2t-3t')]\, d t'=\\ \frac{1}{4}\sin(x) \bigl[\cos (2t-t')+\frac{1}{3}\cos( 2t-3t')\bigr]_{t'=0}^{t'=t}=\\ \frac{1}{3}\sin(x)\bigl[\cos(t)-\cos(2t) \bigr]. \end{multline*} Sure, separation of variables would be simpler here.