
## 2.5. Wave equation reloaded (continued)

We solved IVP with homogeneous ($=0$) initial data for inhomogeneous wave equation. Now consider both inhomogeneous data and equation. So we consider problem (2.4.6)-(2.4.8): \begin{align} &u_{tt}-c^2u_{xx}=f(x,t),\label{eq-2.5.1} \\[3pt] &u|_{t=0}=g(x),\label{eq-2.5.2}\\[3pt] &u_t|_{t=0}=h(x). \label{eq-2.5.3} \end{align} when neither $f$, nor $g,h$ are necessarily equal to $0$.

The good news is that our equation is linear and therefore $u=u_2+u_1$ where $u_1$ satisfies problem with right-hand function $f(x,t)$ but with $g$ and $h$ replaced by $0$ and $u_2$ satisfies the same problem albeit with $f$ replaced by $0$ and original $g,h$: \begin{align*} &u_{1tt}-c^2u_{1xx}=f(x,t), \qquad &&u_{2tt}-c^2u_{2xx}=0, \\[3pt] &u_1|_{t=0}=0, && u_2|_{t=0}=g(x),\\[3pt] &u_{1t}|_{t=0}=0 &&u_{2t}|_{t=0}=h(x). \end{align*}

Exercise 1. Prove it.

Then $u_1$ is given by (2.4.11), and $u_2$ is given by (2.3.14) (with $(f,g)\mapsto (g,h)$)and adding them we arrive to the final \begin{multline} u(x,t)= \underbracket{ \frac{1}{2}\bigl[g(x+ct)+g(x-ct)\bigr]+\frac{1}{2c}\int_{x-ct}^{x+ct} h(y)\,dy}_{=u_2}+\\[3pt] \underbracket{\frac{1}{2c} \iint_{\Delta (x,t)} f(x',t' )\,dx\,d t' }_{=u_1}. \label{eq-2.5.4} \end{multline} where recall that $\Delta(x,t)$ is a characteristic triangle:

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This formula is also called D'Alembert formula.

Remark 1. Note that integral of $h$ in $u_2$ is taken over base of the characteristic triangle $\Delta(x,t)$ and $g$ is taken in the ens of this base.

### Wave equation reloaded (Duhamel integral)

We discuss formula (\ref{eq-2.5.4}) in details in the next lecture and now let us derive it by a completely different method. Again, in virtue of (2.3.14) we need to consider only the case when $g=h=0$.

Let us define an auxillary function $U(x,t,\tau)$ ($0< \tau < t$) as a solution of an auxillary problem \begin{align} &U_{tt}-c^2U_{xx}=0,\label{eq-2.5.5} \\[3pt] &U|_{t=\tau}=0,\label{eq-2.5.6}\\[3pt] &U_t|_{t=\tau}=f(x,\tau). \label{eq-2.5.7} \end{align} We claim that

Proposition 1. $$u(x,t)=\int_0^t U(x,t,\tau)\,d\tau \label{eq-2.5.8}$$ is a required solution.

Proof. Note first that we can differentiate (\ref{eq-2.5.8}) by $x$ easily: $$u_x=\int_0^t U_x (x,t,\tau)\,d\tau,\qquad u_{xx}=\int_0^t U_{xx} (x,t,\tau)\,d\tau \label{eq-2.5.9}$$ and so on. Let us find $u_t$. Note that $u$ depends on $t$ through its upper limit and through integrand. We apply formula \begin{multline} \frac{d\ }{dt}\Bigl(\int_{\alpha(t)}^{\beta(t)} F (t,\tau)\,d\tau\Bigr)=\\ -F(t,\alpha(t))\alpha'(t)+F(t,\beta(t))\beta'(t)+ \int_{\alpha(t)}^{\beta(t)} \frac{\partial F}{\partial t} (t,\tau)\,d\tau\qquad \label{eq-2.5.10} \end{multline} which you should know but we will prove it anyway.

As $\alpha=0$, $\beta=t$ we have $\alpha'=0$, $\beta'=1$ and \begin{equation*} u_t=U(x,t,t)+\int_0^t U_t (x,t,\tau)\,d\tau. \end{equation*} But $U(x,t,t)=0$ due to (\ref{eq-2.5.6}) and therefore $$u_t=\int_0^t U_t (x,t,\tau)\,d\tau. \label{eq-2.5.11}$$ We can differentiate this with respect to $x$ as in (\ref{eq-2.5.9}). Let us differentiate by $t$. Applying the same (\ref{eq-2.5.10}) we get \begin{equation*} u_{tt}=U_t(x,t,t)+\int_0^t U_{tt} (x,t,\tau)\,d\tau. \end{equation*} Due to (\ref{eq-2.5.7}) $U_t(x,t,t)=f(x,t)$: \begin{equation*} u_{tt}=f(x,t)+\int_0^t U_{tt} (x,t,\tau)\,d\tau. \end{equation*} Therefore \begin{equation*} u_{tt}-c^2u_{xx}=f(x,t)+\int_0^t \underbracket{\bigl(U_{tt}-c^2U_{xx}\bigr)}_{=0} (x,t,\tau)\,d\tau =f(x,t) \end{equation*} where integrand vanishes due to (\ref{eq-2.5.5}). So, $u$ really satisfies (\ref{eq-2.5.1}). Due to (\ref{eq-2.5.8}) and (\ref{eq-2.5.11}) $u|_{t=0}=u_t|_{t=0}=0$. QED.

Formula (\ref{eq-2.5.8}) is Duhamel integral formula.

Remark 2. It is not important that it is wave equation: we need assume that equation is linear and has a form $u_{tt} + Lu=f$ where $Lu$ is a linear combination of $u$ and its derivatives but with no more than $1$ differentiation by $t$.

Now we claim that $$U(x,t,\tau)= \frac{1}{2c} \int _{x-c(t-\tau)} ^{x+c(t-\tau)} f(x',\tau)\,dx'. \label{eq-2.5.12}$$ Really, changing variable $t'=t-\tau$ we get the same problem albeit with $t'$ instead of $t$ and with initial data at $t'=0$.

Plugging (\ref{eq-2.5.12}) into (\ref{eq-2.5.8}) we get $$u(x,t)=\int_0^t \frac{1}{2c} \Bigl[\int_{x-c(t-\tau)} ^{x+c(t-\tau)} f(x',\tau)\,dx'\Bigr]\,d\tau. \label{eq-2.5.13}$$

Exercise 2. Rewrite double integral (\ref{eq-2.5.13}) as a $2D$-integral in the right-hand expression of (2.4.11).

Proof of (\ref{eq-2.5.10}). Let us plug $\gamma$ into $F(x,t,\tau)$ instead of $t$. Then integral $I(t)=J(\alpha(t),\beta(t),\gamma(t))$ with $J(\alpha,\beta,\gamma)= \int_\alpha^\beta F(\gamma,\tau)\,d\tau$ and by chain rule $I'(t)=J_\alpha \alpha'+J_\beta \beta'+J_\gamma \gamma'$.

But $J_\alpha=-F(\gamma,\alpha)$, $J_\beta=F(\gamma,\beta)$ (differentiation by lower and upper limits) and $J_\gamma=\int_\alpha^\beta F_\gamma (\gamma,\tau)\,d\tau$. Plugging $\gamma=t$, $\gamma'=1$ we arrive to (\ref{eq-2.5.10}).

### Domains of dependence and influence

Recall formula (\ref{eq-2.5.4}): \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' \end{multline*} where $\Delta (x,t)$ is the same characteristic triangle as above:

Therefore

Proposition 2. 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 1. $\Delta(x,t)$ 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 2. $\Delta^+(\bar{x},0)$ is a triangle of influence for point $(\bar{x},0)$.

Remark 3.

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 3. Solution propagates with the speed not exceeding $c$.
Example 1. \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{eq-2.5.14}) \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.