# Wave Equation with forcing term, vector transformation, method of characteristics

I'm currently doing the exercises in the book Introduction to Partial Differential Equations, (Borthwick). I'm currently on exercise 4.6:

An alternative approach to the one-dimensional wave equation is to recast the PDE as a pair of ODE. Consider the wave equation with forcing term, $$\frac{\partial ^2 u}{\partial t^2} - c^2\frac{\partial ^2 u}{\partial x^2} = f$$

I've completed parts a) and b) where you transform the problem using $$v= \begin{pmatrix} \frac{\partial u}{\partial t}\\ \frac{\partial u}{\partial x}\\ \end{pmatrix}$$ into $$\frac{\partial v}{\partial t} - A\frac{\partial v}{\partial x} = \begin{pmatrix} f\\ 0\\ \end{pmatrix}$$ where $$A = \begin{pmatrix} 0 & c^2 \\ 1 & 0 \\ \end{pmatrix}$$.

Then, using $$T= \begin{pmatrix} 1 & c \\ 1 & -c \\ \end{pmatrix}$$, and the substitution $$w=Tv$$ you can transform the above equation into a pair of ODE's $$\frac{\partial w_1}{\partial t}-c\frac{\partial w_1}{\partial x}=f$$ and $$\frac{\partial w_2}{\partial t}+c\frac{\partial w_2}{\partial x}=f$$.

Part c) asks to translate the initial conditions $$u(0,x)=g(x)$$ and $$\frac{\partial u}{\partial t}(0,x)=h(x)$$ into initial conditions for $$w_1$$ and $$w_2$$, and to then solve the pair of ODEs using the method of characteristics.

I've looked at other examples where they change variables but I can't really wrap my head around this example as to how I transform these initial conditions.

and Part d)

Combine the solutions for $$w_1$$ and $$w_2$$ to compute $$v_1 = \frac{\partial u}{\partial t}$$, and then integrate to solve for $$u$$.

I think once I get part c) i should be able to do this, but obviously at this stage I dont know.

• $\frac{\partial u}{\partial x}(0,x) = g'(x)$ Sep 10 '19 at 1:05
• So then you just set $v_1 = g'(x)$ and $v_2 = h(x)$ at $(0,x)$? Sep 10 '19 at 1:15
• yes, that is correct Sep 10 '19 at 17:22

## 1 Answer

You may have a look at this related post and references therein (including linked posts). To solve the present initial-value problem, one notes that the method of characteristics for $$w_1$$, $$w_2$$ gives $$w_{1,2}(x,t) = w_{1,2}(x\pm ct,0) + \int_0^t f(x\pm c(t-\tau),\tau) \,\text d \tau \, .$$ The functions $$x\mapsto w_{1,2}(x,0)$$ denote the initial conditions for $$w_1$$, $$w_2$$, which are \begin{aligned} w_{1,2}(x,0) &= u_t(x,0) \pm c u_x(x,0) \\ &= h(x) \pm c g'(x) \, . \end{aligned} Hence, $$u_t = \frac12 (w_1+w_2)$$ satisfies $$\frac{\partial u}{\partial t} = \tfrac12(h(x+ct) + h(x-ct)) + \tfrac12 c (g'(x+ct) + g'(x-ct)) + \tfrac12\int_0^t \big( f(x+ c(t-\tau),\tau) + f(x- c(t-\tau),\tau)\big) \text d \tau\, ,$$ which integration in time leads to the generalized d'Alembert's formula (see Generalization for inhomogeneous canonical hyperbolic differential equations in the Wikipedia article).