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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.

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

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