Chapter 20 The method of characteristics: second order PDEs
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Example 20.1. We consider the wave equation (here \(y\) is the time variable and \(x\) the one-dimensional spatial variable):
\[ \partial _{yy}u-c^2\partial _{xx}u=0, \]
where \(c>0\) is called the wave speed. We additionally have the initial conditions
\[ u(x,0)=\phi (x),\qquad \partial _y u(x,0)=\psi (x), \]
where \(\phi \) and \(\psi \) are given. For later use, we denote an anti-derivative of \(\psi \) by \(\Psi \) (it will turn out that the arbitrary constant will cancel, so we don’t have to worry about which anti-derivative to consider). We can factor the wave equation as
\[ (\partial _y+c\partial _x)(\partial _y-c\partial _x)u=0. \]
If we define
\[ w:=(\partial _y-c\partial _x)u, \]
then \(w\) satisfies
\[ (\partial _y+c\partial _x)w=0. \]
We therefore have written the wave equation as a system of equations in the two unknowns \(u\) and \(w\):
\[ (\partial _y+c\partial _x)w=0,\qquad (\partial _y-c\partial _x)u=w. \]
We obtain an initial condition for \(w\) from its definition and the initial conditions for \(u\) as:
\[ w(x,0) =\partial _yu(x,0)-c\partial _xu(x,0) =\psi (x)-c\phi '(x). \]
As a first step we therefore wish to solve
\[ c\partial _xw+\partial _yw=0,\qquad w(x,0)=\psi (x)-c\phi '(x). \]
This is a transport equation which we can solve using the method of characteristics (for first order PDEs). The characteristic equations are
\[ \dot {x}=c,\qquad \dot {y}=1,\qquad \dot {z}=0, \]
the data curve is \(\mR \times \{0\}\) and we have the initial conditions
\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=\psi (r)-c\phi '(r). \]
Solving these initial value problems for the ODEs gives
\[ x(r,t)=ct+r,\qquad y(r,t)=t,\qquad z(r,t)=\psi (r)-c\phi '(r). \]
We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) as \(t=y\), \(r=x-cy\). Substituting in \(z\) gives
\[ w(x,y)=\psi (x-cy)-c\phi '(x-cy). \]
As the second step, now that we have obtained \(w\), we can solve for \(u\). The relevant equation and boundary condition are
\[ -c\partial _xu+\partial _yu=w,\qquad u(x,0)=\phi (x). \]
The characteristic equations are
\[ \dot {x}=-c,\qquad \dot {y}=1,\qquad \dot {z}=w, \]
with the initial conditions
\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=\phi (r). \]
Solving the initial value problems for the ODEs for \(x\) and \(y\) gives
\[ x=-ct+r,\qquad y=t. \]
The equation for \(z\) is
\[ \dot {z}=w=\psi (x-cy)-c\phi '(x-cy)=\psi (-ct+r-ct)-c\phi '(-ct+r-ct)=\psi (r-2ct)-c\phi '(r-2ct). \]
The general solution of this is (recall that \(\Psi \) is an anti-derivative of \(\psi \))
\[ z=\frac {-1}{2c}\Psi (r-2ct)+\frac {1}{2}\phi (r-2ct)+K, \]
for a constant \(K\) (which is constant in \(t\), but may depend on \(r\)). From the initial condition \(z(r,0)=\phi (r)\) we obtain \(K\) and subsequently
\[ z=\frac {-1}{2c}\Psi (r-2ct)+\frac {1}{2}\phi (r-2ct)+\frac {1}{2c}\Psi (r)+\frac {1}{2}\phi (r). \]
We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) as \(t=y\), \(r=x+cy\). Substituting this in \(z\) gives
\[ u(x,y)=\frac {-1}{2c}\Psi (x-cy)+\frac {1}{2}\phi (x-cy)+\frac {1}{2c}\Psi (x+cy)+\frac {1}{2}\phi (x+cy). \]
This is called d’Alembert’s formula.
A second order semi-linear PDE is an equation of the form
\[ a(x,y)\partial _{xx} u+b(x,y)\partial _{xy}u+c(x,y)\partial _{yy}u=d(x,y,u,\partial _xu,\partial _yu). \]
We will first consider the case where the coefficients \(a\), \(b\) and \(c\) are constants. The crucial thing that makes Example 20.1 work is that the expression \(a\partial _{xx} +b\partial _{xy}+c\partial _{yy}\) factors as the product of two linear terms. In general we have
\[ a\partial _{xx} u+b\partial _{xy}u+c\partial _{yy}u =a(\partial _x-\lambda _+\partial _y)(\partial _x-\lambda _-\partial _y)u, \]
where
\[ \lambda _{\pm }=\frac {-b\pm \sqrt {b^2-4ac}}{2a}. \]
A crucial assumption to keep things real is that \(b^2-4ac>0\). If the coefficients \(a,b,c\) are not constant, then a similar procedure still works. This leads to the following classification of second order semi-linear PDEs:
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• hyperbolic if \(b^2-4ac>0\);
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• parabolic if \(b^2-4ac=0\);
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• elliptic if \(b^2-4ac<0\).
Note that in the variable coefficients case, on different parts of \(\mR ^2\), the PDE might have a different classification. The “canonical” example of a hyperbolic equation is the wave equation and that of an elliptic equation is Laplace’s equation.
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Example 20.2. Find the solution of
\[ y\partial _{xx}u+(y-x)\partial _{xy}u-x\partial _{yy}u=0,\qquad u(x,0)=1,\qquad \partial _y u(x,0)=x^2, \]
using the method of characteristics.
The equation factors as
\[ (y\partial _x-x\partial _y)(\partial _x+\partial _y)u=0. \]
We therefore define
\[ w=(\partial _x+\partial _y)u, \]
and obtain
\[ (y\partial _x-x\partial _y)w=0,\qquad w(x,0)=x^2. \]
We already solved this in Example 18.2 and from there obtain
\[ w(x,y)=x^2+y^2. \]
The equation for \(u\) then becomes
\[ \partial _xu+\partial _yu=x^2+y^2,\qquad u(x,0)=1. \]
The characteristic equations are
\[ \dot {x}=1,\qquad \dot {y}=1,\qquad \dot {z}=x^2+y^2. \]
with initial conditions
\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=1. \]
Solving the initial value problems for \(x\) and \(y\) gives
\[ x(r,t)=t+r,\qquad y(r,t)=t. \]
The equation for \(z\) then is
\[ \dot {z}=(t+r)^2+t^2,\qquad z(r,0)=1, \]
which has solution
\[ z(r,t)=\frac {1}{3}(t+r)^3+\frac {1}{3}t^3-\frac {1}{3}r^3+1. \]
We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) as \(t=y\), \(r=x-y\). Substituting in \(z\) gives
\[ u(x,y)=\frac {1}{3}x^3+\frac {1}{3}y^3-\frac {1}{3}(x-y)^3+1. \]