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In this Chapter we first consider first order PDE and then move to \(1\)-dimensional wave equation which we analyze by the method of characteristics.
Recall from ODE cours that an integral line of the vector field is a line, tangent to it in each point.
Often it is called directional derivative but also often then \(\ell\) is normalized, replaced by the unit vector of the same direction \(\ell^0=\ell/|\ell|\).
where \(\phi\) is an arbitrary function.
This is a general solution of our equation.
Consider initial value condition \(u|\_{t=0}=f(x)\). It allows us define \(\phi\): \(\phi(-x/b)=f(x)\implies \phi (x)= f(-bx)\). Plugging in \(u\) we get \[\begin{equation} u=f\bigl( x-ct\bigr)\qquad\text{with } c=b/a. \label{eq-2.1.4} \end{equation}\] It is a solution of IVP \[\begin{equation} \left\\{\begin{aligned} &au\_t+bu\_x=0,\\\\ &u(x,0)=f(x). \end{aligned} \\right. \label{eq-2.1.5} \end{equation}\]Obviously we need to assume that \(a\ne 0\).
If \(a=1\) we can rewrite general solution in the form \(u(x,t)=\phi\_1 (x-bt)\) where \(\phi\_1(x)=\phi(-x/b)\) is another arbitrary function.
Definition 1. Solutions \(u=\chi(x-ct)\) are running waves where \(c\) is a propagation speed.
If \(a\) and/or \(b\) are not constant these integral lines are curves.
Example 1. Consider equation \(u\_t+tu\_x=0\). Then equation of the integral curve is \(\frac{dt}{1}=\frac{dx}{t}\) or equivalently \(tdt-dx=0\) which solves as \(x-\frac{1}{2}t^2=C\) and therefore \(u=\phi (x-\frac{1}{2}t^2)\) is a general solution to this equation.
One can see easily that \(u=f(x-\frac{1}{2}t^2)\) is a solution of IVP.
Example 2. Consider the same equation but let us consider IVP as \(x=0\): \(u(0,t)=g(t)\). However it is not a good problem: first, some integral curves intersect line \(x=0\) more than once and if in different points of intersection of the same curve initial values are different we get a contradiction (therefore problem is not solvable for \(g\) which are not even functions).
On the other hand, if we consider even function \(g\) (or equivalently impose initial condition only for \(t\>0\)) then \(u\) is not defined on the curves which are not intersecting \(x=0\) (which means that \(u\) is not defined for \(x\>\frac{1}{2}t^2\).)
In this example both solvability and unicity are broken.
Example 3. Consider problem \(u\_t+u\_x=x\). Then \(\frac{dx}{1}=\frac{dt}{1}=\frac{du}{x}\). Then \(x-t=C\) and \(u-\frac{1}{2}x^2=D\) and we get \(u-\frac{1}{2}x^2 = \phi (x-t)\) as relation between \(C\) and \(D\) both of which are constants along integral curves. Here \(\phi\) is an arbitrary function. So \(u=\frac{1}{2}x^2 + \phi (x-t)\) is a general solution. Imposing Imposing initial condition \(u|\_{t=0}=0\) (sure, we could impose another condition) we have \(\phi(x)=-\frac{1}{2}x^2\) and plugging into \(u\) we get \(u(x,t)=\frac{1}{2}x^2-\frac{1}{2}(x-t)^2= xt - \frac{1}{2}t^2\).
Example 4.Example 4. Consider \(u\_t+ xu\_x = x t\). Then \(\frac{dt}{1}=\frac{dx}{x}=\frac{du}{xt}\). Solving the first equation \(t-\ln x=-\ln C\implies x =Ce^t\) we get integral curves. Now we have \begin{equation*} =dt du= x t dt= Cte^t dt u=C(t-1)e^t +D = x(t-1)+D \end{equation*} where \(D\) must be constant along integral curves and therefore \(D=\phi (xe^{-t})\) with an arbitrary function \(\phi\). So \(u=x(t-1)+\phi (xe^{-t})\) is a general solution of this equation.
Imposing initial condition \(u|\_{t=0}=0\) (sure, we could impose another condition) we have \(\phi(x)=x\) and then \(u=x(t-1 +e^{-t})\).
Definition 2. If \(a=a(x,t)\) and \(b=b(x,t)\) equation is semilinear.
In this case we first define integral curves which do not depend on \(u\) and then find \(u\) as a solution of ODE along these curves.
Definition 3. Furthermore if \(f\) is a linear function of \(u\): \(f=c(x,t)u + g(x,t)\) original equation is linear.
In this case the last ODE is also linear.
Example 5. Consider \(u\_t+ xu\_x = u\). Then \(\frac{dt}{1}=\frac{dx}{x}=\frac{du}{u}\). Solving the first equation \(t-\ln x=-\ln C\implies x =Ce^t\) we get integral curves. Now we have \begin{equation*} =dt u= t+D u=Det=(xe{-t})e^t \end{equation*} which is a general solution of this equation.
Imposing initial condition \(u|\_{t=0}=x^2\) (sure, we could impose another condition) we have \(\phi(x)= x^2\) and then \(u=x^2 e^{-t}\).
Example 6. Consider \(u\_t+ xu\_x = -u^2\). Then \(\frac{dt}{1}=\frac{dx}{x}=-\frac{du}{u^2}\). Solving the first equation \(x =Ce^t\) we get integral curves. Now we have \begin{equation*} -=dt u^{-1}= t+ D u=(t+ (xe{-t})){-1}. \end{equation*} which is a general solution of this equation.
Definition 4. If \(a\) and/or \(b\) depend on \(u\) this is quasininear equation.
For such equations integral curves depend on the solution which can lead to breaking of solution.
Example 7. Consider Hopf equation \(u\_t+uu\_x=0\) (which is an extremely simplified model of gas dynamics. ) We have \(\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}\) and therefore \(u=\const\) along integral curves and therefore integral curves are \(x-ut=C\).
Consider initial problem \(u(x,0)=f(x)\). We take initial point \((y,0)\), find here \(u=f(y)\), then \(x-f(y)t =y\) (think why?) and we get \(u=f(y)\) where \(y=y(x,t)\) is a solution of equation \(x=f(y)t +y\).
The trouble is that we can define \(y\) for all \(x\) only if \(\frac{\partial }{\partial y}\bigl(f(y)t +y\bigr)\) does not vanish. So, \(f'(y)t +1\ne 0\).
This is possible for all \(t\>0\) if and only if \(f'(y)\ge 0\) i.e. \(f\) is a monotone non-decreasing function.
So, classical solution breaks if \(f\) is not a monotone non-decreasing function. A proper understanding of the global solution for such equation goes well beyond our course.
Example 8. Traffic flow is considered in Appendix
The general solution is \(u=\phi(x-ct)\) and plugging into initial data we get \(\phi(x)=f(x)\) (as \(x\>0\)).
So, \(u(x,t)= f(x-ct)\). Done!–Not so fast. \(f\) is defined only for \(x\>0\) so \(u\) is defined for \(x-ct\>0\) (or \(x\>ct\)). It covers the whole quadrant if \(c\\le 0\) (so waves run to the left) and only in this case we are done.
If \(c\>0\) (waves run to the right) \(u\) is not defined as \(x\< ct\) and to define it here we need a boundary condition at \(x=0\). So we get IBVP (initial-boundary value problem) \[\begin{equation} \left\\{\begin{aligned} &u\_t +cu\_x=0, \qquad &&x\>0, t\>0,\\\\ &u|\_{t=0}= f(x) \qquad &&x\>0,\\\\ &u|\_{x=0}=g(t) \qquad &&t\>0. \end{aligned}\\right. \label{eq-2.1.9} \end{equation}\]Then we get \(\phi(-ct)=g(t)\) as \(t\>0\) which implies \(\phi(x)=g(-\frac{1}{c}x)\) as \(x\<0\) and then \(u(x,t)=g(-\frac{1}{c}(x-ct))=g(t-\frac{1}{c}x)\) as \(x\< ct\).
So solution is \[\begin{equation} u=\left\\{\begin{aligned} &f(x-ct)\qquad &&x\> c t,\\\\ &g(t-\frac{1}{c}x)\qquad && x \< ct. \end{aligned}\right. \label{eq-2.1.10} \end{equation}\]\(\Leftarrow\) \(\Uparrow\) \(\Downarrow\) \(\downarrow\) \(\Rightarrow\)