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Consider the second-order parabolic inhomogeneous second-order PDE $$ u_{tt}+2u_{tx}+u_{xx}=2c $$ I have seen two ways to solve this problem. I would like to know (1) if Solution 1 is correct (2) if Solution 2 is correct and clarification of details (3) what is the connection between Solution 1 and 2 (4) if there are any other ways to solve the PDE and (5) in more complicated second-order PDEs (e.g. hyperbolic, elliptic), could I still use these solution strategies?

$\textbf{Solution 1:}$ We can factor $$ \left( \frac{\partial }{\partial t} + \frac{\partial }{\partial x} \right)^2 u=2c $$ Now, we use a transformation of variables. Let $x=y + z$ and $t=y$ and $v(y,z)=u(t,x)$. Then, \begin{align*} v_{yy} &= \left( \frac{\partial }{\partial t} + \frac{\partial }{\partial x} \right)^2 u=2c\\ v(y,z) &= cy^2+yf_1(z) + f_2(z)\\ u(t,x) &=ct^2+tf_1(x - t) + f_2(x -t) \qquad \textit{(1)} \end{align*} and by symmetry, \begin{align*} u(t,x) &=cx^2+xg_1(t - x) + g_2(t -x) \end{align*}

$\textbf{Solution 2:}$ Again we can factor $$ \left( \frac{\partial }{\partial t} + \frac{\partial }{\partial x} \right)^2 u=2c $$ We solve the PDE in two steps, \begin{align*} \left( \frac{\partial }{\partial t} + \frac{\partial }{\partial x} \right) v &=2c \qquad \textit{(2)}\\ \left( \frac{\partial }{\partial t} + \frac{\partial }{\partial x} \right) u &=v \qquad \textit{(3)} \end{align*} By the method of characteristics for $\textit{(2)}$,we have $\frac{\partial t}{\partial w}=\frac{\partial x}{\partial w}=1$ and $\frac{\partial v}{\partial w}=2c$ so $x=w+x_0$ and $v(?,?)=2cw + \gamma_1(x_0)$.

Also by the method of characteristics for $\textit{(3)}$, we have $\frac{\partial t}{\partial s}=\frac{\partial x}{\partial s}=1$ and $\frac{\partial u}{\partial s}=v=2cs + \gamma_1(r_0)$ so $x=t+r_0$ and $$u(?,?)=cs^2 + s\gamma_1(r_0) + \gamma_2(r_0) =ct^2 + t\gamma_1(x-t) + \gamma_2(x-t) \qquad \textit{(4)}$$ This is the same as $\textit{(1)}$.

I'm a little confused about the details of this proof. Specifically, I'm just not sure what the arguments of the functions are and it feels like I'm substituting in where convenient to get to the solution (e.g. the arguments of $u(?,?)$ are $t$ and $x$ but then they're $s$ and $r_0$ and then I just substitute in for $s$ and $r_0$).

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There is another way to solve this problem, but it involves applied Lie theory and not a lot of people know about it. I picked this up from Dr. Lawrence Dresner in Oak Ridge back in the 90's. Your pde is invariant to Lie group $$ G(x,t,u)=(\lambda x,\lambda^\beta t,\lambda^\alpha u)\lambda_o=1 $$such that $$ \lambda^\alpha u(x,t)=u(\lambda x,\lambda^\beta t) $$Differentiating both sides w.r.t. $\lambda$ and setting $\lambda = \lambda_o=1$, $$ \alpha u=xu_x+\beta tu_t $$The characteristic equation for this expression is $$ \frac{du}{\alpha u}=\frac{dx}{x}=\frac{dt}{\beta t} $$Two independent integrals are $$ \frac{u}{t^{\frac{\alpha}{\beta}}} $$and $$ \frac{x}{t^{\frac{1}{\beta}}} $$The most general solution to your PDE is to take one of these differential invariants and set it equal to a function of the other one, with the result that $$ u=t^{\frac{\alpha}{\beta}}F\bigg(\frac{x}{t^{\frac{1}{\beta}}}\bigg) $$Since $x'=\lambda x$, $t'=\lambda^\beta t$, and $u'=\lambda^\alpha u$, $$ u'_{x'x'}=\lambda^{\alpha -2}u_{xx} $$ $$ u'_{t'x'}=\lambda^{\alpha -\beta -1}u_{tx} $$ $$ u'_{t't'}=\lambda^{\alpha -2\beta}u_{tt} $$and $$ 2C=\lambda^0 2C $$then for invariance to exist, $\alpha -2=\alpha -\beta -1=\alpha -2\beta =0$ which implies $\alpha =2$ and $\beta =1$. Therefore the most general solution to your PDE is $$ u=t^2F(\mu), \mu=\frac{x}{t} $$If you take derivatives of this solution you get $$ u_{xx}=F_{\mu \mu} $$ $$ u_{tx}=F_\mu-\mu F_{\mu \mu} $$ $$ u_{tt}=2F-2\mu F_\mu+\mu^2 F_{\mu \mu} $$Plug this into the PDE and simplify and you get $$ \frac{1}{2}(1-\mu)^2F_{\mu \mu}+(1-\mu)F_\mu+F=C $$a 2nd-order, linear ODE. You can solve this (or hand it to Wolfram Alpha) to get $F(\mu)$. Remember that $\mu=\frac{x}{t}$ and $u=t^2F$ and the answer is yours.

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  • $\begingroup$ This looks very interesting but I couldn't understand some of the details . For example, in the first line, what does it mean $(\lambda x,\lambda^{\beta}t,\lambda^{\alpha}u)\lambda_0=1$ (i.e. the left hand sied looks like a vector and the right hand side a scalar)? Do you know of any online notes / references that I can read up on this method? $\endgroup$
    – user103828
    Aug 10, 2014 at 8:50
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    $\begingroup$ Similarity Solutions of Nonlinear Partial Differential Equations by Lawrence Dresner, ISBN 0-273-08621-9 is a very good reference. I just looked on Amazon.com and it is available, but not cheap. You could get it through a library sharing program and then decide whether or not to buy it. You'll also see similar references to applied Lie theory when you look it up. You can go to the source and get Sophus Lie's 1884 Differential Invariant Paper by Math Sci Press, ISBN 0-915692-13-9, but that is one tough book, my friend, although well worth the effort. $\endgroup$
    – atomteori
    Aug 10, 2014 at 10:59
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    $\begingroup$ The group G is an infinite cyclic group, a "stretching" group. The x-component is stretched to $\lambda x$, t is stretched to $\lambda^2 t$ and u to $\lambda u$. Individual points on the solution curves move around, but the curves themselves do not change. It's a partitioned, bilinear automorphism from R to R. The stabilizers of the group (differential invariants) form an embedded n-1 manifold. By expressing the DEQ and its solution as a function of stabilizers you place it in the kernel with the Lie algebra which can then be used to turn the PDE into an ODE. Lie was a genius. $\endgroup$
    – atomteori
    Aug 10, 2014 at 11:22
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    $\begingroup$ Oh yeah, $\lambda_o=1$ is the identity transformation which maps points on the solution curves back onto themselves. No group transformation is complete without one. It's also needed to find the infinitesimal transformation. $\frac{\partial y'}{\partial \lambda}\bigg|_{\lambda =1}=2y$ $\endgroup$
    – atomteori
    Aug 10, 2014 at 11:28
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I will try to shed some light, let's see if I can help:

  • Q1 and Q2: Your solution given by both the 1st and 2nd methods is completely right. I have checked them both getting your solution $u(x,t) = c t^2 + t f(x-t) + g(x-t)$, where $f$ and $g$ are arbitrary functions of the characteristic(s) $x-t$. This tells you that the solution in absence of forcing (the nonhomogenous term) travels along curves given by $x - t = \xi, \ \xi \in \mathbb{R}$, where $u$ is precisely constant.

  • Q3: You can check that both ways to proceed are exactly the same if you write the original PDE in canonical form. Your 2nd method just take advantage of the operator decomposition: $$(D_{xx} + 2 D_{xt} + D_{tt})u = (D_{x} + D_{t})(D_{x} + D_{t})u = (D_{x} + D_{t})^2 u, \quad D_{x_i} = \partial_{x_i}, $$ which yields to two 1st-order hyperbolic PDEs.

  • Q4: The general way to solve PDEs of the form: $$A u_{xx} + B u_{xt} + C u_{xx} = P(x,t;u,u_x,u_t),$$ for some linear function $P$, is to compute first the discriminant, $\Delta = B^2 -4AC$, in order to determine whether we have a hyperbolic ($\Delta >0$), elliptic ($\Delta <0$), or parabolic ($\Delta =0$) PDE. In your case: $A=C=1, B=2$, so $\Delta = 0$, making your PDE parabolic. The next step is to compute the characteristics of the equation which are given by the equations: $$\begin{align} A \xi_{x}^2 + B \xi_x \xi_t + C\xi_t^2 = 0, \tag{1} \\ A \eta_{x}^2 + B \eta_x \eta_t + C\eta_t^2 = 0. \tag{2}\\ \end{align}$$ Solving $(1)$ for $\xi_x/\xi_t$ we arrive at: $$\frac{\xi_x}{\xi_t} = -1 = \color{blue}{-\frac{dt}{dx}}^*, \tag{3}$$ which after integration leads us to the first characteristic curve, $dx - dt = 0$, or equivalently, $x - t = c \equiv \xi$. Since the equation is parabolic, the other solution (remember $\Delta = 0$) can be arbitrarily chosen so we can make the other characteristic to be $\eta = t$ or $\eta = x$ (symmetry). Once the characteristics are computed you write the equation making the transformation $(x,t) \to (\xi,\eta)$ and using the chain rule in order to come up with the known as canonical form of the PDE which will be (prove it, taking $\eta = t$): $$ u_{\eta \eta}(\xi,\eta) = 2c, $$ which after integrating yields the desired solution: $$ u(\xi,\eta) = c \eta^2 + \eta f(\xi) + g(\xi) \Leftrightarrow u(x,t) = c t^2 + t f(x-t) +g(x-t),$$ as we expected.

  • Q5: As you can see, we can only avoid this tedious work when the PDE is easily factorizable like in the present case. Otherwise, the general method always tries to put your PDE into canonical form which, by the way, is not always simpler.

I hope this helps.

Cheers!

Edit: *, the equation highlighted in blue comes from the fact that given $\xi = \xi(x,t)$, then $d\xi = \xi_x dx + \xi_t dt$, but we know that $\xi$ is a constant (characteristic), therefore $d\xi = 0$ and we have $dt/dx = -\xi_x/\xi_t$.

Edit 2: of course, we have forgotten about boundary value problems, which can be solved using completely different techniques. Here, the shape of the arbitrary functions $f$ and $g$ are either determined by the initial conditions or by any other specification of $u$ as, for example, passing through a given parametrized curve.

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    $\begingroup$ Great. So Solution 1 is basically a shortcut of Solution 2 (i.e. I just guessed the characteristics). What is the point of calculating $B^2-4AC$ if any any case we will then (1) find characteristics and (2) make substitution into the original PDE and turn into canonical form? $\endgroup$
    – user103828
    Aug 9, 2014 at 10:52
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    $\begingroup$ You're right, I guess it's just for anticipating the behaviour of the solution and, obviously, its canonical form. $\endgroup$
    – Dmoreno
    Aug 9, 2014 at 10:55

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