I am trying to use the method from my previous question to solve this PDE: $$ 3u_x + 2u_t = \cos x $$

with initial condition $u(x,0) = x^2$.

So I need to solve these:

\begin{align} \frac{dx}{ds} &= 3 &\implies x &= 3s + x_0 \\ \frac{dt}{ds} &= 2 &\implies t &= 2s + t_0\\ \frac{du}{ds} &= \cos x \end{align}

To integrate the last one, I use the chain rule backwards since $x$ is a function of $s$:

\begin{align} u &= \sin x \frac{ds}{dx} + u_0 \\ &= \frac{1}{3} \sin x + u_0 \end{align}

Which appears to differentiate again properly, but continuing along these lines (AWLOG $t_0 = 0 \implies x_0 = x - \frac{3t}{2}$, and take $u = \frac{1}{3} \sin x + u_0 = \frac{1}{3} \sin x + u(x - \frac{3t}{2}, 0) = \frac{1}{3}\sin x + {\left(x - \frac{3t}{2}\right)}^2$.

But the given answer is actually:

$$ u = {\left(x - \frac{3t}{2}\right)}^2 - \frac{1}{3}\sin{\left(x - \frac{3t}{2}\right)} + \frac{1}{3} \sin x$$

What have I done wrong here?


$$3u_x+2u_t=\cos(x)$$ HINT :

The equations characteristics are : $$\frac{dx}{3}=\frac{dt}{2}=\frac{du}{\cos(x)}$$

First characteristic :

$\frac{dx}{3}=\frac{du}{\cos(x)} \quad\to\quad 3du-\cos(x)dx=0 \quad\to\quad u-\frac{1}{3}\sin(x)=c_1$

Second characteristic :

$\frac{dx}{3}-\frac{dt}{2}=0 \quad\to\quad 2x-3t=c_2$

General solution on implicite form, with any derivable function $\Phi$ of two variables :


With any derivable function $F$, this is equivalent to : $\quad u-\frac{1}{3}\sin(x)=F(2x-3t)$ $$u(x,t)=\frac{1}{3}\sin(x)+F(2x-3t)$$

The boundary condition $u(x,0)=x^2$ implies : $x^2=\frac{1}{3}\sin(x)+F(2x)$ . This defines the function $F(X)= -\frac{1}{3}\sin(\frac{X}{2})+\frac{X^2}{4}$ $$u(x,t)=\frac{1}{3}\sin(x)-\frac{1}{3}\sin\left(\frac{2x-3t}{2}\right)+\left(\frac{2x-3t}{2}\right)^2$$

With comparison with what you did, you certainly will see where is the discrepancy. The same for the second example. If there is still some difficulty, please ask again you are welcome.

In addition, after comments :

The comparison shows the next discrepancies :

enter image description here

enter image description here

$u(x,0)=\frac{1}{3}\sin(x)+u_0\quad$ is true on the characteristic curve, but not elswhere for $x\neq 0\quad $. The correct expression is : $$u(x-\frac{3t}{2}\:,\:0)=\frac{1}{3}\sin(x-\frac{3t}{2})+u_0$$ with $\quad u(x-\frac{3t}{2}\:,\:0)=\left(x-\frac{3t}{2}\right)^2$ $$u_0=\left(x-\frac{3t}{2}\right)^2-\frac{1}{3}\sin(x-\frac{3t}{2})$$ $$u(x,t)=\frac{1}{3}\sin(x) + \left(x-\frac{3t}{2}\right)^2 -\frac{1}{3}\sin\left(x-\frac{3t}{2}\right)$$

Reponse to the question in comments :

The method used in my above answer is equivalent to the method that you use. For more details, see for example : https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf

Copy of the theorem (with change of notations for consistancy) :

enter image description here

Equation (2.40) means that $\Psi_1$ and $\Psi_2$ are related through arbitrary function, for example : $\Psi_1=F(\Psi_2)$ or $\Psi_2=G(\Psi_1)$.

  • $\begingroup$ So I get the integrations, which appear to give the same answer as mine (modulo a scale factor in the second one). I don't get where you pull the implicit general solution from, why it is equivalent to the next line (where you introduce F), or how to relate this to what I did. After that I follow again, so I suspect the discrepancy is in that bit somewhere. Are you able to answer more explicitly in terms of the process I'm using? $\endgroup$ – lvc Feb 5 '16 at 23:20
  • $\begingroup$ See the addition to my first answer. $\endgroup$ – JJacquelin Feb 6 '16 at 8:55
  • $\begingroup$ Oops. That second discrepancy was actually my error in transcribing the given answer - the listed correct answer does match yours. I've corrected this in the question. That being said, I think your discussion below that has finally given me the lightbulb moment, especially since it related back very clearly to other initial value problems that I have done (substitute ICs into the general solution and solve for the constant, rather than the ICs giving the constant directly as $u_0 = u(s=0) = u(x_0, t_0)$ as I had previously assumed). $\endgroup$ – lvc Feb 6 '16 at 9:07
  • $\begingroup$ In order to answer to your question : '' I don't get where you pull the implicit general solution from'', see the new addition to my previous answer. $\endgroup$ – JJacquelin Feb 6 '16 at 9:49

The best way to view this in my opinion is by looking at the actual ODE you get when you solve the characteristic equation. This ODE is what you get along the path $x=x_0+\frac{3}{2}t$. For each $x_0$ you have $v(t)=u(t,x_0+\frac{3}{2}t)$ with $v(0)=f(x_0)$. It satisfies $\frac{dv}{dt}=\frac{1}{2} \cos(x(t))=\frac{1}{2} \cos(x_0+\frac{3}{2}t)$. The solution to this equation is $v(t)=f(x_0)+\frac{1}{3} \sin(x_0+\frac{3}{2}t)-\frac{1}{3} \sin(x_0)$, as you see by integrating.

Now to find $u(t,x)$ you need to solve for $x_0(t,x)$ and plug back in to the formula for $v$. You get $x_0=x-\frac{3}{2}t$, so $u(t,x)=f(x-\frac{3}{2}t)+\frac{1}{3} \sin(x)-\frac{1}{3} \sin(x-\frac{3}{2}t)$.

  • $\begingroup$ How did you find $\frac{dv}{dt} = \frac{1}{2} \cos(x(t))$? $\endgroup$ – lvc Feb 6 '16 at 0:10
  • $\begingroup$ Divide both sides by 2, then it's the usual procedure for a transport equation with a forcing that doesn't depend on $u$. $\endgroup$ – Ian Feb 6 '16 at 0:23
  • $\begingroup$ I'm not actually familiar with transport equations or forcings. Is there a way to find an expression for $\frac{dv}{dt}$ without knowing that theory? $\endgroup$ – lvc Feb 6 '16 at 1:27
  • $\begingroup$ You just use the chain rule on $u$. $\endgroup$ – Ian Feb 6 '16 at 1:43
  • $\begingroup$ To find $\frac{dv}{dt}$ by the chain rule, I would do $\frac{dv}{dt} = \frac{dv}{du} \frac{du}{dt}$, but I don't know $\frac{dv}{du}$? $\endgroup$ – lvc Feb 6 '16 at 1:53

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