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EDIT The issue is with the characteristic equations.

It should be

$$\frac{dt}{1} = \frac{dx}{2} =\frac{dw}{3u}$$

and not

$$\frac{dt}{1} = \frac{dx}{2} =\frac{dw}{3}$$

Solving the following PDE $u_t+2u_x+3u=0$ with $u(x,t=0) = e^{-x^2/l^2}$

So I've got my characteristics:

$$C_1 = x-2t$$ $$C_2 = 3x-2u$$

And I know the general form will be a function of the form

$$f(x-2t, 3x-2u)$$

And we can do the following

$$3x-2u = f(x-2t) \Rightarrow u = \frac12\left(3x-f(x-2t)\right) $$

Now we know that:

$$u(x,t=0) = e^{-x^2/l^2} = \frac12\left(3x-f(x)\right)$$

and

$$f(x) = 3x - 2e^{-x^2/l^2}\Rightarrow f(x-2t) = 3(x-2t) - 2e^{-(x-2t)^2/l^2} $$

Now the next step is where I'm having difficulties. Am I plugging f(x) back into here?

$$u = \frac12\left(3x-f(x-2t)\right)$$

I'd appreciate any hints

  • 1
    i think the solution is given by $$u \left( x,t \right) ={\it \_F1} \left( -x/2+t \right) {{\rm e}^{-3/2 \,x}} $$ – 2017-02-11
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    I'd appreciate some help as to what to do next as opposed to just a solution. – 2017-02-11
  • 0
    You can first perform the substitution $v = u e^{3t}$ to get rid of the $3u$ term. This gives us the equation $v_t + 2v_x = 0$ with the same initial condition for which solving is straight forward. Another way to do it is to notice that if $(x(s),t(s))$ is a characteristic line then $\frac{d}{ds}u(x(s),t(s)) + 3u = \dot{t}u_t + \dot{x} u_x + 3 u = 0$ when $\dot{x} = 2$ and $\dot{t} = 1$ so the $ue^{3s} \propto u e^{3t}$ is constant along these characteristics. – 2017-02-11

3 Answers 3

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$$u_t+2u_x=-3u$$ The system of characteristic differential equations can be summarized as : $$\frac{dt}{1}=\frac{dx}{2}=\frac{du}{-3u}$$ The equation of a first family of characteristic curves comes from : $$\frac{dt}{1}=\frac{dx}{2} \quad\to\quad x-2t=c_1$$ The equation of a second family of characteristic curves comes from : $$\frac{dx}{2}=\frac{du}{-3u} \quad\to\quad ue^{\frac{3}{2}x}=c_2$$ The general solution of the PDE is, expressed on the form of implicit equation, is : $$\Phi\left( (x-2t)\:,\:(ue^{\frac{3}{2}x})\right)=0$$ Where $\Phi$ is any differentiable function of two variables.

In this case, it is possible to separate $u$ to the explicit form : $$ue^{\frac{3}{2}x}=f(x-2t) \quad\to\quad u= e^{-\frac{3}{2}x}f(x-2t)$$ where $f$ is any differentiable function.

With the condition : $$u(x,0)=e^{-\frac{x^2}{L^2}}=e^{-\frac{3}{2}x}f(x-2*0)=e^{-\frac{3}{2}x}f(x)$$ The function $f$ is determined (doesn't matter the symbol of the variable) $$f(X)=e^{\frac{3}{2}X}e^{-\frac{X^2}{L^2}}$$

In the above general solution we have $\quad X=x-2t\quad\to\quad f(x-2t)=e^{\frac{3}{2}(x-2t)}e^{-\frac{(x-2t)^2}{L^2}}\quad$ which leads to : $$u(x,t)= e^{-\frac{3}{2}x} e^{\frac{3}{2}(x-2t)}e^{-\frac{(x-2t)^2}{L^2}}$$ $$u(x,t)= e^{-3t-\frac{(x-2t)^2}{L^2}}$$

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This question can be solved without using the method of characteristics. Suppose we have $au_{t} + bu_{x} + cu = 0$ with initial value condition $u(x,0) = f(x)$. Then $a\frac{u_{t}}{u} + b\frac{u_{x}}{u} = -c \Rightarrow (log(u))_{t} + \frac{b}{a}(log(u))_{x} = -\frac{c}{a}$. Define $v = log(u)$, where our initial value condition is now $v(x,0)= log(f(x))$ - what we have is the transport equation for $v$, which has solution

$$ v(x,t) = log(f(x - \frac{b}{a}t)) + \int_{0}^{t}(-\frac{c}{a})ds = log(f(x - \frac{b}{a}t)) - \frac{c}{a}t$$

$$ \Rightarrow u(x,t) = exp(v(x,t)) = f(x - \frac{b}{a}t)exp(-\frac{c}{a}t)$$

See Evans' Partial Differential Equations, section 2.1 for a reference on the transport equation.

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The simplest way to solve the problem is to consider the equation as a linear. (No need to hurry, the proven way to study first-order PDEs is first to study linear, then semilinear, and only then quasilinear equations; I'd recommend my book on PDEs, in which I follow this approach.)

There is a simple general result stating that if $(a,b) \ne (0,0),$ then the general solution of the linear PDE $$ au_x + bu_y +cu=0 $$ with constant coefficients is given by $$ u(x,y)=e^{-\lambda(ax+by)} f(bx-ay) $$ where $$ \lambda=\frac c{a^2+b^2} $$ and $f \in C^1(R)$ is a continuously differentiable function (it's a result worth remembering). In particular, the general solution of your equation is $$ u(t,x)=e^{-3t/5-6x/5}f(2t-x) $$ where $f \in C^1(\mathbf R)$ is a continuously differentiable function. Now by the initial condition, $$ u(0,x)=e^{-6x/5}f(-x)=e^{-x^2/l^2}, $$ whence $$ f(-x)=e^{-x^2/l^2} e^{6x/5} $$ whence $$ f(x)=e^{-x^2/l^2}e^{-6x/5}, $$ since $f$ is defined everywhere on $\mathbf R.$ It then easily follows that the solution of the Cauchy problem in question is $$ u(t,x)=e^{-3t} e^{-(2t-x)^2/l^2}. $$