Hi, can anyone help solve this equation?
du/dt + u * du/dx - v* (d^2u/dx^2) = 0
with
u(x,0)= 2 for x < 0
u(x,0)= 0 for x >/= 0
The solution is u(x,t) = 2 / (1 + e^((x-t)/v)).
Can anyone show me how to get to this solution please?
First, you have to understand that, this partial differential equation represents diffusive waves moving in fluid. Therefore, we are looking for a solution in the form of:
[imath]u(x,t) = f(x - ct)[/imath], where [imath]c[/imath] is the speed of the wave.
By using our initial guess above and the initial conditions, we want to show that the solution is:
[imath]\displaystyle u(x,t) = \frac{2}{1 + e^{(x-t)/v}}[/imath]
Notice that, the initial conditions are:
[imath]\displaystyle \lim_{x\rightarrow -\infty} u(x,0) = \lim_{x\rightarrow -\infty} \frac{2}{1 + e^{x/v}} = 2[/imath]
AND
[imath]\displaystyle \lim_{x\rightarrow \infty} u(x,0) = \lim_{x\rightarrow \infty} \frac{2}{1 + e^{x/v}} = 0[/imath]
We will call these two constats [imath]y_1[/imath] and [imath]y_2[/imath].
[imath]y_1 = 2[/imath]
[imath]y_2 = 0[/imath]
Let
[imath]\alpha = x - ct[/imath]
AND
[imath]y = f(\alpha)[/imath]
[imath]\displaystyle \frac{\partial u}{\partial t} = \frac{dy}{d\alpha}\frac{\partial \alpha}{\partial t} = -c\frac{dy}{d\alpha}[/imath]
[imath]\displaystyle \frac{\partial u}{\partial x} = \frac{dy}{d\alpha}\frac{\partial \alpha}{\partial x} = \frac{dy}{d\alpha}[/imath]
[imath]\displaystyle \frac{\partial^2 u}{\partial x^2} = \frac{dy^2}{d\alpha^2}[/imath]
Substitute in the original differential equation:
[imath]\displaystyle -c\frac{dy}{d\alpha} + y\frac{dy}{d\alpha} - v\frac{dy^2}{d\alpha^2} = 0[/imath]
Now we can reduce this differential equation one degree by integrating each term with respect to [imath]\alpha.[/imath]
[imath]\displaystyle -cy + \frac{1}{2}y^2 - v\frac{dy}{d\alpha} = A, \ \ \ \ [/imath] where [imath]A[/imath] is the constant of integration.
Rearranging the equation, gives:
[imath]\displaystyle 2v\frac{dy}{d\alpha} = y^2 - 2cy - 2A = (y - y_1)(y - y_2)[/imath],
where [imath]y_1 = c + \sqrt{c^2 + 2A} \ \ \ \ \ [/imath] AND [imath] \ \ \ \ \ y_2 = c - \sqrt{c^2 + 2A}[/imath]
These are the same constants of the initial conditions.
Now we can integrate by partial fraction decomposition.
[imath]\displaystyle \int \frac{2v}{(y - y_1)(y - y_2)} \ dy = \int \ d\alpha[/imath]
[imath]\displaystyle \frac{2v}{y_1 - y_2}\ln\left(\frac{y_1-y}{y - y_2}\right) = \alpha[/imath]
We don't have an integration constant because we know that [imath]y_2 < y < y_1[/imath].
Simplifying the solution and rearrange to solve for [imath]y[/imath], we get:
[imath]\displaystyle y = f(\alpha) = \frac{y_1 + y_2e^{\frac{y_1-y_2}{2v}\alpha}}{1+e^{\frac{y_1-y_2}{2v}\alpha}}[/imath]
We already know that [imath]y_1 = 2 \ \ \ \ [/imath] AND [imath] \ \ \ \ y_2 = 0[/imath]. Therefore, the solution will be:
[imath]f(\alpha) = \frac{2}{1+e^{\frac{\alpha}{v}}}[/imath]
[imath]u(x,t) = f(x - ct) = \frac{2}{1+e^{\frac{x - ct}{v}}}[/imath]
The final trick is to find the speed of the moving wave, [imath]c[/imath].
[imath]c = \frac{1}{2}(y_1 + y_2) = \frac{1}{2}(2 + 0) = 1[/imath]
[imath]u(x,t) = \frac{2}{1+e^{\frac{x - t}{v}}}[/imath]
