1
$\begingroup$

what is the best way to solve a partial differential equation: $ (1-ax)(∂^4 y)/(∂x^4)+2a (∂^3 y)/(∂x^3)=0 $

like in ordinary differential equations I tried the power series method (I'm not very good with differential equations). I got something like: $y= C_1+C_2+C_3 (1+(1/3) ax)+C_4 (1-ax)$

which is difficult subjecting to the boundary conditions:

$ y=0,y''=0,x=0 \\ y=M,y''=(-1-y')/k(1-ax),x=z $

Can anyone help?

  • 0
    That is what I was after. Thank you.2012-06-11

2 Answers 2

3

Since the statement of the problem gives no reason to consider the equation a PDE, a standard method of lowering the order applies: $(1-ax)\frac{d^4y}{dx^4} +2a\frac{d^3y}{dx^3}=0$ $z=\frac{d^3y}{dx^3}$ $(1-ax)z'+2az=0$ $\frac{z'}{z}=-\frac{2a}{1-ax}$ $\ln |z|=\ln[(1-ax)^2]+C_1$ $z=C_1(1-ax)^2$ Now integrate three times and apply boundary conditions as appropriate.

  • 0
    Thank you Valentin. You are a good teacher.2012-06-11
0

The differential equation, $(1-\alpha x)\partial_x^4 y +2\alpha\partial_x^3 y = 0,$ is despite the usage of partial derivatives an ordinary differential equation since the function $y=y(x)$ to be determined depends solely on the variable $x$. Introducing the function, $z=\partial_x^3y,$ the differential equation is recast in the form of a homogeneous first-order ordinary differential equation, $(1-\alpha x)\partial_x z +2\alpha z=0.$ This differential equation can be brought - using physical notation - in the form, $\frac{dz}{z}=-\frac{2\alpha dx}{1-\alpha x}.$ Integrating, one obtains, $\partial_x^3y(x)=z(x)=z(x=0)\exp\left(2\log\vert 1-\alpha x\vert\right)=(\partial_x^3y)_0(1-\alpha x)^2=(\partial_x^3y)_0(\alpha x-1)^2,$ using $\log 1 = 0$ during the integration. Integrating thrice in order to obtain $y(x)$, one has, $y(x)=(\partial^3_x y)_0\dfrac{1}{3\alpha}\frac{1}{4\alpha}\frac{1}{5\alpha}(\alpha x-1)^5+c_2x^2 + c_1x+c_0=\dfrac{(\partial_x^3y)_0(\alpha x-1)^5}{60\alpha^3}+c_2x^2+c_1x^1+c_0.$ From the boundary conditions at $x=0$ one has $c_0=0=c_2$ by inspection. The boundary conditions at $x=z$ could also be applied, but I don't understand why there is a differential equation as a boundary condition. Shall one solve this differential equation first and then evaluate at that point? A convention I am not familiar with?

Although the question is old, perhaps the answer is helpful.

Best regards.