Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
How is this "partial"? I only see derivatives with respect to $x$. – Robert Israel Jun 10 '12 at 23:03
The differential equation (if that's really what you're after) has general solution $c_1 + c_2 (x-1/a) + c_3 (x-1/a)^2 + c_4 (x-1/a)^5$, where $c_1, \ldots, c_4$ are arbitrary constants. – Robert Israel Jun 10 '12 at 23:08
Thank you Mr Israel. that was very helpful. – sani Jun 11 '12 at 8:04
That is what I was after. Thank you. – sani Jun 11 '12 at 8:11
up vote 3 down vote accepted

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.

share|cite|improve this answer
Thank you Valentin. You are a good teacher. – sani Jun 11 '12 at 8:13

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.