# Exact solution of a 4th order DE

• DarkLord777

#### DarkLord777

Hi guys,

I am trying to find the exact solution of:

u'''' + 5u =f on (0,1) f is a constant >0

where

u(0)=u'(0) =0
u''(1) =k constant >0
u'''(1)=0

I assumed I should look at solving the homogeneous equation and got the following

uh = A*exp(Zx)*sin(Zx) + B*exp(ZX)*cos(Zx) + C*exp(-Zx)*sin(Zx) +D*exp(-Zx)*cos(Zx)

Where Z is (sqrt(2)/2)*sqrt(sqrt(5)) and A,B,C,D are constants.

My questions are:

1) Am I on the right track
2) How would I go about solving for the particular solution
3) Does the exact solution actually exist?

Thanks

## Answers and Replies

Just by inspection, u=f/5 is a particular solution right?

surely the particular solution must satisfy the boundary conditions?

You are on the right track and the problem is well posed.
^depends how particular

The solution of the equation can be written as a sum by the superposition principle. The final answer needs to satisfy the differential equation and the boundary conditions (and is a particular solution. The terms we are adding up need not satisfy all (or any) of the conditions. Just let each term satisfy certain conditions and keep track that the sum will satisfy all conditions.

basically either take
u = A*exp(Zx)*sin(Zx) + B*exp(ZX)*cos(Zx) + C*exp(-Zx)*sin(Zx) +D*exp(-Zx)*cos(Zx) +f/5
and fit the boundary conditions or take
u = A*exp(Zx)*sin(Zx) + B*exp(ZX)*cos(Zx) + C*exp(-Zx)*sin(Zx) +D*exp(-Zx)*cos(Zx) +E
and fit the boundary conditions and u'''' + 5u =f

It is the same in the end

What did we do before Mathematica?

Code:
z = (Sqrt[2]/2)*Sqrt[Sqrt[5]];

u[x_] := a*Exp[z*x]*Sin[z*x] + b*Exp[z*x]*Cos[z*x] + c*Exp[(-z)*x]*Sin[z*x] + d*Exp[(-z)*x]*Cos[z*x]+f/5;

myequations = {0 == u[0], 0 == Derivative[1][u][0], k == Derivative[2][u][1], 0 == Derivative[3][u][1]}

Solve[myequations, {a, b, c, d}]

Hmmm and if I don't have mathematica?

I tried wolfram alpha but it can't solve it with all 4 boundary conditions.
If I only use 2 I end up with a massive equation (18+ terms) that I still have to solve for 2 constants.

I tried solving it by hand from scratch and get the wrong answer. (well wrong according to my finite element program - and I know it is correct!)

wolfram alpha can be wonky but I entered

u''''(x) + 5u(x) =f;u(0)=0,u'(0)=0,u''(1)=k,u'''(1)=0

and got a reasonable answer