Exact solution of a 4th order DE

  • Thread starter DarkLord777
  • Start date
In summary, you can solve for the particular solution by solving the sum equation with the superposition principle.
  • #1
DarkLord777
3
0
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
 
Physics news on Phys.org
  • #2
Just by inspection, u=f/5 is a particular solution right?
 
  • #3
surely the particular solution must satisfy the boundary conditions?
 
  • #4
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
 
  • #5
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}]
 
  • #6
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!)
 
  • #7
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
 

1. What is a 4th order differential equation?

A 4th order differential equation is an equation that involves the fourth derivative of an unknown function. It can be written in the form: f(4)(x) = g(x), where f(4)(x) represents the fourth derivative of the function f(x) and g(x) represents a known function.

2. What does it mean to find an exact solution of a 4th order DE?

Finding an exact solution of a 4th order DE means finding a function that satisfies the equation for all values of x. In other words, the function must satisfy the equation and all its derivatives up to the fourth order.

3. How is an exact solution of a 4th order DE found?

An exact solution of a 4th order DE can be found by using various methods such as separation of variables, method of undetermined coefficients, or variation of parameters. These methods involve manipulating the equation to isolate the unknown function and then solving for it.

4. Are there any special cases when finding the exact solution of a 4th order DE?

Yes, there are special cases when finding the exact solution of a 4th order DE. One example is when the equation is a homogeneous equation, meaning the right side of the equation is equal to zero. In this case, the solution can be found by assuming the unknown function is equal to erx and solving for r.

5. What is the importance of finding an exact solution of a 4th order DE?

Finding an exact solution of a 4th order DE is important in many fields of science and engineering. It allows for a better understanding and modeling of complex systems, such as in physics, chemistry, and biology. It also has practical applications in areas such as engineering design and control systems.

Similar threads

  • Differential Equations
Replies
1
Views
1K
  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
7
Views
308
  • Differential Equations
Replies
7
Views
2K
  • Differential Equations
Replies
3
Views
1K
  • Differential Equations
Replies
11
Views
1K
  • Differential Equations
Replies
5
Views
1K
Replies
3
Views
1K
Replies
2
Views
1K
  • Differential Equations
2
Replies
52
Views
417
Back
Top