[?] Solve problem with Sum function?

Questions
,
1

Hi,

I’m trying to find out the value of alpha for which the function w(alpha) is equal to zero. The solve function does not return any value. How can I resolve the issue?

Alpha.PNG
Alpha.PNG462×183 2.92 KB

Any assistance will be highly appreciated.
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Alpha.PNG&userId=1772296

2

In[10]:= Limit[w[[Alpha]], [Alpha] → [Infinity]]

Out[10]= 0

In[11]:= Limit[w[\[Alpha]], \[Alpha] -> -\[Infinity]]

Out[11]= 0
3

For transcendental equation it is better to give bounds for the variables:

Solve[w[\[Alpha]] == 0 && 0 < \[Alpha] < 100, \[Alpha], Reals]

By the way, the delayed evaluation (:=) cancels out with Evaluate. Immediate evaluation (=) is simpler and (I expect) equivalent.

4

Hi, thanks for the input.
However, the Solve function still returns empty curly brackets { }. When I plot the value of w[\Alpha] against Alpha, the plot is asymptotic to both the x and y axes.
This is the entire equation for which I’m trying to find the value of Alpha

3224SSSS.PNG
3224SSSS.PNG1141×432 7.23 KB

x varies from -0.5a to 0.5a, while y varies from 0 to b.
Is there an error in my syntax or what alternatives can I use?

5

You should use the Code Sample icon (first one) when posting code.

6

Apologies. Here’s the code.

q = 15.904;
v = 0.12157;
E1 = 18904;
b = 1830;
a = 1*b;
h = .04*b;

D1 = (E1 h^3)/(12 (1 - v^2)); 
w[x_, y_] = 
  q/(24*D1)*(y^4 - 2* b * y^3 + b^3 * y) + (2*q)/(D1*b) * 
    Evaluate[
     Sum[( ((n Pi)/b * x  *Sinh[(n Pi)/b * x] - 
         Cosh[(n Pi)/b * 
            x]*(2 + (\[Alpha] n Pi)/(2 b)* 
             Tanh[(\[Alpha] n Pi)/(2 b)]))  Sinh[(n Pi)/b* y])/(((
        n Pi)/b)^5 * Cosh[(\[Alpha] n Pi)/(2 b)]), {n, 1, 5}]];
(*Boundary conditions*)
w[0, 0] == 0;
w[0, b] == 0;
w[-0.5*a, b] == 0;
w[0.5*a, b] == 0;
w[-0.5*a, 0] == 0;
w[0.5*a, 0] == 0;
w[-0.5*a, 0.5*b] == 0;
w[0.5*a, 0.5*b] == 0;
7

Empty brackets means no solution exists. For example

Solve[x + 1 == x, x]
8

boundary conditions are for differential equations

9

Hello Matthew,

for me it is by no means clear what you want to do.
What is the w[alpha] you are talking about? In your last post I see only w[ x, y ].

If the “Boundary Conditions” are the equations you want to be fulfilled with a certain alpha that seems to be impossible.

With (giving your w [ x, y ] from above, but as function)

q = 15.904;
v = 0.12157;
E1 = 18904;
b = 1830;
a = 1*b;
h = .04*b;

D1 = (E1 h^3)/(12 (1 - v^2));
w[x_, y_] := 
  q/(24*D1)*(y^4 - 2*b*y^3 + b^3*y) + (2*q)/(D1*b)*
    Evaluate[
     Sum[(((n Pi)/b*x*Sinh[(n Pi)/b*x] - 
           Cosh[(n Pi)/b*
              x]*(2 + (\[Alpha] n Pi)/(2 b)*
               Tanh[(\[Alpha] n Pi)/(2 b)])) Sinh[(n Pi)/b*
           y])/(((n Pi)/b)^5*Cosh[(\[Alpha] n Pi)/(2 b)]), {n, 1, 
       5}]];

some equations are true for any alpha:

w[0, 0]
w[.5 a, 0]

Out[170]= 0

Out[171]= 0

For dealing with other equations I define a new function

f[x_, y_, a_] := w[x, y] /. \[Alpha] -> a

and find

f[-.5 a, b, 1340]
f[-.5 a, b, 1360]

Out[155]= 1.01387*10^6

Out[156]= -459226.

Here we have different signs, therefore I can do a bisection

eps = 10^-6;
xxx = 10;
a1 = 1340;
a2 = 1360;
While[
 xxx > eps,
 a3 = (a1 + a2)/2;
 If[
  Sign[f[.5 a, b, a3]] == Sign[f[.5 a, b, a2]],
  a2 = a3,
  a1 = a3];
 xxx = Abs[a2 - a1]
 ]
a3 // N
f[.5 a, b, a3]

1353.4

0.0386934

0.0386 is not zero, ok, but decreasing eps should give better values.
So w[ .5 a , .5 b ] == 0 is almost fulfilled.

But w [ -.5 a, .5 b ] is not

f[-.5 a, .5 b, a3]

Out[26]= -9041.8

In fact I doubt that you can achieve all your “Boundary Conditions” with a single alpha

10

Thank you for the insight. Let me try to rephrase the equation so that a single value of [Alpha] can be achieved