For the Ritz method, the governing differential equation is such that whenever we try n=4,5 or more, we get back the same polynomial for n=3. Is this correct? I think it is necessary since for n=4,5 we would still have only 3 equations to solve for constants
I have been able to obtain the variational statement for this question but I am having difficulty with the exact solution. Solving exactly means solving the given differential equation? How do we solve the differential equation since it contains a discontinuous function (Q(x))?
When we divide it into 2 domains, and solve the 2nd order ODE separately, we will obtain 4 constants whereas we are only given 2 boundary conditions. I am thinking maybe we can enforce continuity of phi at L/2 to obtain the 3rd condition. I am not able to obtain the 4th condition to solve all the constants completely.
i was substituting the trial functions in the differential equation. I realize that i should have substituted them in the variational statement and then integrated it
When solving using the Ritz method with n=4 and n=5, I can't seem to get a final expression, as if I'm missing some some boundary conditions or perhaps doing a step incorrectly. For n=4, after substituting the trial expression (which satisfies the boundary conditions) into the functional and solving delta I = 0, I get
a3=-1.2*(0.215*Phi_0+1.3*a4). That leaves one coefficient still unknown. Is that true? Or am I missing something?
For Ritz method, after substitute the trial expression into the integration, it is better to take the partial derivative with respect to a_i instead of letting delta I =0. It is more straightforward. For example, for n= 4, after substituting the trial expression (which satisfies the boundary conditions) into the functional, you should have (partial I)/ (partial a3) =0 and (partial I)/ (partial a4) =0
That looks fine to me. Then you need to take variation of I. delta I = ** delta_a3 + ** delta_a4, the coefficient of delta_a3/a4 should vanish to make delta I zero
You are welcome. Actually I'm saying the same thing in my second reply. You do variation, you take partial derivative by a3/a4. They are the same thing. deltaI=delta_a3*(partial I/partial a3)+delta_a4*(partial I/partial a4). Because your delta_a3/delta_a4 is arbitrary, your partial derivatives must be zero. You are not making any mistake when you say taking variation.
Imad, you are welcome. Yes, they should get the same result. Taking partial derivative with respect to coefficients is derived from delta I =0. Thank Haodong for the clarification.
Kunal Samel @ on
For the Ritz method, the governing differential equation is such that whenever we try n=4,5 or more, we get back the same polynomial for n=3. Is this correct? I think it is necessary since for n=4,5 we would still have only 3 equations to solve for constants
thanks
Kunal
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Lohit Gudivada @ on
Dear all,
I have been able to obtain the variational statement for this question but I am having difficulty with the exact solution. Solving exactly means solving the given differential equation? How do we solve the differential equation since it contains a discontinuous function (Q(x))?
thanks,
Lohit
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Lohit Gudivada @ on
When we divide it into 2 domains, and solve the 2nd order ODE separately, we will obtain 4 constants whereas we are only given 2 boundary conditions. I am thinking maybe we can enforce continuity of phi at L/2 to obtain the 3rd condition. I am not able to obtain the 4th condition to solve all the constants completely.
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Lohit Gudivada @ on
Should we use continuity of d(phi)/dx at L/2 as the 4th condition? Isn't anyone else facing this issue?
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Kunal Samel @ on
i was substituting the trial functions in the differential equation. I realize that i should have substituted them in the variational statement and then integrated it
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Imad Hanhan @ on
Hello all,
When solving using the Ritz method with n=4 and n=5, I can't seem to get a final expression, as if I'm missing some some boundary conditions or perhaps doing a step incorrectly. For n=4, after substituting the trial expression (which satisfies the boundary conditions) into the functional and solving delta I = 0, I get
a3=-1.2*(0.215*Phi_0+1.3*a4). That leaves one coefficient still unknown. Is that true? Or am I missing something?
Thanks,
Imad
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Fei Tao @ on — Edited @ @ on
For Ritz method, after substitute the trial expression into the integration, it is better to take the partial derivative with respect to a_i instead of letting delta I =0. It is more straightforward. For example, for n= 4, after substituting the trial expression (which satisfies the boundary conditions) into the functional, you should have (partial I)/ (partial a3) =0 and (partial I)/ (partial a4) =0
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Haodong Du @ on — Edited @ @ on
Hi Imad,
Seems like you missed the boundary condition. You should let your trial function be admissible first.
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Imad Hanhan @ on
Haodong,
For n=4, the function: Phi(x)=Phi_0+a3*(x^2-L*x)+a4*(x^3-L^2*x) satisfies Phi(0)=Phi_0 and Phi(L)=Phi_0.
Which boundary conditions am I missing?
Thanks,
Imad
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Haodong Du @ on — Edited @ @ on
Hi Imad,
That looks fine to me. Then you need to take variation of I. delta I = ** delta_a3 + ** delta_a4, the coefficient of delta_a3/a4 should vanish to make delta I zero
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Imad Hanhan @ on
Thank you Fei and Haodong. Fei thank you for that correction, that's where my mistake was.
-Imad
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Haodong Du @ on
You are welcome. Actually I'm saying the same thing in my second reply. You do variation, you take partial derivative by a3/a4. They are the same thing. deltaI=delta_a3*(partial I/partial a3)+delta_a4*(partial I/partial a4). Because your delta_a3/delta_a4 is arbitrary, your partial derivatives must be zero. You are not making any mistake when you say taking variation.
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Fei Tao @ on
Imad, you are welcome. Yes, they should get the same result. Taking partial derivative with respect to coefficients is derived from delta I =0. Thank Haodong for the clarification.
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