Recent content by jasonmcc
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MHB Integrating factor, initial value problem
$ kxy \frac{dy}{dx} = y^2 - x^2 \quad , \quad y(1) = 0 $ My professor suggests substituting P in for y^2, such that: $ P = y^2 dP = 2y dy $ I am proceeding with an integrating factor method, but unable to use it to separate the variables, may be coming up with the wrong integrating factor ( x )- jasonmcc
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- Initial Initial value problem Value
- Replies: 1
- Forum: Differential Equations
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MHB How to Solve This ODE with Substitution?
I haven't done ODEs in a while nor have a book handing. How do I tackle an equation of the form \[ 2xyy'=-x^2-y^2 \] I tried polar but that didn't seem to work.- jasonmcc
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- Ode Separable
- Replies: 2
- Forum: Differential Equations
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Stress tensor vanishes on cylinder edge
Thanks for all the help and action! I was traveling and away from my computer; great to see the discussion.- jasonmcc
- Post #10
- Forum: Engineering and Comp Sci Homework Help
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Stress tensor vanishes on cylinder edge
Homework Statement Given a cylinder in the Ox1x2x3 coordinate system, such that x1 is in the Length direction and x2 and x3 are in the radial directions. The stress components are given by the tensor $$ [T_{ij}] = \begin{bmatrix}Ax_2 + Bx_3 & Cx_3 & -Cx_2 \\ Cx_3 & 0 & 0 \\ -C_2 & 0 &...- jasonmcc
- Thread
- Cylinder Edge Stress Stress tensor Tensor
- Replies: 9
- Forum: Engineering and Comp Sci Homework Help
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MHB Tensor multiplication 3 dimesnsions
Yes nevermind I was looking at your solution backwards. Breaking up $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)}$ into elements it all canceled out except $$ \frac{1}{2}(A_{ij}B_{ij} - A_{ji}B_{ji}) $$ so if that equals zero we are good...- jasonmcc
- Post #4
- Forum: Linear and Abstract Algebra
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MHB Tensor multiplication 3 dimesnsions
Dustin, I think you basically solved it in line 1, since $$ A_{ij} = A_{(ij)} + A_{[ij]} $$ (decomposition of the tensor into symmetric $(A_{(ij)}$ and antisymmetric $(A_{[ij]})$ parts), so $$ A_{ij}B_{ij} = (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]}) $$ But how does $$ A_{(ij)}B_{(ij)} +...- jasonmcc
- Post #2
- Forum: Linear and Abstract Algebra
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MHB Indicial notation - Levi-Cevita and Tensor
there is an easier way, of course, using indicial. $$ \mathcal{A}_{mi}\varepsilon_{mjk} + \mathcal{A}_{mj}\varepsilon_{imk} + \mathcal{A}_{mj}\varepsilon_{ikm} = \mathcal{A}_{mk}\varepsilon_{ijk}\\ $$ multiplying all by $\varepsilon_{ijk}$ leads to kroniker delta rules, whereupon the expression...- jasonmcc
- Post #2
- Forum: Linear and Abstract Algebra
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MHB Indicial notation - Levi-Cevita and Tensor
Use indicial notation to show that: $$ \mathcal{A}_{mi}\varepsilon_{mjk} + \mathcal{A}_{mj}\varepsilon_{imk} + \mathcal{A}_{mk}\varepsilon_{ijm} = \mathcal{A}_{mm}\varepsilon_{ijk} $$ I'm probably missing an easier way, but my approach is to rearrange and expand on the terms: $$...- jasonmcc
- Thread
- Notation Tensor
- Replies: 1
- Forum: Linear and Abstract Algebra
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MHB How to Expand and Simplify the Expression of Kronecker Delta?
Thanks. What do you think they mean by "Expand", then?- jasonmcc
- Post #3
- Forum: Linear and Abstract Algebra
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MHB How to Expand and Simplify the Expression of Kronecker Delta?
Hi, I'm working on a problem stated as: Expand the following expression and simplify where possible $$ \delta_{ij}\delta_{ij} $$ I'm pretty sure this is correct, but not sure that I am satisfying the expand question. I'm not up to speed in linear algebra (taking a continuum mechanics course) -...- jasonmcc
- Thread
- Delta
- Replies: 3
- Forum: Linear and Abstract Algebra