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MATH 2421 Introduction To Differential Equation
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MATH 2421 Introduction To Differential Equation
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Course Code: MATH2421
University: Douglas College
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Country: Canada
Question:
Write the Introduction to Differential Equation For Positive Definite.
Answer:
x=0 is the equilibrium point for this system. If we introduce a quadratic Lyapunov function
V=x12+ax22
Where a is a positive constant to be determined. v is positive definite on the entire state space ?2. V is also radially unbounded that is| v(x)|→∞ as? x?→∞. The derivative of v along the trajectories is given by
V’ = 2×1 2ax2 -x1-x1x22
-2×2-2x12x2
=2×1(-x1-x12)+2ax2(-2×2-2x12x2)
=-2×12+(-2-4a)x1x22-4ax22
If we choose a =-1/2 then we eliminate x1x2 and v’ becomes
v’=-2(x12-x22) .Therefore x=0 is globally asymptotically stable equilibrium point if x1>x2 or -x1<-x2
since v(x) is a continuous differentiable function i.e.
v'(x)≤-k?x?a for all x??2 and where k, a>0
hence the origin is exponentially stable
b) x1’=x2
x2’=-x1-x23
X=0 is the equilibrium point for this system. If we introduce a quadratic Lyapunov function
V=x12+ax22
Where a is a positive constant to be determined. v is positive definite on the entire state space ?2. V is also radially unbounded that is| v(x)|→∞ as? x?→∞. The derivative of v along the trajectories is given by
V’= 2×1 2ax2 x2
-x1-x23
=2x1x2+2ax2(-x1-x23)
=(2-2a)x1x2-2ax24)
If we choose a=1 we then eliminate the cross terms and we then have v’=-2×24 therefore x=0 is a globally asymptotically stable equilibrium point
since v(x) is a continuous differentiable function i.e.
v'(x)≤-k?x?a for all x??2 and where k,a>0
hence the origin is globally exponentially stable.
c) x1’=-x2-x1(1-x12-x22)
x2’=x1-x2(1-x12-x22)
X=0 is the equilibrium point for this system. If we introduce a quadratic Lyapunov function
V=x12+ax22
Where a is a positive constant to be determined. v is positive definite on the entire state space ?2. V is also radially unbounded that is| v(x)|→∞ as? x?→∞. The derivative of v along the trajectories is given by
V’= 2×1 2ax2 -x2-x1(1-x12-x22)
X1-x2(1-x12-x22)
Multiplying this matrix we obtain
v’ = -2×12+2×14-2ax22+2ax24+(-2+2a)x1x2+(2+2a)x12x22
If we choose a=1 so that we eliminate the cross term we have
v’ = -2×12+2×14-2×22+2×24+4x12x22
Hence x=0 is not a globally asymptotically stable equilibrium point. since v(x) is a continuous differentiable function i.e.
v'(x)≥-k?x?a for all x??2 and where k, a>0
hence the origin is neither exponentially stable nor globally exponentially stable.
2) x’=σ(y-x)
y’=rx-y-xz
z’=xy-bz
where σ, r, b are positive constants and again we have that 0
X2=0
X3=0
h1(x1)-h2(0)=0
Since h2(0)=0 implies that h1(x1)=0 and therefore x1=0
Hence the system has a unique equilibrium point at the origin
b) v(x)= 1(y)dy +x22/22(y)dy
If we define v(x) the way it has been defined above, then it is clear that v(x) is locally positive definite.
c) v'(x) =-h1(x1)2+x2-h2(x3)2
Which is locally negative definite implying that x=0 or the origin is asymptotically stable equilibrium point.
d) The function v(x) should be positive definite on the entire state space and has the property |v(x)|→∞ as ?x?→∞ and again
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