We consider a fractional boundary value problem with various boundary conditions. This boundary value problem has two components, one fractional derivative of alpha degree with alpha between n-1 and n, and a fractional derivative of beta degree with beta between 0 and n-2. We prove existence and uniqueness of solutions, and show some examples that were found using a MATLAB simulation

Yost, Grant

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University of Tennessee at ChattanoogaUTC ScholarHonors Theses Student Research, Creative Works, and Publications5-2016A fractional boundary value problemGrant YostUniversity of Tennessee at Chattanooga, ddw497@mocs.utc.eduFollow this and additional works at: http://scholar.utc.edu/honors-thesesPart of the Mathematics CommonsThis Theses is brought to you for free and open access by the Student Research, Creative Works, and Publications at UTC Scholar. It has been acceptedfor inclusion in Honors Theses by an authorized administrator of UTC Scholar. For more information, please contact scholar@utc.edu.Recommended CitationYost, Grant, "A fractional boundary value problem" (2016). Honors Theses.We consider the boundary value problem consisting of the fractional di↵erentialequation D↵0+z + aD 0+z = w(t)f(t, z), 0 < t < 1, (1)and the boundary conditionsz(0) = z0(0) = ... = z(n 2)(0) = 0, z(1) = k, (2)where k 2 R, a 2 R, n 2 N, 0     n  2, n  1 < ↵ < n, w 2 L[0, 1] with w(t)   0,and f 2 C([0, 1] ⇥ R,R). Here, D 0+h is the  -th left Riemann-Liouville fractionalderivative of h : [0, 1]! R defined by(D 0+h)(t) =1 (l    )dldtlZ t0(t  s)l   1h(s)ds, l = b c+ 1,whenever the right-hand side exists with  (·) being the Gamma function.Fractional di↵erential equation models such as this are relevant in various scienceand engineering fields. Problems like this have been solved by first finding the Green’sfunction for z in similar equations such as D↵0+z = f(t, z).The Green’s function (denoted byG) was developed as a method to solve boundaryvalue problems with linear homogeneous equations (i.e. when f(t, z) = 0). However,when the right side of equation is a function of t, say w(t)f(t, z(t)), the unique solutionto the problem is given byu(t) =Z 10G(t, s)w(s)f(s, z(s))ds,where G is the solution to the similar problem with the right hand side equal to zero.2We try to solve problem (1)-(2) by splitting it into the two following problems:8>><>>: D↵0+u+ aD 0+u = w(t)f(t, u), 0 < t < 1,u(0) = u0(0) = ... = u(n 2)(0) = 0, u(1) = 0,(3)8>><>>: D↵0+v + aD 0+v = 0, 0 < t < 1,v(0) = v0(0) = ... = v(n 2)(0) = 0, v(1) = k.(4)Note that if u(t) is a solution of (3) and v(t) is a solution of (4), then z(t) = u(t)+v(t)is a solution of (1)-(2).The solution of (3) is found by using the Green’s Function: G : [0, 1]⇥ [0, 1]! R,which was obtained in [3]:G(t, s) =8>>>>>>><>>>>>>>:[t(1  s)]↵ 1E↵  ,↵[at↵  ]E↵  ,↵[a(1  s)↵  ]E↵  ,↵[a], 0  t  s  1,[t(1  s)]↵ 1E↵  ,↵[at↵  ]E↵  ,↵[a(1  s)↵  ]E↵  ,↵[a] (t  s)↵ 1E↵  ,↵[a(t  s)↵  ], 0  s  t  1,(5)with |a| <  (↵     + 1). Here, the Green’s Function is given in terms of the knownMittag-Le✏er function E ,  : C ! C:E , [z] =1Xn=0zn ( n+  ).Using this function, we find the solution to (3) to have the form:u(t) =Z 10G(t, s)w(s)f(s, u(s))ds3The solution to (4) is easily seen to be:v(t) =kt↵ 1E↵  ,↵(at↵  )E↵  ,↵(a).Note that by Lemma 4.3 of [3], the Mittag-Le✏er function, E, is increasing, so |v(t)| k for t 2 [0, 1].The solution to (1)-(2) is therefore z(t) = u(t) + v(t).Remark: As found in [3], for G defined in (5), when |a| <  (↵     + 1),G(0, ·) = G(1, ·) = G(·, 0) = G(·, 1) = 0, and |G(t, s)|  G(s) on [0, 1]⇥ [0, 1] whereG(s) =8>><>>:(1  s)↵ 1E↵  ,↵[a(1  s)↵  ], a 2 [0, (↵    + 1)),(1  s)↵ 1E↵  ,↵[a(1  s)↵  ] (↵)E↵  ,↵[a], a 2 (  (↵    + 1), 0).(6)We need to define the constantU =Z 10G(s)w(s)ds (7)with G defined in (6). We take X = C[0, 1] with the norm kuk = maxt2[0,1] |u(t)|,u 2 X, to be our Banach space.We have developed two main theorems on the existance and uniqueness of solutionsfor our problem.Theorem 1: Let |a| <  (↵    + 1). Assume there exists r > 0 such that|f(t, x)|  r   kUfor (t, x) 2 [0, 1]⇥ [ r, r], (8)where k is the constant given in the right hand boundary condition given in (2).Then, BVP (1)-(2) has at least one solution z with kzk  r.Theorem 2: Let |a| <  (↵     + 1). Assume that f satisfies the Lipschitz4condition in x|f(t, x1)  f(t, x2)|  B|x1   x2| for (t, x1), (t, x2) 2 [0, 1]⇥ R,with B 2 (0, 1/U). Then, BVP (1)-(2) has a unique solution.Proofs:For the purpose of these proofs, we need to define an operator T : X ! X by(Tz)(t) =Z 10G(t, s)w(s)f(s, u(s))ds+kt↵ 1E↵  ,↵(at↵  )E↵  ,↵(a),where G is the Green’s function used to solve for u(t). It is clear that z is a solution ofBVP (1)-(2) if and only if z 2 X is a fixed point of T . There is a standard argumentto show that T is completely continuous.Proof of Theorem 1: Let ⌦ ⇢ X be the set defined by⌦ = {z 2 X | kzk  r},where r is given in Theorem 1.For any z 2 ⌦, we have kz(t)k  r on [0, 1], and so|(Tz)(t)| =    Z 10G(t, s)w(s)f(s, u(s))ds+kt↵ 1E↵  ,↵(at↵  )E↵  ,↵(a)        Z 10G(t, s)w(s)f(s, u(s))ds    +     kt↵ 1E↵  ,↵(at↵  )E↵  ,↵(a)    Z 10|G(t, s)|w(s)|f(s, u(s))|ds+ kZ 10G(s)w(s)(r   k)Uds+ k =r   kUU + k=r   k + k = r, t 2 [0, 1].Hence, kTuk  r. Therefore, T⌦ ⇢ ⌦.5By Schauder fixed point theorem, T has a fixed point z in ⌦. Hence, our BVPhas at least one solution z(t) with kzk  r. ⇤Proof of Theorem 2: Let z1, z2 2 X be solutions of (0.1). Then z1 = u1+v1and z2 = u2 + v2, where u1, u2 are solutions of (0.3) and v1, v2 are solutions of (0.4).For t 2 [0, 1],|(Tz1   Tz2)(t)| =    Z 10G(t, s)w(s)f(s, u1(s)))ds+kt↵ 1E↵  ,↵(at↵  )E↵  ,↵(a) Z 10G(t, s)w(s)f(s, u2(s)))ds  kt↵ 1E↵  ,↵(at↵  )E↵  ,↵(a)    =    Z 10G(t, s)w(s)f(s, u1(s)))ds Z 10G(t, s)w(s)f(s, u2(s)))ds    =    Z 10G(t, s)w(s) (f(s, u1(s))  f(s, u2(s))) ds    Z 10G(s)w(s) |f(s, u1(s))  f(s, u2(s))| dsZ 10G(s)w(s)B |u1(s)  u2(s)| ds  BUku1   u2k.Since BU < 1, T is a contraction mapping. By the contraction mapping principle,T has a unique fixed point. Therefore, our BVP has a unique solution. ⇤Examples:Example 1: Consider the BVP8>><>>: D3.60+z + aD1.40+z =⇥cos(20t) + 1 + esin(100t)⇤ ⇥12 sin(z) +zz+1 + et⇤z(0) = z0(0) = · · · = z(n 2) = 0, z(1) = 0.01(9)Here, f(t, z) = 12 sin(z)+zz+1 +et and w(t) = cos(20t)+1+esin(100t). Furthermore, (3.2) = 2.424. The following figure represents the solution to this problem using aniterative process with a = 2.4 shown in green, a = 0.4 shown in red, and a =  2.4shown in blue.6Example 2: Consider the BVP8>><>>: D4.50+z + 1.32D2.70+z =hz2z2+1i[cos(t) + 1]z(0) = z0(0) = · · · = z(n 2) = 0, z(1) = 10 13(10)Here, f(t, z) = z2z2+1 and w(t) = cos(t)+1. Note that |a| = 1.32 < 1.6765 ⇡  (2.8).We claim that this problem has a unique solution. By (0.6) and (0.7), 1/U ⇡ 24.5834.Then, for (t, z1), (t, z2) 2 [0, 1]⇥ R:|f(t, z1)  f(t, z2)| =     z21z21 + 1   z22z22 + 1    =    z21(z22 + 1)  z22(z21 + 1)(z21 + 1)(z22 + 1)    =     z21   z22(z21 + 1)(z22 + 1)    =    (z1   z2)(z1 + z2)(z21 + 1)(z22 + 1)    =|z1   z2|     z1 + z2(z21 + 1)(z22 + 1)    7=|z1   z2|     z1(z21 + 1)(z22 + 1) + z2(z21 + 1)(z22 + 1)    <|z1   z2|     z1z21 + 1 + z2z22 + 1    <|z1   z2||1 + 1|=2|z1   z2|.Therefore, f satisfies the Lipschitz condition with B = 2 2 (0, 1/U). Hence byTheorem 2, BVP (0.10) has a unique solution. 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A fractional boundary value problem

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A fractional boundary value problem

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