Problem Statement
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Consider the L2-ODE-CC (5) p.7b-7 with l o g ( 1 + x ) {\displaystyle log(1+x)\!} as excitation:
y ″ − 3 y ′ + 2 y = r ( x ) {\displaystyle y''-3y'+2y=r(x)\!} (5) p.7b-7
r ( x ) = l o g ( 1 + x ) {\displaystyle r(x)=log(1+x)\!} (1) p.7c-28
and the initial conditions
y ( − 3 4 ) = 1 , y ′ ( − 3 4 ) = 0 {\displaystyle y({\frac {-3}{4}})=1,y'({\frac {-3}{4}})=0\!} .
Project the excitation r ( x ) {\displaystyle r(x)\!} on the polynomial basis
b j ( x ) = x j , j = 0 , 1 , . . . , n {\displaystyle {b_{j}(x)=x^{j},j=0,1,...,n}\!} (1)
i.e., find d j {\displaystyle d_{j}\!} such that
r ( x ) ≈ r n ( x ) = ∑ j = 0 n d j x j {\displaystyle r(x)\approx r_{n}(x)=\sum _{j=0}^{n}d_{j}x^{j}\!} (2)
for x in [ − 3 4 , 3 ] {\displaystyle [{\frac {-3}{4}},3]\!} , and for n = 3, 6, 9.
Plot r ( x ) {\displaystyle r(x)\!} and r n ( x ) {\displaystyle r_{n}(x)\!} to show uniform approximation and convergence.
Note that:
⟨ x i , r ⟩ = ∫ a b x i l o g ( 1 + x ) d x {\displaystyle \left\langle x^{i},r\right\rangle =\int _{a}^{b}x^{i}log(1+x)dx\!} (3)
Solution
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Using Matlab, this is the code that was used to produce the results:
Find y n ( x ) {\displaystyle y_{n}(x)\!} such that:
y n ″ + a y n ′ + b y n = r n ( x ) {\displaystyle y_{n}''+ay_{n}'+by_{n}=r_{n}(x)\!} (1) p.7c-27
with the same initial conditions as in (2) p.7c-28.
Plot y n ( x ) {\displaystyle y_{n}(x)\!} for n = 3, 6, 9, for x in [ − 3 4 , 3 ] {\displaystyle [{\frac {-3}{4}},3]\!} .
In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.
Solution
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Using integration by parts, and then with the help of of
General Binomial Theorem( x + y ) n = ∑ k = 0 n ( n k ) x n − k y k {\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{n-k}y^{k}\!}
Solution
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For n = 0 {\displaystyle n=0\!} :∫ x 0 l o g ( 1 + x ) d x = ∫ l o g ( 1 + x ) d x {\displaystyle \int x^{0}log(1+x)dx=\int log(1+x)dx\!}
For substitution by parts, u = l o g ( 1 + x ) , d u = 1 1 + x , d v = d x , v = x {\displaystyle u=log(1+x),du={\frac {1}{1+x}},dv=dx,v=x\!} ∫ l o g ( 1 + x ) d x = x l o g ( 1 + x ) − ∫ x 1 + x d x {\displaystyle \int log(1+x)dx=xlog(1+x)-\int {\frac {x}{1+x}}dx\!} ∫ l o g ( 1 + x ) d x = x l o g ( 1 + x ) − ∫ ( 1 − 1 1 + x ) d x {\displaystyle \int log(1+x)dx=xlog(1+x)-\int (1-{\frac {1}{1+x}})dx\!} ∫ l o g ( 1 + x ) d x = x l o g ( 1 + x ) − x + l o g ( 1 + x ) + C {\displaystyle \int log(1+x)dx=xlog(1+x)-x+log(1+x)+C\!}
Therefore:
∫ l o g ( 1 + x ) d x = ( x + 1 ) l o g ( 1 + x ) − x + C {\displaystyle \int log(1+x)dx=(x+1)log(1+x)-x+C\!}
Using the General Binomial Theorem:( x + y ) 0 = ∑ k = 0 0 ( 0 k ) x 0 − k y k = 1 {\displaystyle (x+y)^{0}=\sum _{k=0}^{0}{\binom {0}{k}}x^{0-k}y^{k}=1\!}
Therefore:
∫ ( 1 ) l o g ( 1 + x ) d x = ∫ l o g ( 1 + x ) d x {\displaystyle \int (1)log(1+x)dx=\int log(1+x)dx\!}
Which we have previously found that answer as:
∫ l o g ( 1 + x ) d x = ( x + 1 ) l o g ( 1 + x ) − x + C {\displaystyle \int log(1+x)dx=(x+1)log(1+x)-x+C\!}
For n = 1 {\displaystyle n=1\!} :∫ x 1 l o g ( 1 + x ) d x = ∫ x l o g ( 1 + x ) d x {\displaystyle \int x^{1}log(1+x)dx=\int xlog(1+x)dx\!}
Initially we use the following substitutions: t = 1 + x , x = t − 1 , d t = d x {\displaystyle t=1+x,x=t-1,dt=dx\!} ∫ x l o g ( 1 + x ) d x = ∫ ( t − 1 ) l o g ( t ) d t = ∫ ( t l o g ( t ) − log ( t ) ) d t {\displaystyle \int xlog(1+x)dx=\int (t-1)log(t)dt=\int (tlog(t)-\log(t))dt\!}
First let us consider the first term: ∫ t l o g ( t ) d t {\displaystyle \int tlog(t)dt\!}
Next, we use the integration by parts: u = log t , d u = 1 t d t , d v = t d t , v = 1 2 t 2 {\displaystyle u=\log {t},du={\frac {1}{t}}dt,dv=tdt,v={\frac {1}{2}}t^{2}\!} ∫ t l o g ( t ) d t = 1 2 t 2 l o g ( t ) − ∫ 1 2 t 2 ( 1 t d t ) {\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-\int {\frac {1}{2}}t^{2}({\frac {1}{t}}dt)\!} ∫ t l o g ( t ) d t = 1 2 t 2 l o g ( t ) − ∫ 1 2 t d t ) {\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-\int {\frac {1}{2}}tdt)\!} ∫ t l o g ( t ) d t = 1 2 t 2 l o g ( t ) − 1 4 t 2 {\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}\!}
Next let us consider the second term: ∫ l o g ( t ) d t {\displaystyle \int log(t)dt\!}
Again, we will use integration by parts: u = log t , d u = 1 t d t , d v = d t , v = t {\displaystyle u=\log {t},du={\frac {1}{t}}dt,dv=dt,v=t\!} ∫ t l o g ( t ) d t = t l o g ( t ) − ∫ t ( 1 t d t ) {\displaystyle \int tlog(t)dt=tlog(t)-\int t({\frac {1}{t}}dt)\!} ∫ t l o g ( t ) d t = t l o g ( t ) − ∫ d t {\displaystyle \int tlog(t)dt=tlog(t)-\int dt\!} ∫ t l o g ( t ) d t = t l o g ( t ) − t {\displaystyle \int tlog(t)dt=tlog(t)-t\!}
Therefore:∫ ( t l o g ( t ) − log ( t ) ) d t = 1 2 t 2 l o g ( t ) − 1 4 t 2 − ( t l o g ( t ) − t ) {\displaystyle \int (tlog(t)-\log(t))dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}-(tlog(t)-t)\!} ∫ ( t l o g ( t ) − log ( t ) ) d t = 1 2 t 2 l o g ( t ) − 1 4 t 2 − t l o g ( t ) + t {\displaystyle \int (tlog(t)-\log(t))dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}-tlog(t)+t\!}
Re-substituting for t: ∫ x l o g ( 1 + x ) d x = 1 2 ( 1 + x ) 2 l o g ( 1 + x ) − 1 4 ( 1 + x ) 2 − ( 1 + x ) l o g ( 1 + x ) + ( 1 + x ) + C {\displaystyle \int xlog(1+x)dx={\frac {1}{2}}(1+x)^{2}log(1+x)-{\frac {1}{4}}(1+x)^{2}-(1+x)log(1+x)+(1+x)+C\!} ∫ x l o g ( 1 + x ) d x = ( 1 + x ) ( 1 2 ( 1 + x ) l o g ( 1 + x ) − 1 4 ( 1 + x ) − l o g ( 1 + x ) + 1 ) + C {\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}(1+x)log(1+x)-{\frac {1}{4}}(1+x)-log(1+x)+1)+C\!} ∫ x l o g ( 1 + x ) d x = ( 1 + x ) ( 1 2 x l o g ( 1 + x ) − 1 2 l o g ( 1 + x ) − 1 4 x + 3 4 ) + C {\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!}
Therefore:
∫ x l o g ( 1 + x ) d x = ( 1 + x ) ( 1 2 x l o g ( 1 + x ) − 1 2 l o g ( 1 + x ) − 1 4 x + 3 4 ) + C {\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!}
Using the General Binomial Theorem for the integral with t substitution ∫ ( t − 1 ) l o g ( t ) d t {\displaystyle \int (t-1)log(t)dt\!} : ( x + y ) 1 = ( t + ( − 1 ) ) 1 = ( t + ( − 1 ) ) = ∑ k = 0 1 ( 1 k ) x 1 − k y k = ∑ k = 0 1 ( 1 k ) t 1 − k ( − 1 ) k = t − 1 {\displaystyle (x+y)^{1}=(t+(-1))^{1}=(t+(-1))=\sum _{k=0}^{1}{\binom {1}{k}}x^{1-k}y^{k}=\sum _{k=0}^{1}{\binom {1}{k}}t^{1-k}(-1)^{k}=t-1\!}
Therefore:
∫ ( t − 1 ) l o g ( t ) d t = ∫ x l o g ( 1 + x ) d x {\displaystyle \int (t-1)log(t)dt=\int xlog(1+x)dx\!}
Which we have previously found that answer as:
∫ x l o g ( 1 + x ) d x = ( 1 + x ) ( 1 2 x l o g ( 1 + x ) − 1 2 l o g ( 1 + x ) − 1 4 x + 3 4 ) + C {\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!}