Q26E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Existence of Fundamental Solution Sets. By Theorem 1, for each j = 1, 2, . . ., n there is a unique solution \({{\bf{y}}_{\bf{j}}}\left( {\bf{x}} \right)\) to equation (17) satisfying the initial conditions 

\({{\bf{y}}_{\bf{j}}}^{\left( {\bf{k}} \right)}\left( {{{\bf{x}}_{\bf{0}}}} \right){\bf{ = }}\left\{ \begin{array}{l}{\bf{1,}}\,\,\,\,\,{\bf{for}}\,{\bf{k = j - 1,}}\\{\bf{0,}}\,\,\,\,{\bf{for}}\,{\bf{k}} \ne {\bf{j - 1,}}\,\,{\bf{0}} \le {\bf{k}} \le {\bf{n - 1}}\end{array} \right\}\)

(a) Show that \(\left\{ {{{\bf{y}}_{\bf{1}}}{\bf{,}}{{\bf{y}}_{\bf{2}}}{\bf{,}}...{\bf{,}}{{\bf{y}}_{\bf{n}}}} \right\}\) is a fundamental solution set for (17). [Hint: Write out the Wronskian at\({{\bf{x}}_{\bf{0}}}\)].

(b) For given initial values\({\gamma _{\bf{0}}}{\bf{,}}{\gamma _{\bf{1}}}{\bf{,}}{\gamma _{\bf{2}}}{\bf{,}}...{\bf{,}}{\gamma _{{\bf{n - 1}}}}\), express the solution y(x) to (17) satisfying \({{\bf{y}}^{\left( {\bf{k}} \right)}}\left( {{{\bf{x}}_{\bf{0}}}} \right){\bf{ = }}{\gamma _{\bf{k}}},\,\,{\bf{k = 0,1,}}...{\bf{,n - 1}}\),[as in equations (4)] in terms of this fundamental solution set.

Step-by-Step Solution

Answer

1. \(\left\{ {{{\bf{y}}_{\bf{1}}}{\bf{,}}{{\bf{y}}_{\bf{2}}}{\bf{,}}...{\bf{,}}{{\bf{y}}_{\bf{n}}}} \right\}\) is a fundamental solution set for (17) equation.
2. \({\bf{y}}\left( {\bf{x}} \right){\bf{ = }}{\gamma _{\bf{0}}}{{\bf{y}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{ + }}{\gamma _{\bf{1}}}{{\bf{y}}_{\bf{2}}}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{\gamma _{{\bf{n - 1}}}}{{\bf{y}}_{\bf{n}}}\left( {\bf{x}} \right)\)

(a) Step 1: Use the given initial condition to prove that, \(\left\{ {{{\bf{y}}_{\bf{1}}}{\bf{,}}{{\bf{y}}_{\bf{2}}}{\bf{,}}...{\bf{,}}{{\bf{y}}_{\bf{n}}}} \right\}\) is a fundamental solution set for (17)

Equation (17); Let \({{\bf{y}}_{\bf{1}}}{\bf{,}}\,{{\bf{y}}_{\bf{2}}}{\bf{,}}...,\,{{\bf{y}}_{\bf{n}}}\)be n solutions on (a, b) of

\({{\bf{y}}^{\left( {\bf{n}} \right)}}\left( {\bf{x}} \right){\bf{ + }}{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right){{\bf{y}}^{\left( {{\bf{n - 1}}} \right)}}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{{\bf{p}}_{\bf{n}}}\left( {\bf{x}} \right){\bf{y}}\left( {\bf{x}} \right){\bf{ = 0}}\)

Let some point \({{\bf{x}}_{\bf{0}}} \in \left( {{\bf{a,}}\,{\bf{b}}} \right)\)

Given,

\({{\bf{y}}_{\bf{j}}}^{\left( {\bf{k}} \right)}\left( {{{\bf{x}}_{\bf{0}}}} \right){\bf{ = }}\left\{ \begin{array}{l}{\bf{1,}}\,\,\,\,\,{\bf{for}}\,{\bf{k = j - 1,}}\\{\bf{0,}}\,\,\,\,{\bf{for}}\,{\bf{k}} \ne {\bf{j - 1,}}\,\,{\bf{0}} \le {\bf{k}} \le {\bf{n - 1}}\end{array} \right\}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( {\bf{1}} \right)\)

 

The Wronskian at\({{\bf{x}}_{\bf{0}}}\),

 

\({\bf{W}}\left[ {{{\bf{y}}_{\bf{1}}}{\bf{,}}{{\bf{y}}_{\bf{2}}}{\bf{,}}...{\bf{,}}{{\bf{y}}_{\bf{n}}}} \right]{\bf{ = }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{\bf{.}}&{\bf{.}}&{{{\bf{y}}_{\bf{n}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{\bf{.}}&{\bf{.}}&{{{\bf{y}}_{\bf{n}}}{\bf{'}}}\\\begin{array}{l}{\bf{.}}\\{\bf{.}}\end{array}&\begin{array}{l}{\bf{.}}\\{\bf{.}}\end{array}&\begin{array}{l}{\bf{.}}\\{\bf{.}}\end{array}&\begin{array}{l}{\bf{.}}\\{\bf{.}}\end{array}&\begin{array}{l}{\bf{.}}\\{\bf{.}}\end{array}\\{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}\\{{{\bf{y}}_{\bf{1}}}^{\left( {{\bf{n - 1}}} \right)}}&{{{\bf{y}}_{\bf{2}}}^{\left( {{\bf{n - 1}}} \right)}}&{\bf{.}}&{\bf{.}}&{{{\bf{y}}_{\bf{n}}}^{\left( {{\bf{n - 1}}} \right)}}\end{array}} \right|\)

 

Use the equation (1),

 

\(\begin{array}{c}{\bf{W}}\left( {{{\bf{x}}_{\bf{0}}}} \right){\bf{ = }}\left| {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{.}}&{\bf{.}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{.}}&{\bf{.}}&{\bf{0}}\\{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}\\{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}\\{\bf{0}}&{\bf{0}}&{\bf{.}}&{\bf{.}}&{\bf{1}}\end{array}} \right|\\{\bf{ = 1}}\end{array}\)

 

But\({\bf{W}}\left( {{{\bf{x}}_{\bf{0}}}} \right) \ne {\bf{0}}\)

Then one says that,

\(\left\{ {{{\bf{y}}_{\bf{1}}}{\bf{,}}{{\bf{y}}_{\bf{2}}}{\bf{,}}...{\bf{,}}{{\bf{y}}_{\bf{n}}}} \right\}\) is a fundamental solution set of equations (17).

 

(b) Step 2: Use the given condition and find the general solution to equation (17)

Equation (17); Let \({{\bf{y}}_{\bf{1}}}{\bf{,}}\,{{\bf{y}}_{\bf{2}}}{\bf{,}}...,\,{{\bf{y}}_{\bf{n}}}\)be n solutions on (a, b) of

\({{\bf{y}}^{\left( {\bf{n}} \right)}}\left( {\bf{x}} \right){\bf{ + }}{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right){{\bf{y}}^{\left( {{\bf{n - 1}}} \right)}}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{{\bf{p}}_{\bf{n}}}\left( {\bf{x}} \right){\bf{y}}\left( {\bf{x}} \right){\bf{ = 0}}\)

The general solution of the above equation,

\({\bf{y}}\left( {\bf{x}} \right){\bf{ = }}{{\bf{c}}_{\bf{1}}}{{\bf{y}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{ + }}{{\bf{c}}_{\bf{2}}}{{\bf{y}}_{\bf{2}}}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{{\bf{c}}_{\bf{n}}}{{\bf{y}}_{\bf{n}}}\left( {\bf{x}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {\bf{1}} \right)\)

 

Given condition,

\({{\bf{y}}^{\left( {\bf{k}} \right)}}\left( {{{\bf{x}}_{\bf{0}}}} \right){\bf{ = }}{\gamma _{\bf{k}}},\,\,{\bf{k = 0,1,}}...{\bf{,n - 1}}\)

From equation (1)

One has,

\({{\bf{c}}_{\bf{1}}}{{\bf{y}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{ + }}{{\bf{c}}_{\bf{2}}}{{\bf{y}}_{\bf{2}}}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{{\bf{c}}_{\bf{n}}}{{\bf{y}}_{\bf{n}}}\left( {\bf{x}} \right){\bf{ = }}{\gamma _{\bf{0}}}\)

Take the differential n-1 time of the above equation,

\(\begin{array}{c}{{\bf{c}}_{\bf{1}}}{{\bf{y}}_{\bf{1}}}{\bf{'}}\left( {\bf{x}} \right){\bf{ + }}{{\bf{c}}_{\bf{2}}}{{\bf{y}}_{\bf{2}}}{\bf{'}}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{{\bf{c}}_{\bf{n}}}{{\bf{y}}_{\bf{n}}}{\bf{'}}\left( {\bf{x}} \right){\bf{ = }}{\gamma _{\bf{1}}}\\{{\bf{c}}_{\bf{1}}}{{\bf{y}}_{\bf{1}}}{\bf{''}}\left( {\bf{x}} \right){\bf{ + }}{{\bf{c}}_{\bf{2}}}{{\bf{y}}_{\bf{2}}}{\bf{''}}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{{\bf{c}}_{\bf{n}}}{{\bf{y}}_{\bf{n}}}{\bf{''}}\left( {\bf{x}} \right){\bf{ = }}{\gamma _{\bf{2}}}\\.\\.\\.\\{{\bf{c}}_{\bf{1}}}{{\bf{y}}_{\bf{1}}}^{\left( {{\bf{n - 1}}} \right)}\left( {\bf{x}} \right){\bf{ + }}{{\bf{c}}_{\bf{2}}}{{\bf{y}}_{\bf{2}}}^{\left( {{\bf{n - 1}}} \right)}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{{\bf{c}}_{\bf{n}}}{{\bf{y}}_{\bf{n}}}^{\left( {{\bf{n - 1}}} \right)}\left( {\bf{x}} \right){\bf{ = }}{\gamma _{{\bf{n - 1}}}}\\.\end{array}\)

Step 3: Solve the equation of step 2 to find the solution to the equation

One can writes as,

\(\left[ {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{\bf{.}}&{\bf{.}}&{{{\bf{y}}_{\bf{n}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{\bf{.}}&{\bf{.}}&{{{\bf{y}}_{\bf{n}}}{\bf{'}}}\\{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}\\{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}\\{{{\bf{y}}_{\bf{1}}}^{\left( {{\bf{n - 1}}} \right)}}&{{{\bf{y}}_{\bf{2}}}^{\left( {{\bf{n - 1}}} \right)}}&{\bf{.}}&{\bf{.}}&{{{\bf{y}}_{\bf{n}}}^{\left( {{\bf{n - 1}}} \right)}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\.\\.\\{{c_n}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{\gamma _{\bf{0}}}}\\{{\gamma _{\bf{1}}}}\\.\\.\\{{\gamma _{{\bf{n - 1}}}}}\end{array}} \right]\)

 

Therefore,

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{.}}&{\bf{.}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{.}}&{\bf{.}}&{\bf{0}}\\{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}\\{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}&{\bf{.}}\\{\bf{0}}&{\bf{0}}&{\bf{.}}&{\bf{.}}&{\bf{1}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{{\bf{c}}_{\bf{1}}}}\\{{{\bf{c}}_{\bf{2}}}}\\{\bf{.}}\\{\bf{.}}\\{{{\bf{c}}_{\bf{n}}}}\end{array}} \right]{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{{\gamma _{\bf{0}}}}\\{{\gamma _{\bf{1}}}}\\.\\.\\{{\gamma _{{\bf{n - 1}}}}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{{\bf{c}}_{\bf{1}}}}\\{{{\bf{c}}_{\bf{2}}}}\\{\bf{.}}\\{\bf{.}}\\{{{\bf{c}}_{\bf{n}}}}\end{array}} \right]{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{{\gamma _{\bf{0}}}}\\{{\gamma _{\bf{1}}}}\\.\\.\\{{\gamma _{{\bf{n - 1}}}}}\end{array}} \right]\end{array}\)

 

Thus, the solution of the equation is,

 

\({\bf{y}}\left( {\bf{x}} \right){\bf{ = }}{\gamma _{\bf{0}}}{{\bf{y}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{ + }}{\gamma _{\bf{1}}}{{\bf{y}}_{\bf{2}}}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{\gamma _{{\bf{n - 1}}}}{{\bf{y}}_{\bf{n}}}\left( {\bf{x}} \right)\)

 

Therefore, 

 

1. \(\left\{ {{{\bf{y}}_{\bf{1}}}{\bf{,}}{{\bf{y}}_{\bf{2}}}{\bf{,}}...{\bf{,}}{{\bf{y}}_{\bf{n}}}} \right\}\) is a fundamental solution set for (17) equation.

\({\bf{y}}\left( {\bf{x}} \right){\bf{ = }}{\gamma _{\bf{0}}}{{\bf{y}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{ + }}{\gamma _{\bf{1}}}{{\bf{y}}_{\bf{2}}}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{\gamma _{{\bf{n - 1}}}}{{\bf{y}}_{\bf{n}}}\left( {\bf{x}} \right)\)