Higher Order PDE Solutions PDF

# Linear Partial Differential Equations with Constant Coefficients ## Introduction Linear partial differential equations with constant coefficients can be expressed as: $F(D, D')z = f(x,y)$ (1) Where *D* denotes the partial derivative with respect to *x* and *D'* denotes the partial derivative w...

# Linear Partial Differential Equations with Constant Coefficients ## Introduction Linear partial differential equations with constant coefficients can be expressed as: $F(D, D')z = f(x,y)$ (1) Where *D* denotes the partial derivative with respect to *x* and *D'* denotes the partial derivative with respect to *y*. ## Solution Approach The complete solution of (1) consists of two parts: 1. Complementary Function (CF): Solution of the homogeneous equation $F(D,D')z=0$. 2. Particular Integral (PI): A solution to the non-homogeneous equation $F(D,D')z = f(x,y)$. ## Complementary Function The complementary function is obtained by considering the factors of the homogeneous equation $F(D,D')z=0$. For an equation with constant coefficients, the factors will be of the form: $(Dm, D') (D-m₂D') ... (D-m_nD')z = 0$ Each factor leads to a particular solution: $(D-mD')z = 0$ The general solution for this is: $z = \phi(y + mx)$, where $\phi$ is an arbitrary function. ### Notes 1. For repeated factors of the form $(D-mD)^2 = 0$, the general solution becomes: $z = \phi_1 (y + mx) + \phi_2(y + mx)$ 2. For distinct roots $m_1, m_2, ..., m_n$, the general solution for the complementary function is: $$ z = \phi_1 (y + m_1x) + \phi_2 (y + m_2x) + ... + \phi_n (y + m_nx) $$ ## Examples **Example 1:** Solve $(D^3 - 3D^2 D' + 2D D'^2)z =0$ The auxiliary equation is: $m^3 - 3m^2 + 2m = 0$ Solving for *m* gives us the roots: $m = 0, 1, 2$ Therefore, the complementary function is: $z = \phi_1(y) + \phi_2(y + x) + \phi_3(y + 2x)$ **Example 2:** Solve $(D^2 - 4D D' + 4 D D'^2)z = 0$ The auxiliary equation is: $m^2 - 4m + 4 = 0$ Solving for *m* gives us the roots: $m =0, 2, 2$ Therefore, the complementary function is: $z = \phi_1(y) + \phi_2(y + 2x) + x\phi_3(y + 2x)$ **Example 3:** Solve $(5D - 4D')^2 z = 0$ The auxiliary equation is: $(5m - 4) ^2 = 0$ This gives us the repeated root: $m = 4/5$ Therefore, the complementary function is: $z = \phi_1(5y + 4 x) + x\phi_2(5y + 4 x)$ **Example 4:** Solve $(4D^2 + 12DD' + 9D'^2)z = 0$ The auxiliary equation is: $(2D + 3D') ^2 = 0$ This gives us the repeated root: $m = -3/2$ Therefore, the complementary function is: $z = \phi_1(2y + 3x) + x\phi_2(2y + 3x)$ **Example 5:** Solve $(D^2 - 6D D' + 11D D' - 6D'^2)z = 0$ The auxiliary equation is: $m^2 - 6m + 11 - 6 = 0$ Solving for *m* gives us the roots: $m = 1, 2, 3$ Therefore, the complementary function is: $z = \phi_1(y + x) + \phi_2(y + 2x) + \phi_3 (y + 3x)$ ## Particular Integral (PI) The PI is a solution to the non-homogeneous equation $F(D, D')z = f(x, y)$. **General Method:** The general method involves the following steps: 1. **Substitution:** Substitute $y + mx = a$ and $dz = f(x, y) dx$ into the original equation $F(D,D')z = f(x, y)$. 2. **Integration** Integrate the resulting equation with respect to *a* to obtain $z$. 3. **Back Substitution:** Substitute $a = y + mx$ back into the solution. **Shortcut Method:** This is applicable for the right-hand side being a function of a single variable: 1. **Case 1:** If $F(a, b) \ne 0$, then: $$ \frac{1}{F(D, D')}e^{ax + by} = \frac{1}{F(a, b)} e^{ax + by} $$ 2. **Case 2:** If $F(a, b) = 0$, then: $$ F(D, D') = (bD - aD')g(D,D')\text{,} $$ where $g(a, b) \ne 0$ $$ \frac{1}{F(D,D')}e^{ax + by} = \frac{1}{g(a, b)} \frac{1}{(bD-aD')^r}e^{ax + by} $$ where *r* is the order of $F(D,D')$. The PI can be obtained by integrating $r$ times with respect to *x*. ## Examples **Example 1:** Solve $(D - 2D') (D - D') z = e^{x+y}$ The auxiliary equation is: $m - 2 = 0$ , $m - 1=0$ The roots are: $m = 2, 1$ The complementary function is: $z = \phi_1 (y + 2x) + \phi_2(y + x)$ Using the shortcut method, the PI is: $$ \begin{aligned} \frac{1}{(D-2D') (D - D')} e^{x + y} &= \frac{1}{(1-2)(1-1)}e^{x + y} \\ &= \frac{1}{(-1)(-1)}e^{x + y} \\ &= e^{x + y} \end{aligned} $$ Therefore, the complete solution is: $z = \phi_1(y + 2x) + \phi_2(y + x) + e^{x + y}$ **Example 2:** Solve $(D - 2 D') (D - D') z = x e^{x+y}$ Using the shortcut method, the PI is: $$ \begin{aligned} \frac{1}{(D-2D') (D - D')} x e^{x + y} &= \frac{1}{(1-2)(1-1)} \frac{1}{(D-D')^2} x e^{x + y} \\ &= - \int \int \frac{1}{(1-D)^{2}}x e^{x + y} dx dx \\ &= - \int \int x e^{x + y} dx dx \\ &= - \int (x - 1) e^{x + y} dx \\ &= - (x - 2)e^{x + y} \end{aligned} $$ Therefore, the complete solution is: $z = \phi_1(y + 2x) + \phi_2(y + x) - (x - 2)e^{x + y}$ **Example 3:** Solve $(D - 2DD' + D'^2)z = cos(x + 2y)$ The auxiliary equation is: $m^2 - 2m + 1 = 0$ The root is: $m = 1$ The complementary function is: $z = \phi_1(y + x) + x\phi_2(y + x)$ The PI can be calculated using the shortcut method: $$ \begin{aligned} \frac{1}{(D^2 -2DD' + D'^2)} cos(x + 2y) &= \frac{1}{(1-2)^2 } cos(x + 2y) \\ &= cos(x + 2y) \end{aligned} $$ Therefore, the complete solution is: $z = \phi_1(y + x) + x\phi_2(y + x) + cos(x + 2y)$ **Example 4:** Solve $(D^3 - 3D^2 D' + 2D D'^2)z = cos(2x + 3y)$ The auxiliary equation is: $m^3 - 3m^2 + 2m = 0$ The roots are: $m = 0, 1, 2$ The complementary function is: $z = \phi_1(y) + \phi_2(y + x) + \phi_3 (y + 2x)$ The PI can be calculated using the shortcut method: $$ \begin{aligned} \frac{1}{(D^3 - 3D^2 D' + 2D D'^2)}cos(2x + 3y) &= \frac{1}{(2^3 - 3 \cdot 2^2 \cdot 3 + 2 \cdot 2 \cdot 3^2)}cos(2x + 3y) \\ &= \frac{1}{8 - 36 + 36} cos(2x + 3y) \\ &= \frac{1}{8}cos(2x + 3y) \end{aligned} $$ Therefore, the complete solution is: $z = \phi_1(y) + \phi_2(y + x) + \phi_3 (y + 2x) + \frac{1}{8}cos(2x + 3y)$ ## Reducible Linear Equations Reducible linear equations have factors that can be further factored. For instance: $(D-D')(D+D'+1)z = 0$ The solution approach for these equations involves finding solutions corresponding to each factor and then combining them to obtain the general solution. ## Irreducible Factors For equations with irreducible factors, we can solve for the general solution by assuming a trial solution of the form: $z = e^{hx + ky}$ Substituting this into the equation and solving for $h$ and $k$ will give us the solution. ## Non-homogenous Equations with Irreducible Factors **Example 1:** Solve $(D^2 - 2DD' + D'^2 + D - D')z = e^{x + y}$ The auxiliary equation is: $m^2 - 2m + 1 + 1 - 1 = 0$ Simplifying, we get: $m^2 - 2m + 1 = 0$ The root is: $m = 1$ The complementary function is: $z = \phi_1(y + x) + x\phi_2(y + x)$ For the PI, we can use the shortcut method: $$ \begin{aligned} \frac{1}{(D^2 - 2DD' + D'^2 + D - D')}e^{x + y} &= \frac{1}{(1^2 - 2 \cdot 1 \cdot 1 + 1^2 + 1 - 1)}e^{x + y} \\ &= e^{x + y} \end{aligned} $$ Therefore, the complete solution is: $z = \phi_1(y + x) + x\phi_2(y + x) + e^{x + y}$ **Example 2:** Solve $(D^2 - 3DD' + D'^2 + D - D')z = e^{x + 2y}$ The auxiliary equation is the same as the previous example. The complementary function remains identical: $z = \phi_1(y + x) + x\phi_2(y + x)$ For the PI: $$ \begin{aligned} \frac{1}{(D^2 - 3DD' + D'^2 + D - D')}e^{x + 2y} &= \frac{1}{(1 - 3 + 1 + 1 - 1)}e^{x + 2y} \\ &= -\frac{1}{1}e^{x + 2y} \\ &= -e^{x + 2y} \end{aligned} $$ Therefore, the complete solution is: $z = \phi_1(y + x) + x\phi_2(y + x) - e^{x + 2y}$ ## General Method for PI For a non-homogeneous equation with right-hand side $f(x, y)$, the general method for finding the PI involves the following steps: 1. **Substitute:** Use $y + mx = a$ and $dz = f(x, y) dx$. 2. **Solve:** Solve the resulting equation by integrating with respect to *a*. 3. **Back Substitute:** Substitute $a = y + mx$ back into the solution to obtain the PI. ## Homogeneous Equations with Reducible Operators Homogeneous equations with reducible operators are solved by first factoring the operators and then finding the solutions for each factor. The combined solution becomes the general solution. **Example:** Solve $(D^2 - y^2D'^2)z = XY$ The equation can be rewritten as: $(D-yD')(D + yD')z = XY$ We need to solve for $z$ taking into account each factor: $(D - yD')z = 0$ & $(D + yD')z = 0$ Using the substitutions $x = e^X$ and $y = e^Y$, we get: $D = \frac{\partial}{\partial x} = \frac{\partial}{\partial X}e^X = de$ $D' = \frac{\partial}{\partial y} = \frac{\partial}{\partial Y}e^Y = de$ Substituting these into the equations, we get: $(d - d')z = 0$ & $(d + d')z = 0$ The solutions for these equations are: $Z = \phi_1(Y + X)$ & $Z = \phi_2(Y - X)$. Combining these solutions, we get the general solution: $Z = \phi_1(\log y + \log x) + \phi_2(\log y - \log x)$. ## Examples of Non-homogeneous Equations with Trigonometric Functions **Example 1:** Solve $(D^2 - DD' - 2D'^2)z = sin(3x + 4y)$ The auxiliary equation is: $m^2 - m - 2 = 0$ Solving for $m$ we get: $m = 2, -1$ The complementary function is: $Z = \phi_1(y + 2x) + \phi_2(y - x)$ Let's calculate the PI. Since $F(x, y) = sin(3x + 4y)$, we've got a trigonometric function. For trigonometric functions, it's beneficial to use the operators $D^2 = 1 - a^2$ and $DD' = -ab$ for the general solution. Therefore: $(D^2 - DD' - 2D'^2)z = (1 - 9 + 12) z = 4z$ This gives us a PI of: $$ \begin{aligned} z &= \frac{1}{4}sin(3x + 4y) \\ &= \frac{1}{4}[sin(3x + 4y) + 2 cos(3x + 4y)] \end{aligned} $$ The complete solution is: $z = \phi_1(y + 2x) + \phi_2(y - x) + \frac{1}{4}[sin(3x + 4y) + 2 cos(3x + 4y)].$ **Example 2:** Solve $(D - D'^2)z = cos(x - 3y)$ The auxiliary equation is: $m - 1 = 0$ Solving for $m$ we get: $m = 1$ The complementary function is: $Z = \phi(y + x)$ Let's calculate the PI. Since $F(x, y) = cos(x - 3y)$, we've got a trigonometric function. Using the operators $D^2 = 1 - a^2$ and $DD' = -ab$ for the general solution, we get: $(D - D'^2)z = (1 - 9)z = (-8)z$ $$ \begin{aligned} z &= \frac{1}{-8}cos(x - 3y) \\ &= -\frac{1}{8}[cos(x - 3y) - 2 sin(x - 3y)] \end{aligned} $$ The complete solution is: $z = \phi(y + x) - \frac{1}{8}[cos(x - 3y) - 2 sin(x - 3y)]$. These examples showcase the methods and strategies for solving linear partial differential equations with constant coefficients. By understanding these techniques, you can effectively address a wide range of problems in this field.