Use separation of variables to reduce the PDE to two ODEs. Solve the X equation considering all...


Use separation of variables to reduce the PDE to two ODEs. Solve the {eq}X {/eq} equation considering all possible situations.

{eq}\displaystyle \dfrac {\partial^2 u}{\partial x^2} + \dfrac {\partial u}{\partial x} - \dfrac 1 k \dfrac {\partial u}{\partial t} = 0 {/eq}.

Solving PDE by Separation of Variables:

Any Partial Differential Equation (PDE) with independent variables x and t and u as dependent variable, when solved by the method of separation of variables, assumes the solution of the form

{eq}u(x, \ t)=X(x)T(t). {/eq} This form of solution is substituted in the PDE to obtain two ODEs in X(x) and T(t), which can be further solved to obtain the solution of given PDE,.

Answer and Explanation:

Using separation of variables,

{eq}\begin{align} u(x, \ t)=X(x)T(t)\\ \Rightarrow & u_x(x, \ t)=X'(x)T(t)\\ & u_{xx}(x, \ t)=X''(x)T(t)\\ & u_t(x, \ t)=X(x)T'(t)\\ \end{align} {/eq}


{eq}\displaystyle \dfrac {\partial^2 u}{\partial x^2} + \dfrac {\partial u}{\partial x} - \dfrac 1 k \dfrac {\partial u}{\partial t} = 0\\ \displaystyle \Rightarrow X''(x)T(t)+X'(x)T(t)-\frac{1}{k}X(x)T'(t)=0\\ \displaystyle \Rightarrow (X''(x)+X'(x))T(t)=\frac{1}{k}X(x)T'(t)\\ \displaystyle \Rightarrow \frac{(X''(x)+X'(x))}{X(x)}=\frac{1}{k}\frac{T'(t)}{T(t)}=p, {/eq} where p is a constant.

We consider three cases where {eq}p=0\\ p=a^2>0\\ p=-a^2<0\\ {/eq}

Solving the X equation, we get

{eq}\displaystyle \frac{(X''(x)+X'(x))}{X(x)}=p\\ \displaystyle \Rightarrow X''+X'-pX=0\\ {/eq}

Case 1: {eq}p=0\\ {/eq}

{eq}\displaystyle X''+X'-pX=0 \Rightarrow X''+X'=0\\ {/eq}

The characteristic equation becomes

{eq}m^2+m=0\\ \Rightarrow m=0, \ -1. {/eq}

Therefore, {eq}X(x)=C_1+C_2e^{-x} {/eq}

Case 2: {eq}p= \ a^2 \ >0 \\ {/eq}

{eq}\displaystyle X''+X'-pX=0 \Rightarrow X''+X'-a^2X=0\\ {/eq}

The characteristic equation becomes

{eq}m^2+m-a^2X=0\\ \Rightarrow m=\frac{-1+\sqrt{1^2-4.1.(-a^2)}}{2.1}, \ \frac{-1-\sqrt{1^2-4.1.(-a^2)}}{2.1}\\ \Rightarrow m=\frac{-1+\sqrt{1+4a^2}}{2}, \ \frac{-1-\sqrt{1+4a^2}}{2}\\ {/eq}

The roots are real and distinct. Therefore, {eq}X(x)=C_3e^{\frac{(-1+\sqrt{1+4a^2}}{2})x}+C_4e^{\frac{(-1-\sqrt{1+4a^2}}{2})x} {/eq}

Case 3: {eq}p= \ -a^2 \ <0 \\ {/eq}

{eq}\displaystyle X''+X'-pX=0 \Rightarrow X''+X'+a^2X=0\\ {/eq}

The characteristic equation becomes

{eq}m^2+m+a^2X=0\\ \Rightarrow m=\frac{-1+\sqrt{1^2-4.1.(a^2)}}{2.1}, \ \frac{-1-\sqrt{1^2-4.1.(a^2)}}{2.1}\\ \Rightarrow m=\frac{-1+\sqrt{1-4a^2}}{2}, \ \frac{-1-\sqrt{1-4a^2}}{2}\\ {/eq}

Further we can divide into 3 sub-cases where {eq}Sub-case \ 1: 1-4a^2>0 \Rightarrow a^2<\frac{1}{4} \Rightarrow 0>p>-\frac{1}{4}\\ {/eq} The roots will be real and distinct.

{eq}X(x)=C_5e^{\frac{(-1+\sqrt{1-4a^2}}{2})x}+C_6e^{\frac{(-1-\sqrt{1-4a^2}}{2})x} {/eq}

{eq}Sub-case \ 2: 1-4a^2<0 \Rightarrow a^2>\frac{1}{4} \Rightarrow p<-\frac{1}{4}\\ m=\frac{-1+\sqrt{4a^2-1}i}{2}, \ \frac{-1-\sqrt{4a^2-1}i}{2}\\ {/eq}

The roots will be complex.

{eq}X(x)=e^{(\frac{-1}{2})x}(C_7 \cos (\frac{\sqrt{4a^2-1}}{2})x)+C_8 \sin (\frac{\sqrt{4a^2-1}}{2})x) {/eq}

{eq}Sub-case \ 3: 1-4a^2=0 \Rightarrow a^2=\frac{1}{4} \Rightarrow p=-\frac{1}{4}\\ m=\frac{-1}{2}, \ \frac{-1}{2} {/eq}

The roots are real and repeated. Therefore, {eq}X(x)=(C_9+C_{10}x)e^{(\frac{(-1}{2})x} {/eq}

Learn more about this topic:

Solving Partial Derivative Equations

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 1

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