How to solve pde ut+aux=e^(-2x), u(x,0)=f(x)

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Introduction : In this solution we explain step by step the method How to solve pde ut+aux=e^(-2x), u(x,0)=f(x).

Question: solve pde

\[u_{t}+au_{x}=e^{-2x}\]

subject to initial condition \[u(x,0)=f(x)\]

Solution:

Given pde

\[u_{t}+au_{x}=e^{-2x}\]

\[u(x,0)=f(x)\]

So the parametrizing the curve as

\[Γ=(s,0,f(s))\]

Now characteristics equations of the pde are

\[\frac{dx}{dt}=a,\frac{dy}{dt}=1,\frac{dz}{dt}=e^{-2x}\]

integrate first equation

\[x=at+c_{1}\]

using initial condition

\[s=0+c_{1}=>c_{1}=s\]

\[x=at+s\]

And integrating second equation

\[ y=t\]

similarly integrating third equation

\[\frac{dz}{dt}=e^{-2x}=e^{-2\left(at+s\right)}\]

\[z=-\frac{1}{2c}e^{-2(at+s))}+c_{2}\]

using initial condition

\[f(s)=-\frac{1}{2a}e^{-2a}+c_{2}=>c_{2}=f(s)+\frac{1}{2a}\]

\[z(s,t)=-\frac{1}{2a}e^{-2(at+s)}+f(s)+\frac{1}{2a}e^{-2s}\]

\[u(x,y)=-\frac{1}{2a}e^{-2(ay+x-at)}+f(x-at)+\frac{1}{2a}e^{\left(-2(x-at\right))}\]

\[=\frac{1}{2a}(e^{-2(x-at)}-e^{-2(ay+x-at)})+f(x-at)\]

since t= y

\[u(x,y)=\frac{1}{2a}e^{-2(x-ay)}(1-e^{-2ay})+(x-ay)\]

This is the required solution of the the given pde.


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